{ "schema": "cie-topical-past-papers/answers-v1", "subject": "9702", "subject_name": "Physics", "generated": "2026-08-14", "coverage": { "question_records": 436, "available_mark_scheme_entries": 436, "unavailable_mark_scheme_entries": 0 }, "records": [ { "id": "9702-2021-m-42-q01", "question_id": "9702-2021-m-42-q01", "subject": "9702", "year": 2021, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 1, "topic": "Gravitational fields", "topic_slug": "9702-topic-13-gravitational-fields", "marks": 12, "status": "available", "reason": null, "text": "1(a) (gravitational) force is (directly) proportional to product of masses B1\nforce (between point masses) is inversely proportional to the square of their separation B1\n\n1(b) correct read offs from the graph with correct power of ten for R3 C1\n4×π2×1.2×1034 C1\nM =\n6.67×10 −11×2.4×( 365×24×3600 )2\n= 3.0×1030 kg A1\n\n1(c)(i) potential energy is zero at infinity B1\n(gravitational) forces are attractive B1\nwork must be done on the rock to move it to infinity B1\n\n1(c)(ii) GMm mv2 GM GM M1\n= O R v2 = O R v =\nr2 r r r\nGMm A1\nuse of ½ mv2 (e.g. multiplication by ½ m) leading to\n2r\n\n1(c)(iii) −GM −GMm C1\nEp = φ m and φ = or E =\nr p r\nTotal energy = E + E\nk p\nGMm −GMm −GMm A1\nTotal energy= + =\n2r r 2r\n© UCLES 2021 Page 8 of 19", "source_pages": [ 8 ], "source_pdf": "_source-pdfs/2021-March/ms/9702_m21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-March/9702_m21_ms_42.pdf?download=true", "html": "9702-topic-13-gravitational-fields/answers.html", "image_paths": [ "../answer-assets/9702_m21_ms_42-p08.png" ] }, { "id": "9702-2021-m-42-q02", "question_id": "9702-2021-m-42-q02", "subject": "9702", "year": 2021, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 2, "topic": "Ideal gases", "topic_slug": "9702-topic-15-ideal-gases", "marks": 6, "status": "available", "reason": null, "text": "2(a)(i) pV =NkT or pV =nRT and N =nN C1\nA\n2.3×105×3.5×10 −3\nN =\n1.38×10 −23×294\n= 2.0 × 1023 A1\n\n2(a)(ii) 1 C1\npV = Nmc2", "source_pages": [ 9 ], "source_pdf": "_source-pdfs/2021-March/ms/9702_m21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-March/9702_m21_ms_42.pdf?download=true", "html": "9702-topic-15-ideal-gases/answers.html", "image_paths": [ "../answer-assets/9702_m21_ms_42-p09.png" ] }, { "id": "9702-2021-m-42-q03", "question_id": "9702-2021-m-42-q03", "subject": "9702", "year": 2021, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 3, "topic": "Temperature", "topic_slug": "9702-topic-14-temperature", "marks": 8, "status": "available", "reason": null, "text": "3\n3×2.3×105 ×3.5×10 −3\nc2 =\n2.0×1023×40×1.66×10 −27\n= 182 000\nr.m.s. speed = 430 m s–1\nor A1\n1 mc2 =3 kT\n2 2\n3×1.38×10 −23 ×294 (C1)\nc2 =\n40×1.66×10 −27\n= 183 000\nr.m.s.speed = 430 m s–1 (A1)\n© UCLES 2021 Page 9 of 19\n\n3(a) Any 2 from: B2\n• particles / atoms / molecules / ions (very) close together / touching\n• regular, repeating pattern\n• vibrate about a fixed point\n\n3(b) (much) greater increase in spacing of molecules (for vaporisation compared with fusion) B1\n\n3(c)(i) –100 °C B1\n© UCLES 2021 Page 10 of 19\n\n3(c)(ii) time = 8.5 – 3.0 C1\n= 5.5 min\nPt = mL C1\nenergy = power × time = 150 × 5.5 × 60\n= 49 500 J\nE\nL =\nm\n49 500\n=\n0.045\n=1100 kJ kg −1 A1\n\n3(c)(iii) gas has a higher specific heat capacity (than liquid) B1\nQuestion Answer Marks", "source_pages": [ 9, 10, 11 ], "source_pdf": "_source-pdfs/2021-March/ms/9702_m21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-March/9702_m21_ms_42.pdf?download=true", "html": "9702-topic-14-temperature/answers.html", "image_paths": [ "../answer-assets/9702_m21_ms_42-p09.png", "../answer-assets/9702_m21_ms_42-p10.png", "../answer-assets/9702_m21_ms_42-p11.png" ] }, { "id": "9702-2021-m-42-q04", "question_id": "9702-2021-m-42-q04", "subject": "9702", "year": 2021, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 4, "topic": "Oscillations", "topic_slug": "9702-topic-17-oscillations", "marks": 9, "status": "available", "reason": null, "text": "4(a) acceleration and displacement are in opposite directions B1\n\n4(b)(i) F =kx M1\n( ) ( )\n=8.0× 0.060−0.048 or 8.0× 0.060+0.048\nor 8.0×0.012 or 8.0×0.108\n( ) ( )\nΣF = 8.0×0.012 − 8.0×0.108 =0.77 N A1\nor\nΣF =0.864−0.096=0.77 N\n© UCLES 2021 Page 11 of 19\n\n4(b)(ii) F A1\na=\nm\n0.77\n=\n0.25\n=3.1 m s −2\n\n4(b)(iii) a = – ω2x C1\n3.1\nω =\n0.048\nω=8.04\nT = 2 π / ω C1\nT = 2π / 8.04 A1\n= 0.78 s\n\n4(b)(iv) (resultant) force halved and distance halved B1\nsame T B1\nQuestion Answer Marks", "source_pages": [ 11, 12 ], "source_pdf": "_source-pdfs/2021-March/ms/9702_m21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-March/9702_m21_ms_42.pdf?download=true", "html": "9702-topic-17-oscillations/answers.html", "image_paths": [ "../answer-assets/9702_m21_ms_42-p11.png", "../answer-assets/9702_m21_ms_42-p12.png" ] }, { "id": "9702-2021-m-42-q05", "question_id": "9702-2021-m-42-q05", "subject": "9702", "year": 2021, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 5, "topic": "Alternating currents", "topic_slug": "9702-topic-21-alternating-currents", "marks": 7, "status": "available", "reason": null, "text": "5(a)(i) amplitude of the carrier wave varies M1\nin synchrony with the displacement of the (information) signal A1\n\n5(a)(ii) Any 2 from: B2\n• fewer transmitters needed / each transmitter can cover a greater distance\n• more stations can share waveband\n• transmitters and receivers are cheaper\n© UCLES 2021 Page 12 of 19\n\n5(b)(i) v A1\nλ=\nf\n3.0 ×108\n= =200 m\n1.5×106\n\n5(b)(ii) 10 kHz B1\n\n5(c) 1520 kHz B1\nQuestion Answer Marks", "source_pages": [ 12, 13 ], "source_pdf": "_source-pdfs/2021-March/ms/9702_m21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-March/9702_m21_ms_42.pdf?download=true", "html": "9702-topic-21-alternating-currents/answers.html", "image_paths": [ "../answer-assets/9702_m21_ms_42-p12.png", "../answer-assets/9702_m21_ms_42-p13.png" ] }, { "id": "9702-2021-m-42-q06", "question_id": "9702-2021-m-42-q06", "subject": "9702", "year": 2021, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 6, "topic": "Capacitance", "topic_slug": "9702-topic-19-capacitance", "marks": 7, "status": "available", "reason": null, "text": "6(a) (both have) radial field lines B1\n\n6(b)(i) 2.1 cm B1\n\n6(b)(ii) Q C1\nE =\n4πεr2\no\ne.g. r = 2.1 cm, E = 1.30 × 105 V m–1\nQ =4πεr2E\no\n=4×π×8.85×10 −12×0.0212×1.30×105\n=6.4×10 −9C A1\n© UCLES 2021 Page 13 of 19\n\n6(c) Q C1\nC =\nV\neither\nQ\nV = leading to C = 4πεr\n4πεr o\no\nC =4×π×8.85×10 −12×0.021 C1\n( C = ) 2.3×10 −12F A1\nor (C1)\nQ\nV =\n4πεr\no\n6.4×10 −9\n=\n4×π×8.85×10 −12×0.021\n=2740V\n6.4×10 −9\nC =\n2740\n=2.3×10 −12F (A1)\n© UCLES 2021 Page 14 of 19", "source_pages": [ 13, 14 ], "source_pdf": "_source-pdfs/2021-March/ms/9702_m21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-March/9702_m21_ms_42.pdf?download=true", "html": "9702-topic-19-capacitance/answers.html", "image_paths": [ "../answer-assets/9702_m21_ms_42-p13.png", "../answer-assets/9702_m21_ms_42-p14.png" ] }, { "id": "9702-2021-m-42-q07", "question_id": "9702-2021-m-42-q07", "subject": "9702", "year": 2021, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 7, "topic": "Alternating currents", "topic_slug": "9702-topic-21-alternating-currents", "marks": 9, "status": "available", "reason": null, "text": "7(a)(i) non-inverting (amplifier) B1\n\n7(a)(ii) R B1\ngain= f +1\nR\n3.6\ngain= +1=6.0\n0.72\n\n7(a)(iii) straight line from (0,0) to (T / 2, 3) B1\nline from origin to 3.0 V then horizontal line at 3.0 V to T B1\n\n7(a)(iv) ldr / light dependent resistor replaces one of the two resistors B1\n\n7(b)(i) relay coil B1\n\n7(b)(ii) relay coil between op-amp and earth B1\ndiode with correct polarity (pointing away from output) connected between output and device and no other connections B1\nor diode with correct polarity (pointing towards earth) between device and earth and no other connections\nswitch connected to high voltage circuit B1\nQuestion Answer Marks", "source_pages": [ 15 ], "source_pdf": "_source-pdfs/2021-March/ms/9702_m21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-March/9702_m21_ms_42.pdf?download=true", "html": "9702-topic-21-alternating-currents/answers.html", "image_paths": [ "../answer-assets/9702_m21_ms_42-p15.png" ] }, { "id": "9702-2021-m-42-q08", "question_id": "9702-2021-m-42-q08", "subject": "9702", "year": 2021, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 8, "topic": "Medical physics", "topic_slug": "9702-topic-24-medical-physics", "marks": 8, "status": "available", "reason": null, "text": "8(a)(i) at least one anticlockwise arrow and no clockwise arrows B1\n\n8(a)(ii) (force is to the) left B1\n\n8(a)(iii) force is the same B1\nNewton’s third law (of motion) B1\nor force depends on the product of the two currents\n© UCLES 2021 Page 15 of 19\n\n8(b)(i) frequency of radio waves is equal to natural frequency of protons B1\nresonance of protons occurs / protons absorb energy B1\n\n8(b)(ii) in between pulses / when pulse stops B1\nAny 1 from: B1\n• protons de-excite\n• protons emit r.f. pulses\n• emitted (r.f.) pulse (from proton) detected\nQuestion Answer Marks", "source_pages": [ 15, 16 ], "source_pdf": "_source-pdfs/2021-March/ms/9702_m21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-March/9702_m21_ms_42.pdf?download=true", "html": "9702-topic-24-medical-physics/answers.html", "image_paths": [ "../answer-assets/9702_m21_ms_42-p15.png", "../answer-assets/9702_m21_ms_42-p16.png" ] }, { "id": "9702-2021-m-42-q09", "question_id": "9702-2021-m-42-q09", "subject": "9702", "year": 2021, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 9, "topic": "Magnetic fields", "topic_slug": "9702-topic-20-magnetic-fields", "marks": 10, "status": "available", "reason": null, "text": "9(a) (magnetic) flux density × area × number of turns M1\narea is perpendicular to (magnetic) field A1\n\n9(b) use of t = 1.2 s C1\nΔBAN C1\nε=\nΔt\n0.250×π×0.0302×540\n=\n1.2\n=0.32V A1\n\n9(c)(i) light damping B1\n© UCLES 2021 Page 16 of 19\n\n9(c)(ii) sheet cuts (magnetic) flux and causes induced emf B1\n(induced) emf causes (eddy) currents (in sheet) B1\neither currents (in sheet) cause resistive force B1\nor currents (in sheet) dissipate energy\nsmaller currents in Y or larger currents in X, so dashed line is X B1\nQuestion Answer Marks", "source_pages": [ 16, 17 ], "source_pdf": "_source-pdfs/2021-March/ms/9702_m21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-March/9702_m21_ms_42.pdf?download=true", "html": "9702-topic-20-magnetic-fields/answers.html", "image_paths": [ "../answer-assets/9702_m21_ms_42-p16.png", "../answer-assets/9702_m21_ms_42-p17.png" ] }, { "id": "9702-2021-m-42-q10", "question_id": "9702-2021-m-42-q10", "subject": "9702", "year": 2021, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 10, "topic": "Alternating currents", "topic_slug": "9702-topic-21-alternating-currents", "marks": 8, "status": "available", "reason": null, "text": "10(a) 230 V A1\n\n10(b) ω = 100π C1\n2π 2π\nT = =\nω 100π\n=0.020 s A1\n\n10(c)(i) half-wave (rectification) B1\n\n10(c)(ii) sinusoidal half waves in positive V only or negative V only, peak at 320 V B1\nline at zero for second half of cycle B1\ntwo time periods shown, each of 0.020 s B1\n\n10(c)(iii) capacitor added in parallel with resistor B1\n© UCLES 2021 Page 17 of 19", "source_pages": [ 17 ], "source_pdf": "_source-pdfs/2021-March/ms/9702_m21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-March/9702_m21_ms_42.pdf?download=true", "html": "9702-topic-21-alternating-currents/answers.html", "image_paths": [ "../answer-assets/9702_m21_ms_42-p17.png" ] }, { "id": "9702-2021-m-42-q11", "question_id": "9702-2021-m-42-q11", "subject": "9702", "year": 2021, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 11, "topic": "Medical physics", "topic_slug": "9702-topic-24-medical-physics", "marks": 10, "status": "available", "reason": null, "text": "11(a)(i) electrons decelerate (on hitting target) so X-ray photons produced B1\nrange of decelerations B1\nphoton energy depends on (magnitude of) deceleration B1\n\n11(a)(ii) hc C1\neV =\nλ\n6.63×10 −34 ×3.0×108 C1\nλ=\n1.6×10 −19 ×15000\n=8.3×10 −11m A1\nor (C1)\nE = hf and c = fλ and electron energy = eV\nor\nE = hc / λ and electron energy = eV\nelectron energy = 1.6 × 10–19 × 15000\n= 2.4 × 10–15\n6.63×10 −34 ×3.0×108 (C1)\nλ=\n2.4×10 −15\nλ=8.3×10 −11m (A1)\n\n11(b)(i) μ = – gradient or ln (I / I ) = −μx C1\no\n(e.g. 2.08 / 10.0) = 0.21 cm–1 A1\n© UCLES 2021 Page 18 of 19\n\n11(b)(ii) ln 0.05 =−μx C1\nln0.05 A1\nx =\n−μ\ne.g. x =14 cm\nQuestion Answer Marks", "source_pages": [ 18, 19 ], "source_pdf": "_source-pdfs/2021-March/ms/9702_m21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-March/9702_m21_ms_42.pdf?download=true", "html": "9702-topic-24-medical-physics/answers.html", "image_paths": [ "../answer-assets/9702_m21_ms_42-p18.png", "../answer-assets/9702_m21_ms_42-p19.png" ] }, { "id": "9702-2021-m-42-q12", "question_id": "9702-2021-m-42-q12", "subject": "9702", "year": 2021, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 12, "topic": "Nuclear physics", "topic_slug": "9702-topic-23-nuclear-physics", "marks": 6, "status": "available", "reason": null, "text": "12(a) 1 not affected by external factors B1\n2 cannot predict when a (particular) nucleus will decay B1\nor cannot predict which nucleus will decay (next)\n\n12(b)(i) 1.0×10 −9 1.0×10 −9×6.02×1023 C1\nNumber of atoms = or\n90×1.66×10 −27 90×10 −3\n=6.693×1015\nA=λN C1\n5.2×106\nλ=\n6.693×1015\nλ=7.8×10 −10 s–1 A1\n\n12(b)(ii) daughter nucleus is unstable B1\n© UCLES 2021 Page 19 of 19", "source_pages": [ 19 ], "source_pdf": "_source-pdfs/2021-March/ms/9702_m21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-March/9702_m21_ms_42.pdf?download=true", "html": "9702-topic-23-nuclear-physics/answers.html", "image_paths": [ "../answer-assets/9702_m21_ms_42-p19.png" ] }, { "id": "9702-2021-m-52-q01", "question_id": "9702-2021-m-52-q01", "subject": "9702", "year": 2021, "session": "March", "session_code": "m", "paper": 5, "variant": "52", "question_number": 1, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "1 Defining the problem\nMass of cylinder m is the independent variable and period T is the dependent variable, or vary mass of cylinder m and 1\nmeasure period T.\nKeep radius of cylinder constant. 1\nMethods of data collection\nLabelled diagram of workable experiment including: 1\n• beaker with (cooking) oil on a bench or container supported by stand where stand is on a bench\n• cylinder partially submerged in (cooking) oil\n• cylinder and (cooking) oil labelled.\nMethod to determine mass m of cylinder, e.g. use a (top pan) balance. 1\nMethod to determine period or T, e.g. use a stopwatch / timer to time oscillations. 1\nMethod to determine diameter of cylinder, e.g. micrometer or calliper 1\nMethod of Analysis\nPlots a graph of T2 against m. 1\n(Allow other valid graphs, e.g. lg T against lg m)\nRelationship valid if a straight line passing through the origin is produced. 1\n(Allow gradient = 0.5 for log T against log m).\n4π 1\nK =\ngradient×σr2\n4π\n(K = for lg T against lg m).\n102×y-intercept ×σr2\n© UCLES 2021 Page 5 of 9\n\n1 Additional detail including safety considerations 6\nMax 6\nUse gloves to prevent oil contacting skin / slippery hands OR D1\nPerform experiment in a tray to prevent oil spillages.\nKeep density / temperature of the (cooking) oil constant or keep σ constant. D2\nMass of oil = mass of beaker and oil – mass of beaker and D3\nuse a measuring cylinder to determine the volume of the oil.\nDo not accept (calibrated) beaker.\nMethods to measure volume of oil and determine mass of oil and use equation density σ = mass / volume for D4\nmeasurements.\nTime n oscillations and divide nT by n D5\nwhere n ⩾ 5.\nDescription of method of counting oscillations with position of fiducial mark / mark on cylinder / beaker / fixed point shown in D6\ndiagram.\nRepeat experiment for each value of m and average T. D7\nr = diameter / 2 provided diameter measured. D8\nRepeat measurements of diameter in different directions and average. D9\nWait for oscillations to become even / steady. D10\n© UCLES 2021 Page 6 of 9", "source_pages": [ 5, 6 ], "source_pdf": "_source-pdfs/2021-March/ms/9702_m21_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-March/9702_m21_ms_52.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_m21_ms_52-p05.png", "../answer-assets/9702_m21_ms_52-p06.png" ] }, { "id": "9702-2021-m-52-q02", "question_id": "9702-2021-m-52-q02", "subject": "9702", "year": 2021, "session": "March", "session_code": "m", "paper": 5, "variant": "52", "question_number": 2, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "2(a) 1 1\nGradient =\n2uA\n1\n.\ny-intercept =\n2u\n\n2(b)\n0.046\n0.052\n0.062\n0.072\n0.080\n0.088\n1 1\nFirst mark for values of / s cm–1; allow 3sf.\nv\nSecond mark for absolute uncertainties from 1\n± 0.003 to ± 0.004.\n\n2(c)(i) Six points plotted correctly. 1\nMust be accurate to the nearest half small square. Diameter of points must be less than half a small square.\n1 1\nError bars in plotted correctly.\nv\nAll error bars to be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.\n© UCLES 2021 Page 7 of 9\n\n2(c)(ii) Line of best fit drawn. 1\nPoints must be balanced.\nDo not allow line from top plot to bottom plot.\nLine must pass between\n(320, 0.050) and (345, 0.050) and between (795, 0.085) and (815, 0.085).\nWorst acceptable line drawn. 1\nSteepest or shallowest possible line.\nMark scored only if all error bars are plotted.\n\n2(c)(iii) Gradient determined with clear substitution of data points into Δy / Δx; distance between data points must be at least half 1\nthe length of the drawn line.\nGradient of WAL determined and 1\nuncertainty = (gradient of line of best fit – gradient of worst acceptable line)\nor\nuncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)\n\n2(c)(iv) y-intercept determined by substitution of correct point into y = mx + c 1\ny-intercept of worst acceptable line determined by substitution into y = mx + c. 1\nuncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line, or\nuncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)\nDo not accept ecf from false origin method.\n© UCLES 2021 Page 8 of 9\n\n2(d)(i) u determined using y-intercept and 1\nu and A given to 2 or 3 sf.\n1\nu =\n2×y −intercept\nA determined using gradient with correct substitution and 1\nUnits with correct power of ten for u and A.\ny −intercept 1\nA= or A=\ngradient 2×u× gradient\n\n2(d)(ii) Percentage uncertainty in A. 1\nΔgradient Δy-intercept\n%uncert.=\n\n+\n\n×100\n gradient y-intercept \nOR\nΔu clearly determined and\nΔgradient Δu\n%uncert.=\n\n+\n\n×100\n gradient u \nOR\nCorrect substitution for max/min methods.\n\n2(e) Value of m determined from (d)(i) OR (c)(iii) and (c)(iv) with correct number substitution into relevant equation and correct 1\npower of ten.\n2uAt 2uA\ne.g. m= −A= −A, or\nL 10\n t 1 \nm= − ×2uA or\n \nL 2u\nt\n−y-intercept\nm= L .\ngradient\n© UCLES 2021 Page 9 of 9", "source_pages": [ 7, 8, 9 ], "source_pdf": "_source-pdfs/2021-March/ms/9702_m21_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-March/9702_m21_ms_52.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_m21_ms_52-p07.png", "../answer-assets/9702_m21_ms_52-p08.png", "../answer-assets/9702_m21_ms_52-p09.png" ] }, { "id": "9702-2021-mj-41-q01", "question_id": "9702-2021-mj-41-q01", "subject": "9702", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 1, "topic": "Gravitational fields", "topic_slug": "9702-topic-13-gravitational-fields", "marks": 10, "status": "available", "reason": null, "text": "1(a) force per unit mass B1\n\n1(b) GMm / r 2 = mrω 2 and ω = 2π/T C1\nor\nGMm / r 2 = mv2 / r and v = 2πr / T\n6.67 × 10–11 × 6.0 × 1024 = r3 × [2π / (94 × 60)]2 C1\nr = 6.9 × 106 m A1\n\n1(c)(i) r3ω2 = constant or r3 / T2 = constant C1\nr3 / (6.9 × 106)3 = (150 / 94)2 so r = 9.4 × 106 m A1\nor\nGMT2/4π2 = r3 and clear that M is 6.0 × 1024 (C1)\n6.67 × 10–11 × 6.0 × 1024 = r3 × [2π / (150 × 60)]2 (A1)\nso r = 9.4 × 106 m\n\n1(c)(ii) separation increases so (potential energy) increases B1\nor\nmovement is against gravitational force so (potential energy) increases\n\n1(c)(iii) potential energy = (–)GMm / r C1\nΔE = 6.67 × 10–11 × 6.0 × 1024 × 1200 × [(6.9 × 106)–1 – (9.4 × 106)–1] C1\nP\n= 1.9 × 1010 J A1\n© UCLES 2021 Page 8 of 18", "source_pages": [ 8 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9702_s21_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-May-June/9702_s21_ms_41.pdf?download=true", "html": "9702-topic-13-gravitational-fields/answers.html", "image_paths": [ "../answer-assets/9702_s21_ms_41-p08.png" ] }, { "id": "9702-2021-mj-41-q02", "question_id": "9702-2021-mj-41-q02", "subject": "9702", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 2, "topic": "Thermodynamics", "topic_slug": "9702-topic-16-thermodynamics", "marks": 10, "status": "available", "reason": null, "text": "2(a) pV = NkT C1\nN = (1.8 × 10–3 × 3.3 × 105) / (1.38 × 10–23 × 310) = 1.4 × 1023 A1\nor\npV = nRT and nN = N (C1)\nA\nN = (1.8 × 10–3 × 3.3 × 105 × 6.02 × 1023) / (8.31 × 310) = 1.4 × 1023 (A1)\n\n2(b) speed of molecule decreases on impact with moving piston B1\nmean square speed (directly) proportional to (thermodynamic) temperature B1\nor\nmean square speed (directly) proportional to kinetic energy (of molecules)\nor\nkinetic energy (of molecules) (directly) proportional to (thermodynamic) temperature\nkinetic energy (of molecules) decreases (so temperature decreases) B1\n\n2(c)(i) ΔU = 3/2 × k × ΔT × N C1\n= 3/2 × 1.38 × 10–23 × (288 – 310) × 1.4 × 1023 C1\n= – 64 J A1\n\n2(c)(ii) decrease in internal energy is less than work done by gas M1\n(thermal energy is) transferred to the gas (during the expansion) A1\n© UCLES 2021 Page 9 of 18", "source_pages": [ 9 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9702_s21_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-May-June/9702_s21_ms_41.pdf?download=true", "html": "9702-topic-16-thermodynamics/answers.html", "image_paths": [ "../answer-assets/9702_s21_ms_41-p09.png" ] }, { "id": "9702-2021-mj-41-q03", "question_id": "9702-2021-mj-41-q03", "subject": "9702", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 3, "topic": "Oscillations", "topic_slug": "9702-topic-17-oscillations", "marks": 8, "status": "available", "reason": null, "text": "3(a) acceleration (directly) proportional to displacement B1\nacceleration is in opposite direction to displacement B1\n\n3(b) ω2 = 2k / m and ω = 2πf C1\n(2πf)2 = (2 × 130) / 0.84 C1\nf = 2.8 Hz A1\n\n3(c)(i) resonance B1\n\n3(c)(ii) oscillator supplies energy (continuously) B1\nenergy of trolley constant so energy must be dissipated B1\nor\nwithout loss of energy the amplitude would continuously increase\nQuestion Answer Marks", "source_pages": [ 10 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9702_s21_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-May-June/9702_s21_ms_41.pdf?download=true", "html": "9702-topic-17-oscillations/answers.html", "image_paths": [ "../answer-assets/9702_s21_ms_41-p10.png" ] }, { "id": "9702-2021-mj-41-q04", "question_id": "9702-2021-mj-41-q04", "subject": "9702", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 4, "topic": "Medical physics", "topic_slug": "9702-topic-24-medical-physics", "marks": 5, "status": "available", "reason": null, "text": "4 (ultrasound) pulse B1\nreflected at boundaries B1\ngel is used to minimise reflection at skin B1\nor\ngenerated and detected by quartz crystal\ntime delay between generation and detection gives information about depth B1\nintensity (of reflected wave) gives information about nature of boundary B1\n© UCLES 2021 Page 10 of 18", "source_pages": [ 10 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9702_s21_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-May-June/9702_s21_ms_41.pdf?download=true", "html": "9702-topic-24-medical-physics/answers.html", "image_paths": [ "../answer-assets/9702_s21_ms_41-p10.png" ] }, { "id": "9702-2021-mj-41-q05", "question_id": "9702-2021-mj-41-q05", "subject": "9702", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 5, "topic": "Alternating currents", "topic_slug": "9702-topic-21-alternating-currents", "marks": 8, "status": "available", "reason": null, "text": "5(a) amplitude of the carrier wave varies M1\nin synchrony with the displacement of the (information) signal A1\n\n5(b)(i) wavelength = (3.0 × 108) / (300 × 103) A1\n= 1000 m\n\n5(b)(ii) bandwidth = 16 kHz A1\n\n5(b)(iii) frequency = 8 kHz A1\n\n5(c) attenuation = 10 lg (P / P ) C1\n1 2\n73 = 10 lg (P / P ) C1\nT R\n73 = 10 lg (P x2 / 0.082 P ) or x2 / 0.082 = 107.3\nT T\nx = 1300 m A1\n© UCLES 2021 Page 11 of 18", "source_pages": [ 11 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9702_s21_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-May-June/9702_s21_ms_41.pdf?download=true", "html": "9702-topic-21-alternating-currents/answers.html", "image_paths": [ "../answer-assets/9702_s21_ms_41-p11.png" ] }, { "id": "9702-2021-mj-41-q06", "question_id": "9702-2021-mj-41-q06", "subject": "9702", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 6, "topic": "Nuclear physics", "topic_slug": "9702-topic-23-nuclear-physics", "marks": 8, "status": "available", "reason": null, "text": "6(a) from x = 0 to x = r: E = 0 B1\nfrom x = r to x = 3r: curve with negative gradient of decreasing magnitude passing through (r, E ) B1\n0\nline passing through (2r, E / 4) and (3r, E / 9) B1\n0 0\n\n6(b) from p = p / 2 to p = p : curve with negative gradient of decreasing magnitude passing through (p , λ) B1\n0 0 0 0\nline passing through (½p , 2λ) B1\n0 0\n\n6(c) from t = 0 to t = 45 s: curve with positive gradient of decreasing magnitude starting at (0, 0) B1\nline passing through (15, ½N ) B1\n0\nline passing through (30, 0.75N ) and (45, 0.88N ) B1\n0 0\n© UCLES 2021 Page 12 of 18", "source_pages": [ 12 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9702_s21_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-May-June/9702_s21_ms_41.pdf?download=true", "html": "9702-topic-23-nuclear-physics/answers.html", "image_paths": [ "../answer-assets/9702_s21_ms_41-p12.png" ] }, { "id": "9702-2021-mj-41-q07", "question_id": "9702-2021-mj-41-q07", "subject": "9702", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 7, "topic": "Capacitance", "topic_slug": "9702-topic-19-capacitance", "marks": 9, "status": "available", "reason": null, "text": "7(a) charge / potential M1\ncharge is on one plate, potential is p.d. between the plates A1\n\n7(b)(i) I = Q / t M1\ncharge = CV and time = 1 / f leading to I = fCV A1\n\n7(b)(ii) 4.8 × 10–6 = 150 × 60 × C C1\nC = 530 pF A1\n\n7(c) (total) capacitance is halved B1\ncharge (for each cycle/discharge) is halved B1\nor\nsince f and V are constant, current is proportional to capacitance\ncurrent = 2.4 μA B1\n© UCLES 2021 Page 13 of 18", "source_pages": [ 13 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9702_s21_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-May-June/9702_s21_ms_41.pdf?download=true", "html": "9702-topic-19-capacitance/answers.html", "image_paths": [ "../answer-assets/9702_s21_ms_41-p13.png" ] }, { "id": "9702-2021-mj-41-q08", "question_id": "9702-2021-mj-41-q08", "subject": "9702", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 8, "topic": "Temperature", "topic_slug": "9702-topic-14-temperature", "marks": 9, "status": "available", "reason": null, "text": "8(a) V+ = 3.0 × 3.0 / (2.5 + 3.0) C1\n= 1.6 V A1\n\n8(b) V – is +2.0 V B1\nor\nV – > V +\noutput is negative so (LED) does not emit light B1\n\n8(c) at 0 °C, V – = 1.7 V B1\nor\nfor all temperatures above 0 °C, resistance of thermistor < 4.2 kΩ\nV – always greater than V + (so no switching) B1\n\n8(d) (at 20 °C,) R = 1.8 kΩ C1\nT\n2.5 / 3.0 = 1.8 / R C1\nor\n[R / (R + 1.8)] × 3.0 = 1.6\nR = 2.2 kΩ A1\n© UCLES 2021 Page 14 of 18", "source_pages": [ 14 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9702_s21_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-May-June/9702_s21_ms_41.pdf?download=true", "html": "9702-topic-14-temperature/answers.html", "image_paths": [ "../answer-assets/9702_s21_ms_41-p14.png" ] }, { "id": "9702-2021-mj-41-q09", "question_id": "9702-2021-mj-41-q09", "subject": "9702", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 9, "topic": "Magnetic fields", "topic_slug": "9702-topic-20-magnetic-fields", "marks": 9, "status": "available", "reason": null, "text": "9(a) region where there is a force exerted on M1\na current-carrying conductor A1\nor\na moving charge\nor\na magnetic material/magnetic pole\n\n9(b)(i) face PSWV shaded B1\n\n9(b)(ii) accumulating electrons cause an electric field (between the faces) B1\nforce due to electric field opposes force due to magnetic field B1\naccumulation stops when magnetic force equals electric force B1\n\n9(c)(i) number density of charge carriers B1\n\n9(c)(ii) PV or QT or SW B1\n\n9(d) (for semiconductor,) n is (much) smaller so V (much) larger B1\nH\n© UCLES 2021 Page 15 of 18", "source_pages": [ 15 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9702_s21_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-May-June/9702_s21_ms_41.pdf?download=true", "html": "9702-topic-20-magnetic-fields/answers.html", "image_paths": [ "../answer-assets/9702_s21_ms_41-p15.png" ] }, { "id": "9702-2021-mj-41-q10", "question_id": "9702-2021-mj-41-q10", "subject": "9702", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 10, "topic": "Magnetic fields", "topic_slug": "9702-topic-20-magnetic-fields", "marks": 9, "status": "available", "reason": null, "text": "10(a) direction of (induced) e.m.f. M1\nis such as to oppose the change causing it A1\n\n10(b) ring cuts (magnetic) flux and causes induced e.m.f. in ring B1\n(induced) e.m.f. causes (eddy/induced) currents (in ring) B1\ncurrents (in ring) cause magnetic field (around ring) M1\ntwo fields interact to cause resistive/opposing force A1\nor\ncurrent (in ring) is in a magnetic field (M1)\nwhich causes resistive force (A1)\nor\ncurrents (in ring) dissipate thermal energy (M1)\n(thermal) energy comes from energy of oscillations (A1)\n\n10(c) current cannot pass all the way around the ring B1\n(induced) currents smaller B1\nsmaller resistive force (so more oscillations) B1\nor\nsmaller rate of dissipation of energy (so more oscillations)\n© UCLES 2021 Page 16 of 18", "source_pages": [ 16 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9702_s21_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-May-June/9702_s21_ms_41.pdf?download=true", "html": "9702-topic-20-magnetic-fields/answers.html", "image_paths": [ "../answer-assets/9702_s21_ms_41-p16.png" ] }, { "id": "9702-2021-mj-41-q11", "question_id": "9702-2021-mj-41-q11", "subject": "9702", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 11, "topic": "Medical physics", "topic_slug": "9702-topic-24-medical-physics", "marks": 7, "status": "available", "reason": null, "text": "11(a) intensity: vary filament current/p.d. across filament B1\nhardness: vary accelerating potential difference B1\n\n11(b)(i) I = I e –μx C1\n0\nI = I exp(–0.92 × 9.0) A1\nS 0\n= 2.5 × 10–4 I\n0\n\n11(b)(ii) I = [exp(–0.92 × 6.0) × exp(–2.9 × 3.0)] I C1\nC 0\n= 6.7 × 10–7 I A1\n0\n\n11(c) conclusion consistent with values in (b)(i) and (b)(ii) B1\ne.g. I ≫ I so good contrast\nS C\n© UCLES 2021 Page 17 of 18", "source_pages": [ 17 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9702_s21_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-May-June/9702_s21_ms_41.pdf?download=true", "html": "9702-topic-24-medical-physics/answers.html", "image_paths": [ "../answer-assets/9702_s21_ms_41-p17.png" ] }, { "id": "9702-2021-mj-41-q12", "question_id": "9702-2021-mj-41-q12", "subject": "9702", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 12, "topic": "Quantum physics", "topic_slug": "9702-topic-22-quantum-physics", "marks": 8, "status": "available", "reason": null, "text": "12(a) • frequency determines energy of photon B2\n• intensity determines number of photons (per unit time)\n• intensity does not determine energy of a photon\nAny two points, 1 mark each\nkinetic energy (of the electron) depends on the energy of one photon B1\n\n12(b)(i) E = hc / λ C1\nor\nE = hf and c = fλ\nE = (6.63 × 10–34 × 3.00 × 108) / (250 × 10–9) C1\n(= 7.96 × 10–19 J) A1\n= 5.0 eV\n\n12(b)(ii) E = photon energy – work function C1\nMAX\nwork function = 5.0 – 1.4 A1\n= 3.6 eV\n© UCLES 2021 Page 18 of 18", "source_pages": [ 18 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9702_s21_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-May-June/9702_s21_ms_41.pdf?download=true", "html": "9702-topic-22-quantum-physics/answers.html", "image_paths": [ "../answer-assets/9702_s21_ms_41-p18.png" ] }, { "id": "9702-2021-mj-42-q01", "question_id": "9702-2021-mj-42-q01", "subject": "9702", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 1, "topic": "Gravitational fields", "topic_slug": "9702-topic-13-gravitational-fields", "marks": 6, "status": "available", "reason": null, "text": "1(a) (gravitational) force per unit mass B1\n\n1(b)(i) g = GM / r2 C1\n= (6.67 × 10–11 × 6.42 × 1023) / (3.39 × 106)2 A1\n= 3.73 N kg–1\n\n1(b)(ii) a = rω2 and ω = 2π / T C1\nor\na = v2 / r and v = 2πr / T\na = 3.39 × 106 × (2π / (24.6 × 3600))2 A1\n= 0.0171 m s–2\n\n1(b)(iii) force per unit mass = 3.73 – 0.0171 A1\n= 3.71 N kg–1\n© UCLES 2021 Page 8 of 19", "source_pages": [ 8 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9702_s21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-May-June/9702_s21_ms_42.pdf?download=true", "html": "9702-topic-13-gravitational-fields/answers.html", "image_paths": [ "../answer-assets/9702_s21_ms_42-p08.png" ] }, { "id": "9702-2021-mj-42-q02", "question_id": "9702-2021-mj-42-q02", "subject": "9702", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 2, "topic": "Thermodynamics", "topic_slug": "9702-topic-16-thermodynamics", "marks": 9, "status": "available", "reason": null, "text": "2(a) pV = nRT C1\npV = nRT and N = nN C1\nA\nor\npV = NkT\n3.1 × 10–3 × 8.5 × 105 = (N × 290 × 8.31) / (6.02 × 1023) A1\nso N = 6.6 × 1023\nor\n3.1 × 10–3 × 8.5 × 105 = N × 1.38 × 10–23 × 290\nso N = 6.6 × 1023\n\n2(b)(i) (3.1 × 10–3 × 8.5 × 105) / 290 = (6.3 × 10–3 × 2.7 × 105) / T A1\nso T = 190 K\nor\n6.3 × 10–3 × 2.7 × 105 = 6.6 × 1023 × 1.38 × 10–23 × T\nso T = 190 K\n\n2(b)(ii) ΔU = 3/2 × k × ΔT × N C1\n= 3/2 × 1.38 × 10–23 × (190 – 290) × 6.6 × 1023 C1\n= –1400 J A1\n\n2(c) ΔU = q + w M1\nq = 0 so ΔU = w A1\n© UCLES 2021 Page 9 of 19", "source_pages": [ 9 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9702_s21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-May-June/9702_s21_ms_42.pdf?download=true", "html": "9702-topic-16-thermodynamics/answers.html", "image_paths": [ "../answer-assets/9702_s21_ms_42-p09.png" ] }, { "id": "9702-2021-mj-42-q03", "question_id": "9702-2021-mj-42-q03", "subject": "9702", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 3, "topic": "Oscillations", "topic_slug": "9702-topic-17-oscillations", "marks": 9, "status": "available", "reason": null, "text": "3(a) acceleration in opposite direction to displacement shown by – sign B1\ng / L is constant M1\n(so) acceleration is (directly) proportional to displacement A1\n\n3(b) ω2 = g / L C1\nω = 2π / T C1\nor\nω = 2πf and f = 1 / T\n(2π / T)2 = 9.81 / 0.18 A1\nT = 0.85 s\n\n3(c) energy ∝ x 2 C1\n0\n(after 3 cycles,) amplitude = (0.94)3x C1\n0\n= 0.83x\n0\nratio final energy / initial energy = 0.832 A1\n= 0.69\n© UCLES 2021 Page 10 of 19", "source_pages": [ 10 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9702_s21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-May-June/9702_s21_ms_42.pdf?download=true", "html": "9702-topic-17-oscillations/answers.html", "image_paths": [ "../answer-assets/9702_s21_ms_42-p10.png" ] }, { "id": "9702-2021-mj-42-q04", "question_id": "9702-2021-mj-42-q04", "subject": "9702", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 4, "topic": "Alternating currents", "topic_slug": "9702-topic-21-alternating-currents", "marks": 8, "status": "available", "reason": null, "text": "4(a)(i) frequency (modulation) B1\n\n4(a)(ii) 1. zero B1\n2. frequency (of 1.2 MHz) varies by ±50 kHz B1\nfrequency varies (by ±50 kHz) at a rate of 8000 times per second B1\n\n4(b)(i) wavelength = (3.00 × 108) / (240 × 103) C1\n(= 1250 m) A1\n= 1.25 km\n\n4(b)(ii) bandwidth = 30 kHz A1\n\n4(b)(iii) frequency = 15 kHz A1\n© UCLES 2021 Page 11 of 19", "source_pages": [ 11 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9702_s21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-May-June/9702_s21_ms_42.pdf?download=true", "html": "9702-topic-21-alternating-currents/answers.html", "image_paths": [ "../answer-assets/9702_s21_ms_42-p11.png" ] }, { "id": "9702-2021-mj-42-q05", "question_id": "9702-2021-mj-42-q05", "subject": "9702", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 5, "topic": "Nuclear physics", "topic_slug": "9702-topic-23-nuclear-physics", "marks": 9, "status": "available", "reason": null, "text": "5(a) from x = 0 to x = r: horizontal line at V = 1.0V B1\n0\nfrom x = r to x = 3r: curve with negative gradient of decreasing magnitude starting at (r, 1.0V ) B1\n0\nline passing through (2r, ½V ) and (3r, ⅓V ) B1\n0 0\n\n5(b) line with negative gradient from λ = ⅓λ to λ = λ B1\n0 0\nline passing through (λ, 0) B1\n0\ncurve with negative gradient of decreasing magnitude passing through (½λ, E )and (⅓λ, 2E ) B1\n0 MAX 0 MAX\n\n5(c) 1.0T shown at ½N and 2.0T shown at ¼N B1\n½ 0 ½ 0\nline starting at (0, 0) and reaching (T, N –N) B1\n0\nline starting at (0, 0) and reaching original curve at (1.0T , ½N ) B1\n½ 0\n© UCLES 2021 Page 12 of 19", "source_pages": [ 12 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9702_s21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-May-June/9702_s21_ms_42.pdf?download=true", "html": "9702-topic-23-nuclear-physics/answers.html", "image_paths": [ "../answer-assets/9702_s21_ms_42-p12.png" ] }, { "id": "9702-2021-mj-42-q06", "question_id": "9702-2021-mj-42-q06", "subject": "9702", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 6, "topic": "Capacitance", "topic_slug": "9702-topic-19-capacitance", "marks": 8, "status": "available", "reason": null, "text": "6(a) potential difference applied between the plates M1\ncauses charge separation (between the plates) A1\nor\ncauses energy to be stored (between the plates)\n\n6(b)(i) I = Q / t M1\nclear substitution of Q = CV and f = 1 / t, leading to I = fCV A1\n\n6(b)(ii) 2.5 × 10–6 = 50 × C × 180 C1\nC = 280 pF A1\n\n6(c) (total) capacitance increases B1\ngreater charge (for each cycle/discharge) so greater (average) current B1\nor\nV and f are constant so (average) current increases\nor\nI is (directly) proportional to C so (average) current increases\n© UCLES 2021 Page 13 of 19", "source_pages": [ 13 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9702_s21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-May-June/9702_s21_ms_42.pdf?download=true", "html": "9702-topic-19-capacitance/answers.html", "image_paths": [ "../answer-assets/9702_s21_ms_42-p13.png" ] }, { "id": "9702-2021-mj-42-q07", "question_id": "9702-2021-mj-42-q07", "subject": "9702", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 7, "topic": "Temperature", "topic_slug": "9702-topic-14-temperature", "marks": 9, "status": "available", "reason": null, "text": "7(a)(i) no current enters/leaves the input B1\n\n7(a)(ii) gain is the same for all frequencies B1\n\n7(b)(i) V = 1.5 × 400 / (400 + 1100) = 0.40 V A1\nIN\nor\nV = 1.5 – (1.5 × 1100 / 1500) = 0.40 V\nIN\nor\n(1.5 – V ) / 1100 = V / 400 so V = 0.40 V\nIN IN IN\n\n7(b)(ii) gain = (–) R /R C1\nf i\nV /0.40 = (360 + 100) / 96 C1\nOUT\nV = 1.9 V A1\nOUT\n\n7(b)(iii) resistance of thermistor decreases B1\n(magnitude of) gain decreases so reading decreases B1\n\n7(b)(iv) (at gain 12.5) V is 5.0 V, so (above gain 12.5) output becomes saturated B1\nOUT\n© UCLES 2021 Page 14 of 19", "source_pages": [ 14 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9702_s21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-May-June/9702_s21_ms_42.pdf?download=true", "html": "9702-topic-14-temperature/answers.html", "image_paths": [ "../answer-assets/9702_s21_ms_42-p14.png" ] }, { "id": "9702-2021-mj-42-q08", "question_id": "9702-2021-mj-42-q08", "subject": "9702", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 8, "topic": "Magnetic fields", "topic_slug": "9702-topic-20-magnetic-fields", "marks": 10, "status": "available", "reason": null, "text": "8(a) • force per unit length B2\n• force per unit current\n• length/current perpendicular to field\n1 mark for any two points, 2 marks for all three points\n\n8(b) change in potential energy = change in kinetic energy B1\nor\nqV = ½mv2\nv = √(2qV / m) A1\n\n8(c)(i) magnetic force = centripetal force M1\nor\nBqv = mv2 / r\nclear substitution of expression for v and correct algebra leading to q / m = 2V / B2r2 A1\n\n8(c)(ii) q / m = (2 × 230) / [(0.38 × 10–3)2 × 0.142] C1\n= 1.6 × 1011 C kg–1 A1\n\n8(c)(iii) (for α-particle,) q / m is (much) smaller B1\nr would be much larger B1\n© UCLES 2021 Page 15 of 19", "source_pages": [ 15 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9702_s21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-May-June/9702_s21_ms_42.pdf?download=true", "html": "9702-topic-20-magnetic-fields/answers.html", "image_paths": [ "../answer-assets/9702_s21_ms_42-p15.png" ] }, { "id": "9702-2021-mj-42-q09", "question_id": "9702-2021-mj-42-q09", "subject": "9702", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 9, "topic": "Magnetic fields", "topic_slug": "9702-topic-20-magnetic-fields", "marks": 9, "status": "available", "reason": null, "text": "9(a) (particle is) stationary/not moving B1\n(particle is) moving parallel to the (magnetic) field B1\n\n9(b) magnetic field around each coil is circular B1\nor\neach coil is normal to magnetic field due to adjacent coils\ncurrent in coil interacts with (magnetic) field to exert force (on coil) B1\nforce is normal to both coil and magnetic field B1\nor\nforce parallel to axis (of coil)\nforces between coils are attractive so spring contracts B1\n\n9(c) (oscillating) coils cut magnetic flux B1\nor\nas separation of coils changes, magnetic flux changes\ncutting flux causes induced e.m.f. in coils B1\nchanging (induced) e.m.f. causes changing current (in coil) B1\n© UCLES 2021 Page 16 of 19", "source_pages": [ 16 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9702_s21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-May-June/9702_s21_ms_42.pdf?download=true", "html": "9702-topic-20-magnetic-fields/answers.html", "image_paths": [ "../answer-assets/9702_s21_ms_42-p16.png" ] }, { "id": "9702-2021-mj-42-q10", "question_id": "9702-2021-mj-42-q10", "subject": "9702", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 10, "topic": "Alternating currents", "topic_slug": "9702-topic-21-alternating-currents", "marks": 8, "status": "available", "reason": null, "text": "10(a) the steady current M1\nor\nthe direct current\nthat produces the same heating effect (as the alternating current) A1\n\n10(b)(i) peak current = 2.6 A and r.m.s. current = 1.8 A A1\n\n10(b)(ii) peak current = 2.0 A and r.m.s. current = 2.0 A A1\n\n10(c)(i) k = 2πf C1\n= 2π × 50 A1\n= 310 rad s–1\n\n10(c)(ii) power = V 2 / R or power = V 2 / 2R C1\nRMS 0\nR = (240 / √2)2 / 3200 or R = 2402 / (2 × 3200) A1\nR = 9.0 Ω\n© UCLES 2021 Page 17 of 19", "source_pages": [ 17 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9702_s21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-May-June/9702_s21_ms_42.pdf?download=true", "html": "9702-topic-21-alternating-currents/answers.html", "image_paths": [ "../answer-assets/9702_s21_ms_42-p17.png" ] }, { "id": "9702-2021-mj-42-q11", "question_id": "9702-2021-mj-42-q11", "subject": "9702", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 11, "topic": "Medical physics", "topic_slug": "9702-topic-24-medical-physics", "marks": 6, "status": "available", "reason": null, "text": "11(a) to produce a 3-dimensional image of structure/body B1\n\n11(b) X-rays (are used) B1\nscanning in sections B1\nscanning from many angles B1\nimage of each section is 2-dimensional B1\nscanning repeated for many sections B1\nor\nimages of many sections combined together\nQuestion Answer Marks", "source_pages": [ 18 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9702_s21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-May-June/9702_s21_ms_42.pdf?download=true", "html": "9702-topic-24-medical-physics/answers.html", "image_paths": [ "../answer-assets/9702_s21_ms_42-p18.png" ] }, { "id": "9702-2021-mj-42-q12", "question_id": "9702-2021-mj-42-q12", "subject": "9702", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 12, "topic": "Nuclear physics", "topic_slug": "9702-topic-23-nuclear-physics", "marks": 9, "status": "available", "reason": null, "text": "12(a) quantum of energy M1\nof electromagnetic radiation A1\n\n12(b)(i) energy = hc / λ C1\nor\nenergy = hf and f = c / λ\n0.57 × 106 × 1.60 × 10–19 = (6.63 × 10–34 × 3.00 × 108) / λ A1\nλ = 2.2 × 10–12 m\n© UCLES 2021 Page 18 of 19\n\n12(b)(ii) p = h / λ C1\n= (6.63 × 10–34) / (2.2 × 10–12) A1\n= 3.0 × 10–22 N s\nor\np = E / c (C1)\n= (0.57 × 106 × 1.60 × 10–19) / (3.00 × 108) (A1)\n= 3.0 × 10–22 N s\n\n12(c)(i) mass (of Sm-157 nucleus) = 157 × 1.66 × 10–27 C1\nor\nmass (of Sm-157 nucleus) = 0.157 / (6.02 × 1023)\nrecoil speed = (3.00 × 10–22) / (157 × 1.66 × 10–27) A1\n= 1.2 × 103 m s–1\n\n12(c)(ii) (1.2 ×) 103 m s–1 is much less than (3.0 ×) 108 m s–1 B1\n© UCLES 2021 Page 19 of 19", "source_pages": [ 18, 19 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9702_s21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-May-June/9702_s21_ms_42.pdf?download=true", "html": "9702-topic-23-nuclear-physics/answers.html", "image_paths": [ "../answer-assets/9702_s21_ms_42-p18.png", "../answer-assets/9702_s21_ms_42-p19.png" ] }, { "id": "9702-2021-mj-43-q01", "question_id": "9702-2021-mj-43-q01", "subject": "9702", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 1, "topic": "Gravitational fields", "topic_slug": "9702-topic-13-gravitational-fields", "marks": 10, "status": "available", "reason": null, "text": "1(a) force per unit mass B1\n\n1(b) GMm / r 2 = mrω 2 and ω = 2π/T C1\nor\nGMm / r 2 = mv2 / r and v = 2πr / T\n6.67 × 10–11 × 6.0 × 1024 = r3 × [2π / (94 × 60)]2 C1\nr = 6.9 × 106 m A1\n\n1(c)(i) r3ω2 = constant or r3 / T2 = constant C1\nr3 / (6.9 × 106)3 = (150 / 94)2 so r = 9.4 × 106 m A1\nor\nGMT2/4π2 = r3 and clear that M is 6.0 × 1024 (C1)\n6.67 × 10–11 × 6.0 × 1024 = r3 × [2π / (150 × 60)]2 (A1)\nso r = 9.4 × 106 m\n\n1(c)(ii) separation increases so (potential energy) increases B1\nor\nmovement is against gravitational force so (potential energy) increases\n\n1(c)(iii) potential energy = (–)GMm / r C1\nΔE = 6.67 × 10–11 × 6.0 × 1024 × 1200 × [(6.9 × 106)–1 – (9.4 × 106)–1] C1\nP\n= 1.9 × 1010 J A1\n© UCLES 2021 Page 8 of 18", "source_pages": [ 8 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9702_s21_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-May-June/9702_s21_ms_43.pdf?download=true", "html": "9702-topic-13-gravitational-fields/answers.html", "image_paths": [ "../answer-assets/9702_s21_ms_43-p08.png" ] }, { "id": "9702-2021-mj-43-q02", "question_id": "9702-2021-mj-43-q02", "subject": "9702", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 2, "topic": "Thermodynamics", "topic_slug": "9702-topic-16-thermodynamics", "marks": 10, "status": "available", "reason": null, "text": "2(a) pV = NkT C1\nN = (1.8 × 10–3 × 3.3 × 105) / (1.38 × 10–23 × 310) = 1.4 × 1023 A1\nor\npV = nRT and nN = N (C1)\nA\nN = (1.8 × 10–3 × 3.3 × 105 × 6.02 × 1023) / (8.31 × 310) = 1.4 × 1023 (A1)\n\n2(b) speed of molecule decreases on impact with moving piston B1\nmean square speed (directly) proportional to (thermodynamic) temperature B1\nor\nmean square speed (directly) proportional to kinetic energy (of molecules)\nor\nkinetic energy (of molecules) (directly) proportional to (thermodynamic) temperature\nkinetic energy (of molecules) decreases (so temperature decreases) B1\n\n2(c)(i) ΔU = 3/2 × k × ΔT × N C1\n= 3/2 × 1.38 × 10–23 × (288 – 310) × 1.4 × 1023 C1\n= – 64 J A1\n\n2(c)(ii) decrease in internal energy is less than work done by gas M1\n(thermal energy is) transferred to the gas (during the expansion) A1\n© UCLES 2021 Page 9 of 18", "source_pages": [ 9 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9702_s21_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-May-June/9702_s21_ms_43.pdf?download=true", "html": "9702-topic-16-thermodynamics/answers.html", "image_paths": [ "../answer-assets/9702_s21_ms_43-p09.png" ] }, { "id": "9702-2021-mj-43-q03", "question_id": "9702-2021-mj-43-q03", "subject": "9702", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 3, "topic": "Oscillations", "topic_slug": "9702-topic-17-oscillations", "marks": 8, "status": "available", "reason": null, "text": "3(a) acceleration (directly) proportional to displacement B1\nacceleration is in opposite direction to displacement B1\n\n3(b) ω2 = 2k / m and ω = 2πf C1\n(2πf)2 = (2 × 130) / 0.84 C1\nf = 2.8 Hz A1\n\n3(c)(i) resonance B1\n\n3(c)(ii) oscillator supplies energy (continuously) B1\nenergy of trolley constant so energy must be dissipated B1\nor\nwithout loss of energy the amplitude would continuously increase\nQuestion Answer Marks", "source_pages": [ 10 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9702_s21_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-May-June/9702_s21_ms_43.pdf?download=true", "html": "9702-topic-17-oscillations/answers.html", "image_paths": [ "../answer-assets/9702_s21_ms_43-p10.png" ] }, { "id": "9702-2021-mj-43-q04", "question_id": "9702-2021-mj-43-q04", "subject": "9702", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 4, "topic": "Medical physics", "topic_slug": "9702-topic-24-medical-physics", "marks": 5, "status": "available", "reason": null, "text": "4 (ultrasound) pulse B1\nreflected at boundaries B1\ngel is used to minimise reflection at skin B1\nor\ngenerated and detected by quartz crystal\ntime delay between generation and detection gives information about depth B1\nintensity (of reflected wave) gives information about nature of boundary B1\n© UCLES 2021 Page 10 of 18", "source_pages": [ 10 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9702_s21_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-May-June/9702_s21_ms_43.pdf?download=true", "html": "9702-topic-24-medical-physics/answers.html", "image_paths": [ "../answer-assets/9702_s21_ms_43-p10.png" ] }, { "id": "9702-2021-mj-43-q05", "question_id": "9702-2021-mj-43-q05", "subject": "9702", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 5, "topic": "Alternating currents", "topic_slug": "9702-topic-21-alternating-currents", "marks": 8, "status": "available", "reason": null, "text": "5(a) amplitude of the carrier wave varies M1\nin synchrony with the displacement of the (information) signal A1\n\n5(b)(i) wavelength = (3.0 × 108) / (300 × 103) A1\n= 1000 m\n\n5(b)(ii) bandwidth = 16 kHz A1\n\n5(b)(iii) frequency = 8 kHz A1\n\n5(c) attenuation = 10 lg (P / P ) C1\n1 2\n73 = 10 lg (P / P ) C1\nT R\n73 = 10 lg (P x2 / 0.082 P ) or x2 / 0.082 = 107.3\nT T\nx = 1300 m A1\n© UCLES 2021 Page 11 of 18", "source_pages": [ 11 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9702_s21_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-May-June/9702_s21_ms_43.pdf?download=true", "html": "9702-topic-21-alternating-currents/answers.html", "image_paths": [ "../answer-assets/9702_s21_ms_43-p11.png" ] }, { "id": "9702-2021-mj-43-q06", "question_id": "9702-2021-mj-43-q06", "subject": "9702", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 6, "topic": "Nuclear physics", "topic_slug": "9702-topic-23-nuclear-physics", "marks": 8, "status": "available", "reason": null, "text": "6(a) from x = 0 to x = r: E = 0 B1\nfrom x = r to x = 3r: curve with negative gradient of decreasing magnitude passing through (r, E ) B1\n0\nline passing through (2r, E / 4) and (3r, E / 9) B1\n0 0\n\n6(b) from p = p / 2 to p = p : curve with negative gradient of decreasing magnitude passing through (p , λ) B1\n0 0 0 0\nline passing through (½p , 2λ) B1\n0 0\n\n6(c) from t = 0 to t = 45 s: curve with positive gradient of decreasing magnitude starting at (0, 0) B1\nline passing through (15, ½N ) B1\n0\nline passing through (30, 0.75N ) and (45, 0.88N ) B1\n0 0\n© UCLES 2021 Page 12 of 18", "source_pages": [ 12 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9702_s21_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-May-June/9702_s21_ms_43.pdf?download=true", "html": "9702-topic-23-nuclear-physics/answers.html", "image_paths": [ "../answer-assets/9702_s21_ms_43-p12.png" ] }, { "id": "9702-2021-mj-43-q07", "question_id": "9702-2021-mj-43-q07", "subject": "9702", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 7, "topic": "Capacitance", "topic_slug": "9702-topic-19-capacitance", "marks": 9, "status": "available", "reason": null, "text": "7(a) charge / potential M1\ncharge is on one plate, potential is p.d. between the plates A1\n\n7(b)(i) I = Q / t M1\ncharge = CV and time = 1 / f leading to I = fCV A1\n\n7(b)(ii) 4.8 × 10–6 = 150 × 60 × C C1\nC = 530 pF A1\n\n7(c) (total) capacitance is halved B1\ncharge (for each cycle/discharge) is halved B1\nor\nsince f and V are constant, current is proportional to capacitance\ncurrent = 2.4 μA B1\n© UCLES 2021 Page 13 of 18", "source_pages": [ 13 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9702_s21_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-May-June/9702_s21_ms_43.pdf?download=true", "html": "9702-topic-19-capacitance/answers.html", "image_paths": [ "../answer-assets/9702_s21_ms_43-p13.png" ] }, { "id": "9702-2021-mj-43-q08", "question_id": "9702-2021-mj-43-q08", "subject": "9702", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 8, "topic": "Temperature", "topic_slug": "9702-topic-14-temperature", "marks": 9, "status": "available", "reason": null, "text": "8(a) V+ = 3.0 × 3.0 / (2.5 + 3.0) C1\n= 1.6 V A1\n\n8(b) V – is +2.0 V B1\nor\nV – > V +\noutput is negative so (LED) does not emit light B1\n\n8(c) at 0 °C, V – = 1.7 V B1\nor\nfor all temperatures above 0 °C, resistance of thermistor < 4.2 kΩ\nV – always greater than V + (so no switching) B1\n\n8(d) (at 20 °C,) R = 1.8 kΩ C1\nT\n2.5 / 3.0 = 1.8 / R C1\nor\n[R / (R + 1.8)] × 3.0 = 1.6\nR = 2.2 kΩ A1\n© UCLES 2021 Page 14 of 18", "source_pages": [ 14 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9702_s21_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-May-June/9702_s21_ms_43.pdf?download=true", "html": "9702-topic-14-temperature/answers.html", "image_paths": [ "../answer-assets/9702_s21_ms_43-p14.png" ] }, { "id": "9702-2021-mj-43-q09", "question_id": "9702-2021-mj-43-q09", "subject": "9702", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 9, "topic": "Magnetic fields", "topic_slug": "9702-topic-20-magnetic-fields", "marks": 9, "status": "available", "reason": null, "text": "9(a) region where there is a force exerted on M1\na current-carrying conductor A1\nor\na moving charge\nor\na magnetic material/magnetic pole\n\n9(b)(i) face PSWV shaded B1\n\n9(b)(ii) accumulating electrons cause an electric field (between the faces) B1\nforce due to electric field opposes force due to magnetic field B1\naccumulation stops when magnetic force equals electric force B1\n\n9(c)(i) number density of charge carriers B1\n\n9(c)(ii) PV or QT or SW B1\n\n9(d) (for semiconductor,) n is (much) smaller so V (much) larger B1\nH\n© UCLES 2021 Page 15 of 18", "source_pages": [ 15 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9702_s21_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-May-June/9702_s21_ms_43.pdf?download=true", "html": "9702-topic-20-magnetic-fields/answers.html", "image_paths": [ "../answer-assets/9702_s21_ms_43-p15.png" ] }, { "id": "9702-2021-mj-43-q10", "question_id": "9702-2021-mj-43-q10", "subject": "9702", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 10, "topic": "Magnetic fields", "topic_slug": "9702-topic-20-magnetic-fields", "marks": 9, "status": "available", "reason": null, "text": "10(a) direction of (induced) e.m.f. M1\nis such as to oppose the change causing it A1\n\n10(b) ring cuts (magnetic) flux and causes induced e.m.f. in ring B1\n(induced) e.m.f. causes (eddy/induced) currents (in ring) B1\ncurrents (in ring) cause magnetic field (around ring) M1\ntwo fields interact to cause resistive/opposing force A1\nor\ncurrent (in ring) is in a magnetic field (M1)\nwhich causes resistive force (A1)\nor\ncurrents (in ring) dissipate thermal energy (M1)\n(thermal) energy comes from energy of oscillations (A1)\n\n10(c) current cannot pass all the way around the ring B1\n(induced) currents smaller B1\nsmaller resistive force (so more oscillations) B1\nor\nsmaller rate of dissipation of energy (so more oscillations)\n© UCLES 2021 Page 16 of 18", "source_pages": [ 16 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9702_s21_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-May-June/9702_s21_ms_43.pdf?download=true", "html": "9702-topic-20-magnetic-fields/answers.html", "image_paths": [ "../answer-assets/9702_s21_ms_43-p16.png" ] }, { "id": "9702-2021-mj-43-q11", "question_id": "9702-2021-mj-43-q11", "subject": "9702", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 11, "topic": "Medical physics", "topic_slug": "9702-topic-24-medical-physics", "marks": 7, "status": "available", "reason": null, "text": "11(a) intensity: vary filament current/p.d. across filament B1\nhardness: vary accelerating potential difference B1\n\n11(b)(i) I = I e –μx C1\n0\nI = I exp(–0.92 × 9.0) A1\nS 0\n= 2.5 × 10–4 I\n0\n\n11(b)(ii) I = [exp(–0.92 × 6.0) × exp(–2.9 × 3.0)] I C1\nC 0\n= 6.7 × 10–7 I A1\n0\n\n11(c) conclusion consistent with values in (b)(i) and (b)(ii) B1\ne.g. I ≫ I so good contrast\nS C\n© UCLES 2021 Page 17 of 18", "source_pages": [ 17 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9702_s21_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-May-June/9702_s21_ms_43.pdf?download=true", "html": "9702-topic-24-medical-physics/answers.html", "image_paths": [ "../answer-assets/9702_s21_ms_43-p17.png" ] }, { "id": "9702-2021-mj-43-q12", "question_id": "9702-2021-mj-43-q12", "subject": "9702", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 12, "topic": "Quantum physics", "topic_slug": "9702-topic-22-quantum-physics", "marks": 8, "status": "available", "reason": null, "text": "12(a) • frequency determines energy of photon B2\n• intensity determines number of photons (per unit time)\n• intensity does not determine energy of a photon\nAny two points, 1 mark each\nkinetic energy (of the electron) depends on the energy of one photon B1\n\n12(b)(i) E = hc / λ C1\nor\nE = hf and c = fλ\nE = (6.63 × 10–34 × 3.00 × 108) / (250 × 10–9) C1\n(= 7.96 × 10–19 J) A1\n= 5.0 eV\n\n12(b)(ii) E = photon energy – work function C1\nMAX\nwork function = 5.0 – 1.4 A1\n= 3.6 eV\n© UCLES 2021 Page 18 of 18", "source_pages": [ 18 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9702_s21_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-May-June/9702_s21_ms_43.pdf?download=true", "html": "9702-topic-22-quantum-physics/answers.html", "image_paths": [ "../answer-assets/9702_s21_ms_43-p18.png" ] }, { "id": "9702-2021-mj-51-q01", "question_id": "9702-2021-mj-51-q01", "subject": "9702", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "51", "question_number": 1, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "1 Defining the problem\nR is the independent variable and t is the dependent variable or vary R and measure t 1\nkeep the number of turns on the coil/N constant 1\nMethods of data collection\nlabelled diagram or correct symbols including: 1\n• labelled (d.c.) power supply\n• switch in series with power supply, resistor and coil\n• complete workable circuit\ncircuit diagram to measure R, e.g. ammeter and voltmeter correctly positioned or R connected to ohmmeter with no other 1\nconnections (not ohmmeter in main circuit)\nmethod to determine t (of a few milliseconds) e.g. use (storage) oscilloscope or current/voltage sensor connected to 1\ndatalogger/computer\nmethod to determine A, e.g. micrometer/calipers to determine diameter of coil and A = πd2/4 1\nMethod of analysis\nplot a graph of t against 1 / R 1\n(allow log t against log R)\nrelationship valid if a straight line passing through the origin is produced 1\n(allow gradient = –1 for graph of log t against log R)\ngradient×L 1\nK = .\nAN2\n© UCLES 2021 Page 6 of 10\n\n1 Additional detail including safety considerations 6\nD1 open switch/switch off (high voltage) circuit before changing the resistor/touching components or\nensure no bare wires/use shrouded connectors\nD2 wear (insulating) gloves to prevent electric shock/electrocution\nD3 keep A and L constant\nD4 use ruler/calipers to measure L\nD5 repeat measurements of diameter in different directions/at points along the coil and average\nD6 method to determine R e.g. R = V / I linked to correct circuit diagram for ammeter/voltmeter method or measure\nresistance using ohmmeter\nD7 repeat experiment for each value of R and average t\nD8 method to determine t:\nuse of time-base from oscilloscope explained\nor\nuse of time axis of output from data logger/computer explained\nD9 use smaller values of R to increase I\nD10 reduce L or increase N or increase A to increase t\n© UCLES 2021 Page 7 of 10", "source_pages": [ 6, 7 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9702_s21_ms_51.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-May-June/9702_s21_ms_51.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_s21_ms_51-p06.png", "../answer-assets/9702_s21_ms_51-p07.png" ] }, { "id": "9702-2021-mj-51-q02", "question_id": "9702-2021-mj-51-q02", "subject": "9702", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "51", "question_number": 2, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "2(a) 1 1\ngradient =\nuA\n1\ny-intercept =\nu\n\n2(b) 1\n1\n(M + m) / g / s cm–1\nv\n380 0.226 or 0.2262\n480 0.255 or 0.2551\n580 0.294 or 0.2941\n680 0.331 or 0.3311\n830 0.388 or 0.3876\n930 0.429 or 0.4292\n1\nValues of (M + m) and as shown above.\nv\nAbsolute uncertainties in (M + m) from ± (19 or 20) to ± (46.5 or 47 or 50). 1\n\n2(c)(i) Six points plotted correctly. 1\nMust be accurate to the nearest half a small square. Diameter of points must be less than half a small square.\nError bars in (M + m) plotted correctly. 1\nAll error bars must be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.\n© UCLES 2021 Page 8 of 10\n\n2(c)(ii) Line of best fit drawn covers all points. 1\nPoints must be balanced. Do not allow line from top point to bottom point.\nLine must pass between (425, 0.240) and (440, 0.240) and between (850, 0.400) and (865, 0.400).\nWorst acceptable line drawn (steepest or shallowest possible line that passes through all error bars). 1\nAll error bars must be plotted.\n\n2(c)(iii) Gradient determined with clear substitution of data points into Δy / Δx. 1\nDistance between data points must be at least half the length of the drawn line.\nGradient of worst acceptable line determined. 1\nuncertainty = (gradient of line of best fit – gradient of worst acceptable line)\nor\nuncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)\n\n2(c)(iv) y-intercept determined by substitution of correct point into y = mx + c. 1\ny-intercept of worst acceptable line determined by substitution into y = mx + c. 1\nuncertainty = (y-intercept of line of best fit – y-intercept of worst acceptable line)\nor\nuncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)\nDo not allow ECF from false origin method.\n\n2(d)(i) u determined using y-intercept and u and A given to two or three significant figures. 1\n1\nu =\ny-intercept\nA determined using gradient with correct substitution and units with correct power of ten for u and A. 1\ny-intercept 1\nA= or A=\ngradient u× gradient\n© UCLES 2021 Page 9 of 10\n\n2(d)(ii) Percentage uncertainty in A determined, e.g. 1\nΔgradient Δy-intercept\npercentage uncertainty in A=  + \n gradient y-intercept \nor\nΔu clearly determined using the value of u and\nΔgradient Δu\npercentage uncertainty in A=  +  ×100\n gradient u \nor\ncorrect substitution for max/min methods e.g.\n1\nmaxA=\nminu×min gradient\n1\nminA=\nmaxu×max gradient\n\n2(e) Value of m determined from (d)(i) or (c)(iii) and (c)(iv), with correct number substitution and correct power of ten. 1\nA×u\n( )\nm= − 330+A\n\n2\nor\n0.5−y-intercept\nm = −330\ngradient\n© UCLES 2021 Page 10 of 10", "source_pages": [ 8, 9, 10 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9702_s21_ms_51.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-May-June/9702_s21_ms_51.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_s21_ms_51-p08.png", "../answer-assets/9702_s21_ms_51-p09.png", "../answer-assets/9702_s21_ms_51-p10.png" ] }, { "id": "9702-2021-mj-52-q01", "question_id": "9702-2021-mj-52-q01", "subject": "9702", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "52", "question_number": 1, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "1 Defining the problem\nA is the independent variable and t is the dependent variable or vary A and measure t 1\nkeep Δθ constant 1\nMethods of data collection\nlabelled diagram of workable experiment including: 1\n• beaker of water\n• cylinder in water\n• electrical heater in water\n• thermometer in water\n• minimum of three labels from heater, thermometer, cylinder, water, beaker\ncircuit diagram to determine power of the heater e.g. ammeter and voltmeter correctly positioned with a power supply or 1\nwattmeter correctly connected to power supply and heater\nmethod to determine time for temperature of water to increase or t, e.g. use a stopwatch/timer 1\nmethod to determine A, e.g. micrometer/calipers to determine diameter of cylinder and A = πd2/4 1\nMethod of analysis\nplot a graph of t against A (not logarithmic graphs) 1\ngradient×P 1\nW =\nhΔθ\ny-intercept×P 1\nZ =\nΔθ\n© UCLES 2021 Page 6 of 10\n\n1 Additional detail including safety considerations 6\nD1 wear (heat proof) gloves to prevent burns from hot beaker/cylinder/heater/water\nD2 keep P and h constant\nD3 check that/ensure/keep initial temperature of the water constant or volume/mass of water constant\nD4 use calipers/ruler to measure h\nD5 repeat measurements of diameter in different directions/at different positions along cylinder and average\nD6 method to calculate power of heater e.g. P = VI linked to correct circuit diagram for ammeter/voltmeter method\nD7 repeat measurements of t for same A and average t\nD8 ensure heater and cylinder are (totally) submerged/immersed\nor\nstir water (using a glass rod/stirrer)\nD9 relationship valid if a straight line (not passing through the origin)\nD10 method to insulate beaker, e.g. use of a lid on the beaker or foam/insulation around outside of beaker\n© UCLES 2021 Page 7 of 10", "source_pages": [ 6, 7 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9702_s21_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-May-June/9702_s21_ms_52.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_s21_ms_52-p06.png", "../answer-assets/9702_s21_ms_52-p07.png" ] }, { "id": "9702-2021-mj-52-q02", "question_id": "9702-2021-mj-52-q02", "subject": "9702", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "52", "question_number": 2, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "2(a) 1 1\ngradient =\nE\nr\ny-intercept =\nE\n\n2(b) 1\n1\n(R + R ) / Ω / A–1\n1 2\nI\n55 58.1 or 58.14\n69 70.4 or 70.42\n78 78.1 or 78.13\n80 80.6 or 80.65\n89 87.7 or 87.72\n103 99.0 or 99.01\n1\nValues of (R + R ) and as shown above.\n1 2\nI\nAbsolute uncertainties in (R + R ) from ± (2.75 or 2.8 or 3) to ± (5.15 or 5.2 or 5). 1\n1 2\n\n2(c)(i) Six points plotted correctly. 1\nMust be accurate to the nearest half a small square. Diameter of points must be less than half a small square.\nError bars in (R + R ) plotted correctly. 1\n1 2\nAll error bars must be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.\n© UCLES 2021 Page 8 of 10\n\n2(c)(ii) Line of best fit drawn covers all points. 1\nPoints must be balanced. Do not allow line from top point to bottom point.\nLine must pass between (61.0, 65.0) and (63.5, 65.0) and between (96.5, 95.0) and (98.5, 95.0).\nWorst acceptable line drawn (steepest or shallowest possible line that passes through all error bars). 1\nAll error bars must be plotted.\n\n2(c)(iii) Gradient determined with clear substitution of data points into Δy / Δx. 1\nDistance between data points must be at least half the length of the drawn line.\nGradient of worst acceptable line determined. 1\nuncertainty = (gradient of line of best fit – gradient of worst acceptable line)\nor\nuncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)\n\n2(c)(iv) y-intercept determined by substitution of correct point into y = mx + c. 1\ny-intercept of worst acceptable line determined by substitution into y = mx + c. 1\nuncertainty = (y-intercept of line of best fit – y-intercept of worst acceptable line)\nor\nuncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)\nDo not allow ECF from false origin method.\n\n2(d)(i) E determined using gradient and E and r given to two or three significant figures. 1\n1\nE =\ngradient\nr determined using y-intercept with correct substitution and units with correct power of ten for E and r. 1\nr = y-intercept/gradient or r = E × y-intercept\n© UCLES 2021 Page 9 of 10\n\n2(d)(ii) Absolute uncertainty in E determined with method shown e.g. 1\nΔgradient\nΔE = ×E\ngradient\nor\ncorrect substitution for max/min methods e.g.\n1\nΔE = −E\nmingradient\n1\nΔE =E −\nmax gradient\n\n2(e) Value of R determined from (d)(i) or (c)(iii) and (c)(iv), with correct substitution and correct power of ten. 1\n\n2\nE ( )\nR = − 22+r\n\n2 0.0075\nor\nR = 1 −( 22+r )\n\n2 0.0075×gradient\n© UCLES 2021 Page 10 of 10", "source_pages": [ 8, 9, 10 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9702_s21_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-May-June/9702_s21_ms_52.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_s21_ms_52-p08.png", "../answer-assets/9702_s21_ms_52-p09.png", "../answer-assets/9702_s21_ms_52-p10.png" ] }, { "id": "9702-2021-mj-53-q01", "question_id": "9702-2021-mj-53-q01", "subject": "9702", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "53", "question_number": 1, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "1 Defining the problem\nR is the independent variable and t is the dependent variable or vary R and measure t 1\nkeep the number of turns on the coil/N constant 1\nMethods of data collection\nlabelled diagram or correct symbols including: 1\n• labelled (d.c.) power supply\n• switch in series with power supply, resistor and coil\n• complete workable circuit\ncircuit diagram to measure R, e.g. ammeter and voltmeter correctly positioned or R connected to ohmmeter with no other 1\nconnections (not ohmmeter in main circuit)\nmethod to determine t (of a few milliseconds) e.g. use (storage) oscilloscope or current/voltage sensor connected to 1\ndatalogger/computer\nmethod to determine A, e.g. micrometer/calipers to determine diameter of coil and A = πd2/4 1\nMethod of analysis\nplot a graph of t against 1 / R 1\n(allow log t against log R)\nrelationship valid if a straight line passing through the origin is produced 1\n(allow gradient = –1 for graph of log t against log R)\ngradient×L 1\nK = .\nAN2\n© UCLES 2021 Page 6 of 10\n\n1 Additional detail including safety considerations 6\nD1 open switch/switch off (high voltage) circuit before changing the resistor/touching components or\nensure no bare wires/use shrouded connectors\nD2 wear (insulating) gloves to prevent electric shock/electrocution\nD3 keep A and L constant\nD4 use ruler/calipers to measure L\nD5 repeat measurements of diameter in different directions/at points along the coil and average\nD6 method to determine R e.g. R = V / I linked to correct circuit diagram for ammeter/voltmeter method or measure\nresistance using ohmmeter\nD7 repeat experiment for each value of R and average t\nD8 method to determine t:\nuse of time-base from oscilloscope explained\nor\nuse of time axis of output from data logger/computer explained\nD9 use smaller values of R to increase I\nD10 reduce L or increase N or increase A to increase t\n© UCLES 2021 Page 7 of 10", "source_pages": [ 6, 7 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9702_s21_ms_53.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-May-June/9702_s21_ms_53.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_s21_ms_53-p06.png", "../answer-assets/9702_s21_ms_53-p07.png" ] }, { "id": "9702-2021-mj-53-q02", "question_id": "9702-2021-mj-53-q02", "subject": "9702", "year": 2021, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "53", "question_number": 2, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "2(a) 1 1\ngradient =\nuA\n1\ny-intercept =\nu\n\n2(b) 1\n1\n(M + m) / g / s cm–1\nv\n380 0.226 or 0.2262\n480 0.255 or 0.2551\n580 0.294 or 0.2941\n680 0.331 or 0.3311\n830 0.388 or 0.3876\n930 0.429 or 0.4292\n1\nValues of (M + m) and as shown above.\nv\nAbsolute uncertainties in (M + m) from ± (19 or 20) to ± (46.5 or 47 or 50). 1\n\n2(c)(i) Six points plotted correctly. 1\nMust be accurate to the nearest half a small square. Diameter of points must be less than half a small square.\nError bars in (M + m) plotted correctly. 1\nAll error bars must be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.\n© UCLES 2021 Page 8 of 10\n\n2(c)(ii) Line of best fit drawn covers all points. 1\nPoints must be balanced. Do not allow line from top point to bottom point.\nLine must pass between (425, 0.240) and (440, 0.240) and between (850, 0.400) and (865, 0.400).\nWorst acceptable line drawn (steepest or shallowest possible line that passes through all error bars). 1\nAll error bars must be plotted.\n\n2(c)(iii) Gradient determined with clear substitution of data points into Δy / Δx. 1\nDistance between data points must be at least half the length of the drawn line.\nGradient of worst acceptable line determined. 1\nuncertainty = (gradient of line of best fit – gradient of worst acceptable line)\nor\nuncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)\n\n2(c)(iv) y-intercept determined by substitution of correct point into y = mx + c. 1\ny-intercept of worst acceptable line determined by substitution into y = mx + c. 1\nuncertainty = (y-intercept of line of best fit – y-intercept of worst acceptable line)\nor\nuncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)\nDo not allow ECF from false origin method.\n\n2(d)(i) u determined using y-intercept and u and A given to two or three significant figures. 1\n1\nu =\ny-intercept\nA determined using gradient with correct substitution and units with correct power of ten for u and A. 1\ny-intercept 1\nA= or A=\ngradient u× gradient\n© UCLES 2021 Page 9 of 10\n\n2(d)(ii) Percentage uncertainty in A determined, e.g. 1\nΔgradient Δy-intercept\npercentage uncertainty in A=  + \n gradient y-intercept \nor\nΔu clearly determined using the value of u and\nΔgradient Δu\npercentage uncertainty in A=  +  ×100\n gradient u \nor\ncorrect substitution for max/min methods e.g.\n1\nmaxA=\nminu×min gradient\n1\nminA=\nmaxu×max gradient\n\n2(e) Value of m determined from (d)(i) or (c)(iii) and (c)(iv), with correct number substitution and correct power of ten. 1\nA×u\n( )\nm= − 330+A\n\n2\nor\n0.5−y-intercept\nm = −330\ngradient\n© UCLES 2021 Page 10 of 10", "source_pages": [ 8, 9, 10 ], "source_pdf": "_source-pdfs/2021-May-June/ms/9702_s21_ms_53.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-May-June/9702_s21_ms_53.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_s21_ms_53-p08.png", "../answer-assets/9702_s21_ms_53-p09.png", "../answer-assets/9702_s21_ms_53-p10.png" ] }, { "id": "9702-2021-on-41-q01", "question_id": "9702-2021-on-41-q01", "subject": "9702", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 1, "topic": "Motion in a circle", "topic_slug": "9702-topic-12-motion-in-a-circle", "marks": 7, "status": "available", "reason": null, "text": "1(a) constant speed or constant magnitude of velocity B1\nacceleration (always) perpendicular to velocity B1\n\n1(b)(i) F = mv2 / r C1\nor\nv = rω and F = mrω2\nF = 790 × 942 / 318 A1\n= 22000 N\n\n1(b)(ii) centripetal acceleration: same B1\nmaximum speed: greater B1\ntime taken for one lap of the track: greater B1\n© UCLES 2021 Page 7 of 15", "source_pages": [ 7 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9702_w21_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-Oct-Nov/9702_w21_ms_41.pdf?download=true", "html": "9702-topic-12-motion-in-a-circle/answers.html", "image_paths": [ "../answer-assets/9702_w21_ms_41-p07.png" ] }, { "id": "9702-2021-on-41-q02", "question_id": "9702-2021-on-41-q02", "subject": "9702", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 2, "topic": "Gravitational fields", "topic_slug": "9702-topic-13-gravitational-fields", "marks": 9, "status": "available", "reason": null, "text": "2(a) work done per unit mass B1\n(work done in) moving mass from infinity B1\n\n2(b)(i) (gravitational) fields from the Earth and Moon are in opposite directions B1\n(resultant is zero where gravitational) fields are equal (in magnitude) B1\n\n2(b)(ii) g ∝ M / r2 C1\n5.98 × 1024 / x2 = 7.35 × 1022 / (3.84 × 108 – x)2 A1\nleading to x = 3.5 × 108 (m)\n\n2(b)(iii) φ (Earth) = (–)6.67 × 10–11 × (5.98 × 1024 / 3.5 × 108) C1\nand\nφ (Moon) = (–)6.67 × 10–11 × (7.35 × 1022 / 0.38 × 108)\nφ = (–)6.67 × 10–11 × [(5.98 × 1024 / 3.5 × 108) + (7.35 × 1022 / 0.38 × 108)] C1\n= – 1.3 × 106 J kg–1 A1\n© UCLES 2021 Page 8 of 15", "source_pages": [ 8 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9702_w21_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-Oct-Nov/9702_w21_ms_41.pdf?download=true", "html": "9702-topic-13-gravitational-fields/answers.html", "image_paths": [ "../answer-assets/9702_w21_ms_41-p08.png" ] }, { "id": "9702-2021-on-41-q03", "question_id": "9702-2021-on-41-q03", "subject": "9702", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 3, "topic": "Thermodynamics", "topic_slug": "9702-topic-16-thermodynamics", "marks": 13, "status": "available", "reason": null, "text": "3(a) (thermal) energy per unit mass (to cause temperature change) B1\n(thermal) energy per unit change in temperature B1\n\n3(b)(i) (T =) pV / Nk B1\n\n3(b)(ii) (pV =) NkT = ⅓Nm M1\nor\npV = NkT and pV = ⅓Nm\nleading to ½m = (3/2)kT and ½m = E A1\nK\n\n3(b)(iii) internal energy = ΣE (of molecules) + ΣE (of molecules) B1\nK P\nor\nno forces between molecules\npotential energy of molecules is zero B1\n\n3(c)(i) increase in internal energy = Q + work done B1\nconstant volume so no work done B1\n\n3(c)(ii) c = Q / NmΔT C1\n= [N × (3/2)kΔT] / (NmΔT) = 3k / 2m A1\n\n3(d) (as it expands) gas does work (against the atmosphere/external pressure) B1\nfor same temperature rise) more (thermal) energy needed, so larger specific heat capacity B1\n© UCLES 2021 Page 9 of 15", "source_pages": [ 9 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9702_w21_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-Oct-Nov/9702_w21_ms_41.pdf?download=true", "html": "9702-topic-16-thermodynamics/answers.html", "image_paths": [ "../answer-assets/9702_w21_ms_41-p09.png" ] }, { "id": "9702-2021-on-41-q04", "question_id": "9702-2021-on-41-q04", "subject": "9702", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 4, "topic": "Oscillations", "topic_slug": "9702-topic-17-oscillations", "marks": 11, "status": "available", "reason": null, "text": "4(a)(i) 5.0 cm A1\n\n4(a)(ii) ω = 2π / T C1\nor\nω = 2πf and f = 1 / T\nω = 2π / 4.0 A1\n= 1.6 rad s–1\n\n4(a)(iii) v = ωx C1\n0 0\n= 1.57 × 5.0 A1\n= 7.9 cm s–1\n\n4(b) • initial pull was to the right B3\n• distance from X to trolley (at equilibrium) is 20 cm\n• period is 4.0 s\n• initial motion undamped\n• motion becomes damped at/from 12 s\n• damping is light\n• maximum speed at 1s, 3s, etc. / stationary at 2s, 4s, etc.\nAny three points, 1 mark each\n\n4(c) sketch: closed loop encircling (20, 0) B1\nminimum L shown as 15 cm and maximum L shown as 25 cm B1\nminimum v shown as –7.9 cm s–1 and maximum v shown as +7.9 cm s–1 B1\n© UCLES 2021 Page 10 of 15", "source_pages": [ 10 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9702_w21_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-Oct-Nov/9702_w21_ms_41.pdf?download=true", "html": "9702-topic-17-oscillations/answers.html", "image_paths": [ "../answer-assets/9702_w21_ms_41-p10.png" ] }, { "id": "9702-2021-on-41-q05", "question_id": "9702-2021-on-41-q05", "subject": "9702", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 5, "topic": "Alternating currents", "topic_slug": "9702-topic-21-alternating-currents", "marks": 8, "status": "available", "reason": null, "text": "5(a) • noise can be removed/signal can be regenerated B2\n• extra bits can be added for error-checking\n• signal can be encrypted (for increased security)\n• data compression/multiplexing is possible\nAny two points, 1 mark each\n\n5(b)(i) 4ms: 0101 and 8ms: 0100 B1\n\n5(b)(ii) sketch: horizontal line continues to 8ms, then new horizontal line from 8ms to 12ms B1\nlevel of line after 8ms is 4 mV B1\n\n5(c) sketch: series of steps of width 2ms B1\nstep heights at 0, 2, 4, 6, 4, 6 mV B2\n2 marks if all correct, 1 mark if only one incorrect\nQuestion Answer Marks", "source_pages": [ 11 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9702_w21_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-Oct-Nov/9702_w21_ms_41.pdf?download=true", "html": "9702-topic-21-alternating-currents/answers.html", "image_paths": [ "../answer-assets/9702_w21_ms_41-p11.png" ] }, { "id": "9702-2021-on-41-q06", "question_id": "9702-2021-on-41-q06", "subject": "9702", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 6, "topic": "Capacitance", "topic_slug": "9702-topic-19-capacitance", "marks": 5, "status": "available", "reason": null, "text": "6(a) Q = CV and E = ½CV2 B1\n\n6(b)(i) C = CL / (L – D) B1\nN\n\n6(b)(ii) (charge is unchanged by moving the plates so) Q = CV B1\nN\n\n6(b)(iii) V = Q / C B1\nN N N\n= (CV) / [CL / (L – D)]\n= V(L – D) / L\n\n6(c) oppositely charged plates attract, so energy stored decreases B1\n© UCLES 2021 Page 11 of 15", "source_pages": [ 11 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9702_w21_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-Oct-Nov/9702_w21_ms_41.pdf?download=true", "html": "9702-topic-19-capacitance/answers.html", "image_paths": [ "../answer-assets/9702_w21_ms_41-p11.png" ] }, { "id": "9702-2021-on-41-q07", "question_id": "9702-2021-on-41-q07", "subject": "9702", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 7, "topic": "Alternating currents", "topic_slug": "9702-topic-21-alternating-currents", "marks": 9, "status": "available", "reason": null, "text": "7(a) • infinite (open-loop) gain B2\n• infinite slew rate\n• infinite input impedance\n• zero output impedance\n• infinite bandwidth\nAny two points, 1 mark each\n\n7(b) X: thermistor and Y: relay B1\n\n7(c)(i) (any) difference in voltage at the inputs causes output to saturate (because gain is very large) B1\nsaturates positively if V+ > V– and saturates negatively if V+ < V– B1\n\n7(c)(ii) comparator B1\n\n7(c)(iii) temperature M1\nabove a particular value A1\n\n7(c)(iv) to adjust the temperature (at which the lamp illuminates/extinguishes) B1\n© UCLES 2021 Page 12 of 15", "source_pages": [ 12 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9702_w21_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-Oct-Nov/9702_w21_ms_41.pdf?download=true", "html": "9702-topic-21-alternating-currents/answers.html", "image_paths": [ "../answer-assets/9702_w21_ms_41-p12.png" ] }, { "id": "9702-2021-on-41-q08", "question_id": "9702-2021-on-41-q08", "subject": "9702", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 8, "topic": "Magnetic fields", "topic_slug": "9702-topic-20-magnetic-fields", "marks": 6, "status": "available", "reason": null, "text": "8(a) newton per ampere per metre M1\nwhere current/wire is perpendicular to magnetic field A1\n\n8(b)(i) F = BILsinθ C1\nB = 1.0 / (5.0 × 0.060 × sin 50°) A1\n= 4.4 mT\n\n8(b)(ii) (from Fleming’s left-hand rule) force on wire is upwards, so reading decreases B1\n\n8(b)(iii) frame will rotate (so that PQ becomes perpendicular to the field) B1\nQuestion Answer Marks", "source_pages": [ 13 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9702_w21_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-Oct-Nov/9702_w21_ms_41.pdf?download=true", "html": "9702-topic-20-magnetic-fields/answers.html", "image_paths": [ "../answer-assets/9702_w21_ms_41-p13.png" ] }, { "id": "9702-2021-on-41-q09", "question_id": "9702-2021-on-41-q09", "subject": "9702", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 9, "topic": "Ideal gases", "topic_slug": "9702-topic-15-ideal-gases", "marks": 9, "status": "available", "reason": null, "text": "9(a) constant voltage M1\nthat produces/dissipates same power as (the mean power of) the alternating voltage A1\n\n9(b)(i) (maximum) rate of cutting of (magnetic) flux doubles B1\n(peak and hence) r.m.s. induced e.m.f. doubles B1\n\n9(b)(ii) sketch: (sinusoidal) wave of period 10 ms B1\npeak E shown as ± 34V B2\n(1 mark out of 2 awarded if peak E shown as ± 17V or ± 24V)\n\n9(c) current in the coil results in forces that oppose its rotation B1\nor\ncurrent in the resistor dissipates the energy of rotation\ncoil stops rotating B1\n© UCLES 2021 Page 13 of 15", "source_pages": [ 13 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9702_w21_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-Oct-Nov/9702_w21_ms_41.pdf?download=true", "html": "9702-topic-15-ideal-gases/answers.html", "image_paths": [ "../answer-assets/9702_w21_ms_41-p13.png" ] }, { "id": "9702-2021-on-41-q10", "question_id": "9702-2021-on-41-q10", "subject": "9702", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 10, "topic": "Quantum physics", "topic_slug": "9702-topic-22-quantum-physics", "marks": 9, "status": "available", "reason": null, "text": "10(a)(i) photoelectric effect B1\n\n10(a)(ii) electron diffraction B1\n\n10(b)(i) λ = h / p M1\nh is the Planck constant A1\n\n10(b)(ii) de Broglie (wavelength) B1\n\n10(c)(i) ½mv2 = eV C1\n½ × 9.11 × 10–31 × v2 = 1.60 × 10–19 × 4800 so v = 4.1 × 107 m s–1 A1\n\n10(c)(ii) λ = h / mv C1\n= 6.63 × 10–34 / (9.11 × 10–31 × 4.1 × 107)\n= 1.8 × 10–11 m A1\n© UCLES 2021 Page 14 of 15", "source_pages": [ 14 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9702_w21_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-Oct-Nov/9702_w21_ms_41.pdf?download=true", "html": "9702-topic-22-quantum-physics/answers.html", "image_paths": [ "../answer-assets/9702_w21_ms_41-p14.png" ] }, { "id": "9702-2021-on-41-q11", "question_id": "9702-2021-on-41-q11", "subject": "9702", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 11, "topic": "Medical physics", "topic_slug": "9702-topic-24-medical-physics", "marks": 7, "status": "available", "reason": null, "text": "11(a)(i) ease with which edges can be distinguished B1\n\n11(a)(ii) difference in degrees of blackening B1\n\n11(b) I = I exp (–μx) C1\n0\n0.12 = exp (–μ × 2.3) C1\nln 0.12 = –2.3 × μ\nμ = 0.92 cm–1 A1\n\n11(c) advantage: produces 3-dimensional image B1\ndisadvantage: (much) greater exposure to radiation B1\nQuestion Answer Marks", "source_pages": [ 15 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9702_w21_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-Oct-Nov/9702_w21_ms_41.pdf?download=true", "html": "9702-topic-24-medical-physics/answers.html", "image_paths": [ "../answer-assets/9702_w21_ms_41-p15.png" ] }, { "id": "9702-2021-on-41-q12", "question_id": "9702-2021-on-41-q12", "subject": "9702", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 12, "topic": "Nuclear physics", "topic_slug": "9702-topic-23-nuclear-physics", "marks": 7, "status": "available", "reason": null, "text": "12(a) probability of decay (of a nucleus) M1\nper unit time A1\n\n12(b) A = λN C1\nN = mass / (nucleon number × u) C1\n2.92 × 109 = (λ × 5.87 × 10–10) / (131 × 1.66 × 10–27) A1\nλ = 1.08 × 10–6 s–1\n\n12(c) • sample emits radiation in all directions B2\n• some radiation is absorbed by air/detector window\n• self-absorption within the source\n• dead time/inefficiency of detector\nAny two points, 1 mark each\n© UCLES 2021 Page 15 of 15", "source_pages": [ 15 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9702_w21_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-Oct-Nov/9702_w21_ms_41.pdf?download=true", "html": "9702-topic-23-nuclear-physics/answers.html", "image_paths": [ "../answer-assets/9702_w21_ms_41-p15.png" ] }, { "id": "9702-2021-on-42-q01", "question_id": "9702-2021-on-42-q01", "subject": "9702", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 1, "topic": "Motion in a circle", "topic_slug": "9702-topic-12-motion-in-a-circle", "marks": 8, "status": "available", "reason": null, "text": "1(a) acceleration perpendicular to velocity B1\n\n1(b)(i) decreases B1\n\n1(b)(ii) (acceleration of) 9.8ms–2 is caused by weight of car B1\nor\ncentripetal force must be greater than weight of car\n(acceleration > 9.8ms–2) requires contact force from track B1\nor\n(centripetal force > weight) requires contact force from track\n\n1(c) ½mv 2 = ½mv 2 – mgh C1\nY X\na = v2 / r C1\nv 2 = 3.82 – 2 × 9.81 × 0.62 so v = 1.5ms–1 A1\nY Y\na = 1.52 / 0.31 = 7.3ms–2 (which is less than 9.8ms–2) so no\nor\nv = √(9.81 × 0.31) = 1.74 m s–1 so v 2 = 1.742 + 2 × 9.81 × 0.62 (A1)\nY X\nv = 3.9 m s–1 (which is greater than 3.8 m s–1) so no\nX\n\n1(d) acceleration is independent of mass so makes no difference B1\nor\nmass cancels in the equation so makes no difference\n© UCLES 2021 Page 7 of 19", "source_pages": [ 7 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9702_w21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-Oct-Nov/9702_w21_ms_42.pdf?download=true", "html": "9702-topic-12-motion-in-a-circle/answers.html", "image_paths": [ "../answer-assets/9702_w21_ms_42-p07.png" ] }, { "id": "9702-2021-on-42-q02", "question_id": "9702-2021-on-42-q02", "subject": "9702", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 2, "topic": "Gravitational fields", "topic_slug": "9702-topic-13-gravitational-fields", "marks": 10, "status": "available", "reason": null, "text": "2(a) (gravitational) field strength equals (gravitational) potential gradient M1\nreference to minus sign A1\n\n2(b)(i) potential is zero at infinity B1\n(gravitational) force is attractive B1\n(test) mass getting closer (from infinity) loses potential energy B1\n\n2(b)(ii) • potential at (surface of) planet is smaller than at (surface of) moon B2\n• potential gradient at (surface of) planet is smaller than at (surface of) moon\n• magnitude of potential varies inversely with distance from centre near the spheres\n• (point of) maximum potential is nearer to moon than planet\nAny two points, 1 mark each\n\n2(b)(iii) sketch: one curve, starting with gradient of decreasing magnitude at 2R and finishing with gradient of increasing magnitude B1\nat D – R\nfield strength shown as zero (only) near the point of maximum potential B1\nnegative field strength near one sphere and positive field strength near the other B1\n© UCLES 2021 Page 8 of 19", "source_pages": [ 8 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9702_w21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-Oct-Nov/9702_w21_ms_42.pdf?download=true", "html": "9702-topic-13-gravitational-fields/answers.html", "image_paths": [ "../answer-assets/9702_w21_ms_42-p08.png" ] }, { "id": "9702-2021-on-42-q03", "question_id": "9702-2021-on-42-q03", "subject": "9702", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 3, "topic": "Ideal gases", "topic_slug": "9702-topic-15-ideal-gases", "marks": 11, "status": "available", "reason": null, "text": "3(a)(i) no loss of kinetic energy B1\n\n3(a)(ii) • molecules have negligible volume (compared with gas/container) B2\n• no forces between molecules (except during collisions)\n• molecules are in random motion\n• collisions are instantaneous\nAny two points, 1 mark each\n\n3(b)(i) 2mu A1\n\n3(b)(ii) 2L / u A1\n\n3(b)(iii) force = change in momentum / time = 2mu / (2L / u) A1\n= mu2 / L\n\n3(b)(iv) pressure = force / area = (mu2 / L) / L2 A1\n= mu2 / L3\n\n3(c) pV = NkT C1\nNkT = ⅓Nm leading to ½m = (3/2)kT and ½m = E A1\nK\n\n3(d) ½ × 3.34 × 10–27 × = (3/2) × 1.38 × 10–23 × (25 + 273) C1\nr.m.s. speed = 1.9 × 103 m s–1 A1\n© UCLES 2021 Page 9 of 19", "source_pages": [ 9 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9702_w21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-Oct-Nov/9702_w21_ms_42.pdf?download=true", "html": "9702-topic-15-ideal-gases/answers.html", "image_paths": [ "../answer-assets/9702_w21_ms_42-p09.png" ] }, { "id": "9702-2021-on-42-q04", "question_id": "9702-2021-on-42-q04", "subject": "9702", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 4, "topic": "Oscillations", "topic_slug": "9702-topic-17-oscillations", "marks": 7, "status": "available", "reason": null, "text": "4(a) straight line through the origin B1\nnegative gradient B1\n\n4(b) a = (–)ω2x and T = 2π / ω C1\ne.g. ω = √(0.80 / 0.12) (any correct pair of values of a and x) C1\n( = 2.58 rad s–1)\nT = 2π / 2.58 A1\n= 2.4 s\n\n4(c)(i) Point labelled P at one end of the line B1\n\n4(c)(ii) Point labelled Q at displacement with magnitude more than half but less than maximum B1\n© UCLES 2021 Page 10 of 19", "source_pages": [ 10 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9702_w21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-Oct-Nov/9702_w21_ms_42.pdf?download=true", "html": "9702-topic-17-oscillations/answers.html", "image_paths": [ "../answer-assets/9702_w21_ms_42-p10.png" ] }, { "id": "9702-2021-on-42-q05", "question_id": "9702-2021-on-42-q05", "subject": "9702", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 5, "topic": "Alternating currents", "topic_slug": "9702-topic-21-alternating-currents", "marks": 8, "status": "available", "reason": null, "text": "5(a)(i) unmodulated (radio) waves would interfere with each other B1\nor\nnot modulating would require aerials too long (to be practical)\n\n5(a)(ii) advantage: B1\n• can transmit higher frequencies\n• higher quality reproduction\n• less prone to interference\n• same frequency can be used in different areas\n(any one point)\ndisadvantage: B1\n• takes up greater bandwidth\n• shorter range of transmission\n• requires a greater number of transmitting aerials\n(any one point)\n\n5(b) AM amplitude: min. 8 mV and max. 12 mV B1\nAM frequency: min. 100 kHz and max. 100 kHz B1\nFM amplitude: min. 10 mV and max. 10 mV B1\nFM frequency: min. 90 kHz and max. 110 kHz B1\n\n5(c) 8.4 kHz A1\n© UCLES 2021 Page 11 of 19", "source_pages": [ 11 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9702_w21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-Oct-Nov/9702_w21_ms_42.pdf?download=true", "html": "9702-topic-21-alternating-currents/answers.html", "image_paths": [ "../answer-assets/9702_w21_ms_42-p11.png" ] }, { "id": "9702-2021-on-42-q06", "question_id": "9702-2021-on-42-q06", "subject": "9702", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 6, "topic": "Capacitance", "topic_slug": "9702-topic-19-capacitance", "marks": 7, "status": "available", "reason": null, "text": "6(a) work done per unit charge B1\n(work done in) moving positive charge from infinity B1\n\n6(b) C = Q / V C1\nV = Q / (4πεr) and so C = Q / [Q / (4πεr)] = 4πεr A1\n0 0 0\n\n6(c) Q = 4πεrV = 4π × 8.85 × 10–12 × 0.13 × 4500 C1\n0\n( = 6.5 × 10–8 C)\n(Q – q) / 13 = q / 5.2 C1\n5.2Q – 5.2q = 13q, so q = (5.2 / 18.2)Q A1\nq = (5.2 / 18.2) × 6.5 × 10–8\n= 1.9 × 10–8 C\nor\nV = Q / C (C1)\nT T T\n= 6.5 × 10–8 / [4π × 8.85 × 10–12 × (0.13 + 0.052)]\n( = 3210 V)\nq = 4π × 8.85 × 10–12 × 0.052 × 3210 (A1)\n= 1.9 × 10–8 C\n© UCLES 2021 Page 12 of 19", "source_pages": [ 12 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9702_w21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-Oct-Nov/9702_w21_ms_42.pdf?download=true", "html": "9702-topic-19-capacitance/answers.html", "image_paths": [ "../answer-assets/9702_w21_ms_42-p12.png" ] }, { "id": "9702-2021-on-42-q07", "question_id": "9702-2021-on-42-q07", "subject": "9702", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 7, "topic": "Alternating currents", "topic_slug": "9702-topic-21-alternating-currents", "marks": 10, "status": "available", "reason": null, "text": "7(a) output voltage / input voltage M1\ninput (voltage) is difference between (inverting and non-inverting) inputs A1\n\n7(b) • reduces the gain B2\n• greater bandwidth\n• more stable\nAny two points, 1 mark each\n\n7(c)(i) inverting amplifier B1\n\n7(c)(ii) X marked anywhere between right-hand edge of 480Ω resistor, left-hand edge of 1.2kΩ resistor and the inverting input B1\n\n7(c)(iii) gain = (–)R / R C1\nf i\n= (–)1200 / 480 A1\n= –2.5\n\n7(c)(iv) V = 6.5 / (–2.5) A1\nIN\n= –2.6 V\n\n7(c)(v) (–2.5) × (–5.4) = +13.5 V, and so output saturates A1\nV = (+)8.0 V\nOUT\n© UCLES 2021 Page 13 of 19", "source_pages": [ 13 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9702_w21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-Oct-Nov/9702_w21_ms_42.pdf?download=true", "html": "9702-topic-21-alternating-currents/answers.html", "image_paths": [ "../answer-assets/9702_w21_ms_42-p13.png" ] }, { "id": "9702-2021-on-42-q08", "question_id": "9702-2021-on-42-q08", "subject": "9702", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 8, "topic": "Magnetic fields", "topic_slug": "9702-topic-20-magnetic-fields", "marks": 5, "status": "available", "reason": null, "text": "8(a)(i) arrow from Q pointing downwards, labelled B B1\n\n8(a)(ii) arrow from Q pointing towards P, labelled F B1\n\n8(b)(i) force is proportional to product of both currents (I and 2I) B1\nor\nNewton’s third law\nforces are equal B1\n\n8(b)(ii) opposite B1\n© UCLES 2021 Page 14 of 19", "source_pages": [ 14 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9702_w21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-Oct-Nov/9702_w21_ms_42.pdf?download=true", "html": "9702-topic-20-magnetic-fields/answers.html", "image_paths": [ "../answer-assets/9702_w21_ms_42-p14.png" ] }, { "id": "9702-2021-on-42-q09", "question_id": "9702-2021-on-42-q09", "subject": "9702", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 9, "topic": "Quantum physics", "topic_slug": "9702-topic-22-quantum-physics", "marks": 9, "status": "available", "reason": null, "text": "9(a)(i) emission of electrons (from a metal surface) B1\nwhen electromagnetic radiation is incident (on electrons) B1\n9a(ii) minimum energy required for an electron to leave surface B1\n\n9(b)(i) threshold (frequency) B1\n\n9(b)(ii) • photons are (discrete) packets of energy B2\n• energy of photons depends on frequency (of EM radiation)\n• electrons can only absorb a single photon (of energy)\nAny two points, 1 mark each\nemission only possible if photon energy is at least the work function B1\n\n9(b)(iii) work function = hf = 6.63 × 10–34 × 6.93 × 1014 C1\n0\n= 4.59 × 10–19 (J) A1\n= 4.59 × 10–19 / 1.60 × 10–19 (eV)\n= 2.87 eV\n© UCLES 2021 Page 15 of 19", "source_pages": [ 15 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9702_w21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-Oct-Nov/9702_w21_ms_42.pdf?download=true", "html": "9702-topic-22-quantum-physics/answers.html", "image_paths": [ "../answer-assets/9702_w21_ms_42-p15.png" ] }, { "id": "9702-2021-on-42-q10", "question_id": "9702-2021-on-42-q10", "subject": "9702", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 10, "topic": "Alternating currents", "topic_slug": "9702-topic-21-alternating-currents", "marks": 10, "status": "available", "reason": null, "text": "10(a)(i) to increase the magnetic flux linkage (between the coils) B1\n\n10(a)(ii) to reduce energy losses B1\nby reducing induced currents B1\n\n10(b)(i) maximum V = 12 000 × (625 / 25000) A1\nOUT\n= 300 V\n\n10(b)(ii) r.m.s. current = 300 / (640 × √2) A1\n= 0.33 A\n\n10(b)(iii) sketch: sinusoidal shape in positive half of the graph, sitting with ‘minima’ resting on the time-axis (at P = 0) B1\neach ‘cycle’ shown repeating every 20 ms B1\nmaximum P shown as 140 W B1\n\n10(c) power curve is symmetrical about the midpoint (on the power axis) B1\nmean power is half the peak power B1\n© UCLES 2021 Page 16 of 19", "source_pages": [ 16 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9702_w21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-Oct-Nov/9702_w21_ms_42.pdf?download=true", "html": "9702-topic-21-alternating-currents/answers.html", "image_paths": [ "../answer-assets/9702_w21_ms_42-p16.png" ] }, { "id": "9702-2021-on-42-q11", "question_id": "9702-2021-on-42-q11", "subject": "9702", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 11, "topic": "Medical physics", "topic_slug": "9702-topic-24-medical-physics", "marks": 7, "status": "available", "reason": null, "text": "11(a) generates ultrasound B1\ndetects reflected ultrasound B1\napplied p.d. causes crystal to vibrate B1\nor\nvibrations cause crystal to generate an e.m.f.\n\n11(b)(i) product of density and speed M1\nspeed of ultrasound in medium A1\n\n11(b)(ii) difference between (the specific acoustic impedances) C1\n• if similar/same then reflection coefficient is zero/very low A1\n• if very different then reflection coefficient is (nearly) 1\n• the lower the difference means lower the reflection coefficient\n(any one point)\n© UCLES 2021 Page 17 of 19", "source_pages": [ 17 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9702_w21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-Oct-Nov/9702_w21_ms_42.pdf?download=true", "html": "9702-topic-24-medical-physics/answers.html", "image_paths": [ "../answer-assets/9702_w21_ms_42-p17.png" ] }, { "id": "9702-2021-on-42-q12", "question_id": "9702-2021-on-42-q12", "subject": "9702", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 12, "topic": "Nuclear physics", "topic_slug": "9702-topic-23-nuclear-physics", "marks": 8, "status": "available", "reason": null, "text": "12(a)(i) cannot predict when a particular nucleus will decay B1\nor\ncannot predict which nucleus will decay next\n\n12(a)(ii) (decay is) not affected by external (environmental) factors B1\n\n12(b)(i) A = A exp (–λt) and so ln A = ln A – λt C1\n0 0\ngradient of line = (–)λ\nλ = (36.4 – 35.0) / (20 – 0) C1\n( = 0.07(0) min–1)\nhalf-life = ln 2 / λ A1\n= ln 2 / 0.070\n= 10 min\nor\nA = exp (–36.4) = 6.43 × 1015 (Bq) (C1)\n0\nA / 2 = 3.21 × 1015 (Bq), so ln (A / 2) = 35.7 (C1)\n0 0\nread off half-life = 10 min (A1)\nor\n(at one half-life,) ln A = 36.4 – ln 2 (C1)\n= 35.7 (C1)\nread off half-life = 10 min (A1)\n© UCLES 2021 Page 18 of 19\n\n12(b)(ii) A = λN C1\nN = mass / (nucleon number × u) C1\nor\nN = (mass / nucleon number) × N\nA\nexp(36.4) = (1.17 × 10–3 × 5.66 × 10–7) / (nucleon number × 1.66 × 10–27) A1\nor\nexp(36.4) = (1.17 × 10–3 × 5.66 × 10–4 × 6.02 × 1023) / nucleon number\nnucleon number = 62\n© UCLES 2021 Page 19 of 19", "source_pages": [ 18, 19 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9702_w21_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-Oct-Nov/9702_w21_ms_42.pdf?download=true", "html": "9702-topic-23-nuclear-physics/answers.html", "image_paths": [ "../answer-assets/9702_w21_ms_42-p18.png", "../answer-assets/9702_w21_ms_42-p19.png" ] }, { "id": "9702-2021-on-43-q01", "question_id": "9702-2021-on-43-q01", "subject": "9702", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 1, "topic": "Motion in a circle", "topic_slug": "9702-topic-12-motion-in-a-circle", "marks": 7, "status": "available", "reason": null, "text": "1(a) constant speed or constant magnitude of velocity B1\nacceleration (always) perpendicular to velocity B1\n\n1(b)(i) F = mv2 / r C1\nor\nv = rω and F = mrω2\nF = 790 × 942 / 318 A1\n= 22000 N\n\n1(b)(ii) centripetal acceleration: same B1\nmaximum speed: greater B1\ntime taken for one lap of the track: greater B1\n© UCLES 2021 Page 7 of 15", "source_pages": [ 7 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9702_w21_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-Oct-Nov/9702_w21_ms_43.pdf?download=true", "html": "9702-topic-12-motion-in-a-circle/answers.html", "image_paths": [ "../answer-assets/9702_w21_ms_43-p07.png" ] }, { "id": "9702-2021-on-43-q02", "question_id": "9702-2021-on-43-q02", "subject": "9702", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 2, "topic": "Gravitational fields", "topic_slug": "9702-topic-13-gravitational-fields", "marks": 9, "status": "available", "reason": null, "text": "2(a) work done per unit mass B1\n(work done in) moving mass from infinity B1\n\n2(b)(i) (gravitational) fields from the Earth and Moon are in opposite directions B1\n(resultant is zero where gravitational) fields are equal (in magnitude) B1\n\n2(b)(ii) g ∝ M / r2 C1\n5.98 × 1024 / x2 = 7.35 × 1022 / (3.84 × 108 – x)2 A1\nleading to x = 3.5 × 108 (m)\n\n2(b)(iii) φ (Earth) = (–)6.67 × 10–11 × (5.98 × 1024 / 3.5 × 108) C1\nand\nφ (Moon) = (–)6.67 × 10–11 × (7.35 × 1022 / 0.38 × 108)\nφ = (–)6.67 × 10–11 × [(5.98 × 1024 / 3.5 × 108) + (7.35 × 1022 / 0.38 × 108)] C1\n= – 1.3 × 106 J kg–1 A1\n© UCLES 2021 Page 8 of 15", "source_pages": [ 8 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9702_w21_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-Oct-Nov/9702_w21_ms_43.pdf?download=true", "html": "9702-topic-13-gravitational-fields/answers.html", "image_paths": [ "../answer-assets/9702_w21_ms_43-p08.png" ] }, { "id": "9702-2021-on-43-q03", "question_id": "9702-2021-on-43-q03", "subject": "9702", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 3, "topic": "Thermodynamics", "topic_slug": "9702-topic-16-thermodynamics", "marks": 13, "status": "available", "reason": null, "text": "3(a) (thermal) energy per unit mass (to cause temperature change) B1\n(thermal) energy per unit change in temperature B1\n\n3(b)(i) (T =) pV / Nk B1\n\n3(b)(ii) (pV =) NkT = ⅓Nm M1\nor\npV = NkT and pV = ⅓Nm\nleading to ½m = (3/2)kT and ½m = E A1\nK\n\n3(b)(iii) internal energy = ΣE (of molecules) + ΣE (of molecules) B1\nK P\nor\nno forces between molecules\npotential energy of molecules is zero B1\n\n3(c)(i) increase in internal energy = Q + work done B1\nconstant volume so no work done B1\n\n3(c)(ii) c = Q / NmΔT C1\n= [N × (3/2)kΔT] / (NmΔT) = 3k / 2m A1\n\n3(d) (as it expands) gas does work (against the atmosphere/external pressure) B1\nfor same temperature rise) more (thermal) energy needed, so larger specific heat capacity B1\n© UCLES 2021 Page 9 of 15", "source_pages": [ 9 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9702_w21_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-Oct-Nov/9702_w21_ms_43.pdf?download=true", "html": "9702-topic-16-thermodynamics/answers.html", "image_paths": [ "../answer-assets/9702_w21_ms_43-p09.png" ] }, { "id": "9702-2021-on-43-q04", "question_id": "9702-2021-on-43-q04", "subject": "9702", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 4, "topic": "Oscillations", "topic_slug": "9702-topic-17-oscillations", "marks": 11, "status": "available", "reason": null, "text": "4(a)(i) 5.0 cm A1\n\n4(a)(ii) ω = 2π / T C1\nor\nω = 2πf and f = 1 / T\nω = 2π / 4.0 A1\n= 1.6 rad s–1\n\n4(a)(iii) v = ωx C1\n0 0\n= 1.57 × 5.0 A1\n= 7.9 cm s–1\n\n4(b) • initial pull was to the right B3\n• distance from X to trolley (at equilibrium) is 20 cm\n• period is 4.0 s\n• initial motion undamped\n• motion becomes damped at/from 12 s\n• damping is light\n• maximum speed at 1s, 3s, etc. / stationary at 2s, 4s, etc.\nAny three points, 1 mark each\n\n4(c) sketch: closed loop encircling (20, 0) B1\nminimum L shown as 15 cm and maximum L shown as 25 cm B1\nminimum v shown as –7.9 cm s–1 and maximum v shown as +7.9 cm s–1 B1\n© UCLES 2021 Page 10 of 15", "source_pages": [ 10 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9702_w21_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-Oct-Nov/9702_w21_ms_43.pdf?download=true", "html": "9702-topic-17-oscillations/answers.html", "image_paths": [ "../answer-assets/9702_w21_ms_43-p10.png" ] }, { "id": "9702-2021-on-43-q05", "question_id": "9702-2021-on-43-q05", "subject": "9702", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 5, "topic": "Alternating currents", "topic_slug": "9702-topic-21-alternating-currents", "marks": 8, "status": "available", "reason": null, "text": "5(a) • noise can be removed/signal can be regenerated B2\n• extra bits can be added for error-checking\n• signal can be encrypted (for increased security)\n• data compression/multiplexing is possible\nAny two points, 1 mark each\n\n5(b)(i) 4ms: 0101 and 8ms: 0100 B1\n\n5(b)(ii) sketch: horizontal line continues to 8ms, then new horizontal line from 8ms to 12ms B1\nlevel of line after 8ms is 4 mV B1\n\n5(c) sketch: series of steps of width 2ms B1\nstep heights at 0, 2, 4, 6, 4, 6 mV B2\n2 marks if all correct, 1 mark if only one incorrect\nQuestion Answer Marks", "source_pages": [ 11 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9702_w21_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-Oct-Nov/9702_w21_ms_43.pdf?download=true", "html": "9702-topic-21-alternating-currents/answers.html", "image_paths": [ "../answer-assets/9702_w21_ms_43-p11.png" ] }, { "id": "9702-2021-on-43-q06", "question_id": "9702-2021-on-43-q06", "subject": "9702", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 6, "topic": "Capacitance", "topic_slug": "9702-topic-19-capacitance", "marks": 5, "status": "available", "reason": null, "text": "6(a) Q = CV and E = ½CV2 B1\n\n6(b)(i) C = CL / (L – D) B1\nN\n\n6(b)(ii) (charge is unchanged by moving the plates so) Q = CV B1\nN\n\n6(b)(iii) V = Q / C B1\nN N N\n= (CV) / [CL / (L – D)]\n= V(L – D) / L\n\n6(c) oppositely charged plates attract, so energy stored decreases B1\n© UCLES 2021 Page 11 of 15", "source_pages": [ 11 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9702_w21_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-Oct-Nov/9702_w21_ms_43.pdf?download=true", "html": "9702-topic-19-capacitance/answers.html", "image_paths": [ "../answer-assets/9702_w21_ms_43-p11.png" ] }, { "id": "9702-2021-on-43-q07", "question_id": "9702-2021-on-43-q07", "subject": "9702", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 7, "topic": "Alternating currents", "topic_slug": "9702-topic-21-alternating-currents", "marks": 9, "status": "available", "reason": null, "text": "7(a) • infinite (open-loop) gain B2\n• infinite slew rate\n• infinite input impedance\n• zero output impedance\n• infinite bandwidth\nAny two points, 1 mark each\n\n7(b) X: thermistor and Y: relay B1\n\n7(c)(i) (any) difference in voltage at the inputs causes output to saturate (because gain is very large) B1\nsaturates positively if V+ > V– and saturates negatively if V+ < V– B1\n\n7(c)(ii) comparator B1\n\n7(c)(iii) temperature M1\nabove a particular value A1\n\n7(c)(iv) to adjust the temperature (at which the lamp illuminates/extinguishes) B1\n© UCLES 2021 Page 12 of 15", "source_pages": [ 12 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9702_w21_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-Oct-Nov/9702_w21_ms_43.pdf?download=true", "html": "9702-topic-21-alternating-currents/answers.html", "image_paths": [ "../answer-assets/9702_w21_ms_43-p12.png" ] }, { "id": "9702-2021-on-43-q08", "question_id": "9702-2021-on-43-q08", "subject": "9702", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 8, "topic": "Magnetic fields", "topic_slug": "9702-topic-20-magnetic-fields", "marks": 6, "status": "available", "reason": null, "text": "8(a) newton per ampere per metre M1\nwhere current/wire is perpendicular to magnetic field A1\n\n8(b)(i) F = BILsinθ C1\nB = 1.0 / (5.0 × 0.060 × sin 50°) A1\n= 4.4 mT\n\n8(b)(ii) (from Fleming’s left-hand rule) force on wire is upwards, so reading decreases B1\n\n8(b)(iii) frame will rotate (so that PQ becomes perpendicular to the field) B1\nQuestion Answer Marks", "source_pages": [ 13 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9702_w21_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-Oct-Nov/9702_w21_ms_43.pdf?download=true", "html": "9702-topic-20-magnetic-fields/answers.html", "image_paths": [ "../answer-assets/9702_w21_ms_43-p13.png" ] }, { "id": "9702-2021-on-43-q09", "question_id": "9702-2021-on-43-q09", "subject": "9702", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 9, "topic": "Ideal gases", "topic_slug": "9702-topic-15-ideal-gases", "marks": 9, "status": "available", "reason": null, "text": "9(a) constant voltage M1\nthat produces/dissipates same power as (the mean power of) the alternating voltage A1\n\n9(b)(i) (maximum) rate of cutting of (magnetic) flux doubles B1\n(peak and hence) r.m.s. induced e.m.f. doubles B1\n\n9(b)(ii) sketch: (sinusoidal) wave of period 10 ms B1\npeak E shown as ± 34V B2\n(1 mark out of 2 awarded if peak E shown as ± 17V or ± 24V)\n\n9(c) current in the coil results in forces that oppose its rotation B1\nor\ncurrent in the resistor dissipates the energy of rotation\ncoil stops rotating B1\n© UCLES 2021 Page 13 of 15", "source_pages": [ 13 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9702_w21_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-Oct-Nov/9702_w21_ms_43.pdf?download=true", "html": "9702-topic-15-ideal-gases/answers.html", "image_paths": [ "../answer-assets/9702_w21_ms_43-p13.png" ] }, { "id": "9702-2021-on-43-q10", "question_id": "9702-2021-on-43-q10", "subject": "9702", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 10, "topic": "Quantum physics", "topic_slug": "9702-topic-22-quantum-physics", "marks": 9, "status": "available", "reason": null, "text": "10(a)(i) photoelectric effect B1\n\n10(a)(ii) electron diffraction B1\n\n10(b)(i) λ = h / p M1\nh is the Planck constant A1\n\n10(b)(ii) de Broglie (wavelength) B1\n\n10(c)(i) ½mv2 = eV C1\n½ × 9.11 × 10–31 × v2 = 1.60 × 10–19 × 4800 so v = 4.1 × 107 m s–1 A1\n\n10(c)(ii) λ = h / mv C1\n= 6.63 × 10–34 / (9.11 × 10–31 × 4.1 × 107)\n= 1.8 × 10–11 m A1\n© UCLES 2021 Page 14 of 15", "source_pages": [ 14 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9702_w21_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-Oct-Nov/9702_w21_ms_43.pdf?download=true", "html": "9702-topic-22-quantum-physics/answers.html", "image_paths": [ "../answer-assets/9702_w21_ms_43-p14.png" ] }, { "id": "9702-2021-on-43-q11", "question_id": "9702-2021-on-43-q11", "subject": "9702", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 11, "topic": "Medical physics", "topic_slug": "9702-topic-24-medical-physics", "marks": 7, "status": "available", "reason": null, "text": "11(a)(i) ease with which edges can be distinguished B1\n\n11(a)(ii) difference in degrees of blackening B1\n\n11(b) I = I exp (–μx) C1\n0\n0.12 = exp (–μ × 2.3) C1\nln 0.12 = –2.3 × μ\nμ = 0.92 cm–1 A1\n\n11(c) advantage: produces 3-dimensional image B1\ndisadvantage: (much) greater exposure to radiation B1\nQuestion Answer Marks", "source_pages": [ 15 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9702_w21_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-Oct-Nov/9702_w21_ms_43.pdf?download=true", "html": "9702-topic-24-medical-physics/answers.html", "image_paths": [ "../answer-assets/9702_w21_ms_43-p15.png" ] }, { "id": "9702-2021-on-43-q12", "question_id": "9702-2021-on-43-q12", "subject": "9702", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 12, "topic": "Nuclear physics", "topic_slug": "9702-topic-23-nuclear-physics", "marks": 7, "status": "available", "reason": null, "text": "12(a) probability of decay (of a nucleus) M1\nper unit time A1\n\n12(b) A = λN C1\nN = mass / (nucleon number × u) C1\n2.92 × 109 = (λ × 5.87 × 10–10) / (131 × 1.66 × 10–27) A1\nλ = 1.08 × 10–6 s–1\n\n12(c) • sample emits radiation in all directions B2\n• some radiation is absorbed by air/detector window\n• self-absorption within the source\n• dead time/inefficiency of detector\nAny two points, 1 mark each\n© UCLES 2021 Page 15 of 15", "source_pages": [ 15 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9702_w21_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-Oct-Nov/9702_w21_ms_43.pdf?download=true", "html": "9702-topic-23-nuclear-physics/answers.html", "image_paths": [ "../answer-assets/9702_w21_ms_43-p15.png" ] }, { "id": "9702-2021-on-51-q01", "question_id": "9702-2021-on-51-q01", "subject": "9702", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "51", "question_number": 1, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "1 Defining the problem\ndiameter/d is the independent variable and frequency/f is the dependent variable or vary d and measure f 1\nkeep L constant or length (of tube) constant 1\nMethods of data collection\nlabelled diagram of workable experiment including: 1\n• tube supported\n• (loud)speaker positioned in line with the tube\n• (loud)speaker labelled\nlabelled microphone, positioned outside tube in line with tube, connected to labelled oscilloscope or correct circuit symbol 1\nadjust/change frequency until maximum amplitude detected 1\nuse calipers to measure d 1\n© UCLES 2021 Page 6 of 11\n\n1 Method of analysis\n\n1 1 1\nplot a graph of against d or d against\nf f\n(Do not accept logarithmic graphs.)\n\n1 1 1\nfor against d for d against\nf f\nv = gradient × k\nor\n2L or\nv =\ny-intercept v = −\ngradient×2L\ny-intercept\n\n1 1 1\nfor against d for d against\nf f\nk = gradient × v\nor\nor 2L\nk =\ngradient×2L\nk =−\ny-intercept\ny-intercept\n© UCLES 2021 Page 7 of 11\n\n1 Additional detail including safety considerations 6\nD1 wear ear defenders (to prevent damage to hearing/to avoid loud sounds)\nor\nuse a low volume to prevent damage to hearing/to avoid loud sounds\nD2 use a rule to measure L\nD3 increase frequency from a low frequency to the first maximum amplitude\nD4 method to determine f at maximum amplitude, e.g. increase frequency to f, then continue increasing frequency, and\nthen decrease frequency until value of f determined\nD5 method to determine period from oscilloscope, e.g. no. of divisions × time-base\nD6 for frequency/time period determined by oscilloscope, f = 1 / T\nD7 repeat measurements of d and average in different directions/positions or along the tube\nD8 perform experiment in a quiet room\nD9 signal generator connected to (loud)speaker in diagram\nD10 relationship valid if a straight line produced\n(Do not accept through the origin.)\n© UCLES 2021 Page 8 of 11", "source_pages": [ 6, 7, 8 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9702_w21_ms_51.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-Oct-Nov/9702_w21_ms_51.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_w21_ms_51-p06.png", "../answer-assets/9702_w21_ms_51-p07.png", "../answer-assets/9702_w21_ms_51-p08.png" ] }, { "id": "9702-2021-on-51-q02", "question_id": "9702-2021-on-51-q02", "subject": "9702", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "51", "question_number": 2, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "2(a) t 1\ngradient = –\nC\ny-intercept = ln E\n\n2(b) 1\n(R + R ) / kΩ 1\n1 2 / 10–6 Ω\nR +R\n1 2\n55 (± 3) 18 or 18.2 ± 0.9\n69 (± 3 or 4) 14 or 14.5 ± 0.7\n90 (± 4 or 5) 11 or 11.1 ± 0.6\n80 (± 4) 13 or 12.5 ± 0.6\n101 (± 5) 9.9 or 9.90 or 9.901 ± 0.5\n115 (± 6) 8.7 or 8.70 or 8.696 ± 0.4\n1\nValues of (R + R ) and correct as shown above.\n1 2 R +R\n1 2\n1 1\nAbsolute uncertainties in from ± 0.9 or ± 1 to ± 0.4 or ± 0.5.\nR +R\n1 2\n\n2(c)(i) Six points plotted correctly. 1\nMust be accurate to half a small square. Diameter of points must be less than half a small square.\n1 1\nError bars in plotted correctly.\nR +R\n1 2\nAll error bars must be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.\n© UCLES 2021 Page 9 of 11\n\n2(c)(ii) Line of best fit drawn. 1\nPoints must be balanced. Do not accept line from top point to bottom point.\nLine must pass between (10.2, 1.10) and (10.8, 1.10) and between (16.7, 0.40) and (17.2, 0.40).\nWorst acceptable line drawn (steepest or shallowest possible line that passes through all error bars). 1\nAll error bars must be plotted.\n\n2(c)(iii) Negative gradient determined with clear substitution of data points into Δy/Δx. 1\nDistance between data points must be at least half the length of the drawn line.\nGradient of worst acceptable line determined. 1\nuncertainty = (gradient of line of best fit – gradient of worst acceptable line)\nor\nuncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)\n\n2(c)(iv) y-intercept determined by substitution of point on line into y = mx + c. 1\ny-intercept of worst acceptable line determined by substitution of point on line into y = mx + c. 1\nuncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line\nor\nuncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)\nDo not accept ECF from false origin method.\n\n2(d)(i) C determined using gradient and C and E both given to two or three significant figures. 1\nt 60\nC =− =−\ngradient (c)(iii)\nE determined using y-intercept and C and E both given with correct SI unit. 1\nE =ey-intercept\nunit of C: F or C V–1 or s Ω–1\nunit of E: V\n© UCLES 2021 Page 10 of 11\n\n2(d)(ii) Percentage uncertainty determined with method shown. 1\n 1 Δgradient\npercentage uncertainty =  +  ×100\n60 gradient \nClear substitution must be shown for maximum/minimum methods.\n\n2(e) (R + R ) determined to at least two significant figures from (d)(i) or (c)(iii) and (c)(iv) with correct substitution including 1\n1 2\nsigns and correct power of ten(s).\nDo not accept ECF for POT from (c)(iii), (c)(iv) or (d).\n( R +R )=− t × 1 = − 60 × 1\n1 2 C V C lnV −lnE\nln\nE\nor\n( R +R )= gradient = (c)(iii)\n1 2 ln5.0−y-intercept 1.61−(c)(iv)\n© UCLES 2021 Page 11 of 11", "source_pages": [ 9, 10, 11 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9702_w21_ms_51.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-Oct-Nov/9702_w21_ms_51.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_w21_ms_51-p09.png", "../answer-assets/9702_w21_ms_51-p10.png", "../answer-assets/9702_w21_ms_51-p11.png" ] }, { "id": "9702-2021-on-52-q01", "question_id": "9702-2021-on-52-q01", "subject": "9702", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "52", "question_number": 1, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "1 Defining the problem\nθ is the independent variable and x is the dependent variable or vary θ and measure x 1\nkeep (angle) β constant 1\nMethods of data collection\nlabelled diagram of workable experiment including: 1\n• spring attached at both ends e.g. one end connected to a clamp and stand\n• strip free to move\n• at least two labels from: clamp, stand, wire, strip, spring, bench\n(Do not accept extra masses added to strip.)\nuse a rule to measure L and d 1\nuse a protractor to measure θ 1\nor\nuse a rule to measure appropriate distances to determine θ by trigonometry methods\nmeasure original length of spring and new length of spring using rule/calipers 1\nMethod of analysis\nplot a graph of x against cos θ or cos θ against x 1\n(Allow log x against log (cos θ).)\nrelationship is valid if a straight line passing through the origin is produced 1\n(Allow straight line with gradient = 1 for log-log graph.)\nfor x against cos θ for cos θ against x 1\nW =\ngradient×2kdsinβ\nor W =\n2kdsinβ\nL gradient×L\n© UCLES 2021 Page 6 of 10\n\n1 Additional detail including safety considerations 6\nD1 wear goggles to prevent spring/wire/strip entering into eyes\nor\n(retort) stand used to support spring is clamped to bench\nD2 keep distance d constant\nD3 description of (separate) experiment to determine k, e.g. weigh mass and measure extension\nD4 k = weight / extension or mg / extension or gradient of weight–extension graph for candidate’s workable (separate)\nexperiment\nD5 method to prevent strip at point P sliding, e.g. use adhesive putty/hinge\n(Do not accept methods that prevent rotation at point P.)\nD6 use fiducial markers on spring at both ends or measure length of spring on both sides and average\nD7 method to attach wire to strip, e.g. wire wrapped around the strip/(strong) tape/drill hole and tie wire\nD8 determine x by subtracting original length of spring from new length\nD9 adjust support of spring to keep β constant\nD10 protractor correctly positioned on diagram to measure θ\nor\ncorrect trigonometric relationship given for θ\n© UCLES 2021 Page 7 of 10", "source_pages": [ 6, 7 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9702_w21_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-Oct-Nov/9702_w21_ms_52.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_w21_ms_52-p06.png", "../answer-assets/9702_w21_ms_52-p07.png" ] }, { "id": "9702-2021-on-52-q02", "question_id": "9702-2021-on-52-q02", "subject": "9702", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "52", "question_number": 2, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "2(a) gradient = –μ 1\ny-intercept = ln R\n0\n\n2(b) 1\naverage t / mm ln (R / s–1)\n0.16 ± 0.03 3.865 or 3.8649\n0.25 ± 0.03 3.784 or 3.7842\n0.42 ± 0.03 3.643 or 3.6428\n0.56 ± 0.02 3.535 or 3.5351\n0.66 ± 0.02 3.456 or 3.4563\n0.76 ± 0.02 3.391 or 3.3911\nValues of average t and ln R correct as shown above.\nAbsolute uncertainties in average t correct as shown above. 1\n\n2(c)(i) Six points plotted correctly. 1\nMust be accurate to nearest half a small square. Diameter of points must be less than half a small square.\nError bars in average t plotted correctly. 1\nAll error bars must be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.\n© UCLES 2021 Page 8 of 10\n\n2(c)(ii) Line of best fit drawn. 1\nPoints must be balanced. Do not accept line from top point to bottom point.\nLine must pass between (0.22, 3.80) and (0.24, 3.80) and between (0.60, 3.50) and (0.62, 3.50).\nWorst acceptable line drawn (steepest or shallowest possible line that passes through all error bars). 1\nAll error bars must be plotted.\n\n2(c)(iii) Negative gradient determined with clear substitution of data points into Δy / Δx. 1\nDistance between data points must be at least half the length of the drawn line.\nGradient of worst acceptable line determined. 1\nuncertainty = (gradient of line of best fit – gradient of worst acceptable line)\nor\nuncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)\n\n2(c)(iv) y-intercept determined by substitution of point on line into y = mx + c. 1\ny-intercept of worst acceptable line determined by substitution of point on line into y = mx + c. 1\nuncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line\nor\nuncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)\nDo not accept ECF from false origin method.\n© UCLES 2021 Page 9 of 10\n\n2(d) μ = – gradient value 1\nDo not accept negative values (from a negative gradient).\nR determined using y-intercept and μ and R both given with valid SI unit. 1\n0 0\nR =ey-intercept\n0\nunit of μ: mm–1\nunit of R : s–1\n0\nabsolute uncertainty in μ = absolute uncertainty in gradient 1\nand\nabsolute uncertainty in R = ey−intercept of WAL −R\n0 0\nCorrect substitution of numbers must be seen.\n\n2(e) Value of t determined to two or three significant figures from (d) or (c)(iii) and (c)(iv) with correct substitution and correct 1\npower of ten(s).\nDo not accept ECF for POT from (c)(iii), (c)(iv) or (d).\nlnR −lnR ln20−lnR\nt = 0 = 0\n−μ −μ\nor\nln20−y-intercept 2.996−(c)(iv)\nt = =\ngradient (c)(iii)\n© UCLES 2021 Page 10 of 10", "source_pages": [ 8, 9, 10 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9702_w21_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-Oct-Nov/9702_w21_ms_52.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_w21_ms_52-p08.png", "../answer-assets/9702_w21_ms_52-p09.png", "../answer-assets/9702_w21_ms_52-p10.png" ] }, { "id": "9702-2021-on-53-q01", "question_id": "9702-2021-on-53-q01", "subject": "9702", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "53", "question_number": 1, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "1 Defining the problem\ndiameter/d is the independent variable and frequency/f is the dependent variable or vary d and measure f 1\nkeep L constant or length (of tube) constant 1\nMethods of data collection\nlabelled diagram of workable experiment including: 1\n• tube supported\n• (loud)speaker positioned in line with the tube\n• (loud)speaker labelled\nlabelled microphone, positioned outside tube in line with tube, connected to labelled oscilloscope or correct circuit symbol 1\nadjust/change frequency until maximum amplitude detected 1\nuse calipers to measure d 1\n© UCLES 2021 Page 6 of 11\n\n1 Method of analysis\n\n1 1 1\nplot a graph of against d or d against\nf f\n(Do not accept logarithmic graphs.)\n\n1 1 1\nfor against d for d against\nf f\nv = gradient × k\nor\n2L or\nv =\ny-intercept v = −\ngradient×2L\ny-intercept\n\n1 1 1\nfor against d for d against\nf f\nk = gradient × v\nor\nor 2L\nk =\ngradient×2L\nk =−\ny-intercept\ny-intercept\n© UCLES 2021 Page 7 of 11\n\n1 Additional detail including safety considerations 6\nD1 wear ear defenders (to prevent damage to hearing/to avoid loud sounds)\nor\nuse a low volume to prevent damage to hearing/to avoid loud sounds\nD2 use a rule to measure L\nD3 increase frequency from a low frequency to the first maximum amplitude\nD4 method to determine f at maximum amplitude, e.g. increase frequency to f, then continue increasing frequency, and\nthen decrease frequency until value of f determined\nD5 method to determine period from oscilloscope, e.g. no. of divisions × time-base\nD6 for frequency/time period determined by oscilloscope, f = 1 / T\nD7 repeat measurements of d and average in different directions/positions or along the tube\nD8 perform experiment in a quiet room\nD9 signal generator connected to (loud)speaker in diagram\nD10 relationship valid if a straight line produced\n(Do not accept through the origin.)\n© UCLES 2021 Page 8 of 11", "source_pages": [ 6, 7, 8 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9702_w21_ms_53.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-Oct-Nov/9702_w21_ms_53.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_w21_ms_53-p06.png", "../answer-assets/9702_w21_ms_53-p07.png", "../answer-assets/9702_w21_ms_53-p08.png" ] }, { "id": "9702-2021-on-53-q02", "question_id": "9702-2021-on-53-q02", "subject": "9702", "year": 2021, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "53", "question_number": 2, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "2(a) t 1\ngradient = –\nC\ny-intercept = ln E\n\n2(b) 1\n(R + R ) / kΩ 1\n1 2 / 10–6 Ω\nR +R\n1 2\n55 (± 3) 18 or 18.2 ± 0.9\n69 (± 3 or 4) 14 or 14.5 ± 0.7\n90 (± 4 or 5) 11 or 11.1 ± 0.6\n80 (± 4) 13 or 12.5 ± 0.6\n101 (± 5) 9.9 or 9.90 or 9.901 ± 0.5\n115 (± 6) 8.7 or 8.70 or 8.696 ± 0.4\n1\nValues of (R + R ) and correct as shown above.\n1 2 R +R\n1 2\n1 1\nAbsolute uncertainties in from ± 0.9 or ± 1 to ± 0.4 or ± 0.5.\nR +R\n1 2\n\n2(c)(i) Six points plotted correctly. 1\nMust be accurate to half a small square. Diameter of points must be less than half a small square.\n1 1\nError bars in plotted correctly.\nR +R\n1 2\nAll error bars must be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.\n© UCLES 2021 Page 9 of 11\n\n2(c)(ii) Line of best fit drawn. 1\nPoints must be balanced. Do not accept line from top point to bottom point.\nLine must pass between (10.2, 1.10) and (10.8, 1.10) and between (16.7, 0.40) and (17.2, 0.40).\nWorst acceptable line drawn (steepest or shallowest possible line that passes through all error bars). 1\nAll error bars must be plotted.\n\n2(c)(iii) Negative gradient determined with clear substitution of data points into Δy/Δx. 1\nDistance between data points must be at least half the length of the drawn line.\nGradient of worst acceptable line determined. 1\nuncertainty = (gradient of line of best fit – gradient of worst acceptable line)\nor\nuncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)\n\n2(c)(iv) y-intercept determined by substitution of point on line into y = mx + c. 1\ny-intercept of worst acceptable line determined by substitution of point on line into y = mx + c. 1\nuncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line\nor\nuncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)\nDo not accept ECF from false origin method.\n\n2(d)(i) C determined using gradient and C and E both given to two or three significant figures. 1\nt 60\nC =− =−\ngradient (c)(iii)\nE determined using y-intercept and C and E both given with correct SI unit. 1\nE =ey-intercept\nunit of C: F or C V–1 or s Ω–1\nunit of E: V\n© UCLES 2021 Page 10 of 11\n\n2(d)(ii) Percentage uncertainty determined with method shown. 1\n 1 Δgradient\npercentage uncertainty =  +  ×100\n60 gradient \nClear substitution must be shown for maximum/minimum methods.\n\n2(e) (R + R ) determined to at least two significant figures from (d)(i) or (c)(iii) and (c)(iv) with correct substitution including 1\n1 2\nsigns and correct power of ten(s).\nDo not accept ECF for POT from (c)(iii), (c)(iv) or (d).\n( R +R )=− t × 1 = − 60 × 1\n1 2 C V C lnV −lnE\nln\nE\nor\n( R +R )= gradient = (c)(iii)\n1 2 ln5.0−y-intercept 1.61−(c)(iv)\n© UCLES 2021 Page 11 of 11", "source_pages": [ 9, 10, 11 ], "source_pdf": "_source-pdfs/2021-Oct-Nov/ms/9702_w21_ms_53.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2021-Oct-Nov/9702_w21_ms_53.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_w21_ms_53-p09.png", "../answer-assets/9702_w21_ms_53-p10.png", "../answer-assets/9702_w21_ms_53-p11.png" ] }, { "id": "9702-2022-m-42-q01", "question_id": "9702-2022-m-42-q01", "subject": "9702", "year": 2022, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 1, "topic": "Gravitational fields", "topic_slug": "9702-topic-13-gravitational-fields", "marks": 10, "status": "available", "reason": null, "text": "1(a) at least 4 straight radial lines to P B1\nall arrows pointing along the lines towards P B1\n\n1(b) Any 2 from: B2\ngravitational force provides the centripetal force\n(centripetal or gravitational) force has constant magnitude\n(centripetal or gravitational) force is perpendicular to velocity (of moon) / direction of motion (of moon)\n\n1(c)(i) GMm M1\n= mrω2\nr2\nr3ω2 gradient A1\nM= and gradient = r3ω2 hence M=\nG G\nor\ngradient\nr3 = GM × 1/ω2 so gradient = GM hence M=\nG\n\n1(c)(ii) M = 4.1 × 1023 / (6.0 × 107 × 6.67 × 10–11) = 1.0 × 1026 kg B1\n© UCLES 2022 Page 6 of 17\n\n1(c)(iii) GMm mv2 C1\n=\nr2 r\nGM\n= v2\nr\n6.67×10 −11×1.0×1026 C1\nv2=\n1.2×108\nv2 =5.6×107m s −1\nv =7500 m s −1 A1\nQuestion Answer Marks", "source_pages": [ 6, 7 ], "source_pdf": "_source-pdfs/2022-Feb-March/ms/9702_m22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Feb-March/9702_m22_ms_42.pdf?download=true", "html": "9702-topic-13-gravitational-fields/answers.html", "image_paths": [ "../answer-assets/9702_m22_ms_42-p06.png", "../answer-assets/9702_m22_ms_42-p07.png" ] }, { "id": "9702-2022-m-42-q02", "question_id": "9702-2022-m-42-q02", "subject": "9702", "year": 2022, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 2, "topic": "Thermodynamics", "topic_slug": "9702-topic-16-thermodynamics", "marks": 10, "status": "available", "reason": null, "text": "2(a) 0 B1\n\n2(b) pV = nRT C1\n(n =) 1.5 × 105 × 4.2 × 10–3 / 8.31 × 540\n= 0.14 mol A1\n\n2(c) missing pressure 1.5 (× 105) B1\nboth missing volumes 1.8 (× 10–3) B1\n\n2(d)(i) (ΔU:) increase in internal energy (of the system) B1\n(q:) thermal energy supplied to the system B1\n(W:) work done on system B1\n© UCLES 2022 Page 7 of 17\n\n2(d)(ii) volume increases and work is done by the gas B1\ntemperature decreases and internal energy decreases B1\nQuestion Answer Marks", "source_pages": [ 7, 8 ], "source_pdf": "_source-pdfs/2022-Feb-March/ms/9702_m22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Feb-March/9702_m22_ms_42.pdf?download=true", "html": "9702-topic-16-thermodynamics/answers.html", "image_paths": [ "../answer-assets/9702_m22_ms_42-p07.png", "../answer-assets/9702_m22_ms_42-p08.png" ] }, { "id": "9702-2022-m-42-q03", "question_id": "9702-2022-m-42-q03", "subject": "9702", "year": 2022, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 3, "topic": "Oscillations", "topic_slug": "9702-topic-17-oscillations", "marks": 10, "status": "available", "reason": null, "text": "3(a) upthrust, weight B1\n\n3(b) upthrust greater than weight so (resultant force is) upwards B1\n\n3(c)(i) A, g and ρ all constant so F ∝ x B1\nminus sign means F and x are in opposite directions B1\n\n3(c)(ii) F Agρx M1\n(a = so) a = (−)\nm m\nAgρ Agρ A1\nso ω2 = hence ω =\nm m\n\n3(d)(i) damping due to viscous forces B1\n\n3(d)(ii) ( E = )1 mω2x 2 C1\n2 0\nω2 = (–) gradient C1\n( E = )1 mω2(x 2 − x 2) A1\n2 1 2\n= 1 ×0.57 ×(2.3 )(0.0202 −0.0162)\n2 0.020\n= 4.7 ×10 −3 J\n© UCLES 2022 Page 8 of 17", "source_pages": [ 8 ], "source_pdf": "_source-pdfs/2022-Feb-March/ms/9702_m22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Feb-March/9702_m22_ms_42.pdf?download=true", "html": "9702-topic-17-oscillations/answers.html", "image_paths": [ "../answer-assets/9702_m22_ms_42-p08.png" ] }, { "id": "9702-2022-m-42-q04", "question_id": "9702-2022-m-42-q04", "subject": "9702", "year": 2022, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 4, "topic": "Electric fields", "topic_slug": "9702-topic-18-electric-fields", "marks": 6, "status": "available", "reason": null, "text": "4(a) direction of force B1\nforce on a positive charge B1\n\n4(b)(i) Q C1\nV =\n4πε r\no\n4.0×10 −9 −7.2×10 −9\n+ = 0\n4πε x 4πε (0.120 − x)\no o\n( )\n\n4 0.120 − x = 7.2 x\nx = 0.043 m A1\n\n4(b)(ii) fields are in the same direction so no B1\n\n4(b)(iii) straight arrow drawn leftwards from X in direction between extended line joining Q and X and the horizontal B1\n© UCLES 2022 Page 9 of 17", "source_pages": [ 9 ], "source_pdf": "_source-pdfs/2022-Feb-March/ms/9702_m22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Feb-March/9702_m22_ms_42.pdf?download=true", "html": "9702-topic-18-electric-fields/answers.html", "image_paths": [ "../answer-assets/9702_m22_ms_42-p09.png" ] }, { "id": "9702-2022-m-42-q05", "question_id": "9702-2022-m-42-q05", "subject": "9702", "year": 2022, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 5, "topic": "Capacitance", "topic_slug": "9702-topic-19-capacitance", "marks": 10, "status": "available", "reason": null, "text": "5(a) (energy stored =) area under line or ½ QV C1\n= ½ × 8.0 × 1.2 × 10-4\n= 4.8 × 10–4 J A1\n\n5(b)(i) (τ=) RC C1\n(τ=) 220 × 103 × (1.2 × 10-4/8.0) = 3.3 s A1\n\n5(b)(ii) E ∝ V2 C1\n(so time to) V / 3 C1\no\n−t\nV = V e RC\no\nV −t C1\no = V e 3.3\n3 o\n1 −t\n= e 3.3\n3\nt = 3.6 s A1\n\n5(c) (total) capacitance is doubled M1\ntime constant is doubled A1\n© UCLES 2022 Page 10 of 17", "source_pages": [ 10 ], "source_pdf": "_source-pdfs/2022-Feb-March/ms/9702_m22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Feb-March/9702_m22_ms_42.pdf?download=true", "html": "9702-topic-19-capacitance/answers.html", "image_paths": [ "../answer-assets/9702_m22_ms_42-p10.png" ] }, { "id": "9702-2022-m-42-q06", "question_id": "9702-2022-m-42-q06", "subject": "9702", "year": 2022, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 6, "topic": "Magnetic fields", "topic_slug": "9702-topic-20-magnetic-fields", "marks": 7, "status": "available", "reason": null, "text": "6(a) less in smaller solenoid B1\n\n6(b) greater in smaller solenoid B1\n\n6(c)(i) direction of (induced) e.m.f. M1\nsuch as to (produce effects that) oppose the change that caused it A1\n\n6(c)(ii) change of flux (linkage) in smaller solenoid induces e.m.f. in smaller solenoid B1\n(induced) current in smaller solenoid causes field around it B1\nthe two fields (interact to) create an attractive force B1\nQuestion Answer Marks", "source_pages": [ 11 ], "source_pdf": "_source-pdfs/2022-Feb-March/ms/9702_m22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Feb-March/9702_m22_ms_42.pdf?download=true", "html": "9702-topic-20-magnetic-fields/answers.html", "image_paths": [ "../answer-assets/9702_m22_ms_42-p11.png" ] }, { "id": "9702-2022-m-42-q07", "question_id": "9702-2022-m-42-q07", "subject": "9702", "year": 2022, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 7, "topic": "Alternating currents", "topic_slug": "9702-topic-21-alternating-currents", "marks": 6, "status": "available", "reason": null, "text": "7(a)(i) two diodes added in correct directions (Both diodes pointing inwards and upwards), correct symbols only B1\n\n7(a)(ii) ‘+’ anywhere on upper output wire B1\n\n7(b)(i) ω = 2π / T C1\n= 2π / 2.5\n= 0.80 π or 4π / 5 or 2.5\n(V =) 3.5 sin (0.8π t) or 3.5 sin (4π t / 5) or 3.5 sin (2.5 t) A1\n© UCLES 2022 Page 11 of 17\n\n7(b)(ii) V2 V 2 C1\n(P=) or (P=) r.m.s.\n2R R\n3.52 2.472\n= or\n2×12 12\n= 0.51 W A1\nQuestion Answer Marks", "source_pages": [ 11, 12 ], "source_pdf": "_source-pdfs/2022-Feb-March/ms/9702_m22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Feb-March/9702_m22_ms_42.pdf?download=true", "html": "9702-topic-21-alternating-currents/answers.html", "image_paths": [ "../answer-assets/9702_m22_ms_42-p11.png", "../answer-assets/9702_m22_ms_42-p12.png" ] }, { "id": "9702-2022-m-42-q08", "question_id": "9702-2022-m-42-q08", "subject": "9702", "year": 2022, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 8, "topic": "Quantum physics", "topic_slug": "9702-topic-22-quantum-physics", "marks": 7, "status": "available", "reason": null, "text": "8(a) h h M1\nλ = or λ =\np mv\nwhere h is the Planck constant and A1\np is the momentum (of particle) / mv is the momentum (of particle) / m is the mass (of particle) and v is the velocity (of\nparticle)\n\n8(b)(i) (electron) diffraction B1\n\n8(b)(ii) moving electrons behave like waves B1\n\n8(b)(iii) spacing between atoms ≈ wavelength of electron B1\nor\ndiameter of atom ≈ wavelength of electron\n\n8(b)(iv) Any one of: M1\n• wavelength has decreased\n• electron had greater momentum\nso (accelerating) p.d. was increased A1\n© UCLES 2022 Page 12 of 17", "source_pages": [ 12 ], "source_pdf": "_source-pdfs/2022-Feb-March/ms/9702_m22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Feb-March/9702_m22_ms_42.pdf?download=true", "html": "9702-topic-22-quantum-physics/answers.html", "image_paths": [ "../answer-assets/9702_m22_ms_42-p12.png" ] }, { "id": "9702-2022-m-42-q09", "question_id": "9702-2022-m-42-q09", "subject": "9702", "year": 2022, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 9, "topic": "Nuclear physics", "topic_slug": "9702-topic-23-nuclear-physics", "marks": 11, "status": "available", "reason": null, "text": "9(a) 207, 82 for lead B1\n4, 2 for alpha B1\n\n9(b)(i) (half-life found as) 0.52 s or correctly read points substituted into C1\nN =N e −λt\n0\n0.693\nλ=\nt\n1\n2\n0.693\nλ=\n0.52\nλ = 1.3 s–1 A1\n\n9(b)(ii) A=λN A1\n= 1.3 × 24 ×1012\n= 3.1 ×1013 Bq\n\n9(b)(iii) upwards curve of decreasing gradient starting from (0,0) B1\npasses through (0.52, 12) and (1.2, 18.8) B1\n\n9(c)(i) 16 × 1012 and 7.2 × 1012 C1\n6900 × 103 × 1.6 × 10-19 C1\n(16 × 1012 – 7.2 × 1012) × 6900 × 103 × 1.6 × 10-19\n= 9.7 J A1\n© UCLES 2022 Page 13 of 17\n\n9(c)(ii) lead nuclei have kinetic energy B1\nor\ngamma photons are also emitted\nQuestion Answer Marks", "source_pages": [ 13, 14 ], "source_pdf": "_source-pdfs/2022-Feb-March/ms/9702_m22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Feb-March/9702_m22_ms_42.pdf?download=true", "html": "9702-topic-23-nuclear-physics/answers.html", "image_paths": [ "../answer-assets/9702_m22_ms_42-p13.png", "../answer-assets/9702_m22_ms_42-p14.png" ] }, { "id": "9702-2022-m-42-q10", "question_id": "9702-2022-m-42-q10", "subject": "9702", "year": 2022, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 10, "topic": "Temperature", "topic_slug": "9702-topic-14-temperature", "marks": 7, "status": "available", "reason": null, "text": "10(a) energy = mcΔT C1\nenergy = ItV C1\n0.40×0.020×75 000 ×0.95\n(ΔT =)\n0.015×130\n=290 K A1\n\n10(b) I = I e −μt C1\no\n0.20 = e\n−0.22t\nt = 7.3 cm A1\n© UCLES 2022 Page 14 of 17\n\n10(c) either M1\n(linear) attenuation coefficients / μ very different for bone and muscle\n(very) different amounts (of X-rays) absorbed so good contrast A1\nor (very) different intensities transmitted so good contrast\nor (M1)\n(linear) attenuation coefficients / μ similar for blood and muscle\nsimilar amounts (of X-rays) absorbed so poor contrast (A1)\nor similar intensities transmitted so poor contrast\nQuestion Answer Marks", "source_pages": [ 14, 15 ], "source_pdf": "_source-pdfs/2022-Feb-March/ms/9702_m22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Feb-March/9702_m22_ms_42.pdf?download=true", "html": "9702-topic-14-temperature/answers.html", "image_paths": [ "../answer-assets/9702_m22_ms_42-p14.png", "../answer-assets/9702_m22_ms_42-p15.png" ] }, { "id": "9702-2022-m-42-q11", "question_id": "9702-2022-m-42-q11", "subject": "9702", "year": 2022, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 11, "topic": "Medical physics", "topic_slug": "9702-topic-24-medical-physics", "marks": 9, "status": "available", "reason": null, "text": "11(a) substance containing radioactive nuclei that is introduced into the body or B1\nsubstance containing radioactive nuclei that is absorbed by the tissue being studied\n\n11(b)(i) a particle interacting with its antiparticle so that mass is converted into energy B1\n\n11(b)(ii) electron(s) and positron(s) B1\n\n11(c)(i) E = 2mc2 A1\n= 2×9.11×10 −31×3.00×10 −82\n= 1.64×10 −13J\n© UCLES 2022 Page 15 of 17\n\n11(c)(ii) 2hc C1\nλ =\nE\n2×6.63×10 −34 ×3.00×108\n=\n1.64×10 −13\n= 2.43 × 10 −12 m A1\n\n11(d) Any 3 from: B3\n• the two gamma photons travel in opposite directions\n• gamma photons detected (outside body / by detectors)\n• gamma photons arrive (at detector) at different times\n• determine location of production (of gamma)\n• image of tracer concentration in tissue produced\nQuestion Answer Marks", "source_pages": [ 15, 16 ], "source_pdf": "_source-pdfs/2022-Feb-March/ms/9702_m22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Feb-March/9702_m22_ms_42.pdf?download=true", "html": "9702-topic-24-medical-physics/answers.html", "image_paths": [ "../answer-assets/9702_m22_ms_42-p15.png", "../answer-assets/9702_m22_ms_42-p16.png" ] }, { "id": "9702-2022-m-42-q12", "question_id": "9702-2022-m-42-q12", "subject": "9702", "year": 2022, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 12, "topic": "Astronomy and cosmology", "topic_slug": "9702-topic-25-astronomy-and-cosmology", "marks": 7, "status": "available", "reason": null, "text": "12(a) total power of radiation emitted (by the star) B1\n\n12(b) L C1\nF =\n4πd2\n3.83×1026\n=\n4×π×1.51×10112\n= 1340 W m −2 A1\n© UCLES 2022 Page 16 of 17\n\n12(c) E A1\nm=\nc2\n3.83×1026\n=\n3.00×1082\n= 4.26×109 kg\n\n12(d) L = 4πσr2T 4 B1\n3.83×1026 = 4×π×5.67×10 −8 ×6.96×1082×T4 leading to T = 5770 K\n\n12(e) 1 C1\nλ ∝\n(max) T\n5.00×10 −7 9940\n=\nλ 5770\nλ =2.90 × 10 −7m A1\n© UCLES 2022 Page 17 of 17", "source_pages": [ 16, 17 ], "source_pdf": "_source-pdfs/2022-Feb-March/ms/9702_m22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Feb-March/9702_m22_ms_42.pdf?download=true", "html": "9702-topic-25-astronomy-and-cosmology/answers.html", "image_paths": [ "../answer-assets/9702_m22_ms_42-p16.png", "../answer-assets/9702_m22_ms_42-p17.png" ] }, { "id": "9702-2022-m-52-q01", "question_id": "9702-2022-m-52-q01", "subject": "9702", "year": 2022, "session": "March", "session_code": "m", "paper": 5, "variant": "52", "question_number": 1, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "1 Defining the problem 1\nθ is the independent variable and v is the dependent variable, or vary θ and measure v.\nKeep d constant 1\nMethods of data collection 1\nLabelled diagram of workable experiment including:\n• sheet supported by stand / jack\n• light gate positioned at X\n• support, light gate and X labelled.\nLight gate connected to timer / datalogger. 1\nMeasure length (L) (of card) interrupted by beam for single light gate. 1\nMethod to measure θ, e.g. use protractor 1\nor\nMethod to determine θ , e.g. use a rule(r) to measure two appropriate distances to use in a trigonometrical ratio\nMethod of Analysis 1\nPlots a graph of v2on y-axis and sin θ on x-axis.\nAllow other valid graphs, e.g. sin θ against v2 Do not accept log graphs.\ngradient 1\np= for v2 against sin θ\n2d\nor\n\n1\np= for sin θ against v2\n2d×gradient\n© UCLES 2022 Page 6 of 10\n\n1 m×y −intercept 1\nq =− for v2 against sin θ\n2Bd\nor\nmp×y −intercept m×y −intercept\nq = =\nB 2dB×gradient\nfor sin θ against v2\nAdditional detail including safety considerations 6\nAny six from:\nD1 Method to stop the trolley once the trolley passes X, e.g. place a block / stop on the bench near the end of the sheet\nIgnore trolley falls\nD2 Keep B and m constant\nD3 Use a rule(r) to measure d\nD4 Method to keep d constant, e.g. mark distance d on the sheet or the starting position of the trolley on the sheet\nD5 Method to measure mass of trolley (and magnet), e.g. use balance or use newton meter to measure weight and divide\nby g\nand\nMeasure B using a (calibrated) Hall probe\nD6 Additional detail on use of Hall probe, e.g.\nadjust probe until maximum value or\nmeasure B using Hall probe first in one direction, then in the opposite direction and average\nD7 Determine v (the velocity at X) from L / t (for a single light gate)\nD8 Additional detail on measuring θ, e.g. protractor drawn in correct position on diagram, or\nadditional detail on determining θ , e.g. relationship between measured lengths and θ\n© UCLES 2022 Page 7 of 10\n\n1 D9 Relationship valid if a straight line is produced (not passing through the origin)\nD10 Repeat experiment for each θ and average v.\nQuestion Answer Marks", "source_pages": [ 6, 7, 8 ], "source_pdf": "_source-pdfs/2022-Feb-March/ms/9702_m22_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Feb-March/9702_m22_ms_52.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_m22_ms_52-p06.png", "../answer-assets/9702_m22_ms_52-p07.png", "../answer-assets/9702_m22_ms_52-p08.png" ] }, { "id": "9702-2022-m-52-q02", "question_id": "9702-2022-m-52-q02", "subject": "9702", "year": 2022, "session": "March", "session_code": "m", "paper": 5, "variant": "52", "question_number": 2, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "2(a) 1 1\nGradient =\n2πfC\n\n2(b) 1\n1\n/ 10–3 Ω–1 tan θ\nR\n83 or 83.3 6.17 or 6.174\n63 or 62.5 4.51 or 4.511\n45 or 45.5 3.27 or 3.271\n30 or 30.3 2.16 or 2.164\n26 or 25.6 1.86 or 1.857\n23 or 23.3 1.68 or 1.684\n1 1\nAbsolute uncertainties in\nR\nfrom ± 4 to ± 1\n© UCLES 2022 Page 8 of 10\n\n2(c)(i) Six points from (b) plotted correctly. 1\nMust be within half a small square. Diameter of points must be less than half a small square.\n1 1\nError bars in plotted correctly.\nR\nAll error bars to be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.\n\n2(c)(ii) Straight line of best fit drawn. 1\nPoints must be balanced.\nDo not accept line from top plot to bottom plot.\nLine must pass between\n(33.5, 2.5) and (35.0, 2.5) and\n(74.0, 5.5) and (76.0, 5.5)\nWorst acceptable line drawn. 1\nSteepest or shallowest possible line that passes through all the error bars.\nAll error bars must be plotted.\n\n2(c)(iii) Gradient determined with clear substitution of data points into Δy/Δx; distance between data points must be greater than half 1\nthe length of the drawn line.\nGradient determined of WAL with clear substitution of data points into Δy/Δx; 1\nuncertainty = (gradient of line of best fit – gradient of worst acceptable line)\nor\nuncertainty = ½ (steepest worst line gradient –\nshallowest worst line gradient)\n\n2(d) 99 ± 2 (Hz) 1\n\n2(e)(i) C determined using gradient and C given to two or three significant figures. 1\n1 1\nC = =\n2πf ×gradient 2π×(d)×(c)(iii)\nC determined using gradient with correct SI unit and power of ten for C: F or s Ω–1 1\n© UCLES 2022 Page 9 of 10\n\n2(e)(ii) Percentage uncertainty in C determined with method shown. 1\n Δf Δgradient\n%uncertainty =  +  ×100\n f gradient \nOR\nCorrect substitution for max/min methods\n1\nmaxC =\n2π×minf ×mingradient\n1\nminC =\n2π×maxf ×maxgradient\n\n2(f) R determined to at least two significant figures with appropriate power of ten from (c)(iii) OR (d) and (e)(i) with correct 1\nsubstitution seen.\ngradient (c)(iii)\nR = =\ntanθ 0.839\nOR\n1 1\nR = =\n2πfCtanθ 2π×(d)×(e)(i)×0.839\nAbsolute uncertainty in R determined. 1\nMethod must be consistent with determination of R and correct substitution must be seen.\nFor R determined by using the gradient:\nΔgradient\nΔR = ×R\ngradient\nOR\nFor R determined by using (d) and (e)(i):\n\nΔf ΔC\nΔR = + ×R\n \n f C \nOR\nΔR determined by max / min methods.\n© UCLES 2022 Page 10 of 10", "source_pages": [ 8, 9, 10 ], "source_pdf": "_source-pdfs/2022-Feb-March/ms/9702_m22_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Feb-March/9702_m22_ms_52.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_m22_ms_52-p08.png", "../answer-assets/9702_m22_ms_52-p09.png", "../answer-assets/9702_m22_ms_52-p10.png" ] }, { "id": "9702-2022-mj-41-q01", "question_id": "9702-2022-mj-41-q01", "subject": "9702", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 1, "topic": "Gravitational fields", "topic_slug": "9702-topic-13-gravitational-fields", "marks": 10, "status": "available", "reason": null, "text": "1(a)(i) (gravitational) force is (directly) proportional to product of masses B1\nforce (between point masses) is inversely proportional to the square of their separation B1\n\n1(a)(ii) g = F / m C1\nF = GMm / r2 A1\nand so\ng = [GMm / r2] / m = GM / r2\n\n1(b)(i) g = (6.67  10–11  7.35  1022) / (1.74  106)2 = 1.62 N kg–1 A1\n\n1(b)(ii) fields (due to Earth and the Moon) have equal magnitudes B1\nfields (due to Earth and the Moon) are in opposite directions B1\n\n1(b)(iii) distance of X from Earth = (3.84  108 – x) C1\n(G ) 7.35  1022 / x2 = (G ) 5.98  1024 / (3.84  108 – x)2 C1\nx = 3.8  107 m A1\n© UCLES 2022 Page 7 of 16", "source_pages": [ 7 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9702_s22_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-May-June/9702_s22_ms_41.pdf?download=true", "html": "9702-topic-13-gravitational-fields/answers.html", "image_paths": [ "../answer-assets/9702_s22_ms_41-p07.png" ] }, { "id": "9702-2022-mj-41-q02", "question_id": "9702-2022-mj-41-q02", "subject": "9702", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 2, "topic": "Magnetic fields", "topic_slug": "9702-topic-20-magnetic-fields", "marks": 8, "status": "available", "reason": null, "text": "2(a)(i) (vertically) downwards B1\n\n2(a)(ii) magnetic force (on sphere) is perpendicular to its velocity B1\nmagnetic force perpendicular to velocity is the centripetal force B1\nor\nmagnetic force perpendicular to velocity causes centripetal acceleration\nor\nacceleration perpendicular to velocity is centripetal (acceleration)\nor\nmagnetic force does not change the speed of the sphere\nor\nmagnetic force has constant magnitude\n\n2(b) mg = Eq C1\nE = (1.6  10–10  9.81) / (0.27  10–9) A1\n= 5.8 N C–1\n\n2(c) centripetal force = magnetic force B1\nor\nBqv = mv2 / r\nB = mv / qr C1\n= (1.6  10–10  0.78) / (0.27  10–9  3.4) = 0.14 T A1\n© UCLES 2022 Page 8 of 16", "source_pages": [ 8 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9702_s22_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-May-June/9702_s22_ms_41.pdf?download=true", "html": "9702-topic-20-magnetic-fields/answers.html", "image_paths": [ "../answer-assets/9702_s22_ms_41-p08.png" ] }, { "id": "9702-2022-mj-41-q03", "question_id": "9702-2022-mj-41-q03", "subject": "9702", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 3, "topic": "Thermodynamics", "topic_slug": "9702-topic-16-thermodynamics", "marks": 11, "status": "available", "reason": null, "text": "3(a)(i) a gas that obeys pV  T M1\nwhere p = pressure, V = volume, T = thermodynamic temperature A1\n\n3(a)(ii) T = (273 + 17) K C1\nn = pV / RT A1\n= (1.2  105  0.24) / [8.31  (273 + 17)]\n= 12 mol\n\n3(b)(i) work done = pV C1\n= 1.2  105  (0.24 – 0.08) = 19200 J (= 19.2 kJ) A1\n\n3(b)(ii) AB work done correct (19.2) A1\nBC work done correct (0) A1\nCA increase in internal energy correct (0) and CA thermal energy correct (31.6) A1\nAB increase in internal energy calculated correctly from work done – 48.0 A1\nBC increase in internal energy correctly calculated so the final column adds up to zero and BC thermal energy same as A1\nincrease in internal energy\n(Fully correct table:\nAB 19.2 –48.0 –28.8\nBC 0 28.8 28.8\nCA –31.6 31.6 0\n)\n© UCLES 2022 Page 9 of 16", "source_pages": [ 9 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9702_s22_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-May-June/9702_s22_ms_41.pdf?download=true", "html": "9702-topic-16-thermodynamics/answers.html", "image_paths": [ "../answer-assets/9702_s22_ms_41-p09.png" ] }, { "id": "9702-2022-mj-41-q04", "question_id": "9702-2022-mj-41-q04", "subject": "9702", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 4, "topic": "Oscillations", "topic_slug": "9702-topic-17-oscillations", "marks": 8, "status": "available", "reason": null, "text": "4(a) straight line through origin shows that a is proportional to x B1\nnegative gradient shows that a is in opposite direction to x B1\n\n4(b)(i) a =  2x C1\n0 0\nor\na = – 2x\nor\n2 = – gradient\n = (0.40 / 0.050) A1\n= 2.8 rad s–1\n\n4(b)(ii) k =  2L C1\n= 2.82  1.24\n= 9.7 m s–2 A1\n\n4(c) (increasing L causes)  to decrease M1\nor\nenergy (= ½ m2x 2) = ½ mkx 2 / L (and L increases)\n0 0\nso amplitude increases A1\n© UCLES 2022 Page 10 of 16", "source_pages": [ 10 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9702_s22_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-May-June/9702_s22_ms_41.pdf?download=true", "html": "9702-topic-17-oscillations/answers.html", "image_paths": [ "../answer-assets/9702_s22_ms_41-p10.png" ] }, { "id": "9702-2022-mj-41-q05", "question_id": "9702-2022-mj-41-q05", "subject": "9702", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 5, "topic": "Alternating currents", "topic_slug": "9702-topic-21-alternating-currents", "marks": 12, "status": "available", "reason": null, "text": "5(a)(i) conversion (from a.c.) to d.c. B1\n\n5(a)(ii) full-wave (rectification) B1\n\n5(b)(i) P labelled – and Q labelled + B1\n\n5(b)(ii) V scale labelled 4 and 8 on the 2 cm tick marks B1\nOUT\nT = 2 /  C1\n= 2 / 25\n= 0.08 s\nt scale labelled 0.02, 0.04, 0.06, 0.08, 0.10, 0.12 on the 2 cm tick marks A1\n\n5(c)(i) correct symbol used for capacitor and capacitor connected in parallel with the 1.2 k resistor. B1\n\n5(c)(ii) straight lines or curves, with negative decreasing gradients, drawn between adjacent peaks, from top of first peak to meet B1\nline going up to next peak\nlines, from one peak to the line going up to the next peak, show a drop in p.d. of 1½ small squares B1\n\n5(c)(iii) V = 0.90  6.0 (= 5.4 V) C1\nor\ndischarge time (for each cycle) = 0.034 s\nV = V exp (– t / RC) C1\n0\n5.4 = 6.0 exp [– 0.034 / (1.2  103  C)]\nC = 2.7  10–4 F A1\n© UCLES 2022 Page 11 of 16", "source_pages": [ 11 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9702_s22_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-May-June/9702_s22_ms_41.pdf?download=true", "html": "9702-topic-21-alternating-currents/answers.html", "image_paths": [ "../answer-assets/9702_s22_ms_41-p11.png" ] }, { "id": "9702-2022-mj-41-q06", "question_id": "9702-2022-mj-41-q06", "subject": "9702", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 6, "topic": "Magnetic fields", "topic_slug": "9702-topic-20-magnetic-fields", "marks": 11, "status": "available", "reason": null, "text": "6(a) product of (magnetic) flux density and area M1\nwhere area is perpendicular to the (magnetic) field A1\n\n6(b)(i) N = BAN C1\n= 400  10–3  0.122  8 C1\n= 0.046 Wb A1\n\n6(b)(ii) (line is a) straight line B1\n\n6(b)(iii) (induced) e.m.f. = rate of change of flux linkage C1\ne.m.f. = N / t A1\n= 0.046 / 0.60\n= 0.077 V\n\n6(c) (induced e.m.f. causes) current flow (in the coil) B1\neither\ncurrent (in magnetic field) causes forces to act on the coil B1\n(opposite sides of) coil forced inwards B1\nor\ncurrent causes dissipation of energy in the resistance of the coil (B1)\ntemperature of the coil rises (B1)\n© UCLES 2022 Page 12 of 16", "source_pages": [ 12 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9702_s22_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-May-June/9702_s22_ms_41.pdf?download=true", "html": "9702-topic-20-magnetic-fields/answers.html", "image_paths": [ "../answer-assets/9702_s22_ms_41-p12.png" ] }, { "id": "9702-2022-mj-41-q07", "question_id": "9702-2022-mj-41-q07", "subject": "9702", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 7, "topic": "Quantum physics", "topic_slug": "9702-topic-22-quantum-physics", "marks": 10, "status": "available", "reason": null, "text": "7(a) quantum of energy M1\nof electromagnetic radiation A1\n\n7(b)(i) photoelectric effect B1\n\n7(b)(ii)  there is a frequency below which no electrons are emitted B3\nor\nthreshold frequency = 5.4  1014 Hz\n work function of the metal = 3.6  10–19 J (or 2.2 eV)\n E increases (linearly) with (increasing) frequency\nMAX\n gradient of the line is the Planck constant\nor\ngradient of the line is 6.7  10–34 J s\nAny three bullet points, 1 mark each\n\n7(c)(i) different threshold frequency B1\n(line has) same gradient but different intercept B1\n\n7(c)(ii) photons have same energy B1\nline unchanged B1\n© UCLES 2022 Page 13 of 16", "source_pages": [ 13 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9702_s22_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-May-June/9702_s22_ms_41.pdf?download=true", "html": "9702-topic-22-quantum-physics/answers.html", "image_paths": [ "../answer-assets/9702_s22_ms_41-p13.png" ] }, { "id": "9702-2022-mj-41-q08", "question_id": "9702-2022-mj-41-q08", "subject": "9702", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 8, "topic": "Nuclear physics", "topic_slug": "9702-topic-23-nuclear-physics", "marks": 12, "status": "available", "reason": null, "text": "8(a)(i) energy required to separate the nucleons (in the nucleus) M1\nto infinity A1\n\n8(a)(ii) curve starting close to the origin and forming a single peak B1\npeak shown to left of centre, with steep line on LHS of peak and shallow line on RHS of peak B1\n\n8(b)(i) fusion B1\n\n8(b)(ii) both particles have low A values B1\nor\nboth particles are at left-hand end of graph\nHe-3 has higher binding energy (per nucleon) than H-2 B1\n\n8(c) m = [(2  2.014102) – (3.016029 + 1.008665)] u C1\n( = 0.00351 u)\nE = mc2 C1\n= 0.00351  1.66  10–27  (3.00  108)2 C1\n( = 5.24  10–13 J)\n1.00 mol of deuterium forms 0.500 mol of helium-3 C1\ntotal energy = 0.500  6.02  1023  5.24  10–13 A1\n= 1.58  1011 J\n© UCLES 2022 Page 14 of 16", "source_pages": [ 14 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9702_s22_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-May-June/9702_s22_ms_41.pdf?download=true", "html": "9702-topic-23-nuclear-physics/answers.html", "image_paths": [ "../answer-assets/9702_s22_ms_41-p14.png" ] }, { "id": "9702-2022-mj-41-q09", "question_id": "9702-2022-mj-41-q09", "subject": "9702", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 9, "topic": "Medical physics", "topic_slug": "9702-topic-24-medical-physics", "marks": 9, "status": "available", "reason": null, "text": "9(a)(i) electrons are accelerated (by an applied p.d.) B1\nelectrons hit target B1\nX-rays produced when electrons decelerate B1\n\n9(a)(ii) images of the multiple sections are combined to create a 3-D image B1\n\n9(b)(i) I = I exp (– μx) C1\n0\n= I exp (– 0.89  5.6) A1\n0\n= 0.0068 I\n0\n\n9(b)(ii) I = I exp (– 2.4  3.4)  exp (– 0.89  3.2) C1\n0\n= 1.7  10–5 I A1\n0\n\n9(c) comparison of intensities or values in (b) leading to conclusion consistent with these values B1\n© UCLES 2022 Page 15 of 16", "source_pages": [ 15 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9702_s22_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-May-June/9702_s22_ms_41.pdf?download=true", "html": "9702-topic-24-medical-physics/answers.html", "image_paths": [ "../answer-assets/9702_s22_ms_41-p15.png" ] }, { "id": "9702-2022-mj-41-q10", "question_id": "9702-2022-mj-41-q10", "subject": "9702", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 10, "topic": "Astronomy and cosmology", "topic_slug": "9702-topic-25-astronomy-and-cosmology", "marks": 9, "status": "available", "reason": null, "text": "10(a) wavelength of maximum intensity is inversely proportional to (thermodynamic) temperature B1\n\n10(b)(i)  = 0.50 m for A and 0.65 m for B C1\nMAX\nT = 5800  (0.50 / 0.65) A1\n= 4500 K\n\n10(b)(ii) (star B has) greater peak / average wavelength B1\n(star B looks) redder B1\n\n10(c)(i) apparent wavelength is greater B1\nor\nwavelength is greater than known value\n(due to) movement of star away (from observer) B1\n\n10(c)(ii) by examining the (lines in the) spectrum (of light from the star) B1\nand comparing with known spectrum B1\n© UCLES 2022 Page 16 of 16", "source_pages": [ 16 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9702_s22_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-May-June/9702_s22_ms_41.pdf?download=true", "html": "9702-topic-25-astronomy-and-cosmology/answers.html", "image_paths": [ "../answer-assets/9702_s22_ms_41-p16.png" ] }, { "id": "9702-2022-mj-42-q01", "question_id": "9702-2022-mj-42-q01", "subject": "9702", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 1, "topic": "Gravitational fields", "topic_slug": "9702-topic-13-gravitational-fields", "marks": 10, "status": "available", "reason": null, "text": "1(a)(i) work (done) per unit mass B1\nwork (done on mass) in moving mass from infinity (to the point) B1\n\n1(a)(ii) E = ϕm B1\nP\nE = (– GM / r)  m = – GMm / r\nP\nor\nϕ = – GM / r and E = ϕm = – GMm / r\nP\n\n1(b)(i) E = 6.67  10–11  1.99  1030  2.20  1014  [1 / (6.38  1010) – 1 / (8.44  1011)] C1\nP\n= 4.23  1023 J A1\n\n1(b)(ii) (gravitational) force is attractive so decrease B1\nor\n(gravitational) force does work so decrease\n\n1(b)(iii) E = ½m(v 2 – v 2) C1\nP 2 1\n4.23  1023 = ½  2.20  1014  (v2 – 341002) C1\nv (= 70800 m s–1) = 70.8 km s–1 A1\n\n1(c) both PE and KE equations include m, so path is unchanged B1\n© UCLES 2022 Page 7 of 16", "source_pages": [ 7 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9702_s22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-May-June/9702_s22_ms_42.pdf?download=true", "html": "9702-topic-13-gravitational-fields/answers.html", "image_paths": [ "../answer-assets/9702_s22_ms_42-p07.png" ] }, { "id": "9702-2022-mj-42-q02", "question_id": "9702-2022-mj-42-q02", "subject": "9702", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 2, "topic": "Electric fields", "topic_slug": "9702-topic-18-electric-fields", "marks": 13, "status": "available", "reason": null, "text": "2(a) (electric) force is (directly) proportional to product of charges B1\nforce (between point charges) is inversely proportional to the square of their separation B1\n\n2(b)(i) (electric) force is perpendicular to velocity (of particles) B1\nforce (perpendicular to velocity) causes centripetal acceleration B1\nor\nforce does not change the speed of the particles\nor\nforce has constant magnitude\n\n2(b)(ii) F = e2 / 4x2 C1\n0\n= (1.60  10–19)2 / [4  8.85  10–12  (2  1.59  10–10)2] A1\n= 2.28  10–9 N\n\n2(b)(iii) F = mr2 and  = 2 / T C1\nor\nF = mv2 / r and v = 2r / T\nF = 42mr / T2 C1\nT = √ [42  9.11  10–31  1.59  10–10 / (2.28  10–9)]\n= 1.58  10–15 s A1\n\n2(c)(i)  electron and positron interact B2\n positron is anti-particle of electron\n (pair) annihilation occurs\nAny two points, 1 mark each\nmass of the electron and positron converted into photon energy B1\n\n2(c)(ii) PET scanning B1\n© UCLES 2022 Page 8 of 16", "source_pages": [ 8 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9702_s22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-May-June/9702_s22_ms_42.pdf?download=true", "html": "9702-topic-18-electric-fields/answers.html", "image_paths": [ "../answer-assets/9702_s22_ms_42-p08.png" ] }, { "id": "9702-2022-mj-42-q03", "question_id": "9702-2022-mj-42-q03", "subject": "9702", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 3, "topic": "Thermodynamics", "topic_slug": "9702-topic-16-thermodynamics", "marks": 12, "status": "available", "reason": null, "text": "3(a) (thermal) energy per unit mass B1\nenergy to change state between liquid and gas at constant temperature B1\n\n3(b)(i) q = mL = 0.37  2.3  106 A1\n= 8.5  105 J\n\n3(b)(ii) pV = nRT and T = 373 K C1\nn = 370 / 18 C1\nV = [(370 / 18)  8.31  373] / (1.0  105) = 0.64 m3 A1\n\n3(b)(iii) w = pV C1\n= 1.0  105  0.64 A1\n= 6.4  104 J\n\n3(b)(iv) (water does work against atmosphere so) work done on water is negative B1\nincrease in internal energy = (8.5 – 0.64)  105 = 7.9  105 J A1\n\n3(c) valid reasoning of how work done by water is affected M1\ncorrect use of first law to draw conclusion about effect on specific latent heat that is consistent with work done A1\n© UCLES 2022 Page 9 of 16", "source_pages": [ 9 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9702_s22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-May-June/9702_s22_ms_42.pdf?download=true", "html": "9702-topic-16-thermodynamics/answers.html", "image_paths": [ "../answer-assets/9702_s22_ms_42-p09.png" ] }, { "id": "9702-2022-mj-42-q04", "question_id": "9702-2022-mj-42-q04", "subject": "9702", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 4, "topic": "Oscillations", "topic_slug": "9702-topic-17-oscillations", "marks": 8, "status": "available", "reason": null, "text": "4(a) oscillations (of object) at maximum amplitude B1\nwhen driving frequency equals natural frequency (of object) B1\n\n4(b)(i) T = 2 /  C1\n= 2 / 5.0 A1\n= 0.40 s\n\n4(b)(ii) displacement scale labelled –1.0, –0.5, (0), 0.5, 1.0 on the 2 cm tick marks B1\nt scale labelled 0.2, 0.4, 0.6, 0.8, 1.0, 1.2 on the 2 cm tick marks B1\n\n4(b)(iii) ϕ = 2t / T C1\n= 2  0.10 / 0.40 or 2  0.30 / 0.40\n= 1.6 rad or 4.7 rad A1\n© UCLES 2022 Page 10 of 16", "source_pages": [ 10 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9702_s22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-May-June/9702_s22_ms_42.pdf?download=true", "html": "9702-topic-17-oscillations/answers.html", "image_paths": [ "../answer-assets/9702_s22_ms_42-p10.png" ] }, { "id": "9702-2022-mj-42-q05", "question_id": "9702-2022-mj-42-q05", "subject": "9702", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 5, "topic": "Capacitance", "topic_slug": "9702-topic-19-capacitance", "marks": 11, "status": "available", "reason": null, "text": "5(a) charge / potential (difference) M1\ncharge is charge on one plate, and potential is p.d. across the plates A1\n\n5(b) p.d. across both capacitors = E B1\nQ = Q + Q B1\nT 1 2\nC E = C E + C E hence C = C + C B1\nT 1 2 T 1 2\n\n5(c)(i) [(1 / 22) + (1 / 47)]–1 = 15 F A1\n\n5(c)(ii) energy = ½CV 2 C1\n= ½  15  10–6  122 A1\n= 1.1  10–3 J\n\n5(c)(iii) initial p.d. (across 22 F) = 12  (15 / 22) C1\n= 8.2 V\nor\nfinal p.d. across both capacitors = 6.0  (22 / 15)\n= 8.8 V\nV = V exp [– t / (2.7  106  15  10–6)] C1\n0\n6.0 = 8.2 exp [– t / (2.7  106  15  10–6)] A1\nor\n8.8 = 12 exp [– t / (2.7  106  15  10–6)]\nt = 13 s\n© UCLES 2022 Page 11 of 16", "source_pages": [ 11 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9702_s22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-May-June/9702_s22_ms_42.pdf?download=true", "html": "9702-topic-19-capacitance/answers.html", "image_paths": [ "../answer-assets/9702_s22_ms_42-p11.png" ] }, { "id": "9702-2022-mj-42-q06", "question_id": "9702-2022-mj-42-q06", "subject": "9702", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 6, "topic": "Magnetic fields", "topic_slug": "9702-topic-20-magnetic-fields", "marks": 10, "status": "available", "reason": null, "text": "6(a) there must be a current (in the wire) B1\n(wire) must be at a non-zero angle to the magnetic field B1\n\n6(b)(i) arrow from X pointing horizontally to the left B1\narrow from Y pointing diagonally upwards and to the left at about 45° B1\narrow from Z pointing horizontally to the right B1\n\n6(b)(ii) (flux densities at W and X are approximately) equal B1\n(flux density at) Y greater than (flux density at) Z B1\n\n6(c) current in wire creates magnetic field around wire B1\n(each) wire sits in the magnetic field created by the other B1\n(for each wire,) current / wire is perpendicular to magnetic field (due to other wire), (so) experiences a (magnetic) force B1\n© UCLES 2022 Page 12 of 16", "source_pages": [ 12 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9702_s22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-May-June/9702_s22_ms_42.pdf?download=true", "html": "9702-topic-20-magnetic-fields/answers.html", "image_paths": [ "../answer-assets/9702_s22_ms_42-p12.png" ] }, { "id": "9702-2022-mj-42-q07", "question_id": "9702-2022-mj-42-q07", "subject": "9702", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 7, "topic": "Magnetic fields", "topic_slug": "9702-topic-20-magnetic-fields", "marks": 10, "status": "available", "reason": null, "text": "7(a) induced e.m.f. is (directly) proportional to rate M1\nof change of (magnetic) flux (linkage) A1\n\n7(b) V stepped, all at non-zero values, between t = 0 and t = 0.40 s B1\n2\nV shown with same non-zero magnitude up to t = 0.15 s and after t = 0.25 s but with a different magnitude between these B1\n2\ntimes\nV shown with a magnitude between t = 0.15 s and t = 0.25 s that is three times the magnitude before t = 0.15 s and after B1\n2\nt = 0.25 s\nV shown with same sign up to t = 0.15 s and after t = 0.25 s, and opposite sign in between B1\n2\n\n7(c)(i) changing current in coil causes changing (magnetic) field B1\nor\nchanging (magnetic) flux causes induced e.m.f. in ring\ninduced e.m.f. in ring causes current in ring B1\n(magnetic) field due to (induced) current in ring interacts with (coil’s) field to cause upwards force (on ring) B1\nor\n(induced) current in ring perpendicular to (coil’s magnetic) field causes upwards force (on ring)\n\n7(c)(ii) both magnetic fields reverse direction so ring still jumps up B1\nor\ncurrent (in ring) and (coil’s) field both reverse so ring still jumps up\n© UCLES 2022 Page 13 of 16", "source_pages": [ 13 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9702_s22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-May-June/9702_s22_ms_42.pdf?download=true", "html": "9702-topic-20-magnetic-fields/answers.html", "image_paths": [ "../answer-assets/9702_s22_ms_42-p13.png" ] }, { "id": "9702-2022-mj-42-q08", "question_id": "9702-2022-mj-42-q08", "subject": "9702", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 8, "topic": "Nuclear physics", "topic_slug": "9702-topic-23-nuclear-physics", "marks": 8, "status": "available", "reason": null, "text": "8(a)(i) photoelectric effect B1\n\n8(a)(ii) electron diffraction B1\n\n8(b)(i)  = h / p C1\np = 4  1.66  10–27  6.2  107 C1\n( = 4.1  10–19 N s)\n = 6.63  10–34 / 4.1  10–19 A1\n= 1.6  10–15 m\n\n8(b)(ii) line with negative gradient throughout B1\ncurve asymptotic to both axes with non-zero  at v = 6.2  107 m s–1 B1\n\n8(c) (de Broglie) wavelength negligible compared with width of doorway B1\n© UCLES 2022 Page 14 of 16", "source_pages": [ 14 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9702_s22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-May-June/9702_s22_ms_42.pdf?download=true", "html": "9702-topic-23-nuclear-physics/answers.html", "image_paths": [ "../answer-assets/9702_s22_ms_42-p14.png" ] }, { "id": "9702-2022-mj-42-q09", "question_id": "9702-2022-mj-42-q09", "subject": "9702", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 9, "topic": "Astronomy and cosmology", "topic_slug": "9702-topic-25-astronomy-and-cosmology", "marks": 7, "status": "available", "reason": null, "text": "9(a)(i) speed is (directly) proportional to distance M1\nwhere speed is speed of recession of galaxy (from observer) and distance is distance of galaxy away from observer A1\n\n9(a)(ii) wavelengths (of spectral lines) are greater (than their known values) B1\nredshift shows stars (in distant galaxies) moving away from Earth B1\n\n9(b) (all) parts of Universe moving away from each other B1\nmore distant objects are moving away faster B1\nmatter must have been close together / very dense in the past B1\n© UCLES 2022 Page 15 of 16", "source_pages": [ 15 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9702_s22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-May-June/9702_s22_ms_42.pdf?download=true", "html": "9702-topic-25-astronomy-and-cosmology/answers.html", "image_paths": [ "../answer-assets/9702_s22_ms_42-p15.png" ] }, { "id": "9702-2022-mj-42-q10", "question_id": "9702-2022-mj-42-q10", "subject": "9702", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 10, "topic": "Nuclear physics", "topic_slug": "9702-topic-23-nuclear-physics", "marks": 11, "status": "available", "reason": null, "text": "10(a) spontaneous emission of (ionising) radiation B1\nemission from unstable nucleus B1\n\n10(b)(i) curve with decreasing negative gradient passing through (0, N ) B1\n0\ncurve passing through (T, 0.5N ) B1\n0\ncurve passing through (2T, 0.25N ) and (3T, 0.125N ) B1\n0 0\n\n10(b)(ii) line through origin with positive gradient B1\nstraight line passing through (N , A ) B1\n0 0\n\n10(c)(i) activity B1\n\n10(c)(ii) decay constant B1\n\n10(d) N = N exp (– ln 2  1.70T / T) C1\n0\nN / N = 0.31 A1\n0\n© UCLES 2022 Page 16 of 16", "source_pages": [ 16 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9702_s22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-May-June/9702_s22_ms_42.pdf?download=true", "html": "9702-topic-23-nuclear-physics/answers.html", "image_paths": [ "../answer-assets/9702_s22_ms_42-p16.png" ] }, { "id": "9702-2022-mj-43-q01", "question_id": "9702-2022-mj-43-q01", "subject": "9702", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 1, "topic": "Gravitational fields", "topic_slug": "9702-topic-13-gravitational-fields", "marks": 10, "status": "available", "reason": null, "text": "1(a)(i) (gravitational) force is (directly) proportional to product of masses B1\nforce (between point masses) is inversely proportional to the square of their separation B1\n\n1(a)(ii) g = F / m C1\nF = GMm / r2 A1\nand so\ng = [GMm / r2] / m = GM / r2\n\n1(b)(i) g = (6.67  10–11  7.35  1022) / (1.74  106)2 = 1.62 N kg–1 A1\n\n1(b)(ii) fields (due to Earth and the Moon) have equal magnitudes B1\nfields (due to Earth and the Moon) are in opposite directions B1\n\n1(b)(iii) distance of X from Earth = (3.84  108 – x) C1\n(G ) 7.35  1022 / x2 = (G ) 5.98  1024 / (3.84  108 – x)2 C1\nx = 3.8  107 m A1\n© UCLES 2022 Page 7 of 16", "source_pages": [ 7 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9702_s22_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-May-June/9702_s22_ms_43.pdf?download=true", "html": "9702-topic-13-gravitational-fields/answers.html", "image_paths": [ "../answer-assets/9702_s22_ms_43-p07.png" ] }, { "id": "9702-2022-mj-43-q02", "question_id": "9702-2022-mj-43-q02", "subject": "9702", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 2, "topic": "Magnetic fields", "topic_slug": "9702-topic-20-magnetic-fields", "marks": 8, "status": "available", "reason": null, "text": "2(a)(i) (vertically) downwards B1\n\n2(a)(ii) magnetic force (on sphere) is perpendicular to its velocity B1\nmagnetic force perpendicular to velocity is the centripetal force B1\nor\nmagnetic force perpendicular to velocity causes centripetal acceleration\nor\nacceleration perpendicular to velocity is centripetal (acceleration)\nor\nmagnetic force does not change the speed of the sphere\nor\nmagnetic force has constant magnitude\n\n2(b) mg = Eq C1\nE = (1.6  10–10  9.81) / (0.27  10–9) A1\n= 5.8 N C–1\n\n2(c) centripetal force = magnetic force B1\nor\nBqv = mv2 / r\nB = mv / qr C1\n= (1.6  10–10  0.78) / (0.27  10–9  3.4) = 0.14 T A1\n© UCLES 2022 Page 8 of 16", "source_pages": [ 8 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9702_s22_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-May-June/9702_s22_ms_43.pdf?download=true", "html": "9702-topic-20-magnetic-fields/answers.html", "image_paths": [ "../answer-assets/9702_s22_ms_43-p08.png" ] }, { "id": "9702-2022-mj-43-q03", "question_id": "9702-2022-mj-43-q03", "subject": "9702", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 3, "topic": "Thermodynamics", "topic_slug": "9702-topic-16-thermodynamics", "marks": 11, "status": "available", "reason": null, "text": "3(a)(i) a gas that obeys pV  T M1\nwhere p = pressure, V = volume, T = thermodynamic temperature A1\n\n3(a)(ii) T = (273 + 17) K C1\nn = pV / RT A1\n= (1.2  105  0.24) / [8.31  (273 + 17)]\n= 12 mol\n\n3(b)(i) work done = pV C1\n= 1.2  105  (0.24 – 0.08) = 19200 J (= 19.2 kJ) A1\n\n3(b)(ii) AB work done correct (19.2) A1\nBC work done correct (0) A1\nCA increase in internal energy correct (0) and CA thermal energy correct (31.6) A1\nAB increase in internal energy calculated correctly from work done – 48.0 A1\nBC increase in internal energy correctly calculated so the final column adds up to zero and BC thermal energy same as A1\nincrease in internal energy\n(Fully correct table:\nAB 19.2 –48.0 –28.8\nBC 0 28.8 28.8\nCA –31.6 31.6 0\n)\n© UCLES 2022 Page 9 of 16", "source_pages": [ 9 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9702_s22_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-May-June/9702_s22_ms_43.pdf?download=true", "html": "9702-topic-16-thermodynamics/answers.html", "image_paths": [ "../answer-assets/9702_s22_ms_43-p09.png" ] }, { "id": "9702-2022-mj-43-q04", "question_id": "9702-2022-mj-43-q04", "subject": "9702", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 4, "topic": "Oscillations", "topic_slug": "9702-topic-17-oscillations", "marks": 8, "status": "available", "reason": null, "text": "4(a) straight line through origin shows that a is proportional to x B1\nnegative gradient shows that a is in opposite direction to x B1\n\n4(b)(i) a =  2x C1\n0 0\nor\na = – 2x\nor\n2 = – gradient\n = (0.40 / 0.050) A1\n= 2.8 rad s–1\n\n4(b)(ii) k =  2L C1\n= 2.82  1.24\n= 9.7 m s–2 A1\n\n4(c) (increasing L causes)  to decrease M1\nor\nenergy (= ½ m2x 2) = ½ mkx 2 / L (and L increases)\n0 0\nso amplitude increases A1\n© UCLES 2022 Page 10 of 16", "source_pages": [ 10 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9702_s22_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-May-June/9702_s22_ms_43.pdf?download=true", "html": "9702-topic-17-oscillations/answers.html", "image_paths": [ "../answer-assets/9702_s22_ms_43-p10.png" ] }, { "id": "9702-2022-mj-43-q05", "question_id": "9702-2022-mj-43-q05", "subject": "9702", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 5, "topic": "Alternating currents", "topic_slug": "9702-topic-21-alternating-currents", "marks": 12, "status": "available", "reason": null, "text": "5(a)(i) conversion (from a.c.) to d.c. B1\n\n5(a)(ii) full-wave (rectification) B1\n\n5(b)(i) P labelled – and Q labelled + B1\n\n5(b)(ii) V scale labelled 4 and 8 on the 2 cm tick marks B1\nOUT\nT = 2 /  C1\n= 2 / 25\n= 0.08 s\nt scale labelled 0.02, 0.04, 0.06, 0.08, 0.10, 0.12 on the 2 cm tick marks A1\n\n5(c)(i) correct symbol used for capacitor and capacitor connected in parallel with the 1.2 k resistor. B1\n\n5(c)(ii) straight lines or curves, with negative decreasing gradients, drawn between adjacent peaks, from top of first peak to meet B1\nline going up to next peak\nlines, from one peak to the line going up to the next peak, show a drop in p.d. of 1½ small squares B1\n\n5(c)(iii) V = 0.90  6.0 (= 5.4 V) C1\nor\ndischarge time (for each cycle) = 0.034 s\nV = V exp (– t / RC) C1\n0\n5.4 = 6.0 exp [– 0.034 / (1.2  103  C)]\nC = 2.7  10–4 F A1\n© UCLES 2022 Page 11 of 16", "source_pages": [ 11 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9702_s22_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-May-June/9702_s22_ms_43.pdf?download=true", "html": "9702-topic-21-alternating-currents/answers.html", "image_paths": [ "../answer-assets/9702_s22_ms_43-p11.png" ] }, { "id": "9702-2022-mj-43-q06", "question_id": "9702-2022-mj-43-q06", "subject": "9702", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 6, "topic": "Magnetic fields", "topic_slug": "9702-topic-20-magnetic-fields", "marks": 11, "status": "available", "reason": null, "text": "6(a) product of (magnetic) flux density and area M1\nwhere area is perpendicular to the (magnetic) field A1\n\n6(b)(i) N = BAN C1\n= 400  10–3  0.122  8 C1\n= 0.046 Wb A1\n\n6(b)(ii) (line is a) straight line B1\n\n6(b)(iii) (induced) e.m.f. = rate of change of flux linkage C1\ne.m.f. = N / t A1\n= 0.046 / 0.60\n= 0.077 V\n\n6(c) (induced e.m.f. causes) current flow (in the coil) B1\neither\ncurrent (in magnetic field) causes forces to act on the coil B1\n(opposite sides of) coil forced inwards B1\nor\ncurrent causes dissipation of energy in the resistance of the coil (B1)\ntemperature of the coil rises (B1)\n© UCLES 2022 Page 12 of 16", "source_pages": [ 12 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9702_s22_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-May-June/9702_s22_ms_43.pdf?download=true", "html": "9702-topic-20-magnetic-fields/answers.html", "image_paths": [ "../answer-assets/9702_s22_ms_43-p12.png" ] }, { "id": "9702-2022-mj-43-q07", "question_id": "9702-2022-mj-43-q07", "subject": "9702", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 7, "topic": "Quantum physics", "topic_slug": "9702-topic-22-quantum-physics", "marks": 10, "status": "available", "reason": null, "text": "7(a) quantum of energy M1\nof electromagnetic radiation A1\n\n7(b)(i) photoelectric effect B1\n\n7(b)(ii)  there is a frequency below which no electrons are emitted B3\nor\nthreshold frequency = 5.4  1014 Hz\n work function of the metal = 3.6  10–19 J (or 2.2 eV)\n E increases (linearly) with (increasing) frequency\nMAX\n gradient of the line is the Planck constant\nor\ngradient of the line is 6.7  10–34 J s\nAny three bullet points, 1 mark each\n\n7(c)(i) different threshold frequency B1\n(line has) same gradient but different intercept B1\n\n7(c)(ii) photons have same energy B1\nline unchanged B1\n© UCLES 2022 Page 13 of 16", "source_pages": [ 13 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9702_s22_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-May-June/9702_s22_ms_43.pdf?download=true", "html": "9702-topic-22-quantum-physics/answers.html", "image_paths": [ "../answer-assets/9702_s22_ms_43-p13.png" ] }, { "id": "9702-2022-mj-43-q08", "question_id": "9702-2022-mj-43-q08", "subject": "9702", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 8, "topic": "Nuclear physics", "topic_slug": "9702-topic-23-nuclear-physics", "marks": 12, "status": "available", "reason": null, "text": "8(a)(i) energy required to separate the nucleons (in the nucleus) M1\nto infinity A1\n\n8(a)(ii) curve starting close to the origin and forming a single peak B1\npeak shown to left of centre, with steep line on LHS of peak and shallow line on RHS of peak B1\n\n8(b)(i) fusion B1\n\n8(b)(ii) both particles have low A values B1\nor\nboth particles are at left-hand end of graph\nHe-3 has higher binding energy (per nucleon) than H-2 B1\n\n8(c) m = [(2  2.014102) – (3.016029 + 1.008665)] u C1\n( = 0.00351 u)\nE = mc2 C1\n= 0.00351  1.66  10–27  (3.00  108)2 C1\n( = 5.24  10–13 J)\n1.00 mol of deuterium forms 0.500 mol of helium-3 C1\ntotal energy = 0.500  6.02  1023  5.24  10–13 A1\n= 1.58  1011 J\n© UCLES 2022 Page 14 of 16", "source_pages": [ 14 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9702_s22_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-May-June/9702_s22_ms_43.pdf?download=true", "html": "9702-topic-23-nuclear-physics/answers.html", "image_paths": [ "../answer-assets/9702_s22_ms_43-p14.png" ] }, { "id": "9702-2022-mj-43-q09", "question_id": "9702-2022-mj-43-q09", "subject": "9702", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 9, "topic": "Medical physics", "topic_slug": "9702-topic-24-medical-physics", "marks": 9, "status": "available", "reason": null, "text": "9(a)(i) electrons are accelerated (by an applied p.d.) B1\nelectrons hit target B1\nX-rays produced when electrons decelerate B1\n\n9(a)(ii) images of the multiple sections are combined to create a 3-D image B1\n\n9(b)(i) I = I exp (– μx) C1\n0\n= I exp (– 0.89  5.6) A1\n0\n= 0.0068 I\n0\n\n9(b)(ii) I = I exp (– 2.4  3.4)  exp (– 0.89  3.2) C1\n0\n= 1.7  10–5 I A1\n0\n\n9(c) comparison of intensities or values in (b) leading to conclusion consistent with these values B1\n© UCLES 2022 Page 15 of 16", "source_pages": [ 15 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9702_s22_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-May-June/9702_s22_ms_43.pdf?download=true", "html": "9702-topic-24-medical-physics/answers.html", "image_paths": [ "../answer-assets/9702_s22_ms_43-p15.png" ] }, { "id": "9702-2022-mj-43-q10", "question_id": "9702-2022-mj-43-q10", "subject": "9702", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 10, "topic": "Astronomy and cosmology", "topic_slug": "9702-topic-25-astronomy-and-cosmology", "marks": 9, "status": "available", "reason": null, "text": "10(a) wavelength of maximum intensity is inversely proportional to (thermodynamic) temperature B1\n\n10(b)(i)  = 0.50 m for A and 0.65 m for B C1\nMAX\nT = 5800  (0.50 / 0.65) A1\n= 4500 K\n\n10(b)(ii) (star B has) greater peak / average wavelength B1\n(star B looks) redder B1\n\n10(c)(i) apparent wavelength is greater B1\nor\nwavelength is greater than known value\n(due to) movement of star away (from observer) B1\n\n10(c)(ii) by examining the (lines in the) spectrum (of light from the star) B1\nand comparing with known spectrum B1\n© UCLES 2022 Page 16 of 16", "source_pages": [ 16 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9702_s22_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-May-June/9702_s22_ms_43.pdf?download=true", "html": "9702-topic-25-astronomy-and-cosmology/answers.html", "image_paths": [ "../answer-assets/9702_s22_ms_43-p16.png" ] }, { "id": "9702-2022-mj-51-q01", "question_id": "9702-2022-mj-51-q01", "subject": "9702", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "51", "question_number": 1, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "1 Defining the problem\nd is the independent variable and V is the dependent variable or vary d and measure V 1\nkeep A or area (of overlap) of plates constant 1\nMethods of data collection\nlabelled diagram of workable experiment including: 1\n circuit diagram with voltmeter connected in parallel with the capacitor\n capacitor and voltmeter connected to the metal plates with no power supply in discharge part of the circuit\n correct symbols for capacitor and voltmeter\nmethod to charge parallel plates, e.g. separate circuit diagram showing plates connected to a d.c. power supply or combined 1\ncircuit with switches and d.c. power supply\nuse calipers to measure d 1\nor\nuse micrometer/calipers to measure thickness of spacers\nuse rule(r) to measure lengths to determine A and A = length  breadth 1\nMethod of analysis\n\n1 1 1\nplot a graph of against d or equivalent (e.g. d against )\nV V\n(Do not accept log graphs.)\ny-interceptC C 1\nK  or K \ngradientA gradientAW\n\n1 y-interceptC\n(for d against : K  )\nV A\n© UCLES 2022 Page 6 of 10\n\n1 1 1\nW \ny-intercept\n\n1 gradient gradientC\n(for d against : W  or W  )\nV y-intercept AK\nAdditional detail including safety considerations 6\nD1 use gloves to prevent electric shock or do not touch metal plates to avoid shocks\nD2 keep the initial p.d. across plates or initial charge constant\nD3 method to determine the value of C, e.g. description of an experiment to measure p.d. or current against time during\ndischarge through a resistor\nD4 method of operation of circuit(s) using switch(es)\nD5 description of method to fully discharge capacitor, e.g. between experiments, short-circuit the capacitor or use of\nswitch in parallel with capacitor\nD6 repeat measurements of d at different points across plates and average\nD7 repeat measurements of V for same d and average V\nD8 bottom plate resting on insulating material or top plate supported by strings\nD9 use high voltage power supply to increase charge on plates\nor\nuse a very small value of capacitance to increase voltmeter reading\nD10 relationship valid if a straight line is produced (not passing through the origin)\n© UCLES 2022 Page 7 of 10", "source_pages": [ 6, 7 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9702_s22_ms_51.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-May-June/9702_s22_ms_51.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_s22_ms_51-p06.png", "../answer-assets/9702_s22_ms_51-p07.png" ] }, { "id": "9702-2022-mj-51-q02", "question_id": "9702-2022-mj-51-q02", "subject": "9702", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "51", "question_number": 2, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "2(a) gradient = n 1\ny-intercept = lgSZ\n\n2(b) 1\nlg (M / 1030kg) lg (L / 1028W)\n0.68 or 0.681  0.03 or 0.04 0.15 or 0.146\n0.81 or 0.806  0.02 or 0.03 0.49 or 0.491\n1.08 or 1.079  0.07 or 0.08 1.51 or 1.505\n1.36 or 1.362  0.04 2.54 or 2.544\n1.63 or 1.633  0.04 3.56 or 3.556\n1.96 or 1.959  0.02 4.82 or 4.820\nValues of lg (M / 1030kg) and lg (L / 1028W) correct as shown above.\nAbsolute uncertainties in lg (M / 1030kg) correct as shown above. 1\n\n2(c)(i) Six points from (b) plotted correctly. 1\nMust be within half a small square. Diameter of points must be less than half a small square.\nError bars in lg (M / 1030kg) plotted correctly. 1\nAll error bars must be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.\n\n2(c)(ii) Straight line of best fit drawn. 1\nPoints must be balanced. Do not accept line from top point to bottom point.\nLine must pass between (0.92, 1.0) and (0.96, 1.0) and between (1.86, 4.5) and (1.90, 4.5)\nWorst acceptable line drawn (steepest or shallowest possible line that passes through all the error bars). 1\nAll error bars must be plotted.\n© UCLES 2022 Page 8 of 10\n\n2(c)(iii) Gradient determined with clear substitution of data points into y / x. 1\nDistance between data points must be greater than half the length of the drawn line.\nGradient of worst acceptable line determined. 1\nuncertainty = (gradient of line of best fit – gradient of worst acceptable line)\nor\nuncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)\n\n2(c)(iv) y-intercept determined by substitution into y = mx + c. 1\ny-intercept of worst acceptable line determined by substitution into y = mx + c. 1\nuncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line\nor\nuncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)\nDo not allow methods using a false origin.\n© UCLES 2022 Page 9 of 10\n\n2(d) n gradient (c)(iii) and n and Z both given to two or three significant figures. 1\nValue of Z determined using y-intercept. Correct method must be seen. 1\n10y-intercept 1028 10(c)(iv)1028\nZ  \nS 3.851026\nor\nZ 10y-interceptlg S 1028\nor\nZ 10(c)(iv)lg 3.851026 1028\nAbsolute uncertainty in n = absolute uncertainty in gradient 1\nand\n 10y-intercept 10WAL y-intercept 1028\nZ \nS\nCorrect substitution of numbers must be seen.\n\n2(e) L determined from (d) or (c)(iii) and (c)(iv) with correct substitution and correct power of ten(s). 1\nDo not accept incorrect POT for n or Z.\nL = 3.85  1026  (d)  3.0(c)(iii)\nor\nlgL(c)(iii)lg3.0y-intercept\n© UCLES 2022 Page 10 of 10", "source_pages": [ 8, 9, 10 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9702_s22_ms_51.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-May-June/9702_s22_ms_51.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_s22_ms_51-p08.png", "../answer-assets/9702_s22_ms_51-p09.png", "../answer-assets/9702_s22_ms_51-p10.png" ] }, { "id": "9702-2022-mj-52-q01", "question_id": "9702-2022-mj-52-q01", "subject": "9702", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "52", "question_number": 1, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "1 Defining the problem\nL is the independent variable and I is the dependent variable or vary L and measure I 1\nkeep E constant 1\nMethods of data collection\nlabelled diagram of workable experiment including: 1\n circuit diagram with power supply connected to ends C\n ammeter in series with power supply and conductors\n correct symbol for ammeter and power supply\ncircuit diagram with voltmeter correctly positioned to measure E across the power supply 1\nuse a rule(r) to measure L and x 1\nuse a micrometer/calipers to measure y 1\n© UCLES 2022 Page 6 of 11\n\n1 Method of analysis\n\n1 1 1\nplot a graph of against L or equivalent (e.g. L against )\nI I\n(Do not accept log graphs.)\ngradientAE 1\nP ", "source_pages": [ 6, 7 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9702_s22_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-May-June/9702_s22_ms_52.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_s22_ms_52-p06.png", "../answer-assets/9702_s22_ms_52-p07.png" ] }, { "id": "9702-2022-mj-52-q02", "question_id": "9702-2022-mj-52-q02", "subject": "9702", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "52", "question_number": 2, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "2\n1 AE\n(for L against : P  )\nI 2gradient\ny-interceptEy2 1\nQ \nx\n1 y-intercept2Py2 y-interceptEy2\n(for L against : Q  or Q  )\nI Ax gradientx\n© UCLES 2022 Page 7 of 11\n\n9702/52 Cambridge International AS & A Level – Mark Scheme May/June 2022\nPUBLISHED\nQuestion Answer Marks\n1 Additional detail including safety considerations 6\nD1 do not touch/use (heat resistant) gloves to avoid hot conductors/metal bar\nor\nuse a protective resistor/small e.m.f. to reduce the current\nor\nswitch off when not in use/when moving bar\nD2 keep A and y constant\nD3 keep x constant\nD4 use of micrometer/calipers to measure diameter of conductor and A = d2 / 4.\nD5 repeat measurements of diameter along conductors/different (perpendicular) directions/different points and average\nor\nrepeat measurements of y in different (perpendicular) directions/different points/along bar and average\nD6 method to ensure that L is the same for each conductor, e.g. check both lengths\nD7 method to determine L e.g. measure to edge and add y / 2\nor\nmethod to determine x e.g. measure between the conductors and add diameter\nD8 method to keep x constant with reason, e.g. adhesive/plasticine/blocks (one either side of each conductor) to prevent\ncylindrical conductors from moving\nD9 method of ensuring good electrical contact, e.g. clean metal bar/cylindrical conductors or use of solder or crocodile\nclips to connect circuit to the conductors\nD10 relationship valid if a straight line is produced (not passing through the origin)\n© UCLES 2022 Page 8 of 11\n\n2(a) gradient = a 1\ny-intercept = lgSK\n\n2(b) 1\nlg (T / days) lg (L / 1030W)\n1.34 or 1.342 0.46 or 0.462  0.03\n1.51 or 1.505 0.69 or 0.690  0.02\n1.62 or 1.623 0.84 or 0.839  0.01\n1.73 or 1.732 0.99 or 0.991  0.01\n1.89 or 1.892 1.20 or 1.204  0.05 or 0.06\n1.99 or 1.987 1.32 or 1.322  0.04\nValues of lg (T / days) and lg (L / 1030W) correct as shown above.\nAbsolute uncertainties in lg (L / 1030W) correct as shown above. 1\n\n2(c)(i) Six points from (b) plotted correctly. 1\nMust be within half a small square. Diameter of points must be less than half a small square.\nError bars in lg (L / 1030W) plotted correctly. 1\nAll error bars must be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.\n\n2(c)(ii) Straight line of best fit drawn. 1\nPoints must be balanced. Do not accept line from top point to bottom point.\nLine must pass between (1.43, 0.60) and (1.45, 0.60) and between (1.84, 1.15) and (1.86, 1.15).\nWorst acceptable line drawn (steepest or shallowest possible line that passes through all the error bars). 1\nAll error bars must be plotted.\n© UCLES 2022 Page 9 of 11\n\n2(c)(iii) Gradient determined with clear substitution of data points into y / x. 1\nDistance between data points must be greater than half the length of the drawn line.\nGradient of worst acceptable line determined. 1\nuncertainty = (gradient of line of best fit – gradient of worst acceptable line)\nor\nuncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)\n\n2(c)(iv) y-intercept determined by substitution into y = mx + c. 1\ny-intercept of worst acceptable line determined by substitution into y = mx + c. 1\nuncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line\nor\nuncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)\nDo not allow methods using a false origin.\n© UCLES 2022 Page 10 of 11\n\n2(d) a = gradient = (c)(iii) and a and K both given to two or three significant figures. 1\nValue of K determined using y-intercept. Correct method must be seen. 1\n10y-intercept 1030 10(c)(iv)1030\nK  \nS 3.851026\nor\nK 10y-interceptlg S 1030\nor\nK 10(c)(iv)lg 3.851026 1030\nabsolute uncertainty in a = absolute uncertainty in gradient 1\nand\n 10y-intercept 10WAL y-intercept 1030\nK \nS\nCorrect substitution of numbers must be seen.\n\n2(e) L determined from (d) or (c)(iii) and (c)(iv) with correct substitution and correct power of ten(s). 1\nDo not accept incorrect POT for a or K.\nL = 3.85  1026  (d)  5.0(c)(iii)\nor\nlgL(c)(iii)lg5.0y-intercept\n© UCLES 2022 Page 11 of 11", "source_pages": [ 7, 8, 9, 10, 11 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9702_s22_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-May-June/9702_s22_ms_52.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_s22_ms_52-p07.png", "../answer-assets/9702_s22_ms_52-p08.png", "../answer-assets/9702_s22_ms_52-p09.png", "../answer-assets/9702_s22_ms_52-p10.png", "../answer-assets/9702_s22_ms_52-p11.png" ] }, { "id": "9702-2022-mj-53-q01", "question_id": "9702-2022-mj-53-q01", "subject": "9702", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "53", "question_number": 1, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "1 Defining the problem\nd is the independent variable and V is the dependent variable or vary d and measure V 1\nkeep A or area (of overlap) of plates constant 1\nMethods of data collection\nlabelled diagram of workable experiment including: 1\n circuit diagram with voltmeter connected in parallel with the capacitor\n capacitor and voltmeter connected to the metal plates with no power supply in discharge part of the circuit\n correct symbols for capacitor and voltmeter\nmethod to charge parallel plates, e.g. separate circuit diagram showing plates connected to a d.c. power supply or combined 1\ncircuit with switches and d.c. power supply\nuse calipers to measure d 1\nor\nuse micrometer/calipers to measure thickness of spacers\nuse rule(r) to measure lengths to determine A and A = length  breadth 1\nMethod of analysis\n\n1 1 1\nplot a graph of against d or equivalent (e.g. d against )\nV V\n(Do not accept log graphs.)\ny-interceptC C 1\nK  or K \ngradientA gradientAW\n\n1 y-interceptC\n(for d against : K  )\nV A\n© UCLES 2022 Page 6 of 10\n\n1 1 1\nW \ny-intercept\n\n1 gradient gradientC\n(for d against : W  or W  )\nV y-intercept AK\nAdditional detail including safety considerations 6\nD1 use gloves to prevent electric shock or do not touch metal plates to avoid shocks\nD2 keep the initial p.d. across plates or initial charge constant\nD3 method to determine the value of C, e.g. description of an experiment to measure p.d. or current against time during\ndischarge through a resistor\nD4 method of operation of circuit(s) using switch(es)\nD5 description of method to fully discharge capacitor, e.g. between experiments, short-circuit the capacitor or use of\nswitch in parallel with capacitor\nD6 repeat measurements of d at different points across plates and average\nD7 repeat measurements of V for same d and average V\nD8 bottom plate resting on insulating material or top plate supported by strings\nD9 use high voltage power supply to increase charge on plates\nor\nuse a very small value of capacitance to increase voltmeter reading\nD10 relationship valid if a straight line is produced (not passing through the origin)\n© UCLES 2022 Page 7 of 10", "source_pages": [ 6, 7 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9702_s22_ms_53.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-May-June/9702_s22_ms_53.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_s22_ms_53-p06.png", "../answer-assets/9702_s22_ms_53-p07.png" ] }, { "id": "9702-2022-mj-53-q02", "question_id": "9702-2022-mj-53-q02", "subject": "9702", "year": 2022, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "53", "question_number": 2, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "2(a) gradient = n 1\ny-intercept = lgSZ\n\n2(b) 1\nlg (M / 1030kg) lg (L / 1028W)\n0.68 or 0.681  0.03 or 0.04 0.15 or 0.146\n0.81 or 0.806  0.02 or 0.03 0.49 or 0.491\n1.08 or 1.079  0.07 or 0.08 1.51 or 1.505\n1.36 or 1.362  0.04 2.54 or 2.544\n1.63 or 1.633  0.04 3.56 or 3.556\n1.96 or 1.959  0.02 4.82 or 4.820\nValues of lg (M / 1030kg) and lg (L / 1028W) correct as shown above.\nAbsolute uncertainties in lg (M / 1030kg) correct as shown above. 1\n\n2(c)(i) Six points from (b) plotted correctly. 1\nMust be within half a small square. Diameter of points must be less than half a small square.\nError bars in lg (M / 1030kg) plotted correctly. 1\nAll error bars must be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.\n\n2(c)(ii) Straight line of best fit drawn. 1\nPoints must be balanced. Do not accept line from top point to bottom point.\nLine must pass between (0.92, 1.0) and (0.96, 1.0) and between (1.86, 4.5) and (1.90, 4.5)\nWorst acceptable line drawn (steepest or shallowest possible line that passes through all the error bars). 1\nAll error bars must be plotted.\n© UCLES 2022 Page 8 of 10\n\n2(c)(iii) Gradient determined with clear substitution of data points into y / x. 1\nDistance between data points must be greater than half the length of the drawn line.\nGradient of worst acceptable line determined. 1\nuncertainty = (gradient of line of best fit – gradient of worst acceptable line)\nor\nuncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)\n\n2(c)(iv) y-intercept determined by substitution into y = mx + c. 1\ny-intercept of worst acceptable line determined by substitution into y = mx + c. 1\nuncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line\nor\nuncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)\nDo not allow methods using a false origin.\n© UCLES 2022 Page 9 of 10\n\n2(d) n gradient (c)(iii) and n and Z both given to two or three significant figures. 1\nValue of Z determined using y-intercept. Correct method must be seen. 1\n10y-intercept 1028 10(c)(iv)1028\nZ  \nS 3.851026\nor\nZ 10y-interceptlg S 1028\nor\nZ 10(c)(iv)lg 3.851026 1028\nAbsolute uncertainty in n = absolute uncertainty in gradient 1\nand\n 10y-intercept 10WAL y-intercept 1028\nZ \nS\nCorrect substitution of numbers must be seen.\n\n2(e) L determined from (d) or (c)(iii) and (c)(iv) with correct substitution and correct power of ten(s). 1\nDo not accept incorrect POT for n or Z.\nL = 3.85  1026  (d)  3.0(c)(iii)\nor\nlgL(c)(iii)lg3.0y-intercept\n© UCLES 2022 Page 10 of 10", "source_pages": [ 8, 9, 10 ], "source_pdf": "_source-pdfs/2022-May-June/ms/9702_s22_ms_53.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-May-June/9702_s22_ms_53.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_s22_ms_53-p08.png", "../answer-assets/9702_s22_ms_53-p09.png", "../answer-assets/9702_s22_ms_53-p10.png" ] }, { "id": "9702-2022-on-41-q01", "question_id": "9702-2022-on-41-q01", "subject": "9702", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 1, "topic": "Gravitational fields", "topic_slug": "9702-topic-13-gravitational-fields", "marks": 9, "status": "available", "reason": null, "text": "1(a) F = (Gm m ) / r 2 M1\n\n1 2\nwhere G is the gravitational constant A1\n\n1(b) gravitational force provides the centripetal force B1\nmR 2 = GMm / R 2 and  = 2 / T M1\nor\nmv 2 / R = GMm / R 2 and v = 2R / T\nor\n42mR / T 2 = GMm / R 2\ncorrect completion of algebra to get T2 = (42 / GM) R3, with identification of (42 / GM) as k A1\n\n1(c)(i) (24  3600)2 = (42  R3) / (6.67  10–11  6.0  1024) C1\nR = 4.2  107 m A1\n\n1(c)(ii) (orbit) must be above the Equator B1\n(direction) must be from west to east B1\n© UCLES 2022 Page 6 of 15", "source_pages": [ 6 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9702_w22_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Oct-Nov/9702_w22_ms_41.pdf?download=true", "html": "9702-topic-13-gravitational-fields/answers.html", "image_paths": [ "../answer-assets/9702_w22_ms_41-p06.png" ] }, { "id": "9702-2022-on-41-q02", "question_id": "9702-2022-on-41-q02", "subject": "9702", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 2, "topic": "Temperature", "topic_slug": "9702-topic-14-temperature", "marks": 10, "status": "available", "reason": null, "text": "2(a) • resistance of a metal B2\n• volume of a gas at constant pressure\n• e.m.f. of a thermocouple\nAny two points, 1 mark each\n\n2(b)(i) Q = mcT C1\nevidence of realisation that Q lost by water = Q gained by mercury C1\n18.7  4.18  (37.4 – T) = 6.94  0.140  (T – 23.0) C1\nT = 37.2 °C A1\n\n2(b)(ii) use a liquid with a lower (specific) heat capacity (than mercury) B1\nor\nuse a smaller mass of mercury\n\n2(c)(i) depends on properties of a real substance B1\n0 °C is not absolute zero B1\n\n2(c)(ii) ideal gas B1\n© UCLES 2022 Page 7 of 15", "source_pages": [ 7 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9702_w22_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Oct-Nov/9702_w22_ms_41.pdf?download=true", "html": "9702-topic-14-temperature/answers.html", "image_paths": [ "../answer-assets/9702_w22_ms_41-p07.png" ] }, { "id": "9702-2022-on-41-q03", "question_id": "9702-2022-on-41-q03", "subject": "9702", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 3, "topic": "Oscillations", "topic_slug": "9702-topic-17-oscillations", "marks": 11, "status": "available", "reason": null, "text": "3(a) a = –  2x M1\na = acceleration, x = displacement from equilibrium position and  = angular frequency A1\n\n3(b)(i) x = 0.12 m A1\n0\n\n3(b)(ii) v = (x 2 – x2) C1\n0\ntwo (x, v) pairs correctly read from Fig. 3.2 (one may be (x , 0) or value of x from (i))\n0 0\ne.g. 0.20 = (0.122 – 0) leading to  = 1.7 rad s–1 A1\n\n3(b)(iii) E = ½M 2x 2 C1\n0\n0.050 = ½  M  1.672  0.122 A1\nM = 2.5 kg\nor\n(E ) = ½Mv 2 (C1)\nK max 0\n0.050 = ½ M  0.202 (A1)\nM = 2.5 kg\n\n3(c)(i) loss of (total) energy (of system) B1\ndue to resistive forces B1\n\n3(c)(ii) closed loop surrounding the origin with maximum x at ± 0.060 m passing through v = 0 B1\nmaximum velocity shown as ± 0.10 m s–1 passing through x = 0 B1\n© UCLES 2022 Page 8 of 15", "source_pages": [ 8 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9702_w22_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Oct-Nov/9702_w22_ms_41.pdf?download=true", "html": "9702-topic-17-oscillations/answers.html", "image_paths": [ "../answer-assets/9702_w22_ms_41-p08.png" ] }, { "id": "9702-2022-on-41-q04", "question_id": "9702-2022-on-41-q04", "subject": "9702", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 4, "topic": "Nuclear physics", "topic_slug": "9702-topic-23-nuclear-physics", "marks": 12, "status": "available", "reason": null, "text": "4(a) (field line indicates) direction of force B1\nforce on a positive charge B1\n\n4(b)(i) one straight line perpendicular to plates, starting on one plate and finishing on the other B1\nfive straight lines perpendicular to plates between the plates, uniformly spaced B1\ndownwards arrows on lines B1\n\n4(b)(ii) E = V / d C1\n= 2400 / 0.046 A1\n= 5.2  104 N C–1\n\n4(c)(i) smooth curve in region of field and straight line outside field B1\ndirection of deflection shown as downwards in region of field B1\n\n4(c)(ii) helium nucleus has double the charge but four times the mass B1\nvelocity parallel to plates same and acceleration perpendicular to plates smaller (for helium) B1\nfinal speed is lower (for helium) B1\n© UCLES 2022 Page 9 of 15", "source_pages": [ 9 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9702_w22_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Oct-Nov/9702_w22_ms_41.pdf?download=true", "html": "9702-topic-23-nuclear-physics/answers.html", "image_paths": [ "../answer-assets/9702_w22_ms_41-p09.png" ] }, { "id": "9702-2022-on-41-q05", "question_id": "9702-2022-on-41-q05", "subject": "9702", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 5, "topic": "Capacitance", "topic_slug": "9702-topic-19-capacitance", "marks": 11, "status": "available", "reason": null, "text": "5(a)(i) Q = CV C1\nQ = 24  470  10–6 A1\n0\n= 0.011 C\n\n5(a)(ii) I = 24 / 5600 A1\n0\n= 4.3  10–3 A\n\n5(a)(iii)  = RC C1\n= 5600  470  10–6 A1\n= 2.6 s\n\n5(a)(iv) line with negative gradient throughout passing through (0, I ) B1\n0\nexponential decay curve asymptotic to t-axis B1\n\n5(b)(i) current in wire P gives rise to a magnetic field B1\nas current (in P) changes, wire Q cuts (magnetic) flux (of wire P) B1\ncutting magnetic flux causes induced e.m.f. (across Q) B1\n\n5(b)(ii) sketch shows line with a negative gradient throughout B1\n© UCLES 2022 Page 10 of 15", "source_pages": [ 10 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9702_w22_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Oct-Nov/9702_w22_ms_41.pdf?download=true", "html": "9702-topic-19-capacitance/answers.html", "image_paths": [ "../answer-assets/9702_w22_ms_41-p10.png" ] }, { "id": "9702-2022-on-41-q06", "question_id": "9702-2022-on-41-q06", "subject": "9702", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 6, "topic": "Magnetic fields", "topic_slug": "9702-topic-20-magnetic-fields", "marks": 9, "status": "available", "reason": null, "text": "6(a)(i) PQRS and WXYZ B1\n\n6(a)(ii) force on charge carriers is perpendicular to both (magnetic) field and current B1\nas charge carriers are deflected to one side, an electric field is set up B1\n(steady V when) electric and magnetic forces on charge carriers are equal (and opposite) B1\nH\n\n6(b)(i) n: number density of charge carriers B1\nt: distance PW (or SZ or QX or RY) B1\nq: charge on each charge carrier B1\n\n6(b)(ii) V inversely proportional to t B1\nH\n(so t needs to be small for) V to be large enough to measure B1\nH\n© UCLES 2022 Page 11 of 15", "source_pages": [ 11 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9702_w22_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Oct-Nov/9702_w22_ms_41.pdf?download=true", "html": "9702-topic-20-magnetic-fields/answers.html", "image_paths": [ "../answer-assets/9702_w22_ms_41-p11.png" ] }, { "id": "9702-2022-on-41-q07", "question_id": "9702-2022-on-41-q07", "subject": "9702", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 7, "topic": "Alternating currents", "topic_slug": "9702-topic-21-alternating-currents", "marks": 10, "status": "available", "reason": null, "text": "7(a)(i) peak voltage = 4.2  2 B1\n( = 5.9 V)\npower = V2 / R A1\n= 5.92 / 760 = 0.046 W or 46 mW\n\n7(a)(ii) sketch shows peak(s) in power at 46 mW B1\ncorrect shape (sinusoidal wave sitting on t-axis) B1\nfour cycles of repeating pattern shown, with P = 0 at 0, 10, 20, 30, 40 s B1\n\n7(a)(iii) line is symmetrical about 23 mW B1\n\n7(b)(i) (alternating p.d. makes) the crystal vibrate B1\nvibrations (of crystal) causes air to vibrate B1\nfrequency is in ultrasound range B1\n\n7(b)(ii) (air makes) crystal vibrate, which causes an e.m.f. to be generated across the (second) crystal B1\n© UCLES 2022 Page 12 of 15", "source_pages": [ 12 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9702_w22_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Oct-Nov/9702_w22_ms_41.pdf?download=true", "html": "9702-topic-21-alternating-currents/answers.html", "image_paths": [ "../answer-assets/9702_w22_ms_41-p12.png" ] }, { "id": "9702-2022-on-41-q08", "question_id": "9702-2022-on-41-q08", "subject": "9702", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 8, "topic": "Quantum physics", "topic_slug": "9702-topic-22-quantum-physics", "marks": 10, "status": "available", "reason": null, "text": "8(a) photon energy (to remove electron) B1\nminimum energy to remove electron B1\nor\nenergy to remove electron from surface\nor\nenergy to remove electron with zero kinetic energy\n\n8(b)(i) photon energy = hf C1\nnumber per unit time = 8.36  10–3 / (1.36  1015  6.63  10–34) A1\n= 9.27  1015 s–1\n\n8(b)(ii) hf =  + E C1\nMAX\n = (1.36  1015  6.63  10–34) – (3.09  10–19) A1\n= 5.93  10–19 J\n\n8(c)(i) greater photon energy (and same work function) M1\nso maximum kinetic energy is increased A1\n\n8(c)(ii) (greater photon energy and same power so) lower number of photons (per unit time) M1\n(each electron absorbs one photon) so lower rate of emission A1\n© UCLES 2022 Page 13 of 15", "source_pages": [ 13 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9702_w22_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Oct-Nov/9702_w22_ms_41.pdf?download=true", "html": "9702-topic-22-quantum-physics/answers.html", "image_paths": [ "../answer-assets/9702_w22_ms_41-p13.png" ] }, { "id": "9702-2022-on-41-q09", "question_id": "9702-2022-on-41-q09", "subject": "9702", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 9, "topic": "Astronomy and cosmology", "topic_slug": "9702-topic-25-astronomy-and-cosmology", "marks": 8, "status": "available", "reason": null, "text": "9(a) total power of radiation emitted (by the star) B1\n\n9(b)(i) F = L / (4d 2) C1\n= 9.86  1027 / [4  (8.14  1016)2] A1\n= 1.18  10–7 W m–2\n\n9(b)(ii) L = 4 r 2T4 C1\n9.86  1027 = 4    5.67  10–8  r 2  98304\nradius = 1.22  109 m A1\n\n9(c) wavelength of peak intensity determined (from spectrum of star) B1\nwavelength of peak intensity from object of known temperature determined B1\nWien’s displacement law used B1\nor\nwavelength of peak intensity inversely proportional to temperature\n© UCLES 2022 Page 14 of 15", "source_pages": [ 14 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9702_w22_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Oct-Nov/9702_w22_ms_41.pdf?download=true", "html": "9702-topic-25-astronomy-and-cosmology/answers.html", "image_paths": [ "../answer-assets/9702_w22_ms_41-p14.png" ] }, { "id": "9702-2022-on-41-q10", "question_id": "9702-2022-on-41-q10", "subject": "9702", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 10, "topic": "Nuclear physics", "topic_slug": "9702-topic-23-nuclear-physics", "marks": 10, "status": "available", "reason": null, "text": "10(a)(i) cannot predict when a (particular) nucleus will decay B1\nor\ncannot predict which nucleus will decay next\n\n10(a)(ii) not affected by external / environmental factors B1\n\n10(b)(i) line fluctuates B1\nor\ntrend is a straight line\n\n10(b)(ii) straight line of best fit drawn on Fig. 10.1 B1\n\n10(b)(iii) M = M exp (–t) B1\n0\nso ln M = ln M – t so gradient = – (and magnitude of gradient = )\n0\n\n10(b)(iv) gradient = (–) (8.0 – 4.8) / (11.6 – 0) (allow any correct pair of values from Fig. 10.1) C1\n = 0.28 s–1 A1\n\n10(b)(v) half-life = 0.693 /  A1\n= 0.693 / 0.28\n= 2.5 s\n\n10(c) (for reaction to occur,) energy is released B1\nenergy release comes from fall in mass so total mass of products must be less (than mass of carbon-15) B1\n© UCLES 2022 Page 15 of 15", "source_pages": [ 15 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9702_w22_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Oct-Nov/9702_w22_ms_41.pdf?download=true", "html": "9702-topic-23-nuclear-physics/answers.html", "image_paths": [ "../answer-assets/9702_w22_ms_41-p15.png" ] }, { "id": "9702-2022-on-42-q01", "question_id": "9702-2022-on-42-q01", "subject": "9702", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 1, "topic": "Electric fields", "topic_slug": "9702-topic-18-electric-fields", "marks": 10, "status": "available", "reason": null, "text": "1(a) force per unit mass B1\n\n1(b)(i) lines drawn are radial from the surface B1\narrows show pointing towards planet B1\n\n1(b)(ii) field lines show force (on satellite) is towards centre of planet B1\nor\nvelocity of satellite is perpendicular to field lines\n(gravitational) force perpendicular to velocity causes centripetal acceleration B1\n\n1(c)(i) T = 24 hours C1\na = r 2 and  = 2 / T C1\nor\na = v2 / r and v = 2r /T\nor\na = 42r / T 2\na = (42  6.4  106) / (24  60  60)2 A1\n= 0.034 m s–2\n\n1(c)(ii) identification of the two forces acting on the object as gravitational force and (normal) contact force M1\ngravitational force and normal contact force are in opposite directions, and their resultant causes the (centripetal) A1\nacceleration\n© UCLES 2022 Page 6 of 16", "source_pages": [ 6 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9702_w22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Oct-Nov/9702_w22_ms_42.pdf?download=true", "html": "9702-topic-18-electric-fields/answers.html", "image_paths": [ "../answer-assets/9702_w22_ms_42-p06.png" ] }, { "id": "9702-2022-on-42-q02", "question_id": "9702-2022-on-42-q02", "subject": "9702", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 2, "topic": "Thermodynamics", "topic_slug": "9702-topic-16-thermodynamics", "marks": 7, "status": "available", "reason": null, "text": "2(a) (thermal) energy per unit mass (to cause temperature change) B1\n(thermal) energy per unit change in temperature B1\n\n2(b)(i) work done correct (0) B1\nincrease in internal energy correct (+E) B1\n\n2(b)(ii) work done correct (–W) and increase in internal energy same as (b)(i) B1\nthermal energy correct so that it adds to work done to give increase in internal energy B1\n\n2(c) more thermal energy needed so specific heat capacity is greater B1\n© UCLES 2022 Page 7 of 16", "source_pages": [ 7 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9702_w22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Oct-Nov/9702_w22_ms_42.pdf?download=true", "html": "9702-topic-16-thermodynamics/answers.html", "image_paths": [ "../answer-assets/9702_w22_ms_42-p07.png" ] }, { "id": "9702-2022-on-42-q03", "question_id": "9702-2022-on-42-q03", "subject": "9702", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 3, "topic": "Temperature", "topic_slug": "9702-topic-14-temperature", "marks": 10, "status": "available", "reason": null, "text": "3(a) p = pressure (of gas), V = volume (of gas) and k = Boltzmann constant B1\nN = number of molecules B1\nT = thermodynamic temperature B1\n\n3(b) (pV = NkT and pV = ⅓Nm leading to) NkT = ⅓Nm M1\nalgebra leading to (3/2)kT = ½m and use of ½m = E leading to (3/2)kT = E A1\nK K\n\n3(c)(i) T = 296 K C1\n½m = (3/2)kT C1\n½  5.31  10–26  u2 = (3/2)  1.38  10–23  296\nu = 480 m s–1 A1\n\n3(c)(ii) line passing through (P, u) B1\nhorizontal straight line B1\n© UCLES 2022 Page 8 of 16", "source_pages": [ 8 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9702_w22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Oct-Nov/9702_w22_ms_42.pdf?download=true", "html": "9702-topic-14-temperature/answers.html", "image_paths": [ "../answer-assets/9702_w22_ms_42-p08.png" ] }, { "id": "9702-2022-on-42-q04", "question_id": "9702-2022-on-42-q04", "subject": "9702", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 4, "topic": "Oscillations", "topic_slug": "9702-topic-17-oscillations", "marks": 10, "status": "available", "reason": null, "text": "4(a)(i) x = 8.0 cm A1\n0\n\n4(a)(ii)  = 2 / T C1\n= 2 / 4.0 = 1.6 rad s–1 A1\n\n4(a)(iii) E = ½m 2x 2 C1\n0\n= ½  36  1.62  0.0802 C1\n= 0.29 J A1\n\n4(b) dome-shaped curve, starting and ending at E = 0 B1\nK\nmaximum E shown as 0.29 J B1\nK\nposition of peak shown at h = 10.0 cm B1\nline intercepts h-axis at h = 2.0 cm and at h = 18.0 cm B1\n© UCLES 2022 Page 9 of 16", "source_pages": [ 9 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9702_w22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Oct-Nov/9702_w22_ms_42.pdf?download=true", "html": "9702-topic-17-oscillations/answers.html", "image_paths": [ "../answer-assets/9702_w22_ms_42-p09.png" ] }, { "id": "9702-2022-on-42-q05", "question_id": "9702-2022-on-42-q05", "subject": "9702", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 5, "topic": "Electric fields", "topic_slug": "9702-topic-18-electric-fields", "marks": 10, "status": "available", "reason": null, "text": "5(a) work done per unit charge B1\nwork done (on charge) in moving positive charge from infinity (to the point) B1\n\n5(b)(i) radius = 0.060 m A1\n\n5(b)(ii) V = Q / 4x C1\n0\nQ = (–) 850  4  8.85  10–12  0.060\nor\nQ = (–) 850  0.060 / 8.99  109\n(any correct pair of V and x values from curve)\nQ = – 5.7  10–9 C A1\n\n5(c)(i) E = Q2 / 4x C1\nP 0\n= (5.67  10–9)2 / (4  8.85  10–12  0.46)\n= 6.3  10–7 J A1\n\n5(c)(ii) • force is repulsive so spheres move apart B3\n• force in direction of motion so speed increases\n• potential energy converted to kinetic energy so speed increases\n• force decreases with distance so acceleration decreases\n• momentum is conserved (at zero) (and masses are equal) so velocities are always equal and opposite\nAny three points, 1 mark each\n© UCLES 2022 Page 10 of 16", "source_pages": [ 10 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9702_w22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Oct-Nov/9702_w22_ms_42.pdf?download=true", "html": "9702-topic-18-electric-fields/answers.html", "image_paths": [ "../answer-assets/9702_w22_ms_42-p10.png" ] }, { "id": "9702-2022-on-42-q06", "question_id": "9702-2022-on-42-q06", "subject": "9702", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 6, "topic": "Capacitance", "topic_slug": "9702-topic-19-capacitance", "marks": 10, "status": "available", "reason": null, "text": "6(a) • p.d. across resistor = p.d. across capacitor B2\n• current (in resistor) proportional to p.d. across it\n• current causes capacitor to lose charge\n• charge (on capacitor) proportional to p.d. so p.d. decreases\nAny two points, 1 mark each\nrate of change of p.d. decreases as p.d. decreases B1\n\n6(b) Q = 0.90 mC and at t = one time constant, Q = Q exp (–1) B1\n0 0\nat t = one time constant, Q = 0.90 exp (–1) = 0.33 mC M1\nevidence of graph reading: when Q = 0.33 mC, t = 5.5 s A1\nor\nevidence of two correct sets of readings for Q and t from the graph (B1)\ncorrect substitution of Q and t values into Q = Q exp [(t – t ) / ] (M1)\n2 1 1 2\ncalculation to give  = 5.5 s (A1)\nor\nread-off of half-life as 3.75 s (B1)\nuse of Q = Q exp (–t / ) to show that  = half-life / ln 2 (M1)\n0\n = 3.75 / ln 2 = 5.4 s (A1)\n© UCLES 2022 Page 11 of 16\n\n6(b) or\ntangent drawn on Q–t graph and value of Q at exact same time as tangent read from graph (M1)\ngradient of tangent correctly calculated (A1)\n = Q / gradient used to correctly calculate a value for  as 5.5 s (A1)\n\n6(c)(i) C = Q / V C1\n= [(0.90  10–3) / 7.5] = 1.2  10–4 C A1\n= 120 F\n\n6(c)(ii) R = τ / C C1\n= 5.5 / (1.2  10–4) (= 45 800 ) A1\n= 46 k\n© UCLES 2022 Page 12 of 16", "source_pages": [ 11, 12 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9702_w22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Oct-Nov/9702_w22_ms_42.pdf?download=true", "html": "9702-topic-19-capacitance/answers.html", "image_paths": [ "../answer-assets/9702_w22_ms_42-p11.png", "../answer-assets/9702_w22_ms_42-p12.png" ] }, { "id": "9702-2022-on-42-q07", "question_id": "9702-2022-on-42-q07", "subject": "9702", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 7, "topic": "Magnetic fields", "topic_slug": "9702-topic-20-magnetic-fields", "marks": 10, "status": "available", "reason": null, "text": "7(a) force per unit current M1\nforce per unit length M1\ncurrent / wire is perpendicular to (magnetic) field (lines) A1\n\n7(b)(i) current (in coil) is perpendicular to magnetic field (so force on wire) B1\nforce (on wire) is perpendicular to current and field (so is vertical) B1\nor\ncurrent and field are both horizontal (so force is vertical)\n\n7(b)(ii) NBIL = mg C1\nB = (2.16  10–3  9.81) / (40  3.94  0.0300) C1\n= 4.48  10–3 T A1\n\n7(b)(iii) (magnetic) forces (on balance and newton meter) are (equal and) opposite B1\nreading = 0.563 – (2.16  10–3  9.81) A1\n= 0.542 N\n© UCLES 2022 Page 13 of 16", "source_pages": [ 13 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9702_w22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Oct-Nov/9702_w22_ms_42.pdf?download=true", "html": "9702-topic-20-magnetic-fields/answers.html", "image_paths": [ "../answer-assets/9702_w22_ms_42-p13.png" ] }, { "id": "9702-2022-on-42-q08", "question_id": "9702-2022-on-42-q08", "subject": "9702", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 8, "topic": "Magnetic fields", "topic_slug": "9702-topic-20-magnetic-fields", "marks": 11, "status": "available", "reason": null, "text": "8(a) direction of induced e.m.f. M1\nsuch as to (produce effects that) oppose the change that caused it A1\n\n8(b)(i) X = 0.85 A A1\nY = 2 / 0.040 C1\n= 160 rad s–1 A1\n\n8(b)(ii) two cycles of a sinusoidal curve with a period of 0.040 s B1\ncorrect phase (i.e. V max / min at t = 0, 0.02, 0.04, 0.06 and 0.08 s, and V zero at t = 0.01, 0.03, 0.05, 0.07 s) B1\n2 2\nmaximum / minimum V shown (consistently) at ± 6.5 V B1\n2\n\n8(b)(iii) (magnitude of) V is proportional to rate of change of (magnetic) flux B1\n2\n• V is proportional to gradient of I –t curve B2\n2 1\n• V has maximum magnitude when I –t curve is steepest\n2 1\n• V is zero when I –t curve is horizontal / a maximum or minimum\n2 1\n• V changes sign when sign of gradient of I –t curve changes\n2 1\nAny two points, 1 mark each\n© UCLES 2022 Page 14 of 16", "source_pages": [ 14 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9702_w22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Oct-Nov/9702_w22_ms_42.pdf?download=true", "html": "9702-topic-20-magnetic-fields/answers.html", "image_paths": [ "../answer-assets/9702_w22_ms_42-p14.png" ] }, { "id": "9702-2022-on-42-q09", "question_id": "9702-2022-on-42-q09", "subject": "9702", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 9, "topic": "Astronomy and cosmology", "topic_slug": "9702-topic-25-astronomy-and-cosmology", "marks": 13, "status": "available", "reason": null, "text": "9(a)(i) • energy of photon has a corresponding frequency B3\n• change in electron energy level emits a single photon\n• photon energy = difference in energy levels\n• discrete frequencies must have come from discrete energy gaps\n• discrete energy changes imply discrete energy levels\nAny three points, 1 mark each\n\n9(a)(ii) transition (to – 3.400 eV) from X corresponds to 658 nm line C1\nE – E = hc /  C1\n1 2\nE – (– 3.400) = (6.63  10–34  3.00  108) / (658  10–9  1.60  10–19) A1\n1\nand so E = –1.51 eV (full substitution and answer needed)\n1\n\n9(b)(i) redshift B1\n\n9(b)(ii) moving away (from observer) B1\n\n9(b)(iii)  /  = v / c C1\ne.g. for 658 nm line:  = 686 – 658\n( = 28 nm) (other lines may be used)\n28 / 658 = v / (3.00  108) (other lines may be used) C1\nv = 1.3  107 m s–1 A1\n\n9(c) v = H d C1\n0\nH = (1.3  107) / (5.7  1024) A1\n0\n= 2.3  10–18 s–1\n© UCLES 2022 Page 15 of 16", "source_pages": [ 15 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9702_w22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Oct-Nov/9702_w22_ms_42.pdf?download=true", "html": "9702-topic-25-astronomy-and-cosmology/answers.html", "image_paths": [ "../answer-assets/9702_w22_ms_42-p15.png" ] }, { "id": "9702-2022-on-42-q10", "question_id": "9702-2022-on-42-q10", "subject": "9702", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 10, "topic": "Medical physics", "topic_slug": "9702-topic-24-medical-physics", "marks": 9, "status": "available", "reason": null, "text": "10(a)(i) introduction of tracer (into the body) M1\ncontaining a + emitter A1\n\n10(a)(ii) positron interacts with electron B1\n(pair) annihilation occurs B1\nmass of particles converted into gamma photons B1\n\n10(b) (annihilation of electron and positron) produces two photons B1\nE = ()mc2 B1\nE = hf and f = c /  B1\nor\nE = hc / \n = {[2] 6.63  10–34  3.00  108} / {[2] 9.11  10–31  (3.00  108)2} B1\n= 2.4(3)  10–12 m or 2.4(3) pm (full substitution and answer with unit needed)\n© UCLES 2022 Page 16 of 16", "source_pages": [ 16 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9702_w22_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Oct-Nov/9702_w22_ms_42.pdf?download=true", "html": "9702-topic-24-medical-physics/answers.html", "image_paths": [ "../answer-assets/9702_w22_ms_42-p16.png" ] }, { "id": "9702-2022-on-43-q01", "question_id": "9702-2022-on-43-q01", "subject": "9702", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 1, "topic": "Gravitational fields", "topic_slug": "9702-topic-13-gravitational-fields", "marks": 9, "status": "available", "reason": null, "text": "1(a) F = (Gm m ) / r 2 M1\n\n1 2\nwhere G is the gravitational constant A1\n\n1(b) gravitational force provides the centripetal force B1\nmR 2 = GMm / R 2 and  = 2 / T M1\nor\nmv 2 / R = GMm / R 2 and v = 2R / T\nor\n42mR / T 2 = GMm / R 2\ncorrect completion of algebra to get T2 = (42 / GM) R3, with identification of (42 / GM) as k A1\n\n1(c)(i) (24  3600)2 = (42  R3) / (6.67  10–11  6.0  1024) C1\nR = 4.2  107 m A1\n\n1(c)(ii) (orbit) must be above the Equator B1\n(direction) must be from west to east B1\n© UCLES 2022 Page 6 of 15", "source_pages": [ 6 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9702_w22_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Oct-Nov/9702_w22_ms_43.pdf?download=true", "html": "9702-topic-13-gravitational-fields/answers.html", "image_paths": [ "../answer-assets/9702_w22_ms_43-p06.png" ] }, { "id": "9702-2022-on-43-q02", "question_id": "9702-2022-on-43-q02", "subject": "9702", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 2, "topic": "Temperature", "topic_slug": "9702-topic-14-temperature", "marks": 10, "status": "available", "reason": null, "text": "2(a) • resistance of a metal B2\n• volume of a gas at constant pressure\n• e.m.f. of a thermocouple\nAny two points, 1 mark each\n\n2(b)(i) Q = mcT C1\nevidence of realisation that Q lost by water = Q gained by mercury C1\n18.7  4.18  (37.4 – T) = 6.94  0.140  (T – 23.0) C1\nT = 37.2 °C A1\n\n2(b)(ii) use a liquid with a lower (specific) heat capacity (than mercury) B1\nor\nuse a smaller mass of mercury\n\n2(c)(i) depends on properties of a real substance B1\n0 °C is not absolute zero B1\n\n2(c)(ii) ideal gas B1\n© UCLES 2022 Page 7 of 15", "source_pages": [ 7 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9702_w22_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Oct-Nov/9702_w22_ms_43.pdf?download=true", "html": "9702-topic-14-temperature/answers.html", "image_paths": [ "../answer-assets/9702_w22_ms_43-p07.png" ] }, { "id": "9702-2022-on-43-q03", "question_id": "9702-2022-on-43-q03", "subject": "9702", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 3, "topic": "Oscillations", "topic_slug": "9702-topic-17-oscillations", "marks": 11, "status": "available", "reason": null, "text": "3(a) a = –  2x M1\na = acceleration, x = displacement from equilibrium position and  = angular frequency A1\n\n3(b)(i) x = 0.12 m A1\n0\n\n3(b)(ii) v = (x 2 – x2) C1\n0\ntwo (x, v) pairs correctly read from Fig. 3.2 (one may be (x , 0) or value of x from (i))\n0 0\ne.g. 0.20 = (0.122 – 0) leading to  = 1.7 rad s–1 A1\n\n3(b)(iii) E = ½M 2x 2 C1\n0\n0.050 = ½  M  1.672  0.122 A1\nM = 2.5 kg\nor\n(E ) = ½Mv 2 (C1)\nK max 0\n0.050 = ½ M  0.202 (A1)\nM = 2.5 kg\n\n3(c)(i) loss of (total) energy (of system) B1\ndue to resistive forces B1\n\n3(c)(ii) closed loop surrounding the origin with maximum x at ± 0.060 m passing through v = 0 B1\nmaximum velocity shown as ± 0.10 m s–1 passing through x = 0 B1\n© UCLES 2022 Page 8 of 15", "source_pages": [ 8 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9702_w22_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Oct-Nov/9702_w22_ms_43.pdf?download=true", "html": "9702-topic-17-oscillations/answers.html", "image_paths": [ "../answer-assets/9702_w22_ms_43-p08.png" ] }, { "id": "9702-2022-on-43-q04", "question_id": "9702-2022-on-43-q04", "subject": "9702", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 4, "topic": "Nuclear physics", "topic_slug": "9702-topic-23-nuclear-physics", "marks": 12, "status": "available", "reason": null, "text": "4(a) (field line indicates) direction of force B1\nforce on a positive charge B1\n\n4(b)(i) one straight line perpendicular to plates, starting on one plate and finishing on the other B1\nfive straight lines perpendicular to plates between the plates, uniformly spaced B1\ndownwards arrows on lines B1\n\n4(b)(ii) E = V / d C1\n= 2400 / 0.046 A1\n= 5.2  104 N C–1\n\n4(c)(i) smooth curve in region of field and straight line outside field B1\ndirection of deflection shown as downwards in region of field B1\n\n4(c)(ii) helium nucleus has double the charge but four times the mass B1\nvelocity parallel to plates same and acceleration perpendicular to plates smaller (for helium) B1\nfinal speed is lower (for helium) B1\n© UCLES 2022 Page 9 of 15", "source_pages": [ 9 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9702_w22_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Oct-Nov/9702_w22_ms_43.pdf?download=true", "html": "9702-topic-23-nuclear-physics/answers.html", "image_paths": [ "../answer-assets/9702_w22_ms_43-p09.png" ] }, { "id": "9702-2022-on-43-q05", "question_id": "9702-2022-on-43-q05", "subject": "9702", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 5, "topic": "Capacitance", "topic_slug": "9702-topic-19-capacitance", "marks": 11, "status": "available", "reason": null, "text": "5(a)(i) Q = CV C1\nQ = 24  470  10–6 A1\n0\n= 0.011 C\n\n5(a)(ii) I = 24 / 5600 A1\n0\n= 4.3  10–3 A\n\n5(a)(iii)  = RC C1\n= 5600  470  10–6 A1\n= 2.6 s\n\n5(a)(iv) line with negative gradient throughout passing through (0, I ) B1\n0\nexponential decay curve asymptotic to t-axis B1\n\n5(b)(i) current in wire P gives rise to a magnetic field B1\nas current (in P) changes, wire Q cuts (magnetic) flux (of wire P) B1\ncutting magnetic flux causes induced e.m.f. (across Q) B1\n\n5(b)(ii) sketch shows line with a negative gradient throughout B1\n© UCLES 2022 Page 10 of 15", "source_pages": [ 10 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9702_w22_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Oct-Nov/9702_w22_ms_43.pdf?download=true", "html": "9702-topic-19-capacitance/answers.html", "image_paths": [ "../answer-assets/9702_w22_ms_43-p10.png" ] }, { "id": "9702-2022-on-43-q06", "question_id": "9702-2022-on-43-q06", "subject": "9702", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 6, "topic": "Magnetic fields", "topic_slug": "9702-topic-20-magnetic-fields", "marks": 9, "status": "available", "reason": null, "text": "6(a)(i) PQRS and WXYZ B1\n\n6(a)(ii) force on charge carriers is perpendicular to both (magnetic) field and current B1\nas charge carriers are deflected to one side, an electric field is set up B1\n(steady V when) electric and magnetic forces on charge carriers are equal (and opposite) B1\nH\n\n6(b)(i) n: number density of charge carriers B1\nt: distance PW (or SZ or QX or RY) B1\nq: charge on each charge carrier B1\n\n6(b)(ii) V inversely proportional to t B1\nH\n(so t needs to be small for) V to be large enough to measure B1\nH\n© UCLES 2022 Page 11 of 15", "source_pages": [ 11 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9702_w22_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Oct-Nov/9702_w22_ms_43.pdf?download=true", "html": "9702-topic-20-magnetic-fields/answers.html", "image_paths": [ "../answer-assets/9702_w22_ms_43-p11.png" ] }, { "id": "9702-2022-on-43-q07", "question_id": "9702-2022-on-43-q07", "subject": "9702", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 7, "topic": "Alternating currents", "topic_slug": "9702-topic-21-alternating-currents", "marks": 10, "status": "available", "reason": null, "text": "7(a)(i) peak voltage = 4.2  2 B1\n( = 5.9 V)\npower = V2 / R A1\n= 5.92 / 760 = 0.046 W or 46 mW\n\n7(a)(ii) sketch shows peak(s) in power at 46 mW B1\ncorrect shape (sinusoidal wave sitting on t-axis) B1\nfour cycles of repeating pattern shown, with P = 0 at 0, 10, 20, 30, 40 s B1\n\n7(a)(iii) line is symmetrical about 23 mW B1\n\n7(b)(i) (alternating p.d. makes) the crystal vibrate B1\nvibrations (of crystal) causes air to vibrate B1\nfrequency is in ultrasound range B1\n\n7(b)(ii) (air makes) crystal vibrate, which causes an e.m.f. to be generated across the (second) crystal B1\n© UCLES 2022 Page 12 of 15", "source_pages": [ 12 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9702_w22_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Oct-Nov/9702_w22_ms_43.pdf?download=true", "html": "9702-topic-21-alternating-currents/answers.html", "image_paths": [ "../answer-assets/9702_w22_ms_43-p12.png" ] }, { "id": "9702-2022-on-43-q08", "question_id": "9702-2022-on-43-q08", "subject": "9702", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 8, "topic": "Quantum physics", "topic_slug": "9702-topic-22-quantum-physics", "marks": 10, "status": "available", "reason": null, "text": "8(a) photon energy (to remove electron) B1\nminimum energy to remove electron B1\nor\nenergy to remove electron from surface\nor\nenergy to remove electron with zero kinetic energy\n\n8(b)(i) photon energy = hf C1\nnumber per unit time = 8.36  10–3 / (1.36  1015  6.63  10–34) A1\n= 9.27  1015 s–1\n\n8(b)(ii) hf =  + E C1\nMAX\n = (1.36  1015  6.63  10–34) – (3.09  10–19) A1\n= 5.93  10–19 J\n\n8(c)(i) greater photon energy (and same work function) M1\nso maximum kinetic energy is increased A1\n\n8(c)(ii) (greater photon energy and same power so) lower number of photons (per unit time) M1\n(each electron absorbs one photon) so lower rate of emission A1\n© UCLES 2022 Page 13 of 15", "source_pages": [ 13 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9702_w22_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Oct-Nov/9702_w22_ms_43.pdf?download=true", "html": "9702-topic-22-quantum-physics/answers.html", "image_paths": [ "../answer-assets/9702_w22_ms_43-p13.png" ] }, { "id": "9702-2022-on-43-q09", "question_id": "9702-2022-on-43-q09", "subject": "9702", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 9, "topic": "Astronomy and cosmology", "topic_slug": "9702-topic-25-astronomy-and-cosmology", "marks": 8, "status": "available", "reason": null, "text": "9(a) total power of radiation emitted (by the star) B1\n\n9(b)(i) F = L / (4d 2) C1\n= 9.86  1027 / [4  (8.14  1016)2] A1\n= 1.18  10–7 W m–2\n\n9(b)(ii) L = 4 r 2T4 C1\n9.86  1027 = 4    5.67  10–8  r 2  98304\nradius = 1.22  109 m A1\n\n9(c) wavelength of peak intensity determined (from spectrum of star) B1\nwavelength of peak intensity from object of known temperature determined B1\nWien’s displacement law used B1\nor\nwavelength of peak intensity inversely proportional to temperature\n© UCLES 2022 Page 14 of 15", "source_pages": [ 14 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9702_w22_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Oct-Nov/9702_w22_ms_43.pdf?download=true", "html": "9702-topic-25-astronomy-and-cosmology/answers.html", "image_paths": [ "../answer-assets/9702_w22_ms_43-p14.png" ] }, { "id": "9702-2022-on-43-q10", "question_id": "9702-2022-on-43-q10", "subject": "9702", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 10, "topic": "Nuclear physics", "topic_slug": "9702-topic-23-nuclear-physics", "marks": 10, "status": "available", "reason": null, "text": "10(a)(i) cannot predict when a (particular) nucleus will decay B1\nor\ncannot predict which nucleus will decay next\n\n10(a)(ii) not affected by external / environmental factors B1\n\n10(b)(i) line fluctuates B1\nor\ntrend is a straight line\n\n10(b)(ii) straight line of best fit drawn on Fig. 10.1 B1\n\n10(b)(iii) M = M exp (–t) B1\n0\nso ln M = ln M – t so gradient = – (and magnitude of gradient = )\n0\n\n10(b)(iv) gradient = (–) (8.0 – 4.8) / (11.6 – 0) (allow any correct pair of values from Fig. 10.1) C1\n = 0.28 s–1 A1\n\n10(b)(v) half-life = 0.693 /  A1\n= 0.693 / 0.28\n= 2.5 s\n\n10(c) (for reaction to occur,) energy is released B1\nenergy release comes from fall in mass so total mass of products must be less (than mass of carbon-15) B1\n© UCLES 2022 Page 15 of 15", "source_pages": [ 15 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9702_w22_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Oct-Nov/9702_w22_ms_43.pdf?download=true", "html": "9702-topic-23-nuclear-physics/answers.html", "image_paths": [ "../answer-assets/9702_w22_ms_43-p15.png" ] }, { "id": "9702-2022-on-51-q01", "question_id": "9702-2022-on-51-q01", "subject": "9702", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "51", "question_number": 1, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "1 Defining the problem\nA is the independent variable and s is the dependent variable or vary A and measure s 1\nkeep B and t constant 1\nMethods of data collection\nlabelled diagram of workable experiment including: 1\n• pin / rod through hole\n• supported by a stand\n• sheet able to oscillate freely\n• at least one label from copper/sheet, hole, clamp, stand, rod, pin.\ndrawn clamped rule(r) parallel to the direction of the oscillations (by eye) (to measure s) 1\nuse rule(r) to measure lengths to determine A 1\nand\nA = length  breadth\nuse of micrometer to measure t 1\nMethod of Analysis\nplot a graph of ln s against A or equivalent 1\nrelationship valid if a straight line (with y-intercept = ln s ) is produced 1\n0\ngradient 1\nK =−\nBt\n\n1\n(K =− for A against ln s)\nBtgradient\n© UCLES 2022 Page 5 of 9\n\n1 Additional detail including safety considerations 6\nD1 use of cushion/sand box in case sheet falls\nor\nuse gloves to protect hands from cuts / sharp edges\nD2 keep (initial) distance between (copper) sheet and (poles of) magnet constant\nor\nkeep (initial) distance between (copper) sheet and coil(s) constant\nD3 keep s constant\n0\nD4 method to ensure s is constant, e.g. initially line up (corner of) plate with fiducial marker / vertical pin to keep s\n0 0\nconstant\nD5 method to determine s using video camera:\n• rule(r) in a position to measure s in the diagram\n• video camera shown in diagram or description of use of video camera\n• playback video recording by frame by frame / slow motion (to measure s)\nD6 repeat measurements of t in different positions and average t\nD7 measure B/magnetic flux density using a (calibrated) Hall probe\nD8 additional detail on use of Hall probe, e.g. adjust probe until maximum value\nor\nmeasure B using Hall probe first in one direction and then in the opposite direction and average\nD9 drawn method to create a magnetic field perpendicular to the area of the sheet, e.g. pair of magnets/horseshoe\nmagnet/pair of coils connected to a (d.c.) supply\nD10 repeat experiment for each A and average s\n© UCLES 2022 Page 6 of 9", "source_pages": [ 5, 6 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9702_w22_ms_51.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Oct-Nov/9702_w22_ms_51.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_w22_ms_51-p05.png", "../answer-assets/9702_w22_ms_51-p06.png" ] }, { "id": "9702-2022-on-51-q02", "question_id": "9702-2022-on-51-q02", "subject": "9702", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "51", "question_number": 2, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "2(a) YZd2 1\ngradient =\n4\n\n2(b) 1\n1\n/ 10–3 –1\nR\n45 or 45.5\n37 or 37.0\n30 or 30.3\n26 or 25.6\n21 or 21.3\n19 or 18.5\n1 1\nAbsolute uncertainties in from ± 2 to ± 0.9 or ± 1.\nR\n\n2(c)(i) Six points from (b) plotted correctly. 1\nMust be within half a small square. Diameter of points must be less than half a small square.\n1 1\nError bars in plotted correctly.\nR\nAll error bars to be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.\n\n2(c)(ii) Straight line of best fit drawn. 1\nPoints must be balanced. Do not accept line from top point to bottom point.\nLine must pass between (22.0, 30.0) and (23.0, 30.0) and (40.5, 65.0) and (42.0, 65.0).\nWorst acceptable line drawn (steepest or shallowest possible line that passes through all the error bars). 1\nAll error bars must be plotted.\n© UCLES 2022 Page 7 of 9\n\n2(c)(iii) Gradient determined with clear substitution of data points into y / x. 1\nDistance between data points must be greater than half the length of the drawn line.\nGradient of worst acceptable line determined with clear substitution of data points into y / x. 1\nuncertainty = (gradient of line of best fit – gradient of worst acceptable line)\nor\nuncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)\n\n2(d) 0.261 ± 0.003 (mm) 1\n\n2(e)(i)  determined using gradient and  given to two or three significant figures. 1\nYZd2 2222(d)2\n= =\n4gradient 4(c)(iii)\n determined using gradient and given with correct SI unit ( m) and correct power of ten 1\n\n2(e)(ii) percentage uncertainty in : 1\n 2d gradient \npercentage uncertainty= + +0.05+0.05100\n d gradient \nor\ncorrect substitution for max/min methods\n(1.0522)(1.0522)(d +d)2\nmax=\n4mingradient\n(0.9522)(0.9522)(d −d)2\nmin=\n4maxgradient\n© UCLES 2022 Page 8 of 9\n\n2(f) R determined to at least two significant figures from (c)(iii) or (d) and (e)(i) with correct substitution seen. 1\ngradient\nR =\n0.950\nor\nYZd2 2222(d)2\nR = =\n4L 4(e)(i)0.950\nAbsolute uncertainty in R determined. 1\nMethod must be consistent with determination of R and correct substitution must be seen.\nfor R determined using the gradient:\ngradient\nR = R\ngradient\nor\nfor R determined using (d) and (e)(i):\n 2d  \nR= + +0.05+0.05 R\n d  \nor\ncorrect substitution for max/min methods:\n(1.0522)(1.0522)(d +d)2\nmaxR =\n4min0.950\n(0.9522)(0.9522)(d −d)2\nmin R =\n4max0.950\n© UCLES 2022 Page 9 of 9", "source_pages": [ 7, 8, 9 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9702_w22_ms_51.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Oct-Nov/9702_w22_ms_51.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_w22_ms_51-p07.png", "../answer-assets/9702_w22_ms_51-p08.png", "../answer-assets/9702_w22_ms_51-p09.png" ] }, { "id": "9702-2022-on-52-q01", "question_id": "9702-2022-on-52-q01", "subject": "9702", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "52", "question_number": 1, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "1 Defining the problem\nz is the independent variable and t is the dependent variable or vary z and measure t 1\nkeep B and A constant 1\nMethods of data collection\nlabelled diagram of workable experiment including: 1\n• pin/rod though hole\n• supported by a stand\n• sheet able to oscillate freely\n• at least one label from copper/sheet, hole, clamp stand, rod, pin\nuse of stop-watch/timer to measure t (from release to stopping) 1\nor\nuse of stop-watch/timer to measure time for the sheet (to stop) oscillating\nuse of micrometer to measure z 1\nuse of rule(r) to measure lengths to determine A 1\nand\nA = length  breadth\nMethod of Analysis\nplot a graph of lg t against lg z or equivalent (e.g. ln t against ln z) 1\nq=gradient 1\nK = AB10y-intercept 1\n(K = ABey-intercept for ln t against ln z)\n© UCLES 2022 Page 5 of 9\n\n1 Additional detail including safety considerations 6\nD1 use of cushion / sand box in case sheet falls\nor\nuse gloves to protect hands from cuts / sharp edges\nD2 keep (initial) distance between (copper) sheet and (poles of) magnet constant\nor\nkeep (initial) distance between (copper) sheet and coil(s) constant\nD3 keep initial displacement (of copper sheet) constant\nD4 method to ensure initial displacement (of copper sheet) is constant, e.g. initially line up (corner of) plate with fiducial\nmarker/vertical pin\n K \nD5 relationship valid if a straight line (with y-intercept = log   ) is produced\nAB\nD6 repeat measurements of z in different positions and average z\nD7 measure B / magnetic flux density using a (calibrated) Hall probe\nD8 additional detail on use of Hall probe, e.g. adjust (position of) probe until maximum value\nor\nmeasure B using Hall probe first in one direction and then in the opposite direction and average\nD9 drawn method to create a magnetic field perpendicular to the area of the sheet, e.g. pair of magnets/horseshoe\nmagnet/pair of coils connected to a (d.c.) supply\nD10 repeat experiment for each z and average t\nD11 method to determine , e.g. measure mass with balance and volume = Az and density = mass / volume\n© UCLES 2022 Page 6 of 9", "source_pages": [ 5, 6 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9702_w22_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Oct-Nov/9702_w22_ms_52.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_w22_ms_52-p05.png", "../answer-assets/9702_w22_ms_52-p06.png" ] }, { "id": "9702-2022-on-52-q02", "question_id": "9702-2022-on-52-q02", "subject": "9702", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "52", "question_number": 2, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "2(a) c 1\ngradient =\n2h\nc\ny-intercept = −\n4h\n\n2(b) 1\nT / ms f / Hz\n7.0 or 7.00 ± 1 140 or 143 ± (10–30)\n2.9 or 2.90 ± 0.2 340 or 345 ± (20–30)\n1.8 or 1.80 ± 0.1 560 or 556 ± 30\n1.4 or 1.35 ± 0.1 710 or 714 ± (40–60)\nor\n740 or 741\n1.1 or 1.05 ± 0.1 910 or 909 ± (80–100)\nor\n950or952\n0.88 or 0.880 ± 0.02 1100 or 1140 ± (30–60)\nValues of T and f correct as shown above.\nAbsolute uncertainties in T and f correct as shown above. 1\n\n2(c)(i) Six points from (b) plotted correctly. 1\nMust be within half a small square. Diameter of points must be less than half a small square.\nError bars in f plotted correctly. 1\nAll error bars to be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.\n© UCLES 2022 Page 7 of 9\n\n2(c)(ii) Straight line of best fit drawn. 1\nPoints must be balanced. Do not accept line from top point to bottom point.\nLine must pass between (2.20, 400) and (2.40, 400) and (5.20, 1000) and (5.60, 1000).\nWorst acceptable line drawn (steepest or shallowest possible line that passes through all the error bars). 1\nAll error bars must be plotted.\n\n2(c)(iii) Gradient determined with clear substitution of data points into y / x. 1\nDistance between data points must be greater than half the length of the drawn line.\nGradient of worst acceptable line determined with clear substitution of data points into y / x. 1\nuncertainty = (gradient of line of best fit – gradient of worst acceptable line)\nor\nuncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)\n\n2(d) 83.2 ± 0.3 (cm) 1\n\n2(e)(i) c determined using gradient and c given to two or three significant figures. 1\nc = 2  h  gradient = 2  (d)  (c)(iii)\nc determined using gradient and given with correct SI unit and correct power of ten: m s–1 or cm s–1. 1\n\n2(e)(ii) Percentage uncertainty in c from (c)(iii) and (d) with method shown. 1\n h gradient\npercentage uncertainty= + 100\n h gradient \nor\ncorrect substitution for max/min methods:\nmax c = 2  max h  max gradient\nmin c = 2  min h  min gradient\n© UCLES 2022 Page 8 of 9\n\n2(f) h determined to at least two significant figures from (e)(i) with correct substitution. 1\n3(e)(i)\nh=\n4130\nAbsolute uncertainty in h determined. Correct substitution must be seen. 1\n f c  5 c\nh= + h= + h\n   \n f c  130 c \nor\ncorrect substitution for max/min methods:\n3 max c 3 max(e)(i)\nmaxh= =\n4min f 4125\n3min c 3min(e)(i)\nminh= =\n4max f 4135\n© UCLES 2022 Page 9 of 9", "source_pages": [ 7, 8, 9 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9702_w22_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Oct-Nov/9702_w22_ms_52.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_w22_ms_52-p07.png", "../answer-assets/9702_w22_ms_52-p08.png", "../answer-assets/9702_w22_ms_52-p09.png" ] }, { "id": "9702-2022-on-53-q01", "question_id": "9702-2022-on-53-q01", "subject": "9702", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "53", "question_number": 1, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "1 Defining the problem\nA is the independent variable and s is the dependent variable or vary A and measure s 1\nkeep B and t constant 1\nMethods of data collection\nlabelled diagram of workable experiment including: 1\n• pin / rod through hole\n• supported by a stand\n• sheet able to oscillate freely\n• at least one label from copper/sheet, hole, clamp, stand, rod, pin.\ndrawn clamped rule(r) parallel to the direction of the oscillations (by eye) (to measure s) 1\nuse rule(r) to measure lengths to determine A 1\nand\nA = length  breadth\nuse of micrometer to measure t 1\nMethod of Analysis\nplot a graph of ln s against A or equivalent 1\nrelationship valid if a straight line (with y-intercept = ln s ) is produced 1\n0\ngradient 1\nK =−\nBt\n\n1\n(K =− for A against ln s)\nBtgradient\n© UCLES 2022 Page 5 of 9\n\n1 Additional detail including safety considerations 6\nD1 use of cushion/sand box in case sheet falls\nor\nuse gloves to protect hands from cuts / sharp edges\nD2 keep (initial) distance between (copper) sheet and (poles of) magnet constant\nor\nkeep (initial) distance between (copper) sheet and coil(s) constant\nD3 keep s constant\n0\nD4 method to ensure s is constant, e.g. initially line up (corner of) plate with fiducial marker / vertical pin to keep s\n0 0\nconstant\nD5 method to determine s using video camera:\n• rule(r) in a position to measure s in the diagram\n• video camera shown in diagram or description of use of video camera\n• playback video recording by frame by frame / slow motion (to measure s)\nD6 repeat measurements of t in different positions and average t\nD7 measure B/magnetic flux density using a (calibrated) Hall probe\nD8 additional detail on use of Hall probe, e.g. adjust probe until maximum value\nor\nmeasure B using Hall probe first in one direction and then in the opposite direction and average\nD9 drawn method to create a magnetic field perpendicular to the area of the sheet, e.g. pair of magnets/horseshoe\nmagnet/pair of coils connected to a (d.c.) supply\nD10 repeat experiment for each A and average s\n© UCLES 2022 Page 6 of 9", "source_pages": [ 5, 6 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9702_w22_ms_53.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Oct-Nov/9702_w22_ms_53.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_w22_ms_53-p05.png", "../answer-assets/9702_w22_ms_53-p06.png" ] }, { "id": "9702-2022-on-53-q02", "question_id": "9702-2022-on-53-q02", "subject": "9702", "year": 2022, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "53", "question_number": 2, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "2(a) YZd2 1\ngradient =\n4\n\n2(b) 1\n1\n/ 10–3 –1\nR\n45 or 45.5\n37 or 37.0\n30 or 30.3\n26 or 25.6\n21 or 21.3\n19 or 18.5\n1 1\nAbsolute uncertainties in from ± 2 to ± 0.9 or ± 1.\nR\n\n2(c)(i) Six points from (b) plotted correctly. 1\nMust be within half a small square. Diameter of points must be less than half a small square.\n1 1\nError bars in plotted correctly.\nR\nAll error bars to be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.\n\n2(c)(ii) Straight line of best fit drawn. 1\nPoints must be balanced. Do not accept line from top point to bottom point.\nLine must pass between (22.0, 30.0) and (23.0, 30.0) and (40.5, 65.0) and (42.0, 65.0).\nWorst acceptable line drawn (steepest or shallowest possible line that passes through all the error bars). 1\nAll error bars must be plotted.\n© UCLES 2022 Page 7 of 9\n\n2(c)(iii) Gradient determined with clear substitution of data points into y / x. 1\nDistance between data points must be greater than half the length of the drawn line.\nGradient of worst acceptable line determined with clear substitution of data points into y / x. 1\nuncertainty = (gradient of line of best fit – gradient of worst acceptable line)\nor\nuncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)\n\n2(d) 0.261 ± 0.003 (mm) 1\n\n2(e)(i)  determined using gradient and  given to two or three significant figures. 1\nYZd2 2222(d)2\n= =\n4gradient 4(c)(iii)\n determined using gradient and given with correct SI unit ( m) and correct power of ten 1\n\n2(e)(ii) percentage uncertainty in : 1\n 2d gradient \npercentage uncertainty= + +0.05+0.05100\n d gradient \nor\ncorrect substitution for max/min methods\n(1.0522)(1.0522)(d +d)2\nmax=\n4mingradient\n(0.9522)(0.9522)(d −d)2\nmin=\n4maxgradient\n© UCLES 2022 Page 8 of 9\n\n2(f) R determined to at least two significant figures from (c)(iii) or (d) and (e)(i) with correct substitution seen. 1\ngradient\nR =\n0.950\nor\nYZd2 2222(d)2\nR = =\n4L 4(e)(i)0.950\nAbsolute uncertainty in R determined. 1\nMethod must be consistent with determination of R and correct substitution must be seen.\nfor R determined using the gradient:\ngradient\nR = R\ngradient\nor\nfor R determined using (d) and (e)(i):\n 2d  \nR= + +0.05+0.05 R\n d  \nor\ncorrect substitution for max/min methods:\n(1.0522)(1.0522)(d +d)2\nmaxR =\n4min0.950\n(0.9522)(0.9522)(d −d)2\nmin R =\n4max0.950\n© UCLES 2022 Page 9 of 9", "source_pages": [ 7, 8, 9 ], "source_pdf": "_source-pdfs/2022-Oct-Nov/ms/9702_w22_ms_53.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2022-Oct-Nov/9702_w22_ms_53.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_w22_ms_53-p07.png", "../answer-assets/9702_w22_ms_53-p08.png", "../answer-assets/9702_w22_ms_53-p09.png" ] }, { "id": "9702-2023-m-42-q01", "question_id": "9702-2023-m-42-q01", "subject": "9702", "year": 2023, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 1, "topic": "Electric fields", "topic_slug": "9702-topic-18-electric-fields", "marks": 12, "status": "available", "reason": null, "text": "1(a) work done per unit mass B1\nwork (done on mass) moving mass from infinity (to the point) B1\n\n1(b)(i) –3.55  107 J kg–1 B1\n\n1(b)(ii) GM B1\n =−\nr\n−3.55 1 07  4 800 000\nM = –\n6.6710−11\n= 2.55  1024 kg\n\n1(b)(iii) GM  C1\ng = or g =−\nr2 r\n6.67 1 0−11  2.55 1 024 3.55 1 07 A1\n= or =\n48000002 4800000\n= 7.4 N kg–1\n\n1(b)(iv) r in range 2.60  107 to 2.65  107m C1\nmv2 GMm 2r GMm 2 C1\n= and v = or mr 2 = and  =\nr r2 T r2 T\n42r 3 42  ( 2.65 1 07)3 C1\nT2 = = = 4.20  109\nGM 6.67 1 0−11 2.55 1 024\nT = 64 800s A1\n= 18 hours\n© UCLES 2023 Page 6 of 18\n\n1(c) similarity – any one point from B1\n• inversely proportional to distance (from point)\n• points of equal potential lie on concentric spheres\n• zero at infinite distance\ndifference – any one point from B1\n• gravitational potential is (always) negative\n• electric potential can be positive or negative\nQuestion Answer Marks", "source_pages": [ 6, 7 ], "source_pdf": "_source-pdfs/2023-March/ms/9702_m23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-March/9702_m23_ms_42.pdf?download=true", "html": "9702-topic-18-electric-fields/answers.html", "image_paths": [ "../answer-assets/9702_m23_ms_42-p06.png", "../answer-assets/9702_m23_ms_42-p07.png" ] }, { "id": "9702-2023-m-42-q02", "question_id": "9702-2023-m-42-q02", "subject": "9702", "year": 2023, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 2, "topic": "Thermodynamics", "topic_slug": "9702-topic-16-thermodynamics", "marks": 12, "status": "available", "reason": null, "text": "2(a) gas for which pV  T M1\nwhere T is thermodynamic temperature A1\n\n2(b)(i) evidence of two temperature conversions between C and K B1\ntwo calculations shown, one for each state e.g. A1\n1.1010554010−6 6.701063010−6\n= 0.198 and = 0.198\n(273+27) (273+742)\n\n2(b)(ii) work is done on the gas M1\ninternal energy increases (so temperature increases) A1\n© UCLES 2023 Page 7 of 18\n\n2(b)(iii) pV = NkT e.g. C1\n1.1010554010−6\nN =\n1.3810−23300\n= 1.435  1022\nE = (3 / 2) kTN\nk\n1.1010554010−6 C1\n= (3 / 2)  1.38  1023  (742 – 27) \n1.3810−23300\n= 212 J A1\n\n2(c) E = mc and E = mL C1\n = (27 + 196) or 223 C1\nE = 0.0240  1.04  (27 + 196) + 0.0240  199 A1\n= 10.3 kJ\n© UCLES 2023 Page 8 of 18", "source_pages": [ 7, 8 ], "source_pdf": "_source-pdfs/2023-March/ms/9702_m23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-March/9702_m23_ms_42.pdf?download=true", "html": "9702-topic-16-thermodynamics/answers.html", "image_paths": [ "../answer-assets/9702_m23_ms_42-p07.png", "../answer-assets/9702_m23_ms_42-p08.png" ] }, { "id": "9702-2023-m-42-q03", "question_id": "9702-2023-m-42-q03", "subject": "9702", "year": 2023, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 3, "topic": "Oscillations", "topic_slug": "9702-topic-17-oscillations", "marks": 10, "status": "available", "reason": null, "text": "3(a) P: total energy B2\nQ: potential energy\nR: kinetic energy\n\n3(b) E = ½m2x 2 or E = ½mv 2 and v = x C1\n0 0 0 0\n6.410−3 = 1 0.13020.0152 C1\n2\n(2 = 438)\n( = 20.9)\nT = 2 /  C1\n= 2 / 20.9 A1\n= 0.30 s\n\n3(c)(i) resistive forces B1\n\n3(c)(ii) 0.926 C1\ndecrease in energy = 6.4 – (6.4  0.926) A1\n= 2.5 mJ\n\n3(c)(iii) light damping because the amplitude of oscillations gradually reduces B1\nor\nlight damping because the system still oscillates\n© UCLES 2023 Page 9 of 18", "source_pages": [ 9 ], "source_pdf": "_source-pdfs/2023-March/ms/9702_m23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-March/9702_m23_ms_42.pdf?download=true", "html": "9702-topic-17-oscillations/answers.html", "image_paths": [ "../answer-assets/9702_m23_ms_42-p09.png" ] }, { "id": "9702-2023-m-42-q04", "question_id": "9702-2023-m-42-q04", "subject": "9702", "year": 2023, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 4, "topic": "Electric fields", "topic_slug": "9702-topic-18-electric-fields", "marks": 12, "status": "available", "reason": null, "text": "4(a) (electric) force is (directly) proportional to product of charges B1\nforce (between point charges) is inversely proportional to the square of their separation B1\n\n4(b)(i) arrows showing tension upwards in direction of string, electric force horizontally to the right and weight vertically B1\ndownwards and all three labelled\n\n4(b)(ii) 9610−96410−9 C1\nF =\nE\n48.8510−120.0802\n( = 8.63  10–3 N)\neither angle to vertical = sin–1 0.080 / 1.2 C1\n( = 3.82°)\nweight = F / tan 3.82 = 8.63  10–3 / tan 3.82 C1\nE\n( = 0.129 N)\nmass = 0.129 / 9.81 A1\n= 0.013 kg\nor T sin  = mg and T cos  = F or tan  = mg / F (C1)\nE E\ntan  = 1.2 / 0.080 (C1)\nm = (1.2  8.63  10–3) / (0.080  9.81) (A1)\n= 0.013 kg\n© UCLES 2023 Page 10 of 18\n\n4(b)(iii) QQ 9610−96410−9 A1\nE = 1 2 =\np 4r 48.8510−120.080\no\n= 6.9  10–4 J\n\n4(c)(i) towards the top of the page / towards plate P B1\n\n4(c)(ii) F = QE and E = V / d C1\nF = 1.6  10–19  250 / 0.018 A1\n= 2.2  10–15 N\n\n4(c)(iii) either the force is not (always) perpendicular to the velocity B1\nor the force is always in the same direction\n© UCLES 2023 Page 11 of 18", "source_pages": [ 10, 11 ], "source_pdf": "_source-pdfs/2023-March/ms/9702_m23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-March/9702_m23_ms_42.pdf?download=true", "html": "9702-topic-18-electric-fields/answers.html", "image_paths": [ "../answer-assets/9702_m23_ms_42-p10.png", "../answer-assets/9702_m23_ms_42-p11.png" ] }, { "id": "9702-2023-m-42-q05", "question_id": "9702-2023-m-42-q05", "subject": "9702", "year": 2023, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 5, "topic": "Capacitance", "topic_slug": "9702-topic-19-capacitance", "marks": 10, "status": "available", "reason": null, "text": "5(a) from graph ln Q = 2.9 B1\n(so Q = 18.2 C)\nC = Q / V C1\n= 18.2 / 12 = 1.5 F A1\n\n5(b) gradient = –0.25 C1\ngradient = –1 / RC C1\nR = 1 / (0.25  1.5  10–6) A1\n= 2.7  106 \nQ −t −t (C1)\nor = e CR or ln Q – ln Q =\n0\nQ CR\n0\n−5.2 (C1)\n4.95 ( 1.5 10−6R )\ne.g. = e or 1.6 – 2.9 = 5.2 / (1.5 ×10–6R)\n18.2\nR = 2.7  106 (A1)\n© UCLES 2023 Page 12 of 18\n\n5(c) W = ½ QV C1\n= ½  18.2  10–6  12 A1\n= 1.1  10–4 J\nor W = ½ CV2 (C1)\n= ½  1.5  10–6  122 (A1)\n= 1.1  10–4 J\nor W = ½ Q2 / C (C1)\n= ½  (18.2  10–6)2 / 1.5  10–6 (A1)\n= 1.1  10–4 J\n\n5(d) straight line with different negative gradient starting from (0, 2.9) M1\nstraight line between t = 0 and at least t = 5.0s with twice the gradient of the original line A1\n© UCLES 2023 Page 13 of 18", "source_pages": [ 12, 13 ], "source_pdf": "_source-pdfs/2023-March/ms/9702_m23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-March/9702_m23_ms_42.pdf?download=true", "html": "9702-topic-19-capacitance/answers.html", "image_paths": [ "../answer-assets/9702_m23_ms_42-p12.png", "../answer-assets/9702_m23_ms_42-p13.png" ] }, { "id": "9702-2023-m-42-q06", "question_id": "9702-2023-m-42-q06", "subject": "9702", "year": 2023, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 6, "topic": "Magnetic fields", "topic_slug": "9702-topic-20-magnetic-fields", "marks": 8, "status": "available", "reason": null, "text": "6(a) it is zero when (plane of) probe is parallel to the (magnetic) field (lines) B1\nit is maximum when (plane of) probe is perpendicular to (magnetic) field (lines) B1\n\n6(b)(i) number density of charge carriers B1\n\n6(b)(ii) smaller value of n so greater Hall voltage / V B1\nH\n\n6(c) (36mV corresponds to) 48 mT C1\nuse of 1.4 s or (8.6 – 7.2) s C1\nE = BAN / t C1\n4810−30.0182780 A1\n=\n1.4\n= 0.027V\n© UCLES 2023 Page 14 of 18", "source_pages": [ 14 ], "source_pdf": "_source-pdfs/2023-March/ms/9702_m23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-March/9702_m23_ms_42.pdf?download=true", "html": "9702-topic-20-magnetic-fields/answers.html", "image_paths": [ "../answer-assets/9702_m23_ms_42-p14.png" ] }, { "id": "9702-2023-m-42-q07", "question_id": "9702-2023-m-42-q07", "subject": "9702", "year": 2023, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 7, "topic": "Quantum physics", "topic_slug": "9702-topic-22-quantum-physics", "marks": 7, "status": "available", "reason": null, "text": "7(a) photon absorbed (by electron) and electron excited B1\nphoton energy equal to difference in (energy of two) energy levels B1\nphoton energy relates to a single wavelength / single frequency B1\nelectron de-excites and emits photon in any direction B1\n\n7(b) hc C1\n=E\n\nuses 658nm C1\n6.63 1 0–34  3.00 1 08 A1\n= – E – (–3.40 × 1.60 × 10–19)\n1\n658 1 0–9\nE = –2.42  10–19J\n1\n© UCLES 2023 Page 15 of 18", "source_pages": [ 15 ], "source_pdf": "_source-pdfs/2023-March/ms/9702_m23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-March/9702_m23_ms_42.pdf?download=true", "html": "9702-topic-22-quantum-physics/answers.html", "image_paths": [ "../answer-assets/9702_m23_ms_42-p15.png" ] }, { "id": "9702-2023-m-42-q08", "question_id": "9702-2023-m-42-q08", "subject": "9702", "year": 2023, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 8, "topic": "Nuclear physics", "topic_slug": "9702-topic-23-nuclear-physics", "marks": 11, "status": "available", "reason": null, "text": "8(a) 234, 92 for the uranium nucleus B1\n4, 2 for the alpha particle B1\n\n8(b)(i) N = 0.874 / (238  1.66  10–27) A1\n0\n= 2.21  1024\n\n8(b)(ii) A = N C1\nln2 A1\n= 2.211024\n87.7365243600\n= 5.54  1014Bq\n\n8(b)(iii) power = 5.54  1014  5.59  106  1.60  10–19 C1\n= 496 W A1\n\n8(b)(iv) ln2 C1\n− t\n65.3=100e 87.7\nln 0.653 = – (ln 2 / 87.7) t A1\nt = 53.9 years\n\n8(c) advantage: less mass so less energy needed to launch probe B1\ndisadvantage: half-life shorter so will not provide power for as long B1\n© UCLES 2023 Page 16 of 18", "source_pages": [ 16 ], "source_pdf": "_source-pdfs/2023-March/ms/9702_m23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-March/9702_m23_ms_42.pdf?download=true", "html": "9702-topic-23-nuclear-physics/answers.html", "image_paths": [ "../answer-assets/9702_m23_ms_42-p16.png" ] }, { "id": "9702-2023-m-42-q09", "question_id": "9702-2023-m-42-q09", "subject": "9702", "year": 2023, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 9, "topic": "Ideal gases", "topic_slug": "9702-topic-15-ideal-gases", "marks": 9, "status": "available", "reason": null, "text": "9(a) piezo-electric crystal B1\n(ultrasound) wave causes shape change / vibrations (of crystal) B1\nshape change / vibrations causes e.m.f. (which is detected) B1\n\n9(b)(i) 93V A1\n\n9(b)(ii) 2.7  107rads–1 A1\n\n9(c)(i) kgm–2s–1 B1\n\n9(c)(ii)  = Z / c = 1.7  106 / 1600 A1\n= 1100 kg m–3\n\n9(c)(iii) intensity reflection coefficient ≈ 1 or Z and Z are very different B1\n1 2\nalmost no / no ultrasound transmitted (into air filled cavity) B1\n© UCLES 2023 Page 17 of 18", "source_pages": [ 17 ], "source_pdf": "_source-pdfs/2023-March/ms/9702_m23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-March/9702_m23_ms_42.pdf?download=true", "html": "9702-topic-15-ideal-gases/answers.html", "image_paths": [ "../answer-assets/9702_m23_ms_42-p17.png" ] }, { "id": "9702-2023-m-42-q10", "question_id": "9702-2023-m-42-q10", "subject": "9702", "year": 2023, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 10, "topic": "Astronomy and cosmology", "topic_slug": "9702-topic-25-astronomy-and-cosmology", "marks": 9, "status": "available", "reason": null, "text": "10(a) brighter star could be closer (to Earth) B1\nbrighter star could have a greater luminosity (in the visible wavelengths) B1\n\n10(b) object with known luminosity B1\n\n10(c)(i) 660.9−656.3 v B1\n leading to 2.1 106 m s–1\n656.3 3.0108\n\n10(c)(ii) v = H d C1\no\nd = 2.1  106 / 2.3  10–18 A1\n= 9.1  1023 m\n\n10(c)(iii) wavelength has increased / light is redshifted B1\nstar within galaxy is moving away / receding (from Earth) B1\nUniverse is expanding B1\n© UCLES 2023 Page 18 of 18", "source_pages": [ 18 ], "source_pdf": "_source-pdfs/2023-March/ms/9702_m23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-March/9702_m23_ms_42.pdf?download=true", "html": "9702-topic-25-astronomy-and-cosmology/answers.html", "image_paths": [ "../answer-assets/9702_m23_ms_42-p18.png" ] }, { "id": "9702-2023-m-52-q01", "question_id": "9702-2023-m-52-q01", "subject": "9702", "year": 2023, "session": "March", "session_code": "m", "paper": 5, "variant": "52", "question_number": 1, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "1 Defining the problem\nf is the independent variable and Q is the dependent variable, or vary f and measure Q. 1\nKeep h constant 1\nMethods of data collection\nLabelled diagram of workable experiment including: 1\n• fan is positioned in line with the turbine so that blades of both fan and turbine overlap\n• base of fan on same bench as turbine\n• fan labelled and one other label from bench, (wind) turbine, cable, pump, pipe, liquid\nLabelled apparatus showing workable method to collect the liquid from the top of the pipe, e.g. hose / pipe / tube connected 1\nto top of pipe with the other end over a beaker / measuring cylinder below top of pipe.\nAt least one label related to collection of liquid.\nUse of stop-watch / timer to measure time to collect liquid or to measure time for blades to rotate. 1\nUse of (top pan) balance to measure mass of liquid leaving the pipe. 1\nMethod of Analysis\nPlots a graph of Q against f3or equivalent (e.g. f3 against Q). 1\nDo not accept logarithmic graphs.\nC =ghy-intercept(for f3 against Q: C =−Dy-intercept) 1\nD=ghgradient 1\ngh\n(for f3 against Q: D = )\ngradient\n© UCLES 2023 Page 5 of 9\n\n1 Additional detail including safety considerations 6\nAny six from:\nD1 Precaution with reason linked to prevent liquid spilling (on bench / floor) e.g. use of large bucket / bowl / tray to\ncontain any spilled liquid\nor\nPrecaution with reason linked to prevent air / dust particles in eye, e.g. use of goggles\nor\nPrecaution with reason linked to turbine falling, e.g. clamp turbine to bench.\nD2 Use rule to measure h.\nD3 Method to determine mass of liquid, e.g.\nmass of beaker + liquid – mass of empty beaker\nor\nmass of container / pipe before – mass of container / pipe after.\nD4 Method to determine f, e.g. measure time t for many rotations / revolutions N and period T = t /N and f = 1/T\nor\nmeasure time t for many rotations / revolutions N and f = N/t\nor\nvideo rotating blades, playback frame by frame and use a time stamp to determine period T and f = 1/T.\nD5 Mark one of the blades to assist in counting number of rotations.\nD6 Method to vary f, e.g. change speed of fan / change distance between fan and blades / vary current in fan.\nD7 Wait for steady air flow before starting timing and / or collecting liquid.\nmass (of liquid)\nD8 Q =\ntime (to collect liquid)\nor\nmethod and explanation to reduce uncertainty in Q, e.g. use large value of time or mass of liquid collected.\nD9 Repeat measurements of Q for the same value of f and average Q.\n© UCLES 2023 Page 6 of 9\n\n1 D10 Relationship valid if a straight line is produced (not passing through the origin).\nDo not accept passing through the origin.\nQuestion Answer Marks", "source_pages": [ 5, 6, 7 ], "source_pdf": "_source-pdfs/2023-March/ms/9702_m23_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-March/9702_m23_ms_52.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_m23_ms_52-p05.png", "../answer-assets/9702_m23_ms_52-p06.png", "../answer-assets/9702_m23_ms_52-p07.png" ] }, { "id": "9702-2023-m-52-q02", "question_id": "9702-2023-m-52-q02", "subject": "9702", "year": 2023, "session": "March", "session_code": "m", "paper": 5, "variant": "52", "question_number": 2, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "2(a) v 1\ngradient =\n4\ny-intercept = −k\n\n2(b) 2\n1/f / 10–3 Hz–1 d / cm\n0.67 or 0.667 24.7  0.2\n0.48 or 0.476 17.4  0.2\n0.36 or 0.357 12.7  0.3\n0.24 or 0.244 8.4  0.3\n0.19 or 0.192 6.6  0.4\n0.13 or 0.132 4.6  0.4\nFirst mark: values of 1 / f and d correct as shown.\nSecond mark: uncertainties in d correct as shown.\n\n2(c)(i) Six points from (b) plotted correctly. 1\nMust be within half a small square. Diameter of points must be less than half a small square.\nError bars in d / cm plotted correctly. 1\nAll error bars to be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.\n© UCLES 2023 Page 7 of 9\n\n2(c)(ii) Straight line of best fit drawn. 1\nDo not accept line from top plot to bottom plot.\nPoints must be balanced.\nLine must pass between (0.170, 6.0) and (0.185, 6.0) and between (0.590, 22.0) and (0.610, 22.0)\nWorst acceptable line drawn. 1\nSteepest or shallowest possible line that passes through all the error bars.\nAll error bars must be plotted.\n\n2(c)(iii) Gradient determined with clear substitution of data points into y / x; distance between data points must be greater than 1\nhalf the length of the drawn line.\nGradient determined of worst acceptable line uncertainty = (gradient of line of best fit – gradient of worst acceptable line) 1\nor\nuncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)\n\n2(c)(iv) y-intercept determined by substitution of correct point with consistent power of ten in m and x into y = mx + c. 1\nExpect y-intercept to be negative.\ny-intercept of worst acceptable line determined by substitution into y = mx + c. 1\nuncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line, or\nuncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)\n© UCLES 2023 Page 8 of 9\n\n2(d) v determined using gradient and v and k given to 2 or 3 sf. 1\nv =4gradient=4(c)(iii)\nk determined using y-intercept and units for v and k 1\nk =−y-intercept=−(c)(iv)\nUnits:\nv: m s–1, cm s–1\nk: m, cm\nAbsolute uncertainties in v and k. 1\n gradient\nv: v with correct substitution or v: 4  uncertainty in gradient\ngradient\nand\nk: uncertainty in y-intercept\n\n2(e) f determined to a minimum of 2 significant figures from (c)(iii) and (c)(iv) OR (d) with correct substitution and correct 1\npowers of ten used for all quantities.\nv\nf =\n4(d +k)\nor\ngradient\nf =\nd −(y-intercept)\n© UCLES 2023 Page 9 of 9", "source_pages": [ 7, 8, 9 ], "source_pdf": "_source-pdfs/2023-March/ms/9702_m23_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-March/9702_m23_ms_52.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_m23_ms_52-p07.png", "../answer-assets/9702_m23_ms_52-p08.png", "../answer-assets/9702_m23_ms_52-p09.png" ] }, { "id": "9702-2023-mj-41-q01", "question_id": "9702-2023-mj-41-q01", "subject": "9702", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 1, "topic": "Electric fields", "topic_slug": "9702-topic-18-electric-fields", "marks": 11, "status": "available", "reason": null, "text": "1(a)(i) force per unit mass B1\n\n1(a)(ii) force per unit positive charge B1\n\n1(a)(iii) similarity: B1\n inversely proportional to distance (from point)\n points of equal potential lie on concentric spheres\n zero at infinite distance\nAny point, 1 mark\ndifference: B1\n gravitational potential is (always) negative\n electric potential can be positive or negative\nAny point, 1 mark\n\n1(b)(i) g = GM / r2 M1\nE = Q / 4r2 M1\n0\nalgebra showing the elimination of r leading to M / Q = (1 / 4G) (g / E) A1\n0\n\n1(b)(ii)  = 1 / (4  6.67  10–11  8.85  10–12) = 1.35  1020 (kg2 C–2) A1\nor\n = (8.99  109) / (6.67  10–11) = 1.35  1020 (kg2 C–2)\n\n1(c)(i) E = gQ / M C1\n= (1.35  1020  9.81  4.80  105) / (5.98  1024)\n= 106 N C–1 or 106 V m–1 A1\n\n1(c)(ii) same (direction) B1\n© UCLES 2023 Page 6 of 16", "source_pages": [ 6 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9702_s23_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-May-June/9702_s23_ms_41.pdf?download=true", "html": "9702-topic-18-electric-fields/answers.html", "image_paths": [ "../answer-assets/9702_s23_ms_41-p06.png" ] }, { "id": "9702-2023-mj-41-q02", "question_id": "9702-2023-mj-41-q02", "subject": "9702", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 2, "topic": "Oscillations", "topic_slug": "9702-topic-17-oscillations", "marks": 12, "status": "available", "reason": null, "text": "2(a) horizontal force on sphere causes centripetal acceleration B1\nweight of sphere is (now) equal to vertical component of tension B1\nor\nhorizontal and vertical components (of force) (now) combine to give greater tension (in spring)\ngreater tension in spring so greater extension of spring B1\n\n2(b)(i) r = 10.8  sin 27° = 4.9 cm A1\n\n2(b)(ii) T cos  = mg C1\nor\nT cos  = W and W = mg\nT cos 27° = 0.29  9.81 leading to T = 3.2 N A1\n\n2(b)(iii) T = 3.2 – (0.29  9.81) C1\nk = T / x A1\n= [3.2 – (0.29  9.81)] / [10.8 – 8.5]\n= 0.15 N cm–1\n\n2(c)(i) centripetal acceleration = (T sin ) / m C1\n= (3.2  sin 27°) / 0.29\n= 5.0 m s–2 A1\n© UCLES 2023 Page 7 of 16\n\n2(c)(ii) a = r2 and  = 2 / T C1\nor\na = v2 / r and v = 2r / T\nT = 2  √(0.049 / 5.0) A1\n= 0.62 s\n© UCLES 2023 Page 8 of 16", "source_pages": [ 7, 8 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9702_s23_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-May-June/9702_s23_ms_41.pdf?download=true", "html": "9702-topic-17-oscillations/answers.html", "image_paths": [ "../answer-assets/9702_s23_ms_41-p07.png", "../answer-assets/9702_s23_ms_41-p08.png" ] }, { "id": "9702-2023-mj-41-q03", "question_id": "9702-2023-mj-41-q03", "subject": "9702", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 3, "topic": "Temperature", "topic_slug": "9702-topic-14-temperature", "marks": 11, "status": "available", "reason": null, "text": "3(a) no net thermal energy is transferred (between them) B1\n\n3(b)(i) variation (of density with temperature) is linear B1\nor\neach temperature has a unique value of density\n\n3(b)(ii)  variation (of density with temperature) is not linear B2\n region where the density does not vary with temperature\n different temperatures have the same density\nAny two points, 1 mark each\n\n3(c)(i) boiling point = 80 °C A1\n\n3(c)(ii) Q = Pt and t = 21 s C1\n(thermal energy supplied = 810  21 = 17000 J)\nc = Q / m C1\nthermal energy absorbed by beaker = 42  0.84  (80 – 25) C1\n( = 1940 J)\ns.h.c. of liquid = [(810  21) – (42  0.84  (80 – 25))] / [120  (80 – 25)] A1\n= 2.3 J g–1 K–1\n\n3(d) sketch: straight diagonal line from 25 °C to 100 °C and then horizontal at 100 °C B1\nstraight diagonal line starting at 25 °C with gradient approximately half that of the original line B1\n© UCLES 2023 Page 9 of 16", "source_pages": [ 9 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9702_s23_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-May-June/9702_s23_ms_41.pdf?download=true", "html": "9702-topic-14-temperature/answers.html", "image_paths": [ "../answer-assets/9702_s23_ms_41-p09.png" ] }, { "id": "9702-2023-mj-41-q04", "question_id": "9702-2023-mj-41-q04", "subject": "9702", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 4, "topic": "Thermodynamics", "topic_slug": "9702-topic-16-thermodynamics", "marks": 12, "status": "available", "reason": null, "text": "4(a)  particles are in (continuous) random motion B2\n particles have negligible volume (compared with the gas)\n negligible forces between particles (except during collisions)\n (all) collisions (perfectly) elastic\n time of collision negligible (in comparison with time between collisions)\nAny two points, 1 mark each\n\n4(b)(i) (general starting equation) pV = nRT C1\nT = (2pV / nR) where R is the (molar) gas constant A1\n\n4(b)(ii) sketch: straight vertical line XY from (V, 2p) to (V, p) B1\nstraight horizontal line YZ from (V, p) to (2V, p) B1\ncurve with gradient increasing from Z to X from (2V, p) to (V, 2p) B1\n\n4(b)(iii) XY work done on gas correct (= 0) B1\nZX increase in internal energy correct (= 0) B1\nYZ work done on gas correct (= –pV) B1\nXY increase in internal energy such that the increase in internal energy column adds up to zero B1\nall three thermal energies transferred such that U = q + w in each row B1\n(completely correct answer:\nchange U Q w\nX to Y –U –U 0\nY to Z [ +U ] U + pV –pV\nZ to X 0 –W [ +W ]\n)\n© UCLES 2023 Page 10 of 16", "source_pages": [ 10 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9702_s23_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-May-June/9702_s23_ms_41.pdf?download=true", "html": "9702-topic-16-thermodynamics/answers.html", "image_paths": [ "../answer-assets/9702_s23_ms_41-p10.png" ] }, { "id": "9702-2023-mj-41-q05", "question_id": "9702-2023-mj-41-q05", "subject": "9702", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 5, "topic": "Alternating currents", "topic_slug": "9702-topic-21-alternating-currents", "marks": 9, "status": "available", "reason": null, "text": "5(a)(i) correct circuit symbol for a diode shown correctly connected in series with the wires leading into and out of the dotted box B1\n\n5(a)(ii) smoothing / V is smoothed B1\nOUT\n\n5(b)(i) frequency = 1 / 0.04 A1\n= 25 Hz\n\n5(b)(ii) V = V exp (– t / RC) and  = RC C1\n0\nor\nV = V exp (– t / )\n0\n3.25 = 5.50 exp (– 0.020 / ) leading to  = 0.038 s A1\n\n5(b)(iii)  = RC C1\ncapacitance = 0.038 / 14000 A1\n= 2.7  10–6 F\n\n5(c) V has constant magnitude in both positive and negative directions B1\nIN\n(so) V is (now) constant / V does not vary with time B1\nOUT OUT\n© UCLES 2023 Page 11 of 16", "source_pages": [ 11 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9702_s23_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-May-June/9702_s23_ms_41.pdf?download=true", "html": "9702-topic-21-alternating-currents/answers.html", "image_paths": [ "../answer-assets/9702_s23_ms_41-p11.png" ] }, { "id": "9702-2023-mj-41-q06", "question_id": "9702-2023-mj-41-q06", "subject": "9702", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 6, "topic": "Magnetic fields", "topic_slug": "9702-topic-20-magnetic-fields", "marks": 10, "status": "available", "reason": null, "text": "6(a) a region where a force acts on M1\na current-carrying conductor A1\nor\na moving charge\nor\na magnetic material / magnetic pole\n\n6(b) concentric circles around the wire B1\nspacing between circles increases with distance from wire B1\narrows showing direction of field is clockwise B1\n\n6(c)(i) F = BIL C1\nforce per unit length = BI A1\n= 2.6  10–3  5.0\n= 0.013 N m–1\n\n6(c)(ii) to the right B1\n\n6(c)(iii) force (per unit length) has the same magnitude due to Newton’s 3rd law B1\n0.013 = 1.5  10–3  I A1\ncurrent = 8.7 A\n© UCLES 2023 Page 12 of 16", "source_pages": [ 12 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9702_s23_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-May-June/9702_s23_ms_41.pdf?download=true", "html": "9702-topic-20-magnetic-fields/answers.html", "image_paths": [ "../answer-assets/9702_s23_ms_41-p12.png" ] }, { "id": "9702-2023-mj-41-q07", "question_id": "9702-2023-mj-41-q07", "subject": "9702", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 7, "topic": "Quantum physics", "topic_slug": "9702-topic-22-quantum-physics", "marks": 9, "status": "available", "reason": null, "text": "7(a) wavelength associated with a moving particle B1\n\n7(b)(i) (electron) diffraction B1\n\n7(b)(ii) beam spreads out indicating diffraction B1\nor\nlight and dark regions indicate an interference pattern\nelectron beam is behaving as a wave B1\n\n7(c)(i) central blob and concentric rings B1\nrings closer together (than previously) B1\n\n7(c)(ii) (greater p.d. so) electrons to have greater momentum B1\ngreater momentum so decrease in (de Broglie) wavelength B1\nlower (de Broglie) wavelength (for same grating spacing in crystal) causes: B1\nsmaller diffraction angle\nor\nsmaller angle of intensity maxima (for each order)\nor\ndecrease in fringe spacing in diffraction pattern\n© UCLES 2023 Page 13 of 16", "source_pages": [ 13 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9702_s23_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-May-June/9702_s23_ms_41.pdf?download=true", "html": "9702-topic-22-quantum-physics/answers.html", "image_paths": [ "../answer-assets/9702_s23_ms_41-p13.png" ] }, { "id": "9702-2023-mj-41-q08", "question_id": "9702-2023-mj-41-q08", "subject": "9702", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 8, "topic": "Medical physics", "topic_slug": "9702-topic-24-medical-physics", "marks": 8, "status": "available", "reason": null, "text": "8(a)(i) specific acoustic impedance = 1200  1400 = 1.68  106 kg m–2 s–1 A1\n\n8(a)(ii) density of air shown in table as 1.29 A1\nspeed of sound in tissue shown in table as 1540 A1\n\n8(b)(i) intensity reflection coefficient = (Z – Z )2 / (Z + Z )2 C1\n1 2 1 2\n= (1680000 – 440)2 / (1680000 + 440)2\n= 0.999 A1\n\n8(b)(ii) intensity reflection coefficient = (Z – Z )2 / (Z + Z )2 A1\n1 2 1 2\n= (1680000 – 1680000)2 / (1680000 + 1680000)2\n= 0\n\n8(c) without gel, (almost) all of the (incident) ultrasound is reflected (from skin) B1\nwith gel, (almost) all of the (incident) ultrasound is transmitted (into the body) B1\n© UCLES 2023 Page 14 of 16", "source_pages": [ 14 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9702_s23_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-May-June/9702_s23_ms_41.pdf?download=true", "html": "9702-topic-24-medical-physics/answers.html", "image_paths": [ "../answer-assets/9702_s23_ms_41-p14.png" ] }, { "id": "9702-2023-mj-41-q09", "question_id": "9702-2023-mj-41-q09", "subject": "9702", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 9, "topic": "Nuclear physics", "topic_slug": "9702-topic-23-nuclear-physics", "marks": 9, "status": "available", "reason": null, "text": "9(a) time for activity (of sample) to halve B1\n\n9(b) sketch: line with positive gradient starting at (0,0) and extending to t = 80 min B1\nexponential curve, extending from t = 0 to t = 80 min, with gradient of steadily decreasing magnitude B1\nline passing through (0,0), (20, 0.5N ) and (40, 0.75 N ) B1\n0 0\n\n9(c)(i) every (undecayed) nucleus has the same probability of decay M1\nfewer (undecayed) nuclei remaining (with time), so fewer will decay (in a given time interval) A1\n\n9(c)(ii)  sample emits in all directions but detector only captures emissions in one direction B2\n some emissions are absorbed before reaching detector\n some emissions are scattered within the sample\n simultaneous arrival of multiple particles only registers once\n some particles may reach detector but not cause ionisation\nAny two points, 1 mark each\nmeasured count rate is less than the activity B1\n© UCLES 2023 Page 15 of 16", "source_pages": [ 15 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9702_s23_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-May-June/9702_s23_ms_41.pdf?download=true", "html": "9702-topic-23-nuclear-physics/answers.html", "image_paths": [ "../answer-assets/9702_s23_ms_41-p15.png" ] }, { "id": "9702-2023-mj-41-q10", "question_id": "9702-2023-mj-41-q10", "subject": "9702", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 10, "topic": "Astronomy and cosmology", "topic_slug": "9702-topic-25-astronomy-and-cosmology", "marks": 9, "status": "available", "reason": null, "text": "10(a) speed is (directly) proportional to distance M1\nspeed is speed of recession of galaxy from an observer, and distance is the distance of the galaxy from the observer A1\n\n10(b) F = L / (4d2) C1\n= (3.8  1031) / [4  (1.8  1024)2] A1\n= 9.3  10–19 W m–2\n\n10(c)(i) galaxy is moving away (from the Earth) B1\nwavelength (of light from the galaxy) increased by the Doppler effect / due to redshift B1\n\n10(c)(ii)  /  = v / c C1\nv = [(492 – 486)  3.00  108] / 486\n(v = 3.7  106 m s–1)\nH = v / d C1\n0\n= (3.7  106) / (1.8  1024) A1\n= 2.1  10–18 s–1\n© UCLES 2023 Page 16 of 16", "source_pages": [ 16 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9702_s23_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-May-June/9702_s23_ms_41.pdf?download=true", "html": "9702-topic-25-astronomy-and-cosmology/answers.html", "image_paths": [ "../answer-assets/9702_s23_ms_41-p16.png" ] }, { "id": "9702-2023-mj-42-q01", "question_id": "9702-2023-mj-42-q01", "subject": "9702", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 1, "topic": "Gravitational fields", "topic_slug": "9702-topic-13-gravitational-fields", "marks": 11, "status": "available", "reason": null, "text": "1(a) (gravitational) force is (directly) proportional to product of masses B1\nforce (between point masses) is inversely proportional to the square of their separation B1\n\n1(b) GMm / R2 = mR2 M1\n = 2 / T and algebra leading to 42R3 = GMT2 A1\nor\nGMm / R2 = mv2 / R (M1)\nv = 2R / T and algebra leading to 42R3 = GMT2 (A1)\n\n1(c) 42  R3 = 6.67  10–11  5.98  1024  (24  60  60)2 C1\n(R = 4.22  107 m)\nh = R – (6.37  106) C1\nh = (4.22  107) – (6.37  106) A1\n= 3.6  107 m\n\n1(d)(i)  = 2 / T C1\n= 2 / (24  60  60) A1\n= 7.3  10–5 rad s–1\n\n1(d)(ii) orbit is from east to west B1\norbit is not equatorial / orbit is polar B1\n© UCLES 2023 Page 6 of 15", "source_pages": [ 6 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9702_s23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-May-June/9702_s23_ms_42.pdf?download=true", "html": "9702-topic-13-gravitational-fields/answers.html", "image_paths": [ "../answer-assets/9702_s23_ms_42-p06.png" ] }, { "id": "9702-2023-mj-42-q02", "question_id": "9702-2023-mj-42-q02", "subject": "9702", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 2, "topic": "Temperature", "topic_slug": "9702-topic-14-temperature", "marks": 12, "status": "available", "reason": null, "text": "2(a)(i) (gas that obeys) pV  T (for all values of p,V and T) M1\nwhere T is thermodynamic temperature A1\n\n2(a)(ii) temperature = –273.15 °C A1\n\n2(b)(i) pV = NkT C1\nN = (1.37  105  0.640) / (1.38  10–23  (227 + 273)) C1\n= 1.27  1025 A1\n\n2(b)(ii) mass = 0.0424 / (1.27  1025) A1\n= 3.34  10–27 kg\n\n2(b)(iii) ½m = (3 / 2)kT C1\n3.34  10–27  v2 = 3  1.38  10–23  500 C1\nv = 2490 m s–1 A1\nor\npV = ⅓(Nm) and Nm = mass of gas (C1)\n0.0424  v2 = 3  1.37  105  0.640 (C1)\nv = 2490 m s–1 (A1)\n\n2(c) sketch: line from (0, 0) to (500, v) B1\nline with decreasing positive gradient throughout B1\n© UCLES 2023 Page 7 of 15", "source_pages": [ 7 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9702_s23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-May-June/9702_s23_ms_42.pdf?download=true", "html": "9702-topic-14-temperature/answers.html", "image_paths": [ "../answer-assets/9702_s23_ms_42-p07.png" ] }, { "id": "9702-2023-mj-42-q03", "question_id": "9702-2023-mj-42-q03", "subject": "9702", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 3, "topic": "Thermodynamics", "topic_slug": "9702-topic-16-thermodynamics", "marks": 9, "status": "available", "reason": null, "text": "3(a) change in internal energy = work done + energy transfer by heating C1\nincrease in internal energy = work done on system + energy transferred to the system by heating A1\n\n3(b)(i) AB change in internal energy: decrease B1\nAB work done on gas: positive B1\nBC change in internal energy: increase B1\nBC work done on gas: zero B1\n\n3(b)(ii) more work done by gas in CD than is done on gas in AB B1\nor\n(no work done on gas in BC and DA so) (overall) gas does work\n(overall) change in internal energy is zero B1\n(must be an overall) input of thermal energy B1\n© UCLES 2023 Page 8 of 15", "source_pages": [ 8 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9702_s23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-May-June/9702_s23_ms_42.pdf?download=true", "html": "9702-topic-16-thermodynamics/answers.html", "image_paths": [ "../answer-assets/9702_s23_ms_42-p08.png" ] }, { "id": "9702-2023-mj-42-q04", "question_id": "9702-2023-mj-42-q04", "subject": "9702", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 4, "topic": "Oscillations", "topic_slug": "9702-topic-17-oscillations", "marks": 11, "status": "available", "reason": null, "text": "4(a)(i)  = 2f C1\nf = 9.7 / 2 A1\n= 1.5 Hz\n\n4(a)(ii) amplitude = √(11.6) = 3.4 cm A1\n\n4(a)(iii) a = 2x C1\n0 0\n= 9.72  3.4  10–2 A1\n= 3.2 m s–2\n\n4(b) sketch: straight line through the origin with negative gradient B1\nline with negative gradient passing through (+3.4, –a ) and (–3.4, +a ) B1\n0 0\nline with ends at x =  3.4 cm and a =  a B1\n0\n\n4(c) sum of potential energy and kinetic energy is constant B1\nat maximum displacement, kinetic energy is zero B1\nor\nat maximum displacement, potential energy is maximum\nat zero displacement, kinetic energy is maximum B1\nor\nat zero displacement, potential energy is minimum\n© UCLES 2023 Page 9 of 15", "source_pages": [ 9 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9702_s23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-May-June/9702_s23_ms_42.pdf?download=true", "html": "9702-topic-17-oscillations/answers.html", "image_paths": [ "../answer-assets/9702_s23_ms_42-p09.png" ] }, { "id": "9702-2023-mj-42-q05", "question_id": "9702-2023-mj-42-q05", "subject": "9702", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 5, "topic": "Capacitance", "topic_slug": "9702-topic-19-capacitance", "marks": 9, "status": "available", "reason": null, "text": "5(a)(i) Q = CV A1\nA\n\n5(a)(ii) E = ½CV2 A1\nA\n\n5(b)(i) some of the charge transfers to (the plates of) capacitor B B1\ntransfer is because the p.d.s across the capacitors are not equal B1\nor\ntransfer stops when the p.d.s across the capacitors become equal\n\n5(b)(ii) V = V M1\nA B\ncharge on A + charge on B = CV M1\nCV + 3CV = CV leading to V = V / 4 A1\nB B B\nor\nC = 4C (M1)\nT\nQ = CV (M1)\nT\nV = CV / 4C = V / 4 (A1)\nB\n\n5(b)(iii) E = ½CV2 – nCV2, where n is a multiple that is less than ½ C1\nor\ntotal final energy = ½  4C  (V / 4)2\n= ⅛CV2\nE = ½CV2 – ⅛CV2 A1\n= ⅜CV2\n© UCLES 2023 Page 10 of 15", "source_pages": [ 10 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9702_s23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-May-June/9702_s23_ms_42.pdf?download=true", "html": "9702-topic-19-capacitance/answers.html", "image_paths": [ "../answer-assets/9702_s23_ms_42-p10.png" ] }, { "id": "9702-2023-mj-42-q06", "question_id": "9702-2023-mj-42-q06", "subject": "9702", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 6, "topic": "Magnetic fields", "topic_slug": "9702-topic-20-magnetic-fields", "marks": 10, "status": "available", "reason": null, "text": "6(a)(i) product of (magnetic) flux density and area M1\narea perpendicular to the (magnetic) field A1\n\n6(a)(ii) flux = B  r2 C1\n= 0.17    0.362\n= 6.9  10–2 Wb A1\n\n6(b) time for one revolution = 1 / 25 s C1\ne.m.f. = rate of cutting flux or  / t C1\n= 0.069  25 A1\n= 1.7 V\n\n6(c) current (in disc) is perpendicular to magnetic field B1\nor\ncurrent causes force to act on disc\nforce opposes rotation of disc B1\nleft-hand rule indicates current is from rim to axle B1\n© UCLES 2023 Page 11 of 15", "source_pages": [ 11 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9702_s23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-May-June/9702_s23_ms_42.pdf?download=true", "html": "9702-topic-20-magnetic-fields/answers.html", "image_paths": [ "../answer-assets/9702_s23_ms_42-p11.png" ] }, { "id": "9702-2023-mj-42-q07", "question_id": "9702-2023-mj-42-q07", "subject": "9702", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 7, "topic": "Alternating currents", "topic_slug": "9702-topic-21-alternating-currents", "marks": 10, "status": "available", "reason": null, "text": "7(a)(i) full-wave (rectification) B1\n\n7(a)(ii) lower left diode shown pointing left B1\nlower right and upper left diodes shown pointing left B1\n\n7(a)(iii) arrow indicating current direction in resistor to the right B1\n\n7(b)(i) sketch: periodic line showing minimum V = 0 and maximum V = +V B1\nOUT OUT 0\nline showing peak V at t = 0, 0.5T, 1.0T, 1.5T and 2.0T, with V going to zero half-way in between each peak B1\nOUT OUT\nline showing correct modulated sine shape B1\n\n7(b)(ii) sketch: sinusoidal curve with troughs sitting on the time axis B1\npeak power at t = 0, 0.5T, 1.0T, 1.5T and 2.0T and zero power half-way in between each peak B1\n\n7(b)(iii) same power-time graph with or without rectification, so same V B1\nrms\nor\nV2-time graph is same for both V and V , so same V\nOUT IN rms\nor\npower does not depend on sign of V, so same V\nrms\n© UCLES 2023 Page 12 of 15", "source_pages": [ 12 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9702_s23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-May-June/9702_s23_ms_42.pdf?download=true", "html": "9702-topic-21-alternating-currents/answers.html", "image_paths": [ "../answer-assets/9702_s23_ms_42-p12.png" ] }, { "id": "9702-2023-mj-42-q08", "question_id": "9702-2023-mj-42-q08", "subject": "9702", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 8, "topic": "Quantum physics", "topic_slug": "9702-topic-22-quantum-physics", "marks": 8, "status": "available", "reason": null, "text": "8(a) transition (emits) (one) photon with energy equal to the difference in energy between the two levels B1\nfrequency of radiation corresponds to energy of photon B1\n\n8(b)(i) line to the left of the pair in Fig. 8.2, labelled A B1\nlarger gap between line A and the nearest of the pair in Fig. 8.2 than between the lines in the pair B1\n\n8(b)(ii) line to the left of both the pair in Fig. 8.2 and line A, labelled B B1\nlarger gap between line B and line A than between line A and the nearest one of the pair in Fig. 8.2 B1\n\n8(c) E = hf C1\nE = E + h(f + f ) A1\n3 1 A B\n© UCLES 2023 Page 13 of 15", "source_pages": [ 13 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9702_s23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-May-June/9702_s23_ms_42.pdf?download=true", "html": "9702-topic-22-quantum-physics/answers.html", "image_paths": [ "../answer-assets/9702_s23_ms_42-p13.png" ] }, { "id": "9702-2023-mj-42-q09", "question_id": "9702-2023-mj-42-q09", "subject": "9702", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 9, "topic": "Astronomy and cosmology", "topic_slug": "9702-topic-25-astronomy-and-cosmology", "marks": 11, "status": "available", "reason": null, "text": "9(a) difference between mass of nucleus and (total) mass of nucleons M1\nwhen infinitely separated A1\n\n9(b)(i) neutron B1\n\n9(b)(ii) E = m c2 C1\nm = (0.030377 – 0.002388 – 0.009105)u C1\n( = 0.018884u)\nenergy release = (0.030377 – 0.002388 – 0.009105)  1.66  10–27  (3.00  108)2 = 2.8  10–12 J A1\n\n9(c)(i) number of atoms per unit time = (1.4  1028) / (2.8  10–12) C1\n( = 5.0  1039 s–1)\nmass of one atom = 4  1.66  10–27 or (4  10–3) / (6.02  1023) C1\n( = 6.64  10–27 kg)\nmass per unit time = 6.64  10–27  5.0  1039 A1\n= 3.3  1013 kg s–1\n\n9(c)(ii) L = 4σr2T4 C1\n1.4  1028 = 4  5.67  10–8  (2.3  109)2  T4\nT = 7800 K A1\n© UCLES 2023 Page 14 of 15", "source_pages": [ 14 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9702_s23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-May-June/9702_s23_ms_42.pdf?download=true", "html": "9702-topic-25-astronomy-and-cosmology/answers.html", "image_paths": [ "../answer-assets/9702_s23_ms_42-p14.png" ] }, { "id": "9702-2023-mj-42-q10", "question_id": "9702-2023-mj-42-q10", "subject": "9702", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 10, "topic": "Medical physics", "topic_slug": "9702-topic-24-medical-physics", "marks": 9, "status": "available", "reason": null, "text": "10(a)(i) electrons B1\n\n10(a)(ii) electrons are decelerated / stopped on impact with the target B1\n(kinetic) energy lost by electrons emitted as (X-ray) photons B1\n\n10(a)(iii) eV = hc /  C1\n = (6.63  10–34  3.00  108) / (1.60  10–19  5800) C1\n= 2.14  10–10 m A1\n\n10(b) I = I exp (–x) C1\n0\nI / I = exp (–(1.4  2.8)) C1\nT 0\n= 0.020\n% absorbed = (1.000 – 0.0198)  100 A1\n= 98%\n© UCLES 2023 Page 15 of 15", "source_pages": [ 15 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9702_s23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-May-June/9702_s23_ms_42.pdf?download=true", "html": "9702-topic-24-medical-physics/answers.html", "image_paths": [ "../answer-assets/9702_s23_ms_42-p15.png" ] }, { "id": "9702-2023-mj-43-q01", "question_id": "9702-2023-mj-43-q01", "subject": "9702", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 1, "topic": "Electric fields", "topic_slug": "9702-topic-18-electric-fields", "marks": 11, "status": "available", "reason": null, "text": "1(a)(i) force per unit mass B1\n\n1(a)(ii) force per unit positive charge B1\n\n1(a)(iii) similarity: B1\n inversely proportional to distance (from point)\n points of equal potential lie on concentric spheres\n zero at infinite distance\nAny point, 1 mark\ndifference: B1\n gravitational potential is (always) negative\n electric potential can be positive or negative\nAny point, 1 mark\n\n1(b)(i) g = GM / r2 M1\nE = Q / 4r2 M1\n0\nalgebra showing the elimination of r leading to M / Q = (1 / 4G) (g / E) A1\n0\n\n1(b)(ii)  = 1 / (4  6.67  10–11  8.85  10–12) = 1.35  1020 (kg2 C–2) A1\nor\n = (8.99  109) / (6.67  10–11) = 1.35  1020 (kg2 C–2)\n\n1(c)(i) E = gQ / M C1\n= (1.35  1020  9.81  4.80  105) / (5.98  1024)\n= 106 N C–1 or 106 V m–1 A1\n\n1(c)(ii) same (direction) B1\n© UCLES 2023 Page 6 of 16", "source_pages": [ 6 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9702_s23_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-May-June/9702_s23_ms_43.pdf?download=true", "html": "9702-topic-18-electric-fields/answers.html", "image_paths": [ "../answer-assets/9702_s23_ms_43-p06.png" ] }, { "id": "9702-2023-mj-43-q02", "question_id": "9702-2023-mj-43-q02", "subject": "9702", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 2, "topic": "Oscillations", "topic_slug": "9702-topic-17-oscillations", "marks": 12, "status": "available", "reason": null, "text": "2(a) horizontal force on sphere causes centripetal acceleration B1\nweight of sphere is (now) equal to vertical component of tension B1\nor\nhorizontal and vertical components (of force) (now) combine to give greater tension (in spring)\ngreater tension in spring so greater extension of spring B1\n\n2(b)(i) r = 10.8  sin 27° = 4.9 cm A1\n\n2(b)(ii) T cos  = mg C1\nor\nT cos  = W and W = mg\nT cos 27° = 0.29  9.81 leading to T = 3.2 N A1\n\n2(b)(iii) T = 3.2 – (0.29  9.81) C1\nk = T / x A1\n= [3.2 – (0.29  9.81)] / [10.8 – 8.5]\n= 0.15 N cm–1\n\n2(c)(i) centripetal acceleration = (T sin ) / m C1\n= (3.2  sin 27°) / 0.29\n= 5.0 m s–2 A1\n© UCLES 2023 Page 7 of 16\n\n2(c)(ii) a = r2 and  = 2 / T C1\nor\na = v2 / r and v = 2r / T\nT = 2  √(0.049 / 5.0) A1\n= 0.62 s\n© UCLES 2023 Page 8 of 16", "source_pages": [ 7, 8 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9702_s23_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-May-June/9702_s23_ms_43.pdf?download=true", "html": "9702-topic-17-oscillations/answers.html", "image_paths": [ "../answer-assets/9702_s23_ms_43-p07.png", "../answer-assets/9702_s23_ms_43-p08.png" ] }, { "id": "9702-2023-mj-43-q03", "question_id": "9702-2023-mj-43-q03", "subject": "9702", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 3, "topic": "Temperature", "topic_slug": "9702-topic-14-temperature", "marks": 11, "status": "available", "reason": null, "text": "3(a) no net thermal energy is transferred (between them) B1\n\n3(b)(i) variation (of density with temperature) is linear B1\nor\neach temperature has a unique value of density\n\n3(b)(ii)  variation (of density with temperature) is not linear B2\n region where the density does not vary with temperature\n different temperatures have the same density\nAny two points, 1 mark each\n\n3(c)(i) boiling point = 80 °C A1\n\n3(c)(ii) Q = Pt and t = 21 s C1\n(thermal energy supplied = 810  21 = 17000 J)\nc = Q / m C1\nthermal energy absorbed by beaker = 42  0.84  (80 – 25) C1\n( = 1940 J)\ns.h.c. of liquid = [(810  21) – (42  0.84  (80 – 25))] / [120  (80 – 25)] A1\n= 2.3 J g–1 K–1\n\n3(d) sketch: straight diagonal line from 25 °C to 100 °C and then horizontal at 100 °C B1\nstraight diagonal line starting at 25 °C with gradient approximately half that of the original line B1\n© UCLES 2023 Page 9 of 16", "source_pages": [ 9 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9702_s23_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-May-June/9702_s23_ms_43.pdf?download=true", "html": "9702-topic-14-temperature/answers.html", "image_paths": [ "../answer-assets/9702_s23_ms_43-p09.png" ] }, { "id": "9702-2023-mj-43-q04", "question_id": "9702-2023-mj-43-q04", "subject": "9702", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 4, "topic": "Thermodynamics", "topic_slug": "9702-topic-16-thermodynamics", "marks": 12, "status": "available", "reason": null, "text": "4(a)  particles are in (continuous) random motion B2\n particles have negligible volume (compared with the gas)\n negligible forces between particles (except during collisions)\n (all) collisions (perfectly) elastic\n time of collision negligible (in comparison with time between collisions)\nAny two points, 1 mark each\n\n4(b)(i) (general starting equation) pV = nRT C1\nT = (2pV / nR) where R is the (molar) gas constant A1\n\n4(b)(ii) sketch: straight vertical line XY from (V, 2p) to (V, p) B1\nstraight horizontal line YZ from (V, p) to (2V, p) B1\ncurve with gradient increasing from Z to X from (2V, p) to (V, 2p) B1\n\n4(b)(iii) XY work done on gas correct (= 0) B1\nZX increase in internal energy correct (= 0) B1\nYZ work done on gas correct (= –pV) B1\nXY increase in internal energy such that the increase in internal energy column adds up to zero B1\nall three thermal energies transferred such that U = q + w in each row B1\n(completely correct answer:\nchange U Q w\nX to Y –U –U 0\nY to Z [ +U ] U + pV –pV\nZ to X 0 –W [ +W ]\n)\n© UCLES 2023 Page 10 of 16", "source_pages": [ 10 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9702_s23_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-May-June/9702_s23_ms_43.pdf?download=true", "html": "9702-topic-16-thermodynamics/answers.html", "image_paths": [ "../answer-assets/9702_s23_ms_43-p10.png" ] }, { "id": "9702-2023-mj-43-q05", "question_id": "9702-2023-mj-43-q05", "subject": "9702", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 5, "topic": "Alternating currents", "topic_slug": "9702-topic-21-alternating-currents", "marks": 9, "status": "available", "reason": null, "text": "5(a)(i) correct circuit symbol for a diode shown correctly connected in series with the wires leading into and out of the dotted box B1\n\n5(a)(ii) smoothing / V is smoothed B1\nOUT\n\n5(b)(i) frequency = 1 / 0.04 A1\n= 25 Hz\n\n5(b)(ii) V = V exp (– t / RC) and  = RC C1\n0\nor\nV = V exp (– t / )\n0\n3.25 = 5.50 exp (– 0.020 / ) leading to  = 0.038 s A1\n\n5(b)(iii)  = RC C1\ncapacitance = 0.038 / 14000 A1\n= 2.7  10–6 F\n\n5(c) V has constant magnitude in both positive and negative directions B1\nIN\n(so) V is (now) constant / V does not vary with time B1\nOUT OUT\n© UCLES 2023 Page 11 of 16", "source_pages": [ 11 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9702_s23_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-May-June/9702_s23_ms_43.pdf?download=true", "html": "9702-topic-21-alternating-currents/answers.html", "image_paths": [ "../answer-assets/9702_s23_ms_43-p11.png" ] }, { "id": "9702-2023-mj-43-q06", "question_id": "9702-2023-mj-43-q06", "subject": "9702", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 6, "topic": "Magnetic fields", "topic_slug": "9702-topic-20-magnetic-fields", "marks": 10, "status": "available", "reason": null, "text": "6(a) a region where a force acts on M1\na current-carrying conductor A1\nor\na moving charge\nor\na magnetic material / magnetic pole\n\n6(b) concentric circles around the wire B1\nspacing between circles increases with distance from wire B1\narrows showing direction of field is clockwise B1\n\n6(c)(i) F = BIL C1\nforce per unit length = BI A1\n= 2.6  10–3  5.0\n= 0.013 N m–1\n\n6(c)(ii) to the right B1\n\n6(c)(iii) force (per unit length) has the same magnitude due to Newton’s 3rd law B1\n0.013 = 1.5  10–3  I A1\ncurrent = 8.7 A\n© UCLES 2023 Page 12 of 16", "source_pages": [ 12 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9702_s23_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-May-June/9702_s23_ms_43.pdf?download=true", "html": "9702-topic-20-magnetic-fields/answers.html", "image_paths": [ "../answer-assets/9702_s23_ms_43-p12.png" ] }, { "id": "9702-2023-mj-43-q07", "question_id": "9702-2023-mj-43-q07", "subject": "9702", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 7, "topic": "Quantum physics", "topic_slug": "9702-topic-22-quantum-physics", "marks": 9, "status": "available", "reason": null, "text": "7(a) wavelength associated with a moving particle B1\n\n7(b)(i) (electron) diffraction B1\n\n7(b)(ii) beam spreads out indicating diffraction B1\nor\nlight and dark regions indicate an interference pattern\nelectron beam is behaving as a wave B1\n\n7(c)(i) central blob and concentric rings B1\nrings closer together (than previously) B1\n\n7(c)(ii) (greater p.d. so) electrons to have greater momentum B1\ngreater momentum so decrease in (de Broglie) wavelength B1\nlower (de Broglie) wavelength (for same grating spacing in crystal) causes: B1\nsmaller diffraction angle\nor\nsmaller angle of intensity maxima (for each order)\nor\ndecrease in fringe spacing in diffraction pattern\n© UCLES 2023 Page 13 of 16", "source_pages": [ 13 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9702_s23_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-May-June/9702_s23_ms_43.pdf?download=true", "html": "9702-topic-22-quantum-physics/answers.html", "image_paths": [ "../answer-assets/9702_s23_ms_43-p13.png" ] }, { "id": "9702-2023-mj-43-q08", "question_id": "9702-2023-mj-43-q08", "subject": "9702", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 8, "topic": "Medical physics", "topic_slug": "9702-topic-24-medical-physics", "marks": 8, "status": "available", "reason": null, "text": "8(a)(i) specific acoustic impedance = 1200  1400 = 1.68  106 kg m–2 s–1 A1\n\n8(a)(ii) density of air shown in table as 1.29 A1\nspeed of sound in tissue shown in table as 1540 A1\n\n8(b)(i) intensity reflection coefficient = (Z – Z )2 / (Z + Z )2 C1\n1 2 1 2\n= (1680000 – 440)2 / (1680000 + 440)2\n= 0.999 A1\n\n8(b)(ii) intensity reflection coefficient = (Z – Z )2 / (Z + Z )2 A1\n1 2 1 2\n= (1680000 – 1680000)2 / (1680000 + 1680000)2\n= 0\n\n8(c) without gel, (almost) all of the (incident) ultrasound is reflected (from skin) B1\nwith gel, (almost) all of the (incident) ultrasound is transmitted (into the body) B1\n© UCLES 2023 Page 14 of 16", "source_pages": [ 14 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9702_s23_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-May-June/9702_s23_ms_43.pdf?download=true", "html": "9702-topic-24-medical-physics/answers.html", "image_paths": [ "../answer-assets/9702_s23_ms_43-p14.png" ] }, { "id": "9702-2023-mj-43-q09", "question_id": "9702-2023-mj-43-q09", "subject": "9702", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 9, "topic": "Nuclear physics", "topic_slug": "9702-topic-23-nuclear-physics", "marks": 9, "status": "available", "reason": null, "text": "9(a) time for activity (of sample) to halve B1\n\n9(b) sketch: line with positive gradient starting at (0,0) and extending to t = 80 min B1\nexponential curve, extending from t = 0 to t = 80 min, with gradient of steadily decreasing magnitude B1\nline passing through (0,0), (20, 0.5N ) and (40, 0.75 N ) B1\n0 0\n\n9(c)(i) every (undecayed) nucleus has the same probability of decay M1\nfewer (undecayed) nuclei remaining (with time), so fewer will decay (in a given time interval) A1\n\n9(c)(ii)  sample emits in all directions but detector only captures emissions in one direction B2\n some emissions are absorbed before reaching detector\n some emissions are scattered within the sample\n simultaneous arrival of multiple particles only registers once\n some particles may reach detector but not cause ionisation\nAny two points, 1 mark each\nmeasured count rate is less than the activity B1\n© UCLES 2023 Page 15 of 16", "source_pages": [ 15 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9702_s23_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-May-June/9702_s23_ms_43.pdf?download=true", "html": "9702-topic-23-nuclear-physics/answers.html", "image_paths": [ "../answer-assets/9702_s23_ms_43-p15.png" ] }, { "id": "9702-2023-mj-43-q10", "question_id": "9702-2023-mj-43-q10", "subject": "9702", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 10, "topic": "Astronomy and cosmology", "topic_slug": "9702-topic-25-astronomy-and-cosmology", "marks": 9, "status": "available", "reason": null, "text": "10(a) speed is (directly) proportional to distance M1\nspeed is speed of recession of galaxy from an observer, and distance is the distance of the galaxy from the observer A1\n\n10(b) F = L / (4d2) C1\n= (3.8  1031) / [4  (1.8  1024)2] A1\n= 9.3  10–19 W m–2\n\n10(c)(i) galaxy is moving away (from the Earth) B1\nwavelength (of light from the galaxy) increased by the Doppler effect / due to redshift B1\n\n10(c)(ii)  /  = v / c C1\nv = [(492 – 486)  3.00  108] / 486\n(v = 3.7  106 m s–1)\nH = v / d C1\n0\n= (3.7  106) / (1.8  1024) A1\n= 2.1  10–18 s–1\n© UCLES 2023 Page 16 of 16", "source_pages": [ 16 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9702_s23_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-May-June/9702_s23_ms_43.pdf?download=true", "html": "9702-topic-25-astronomy-and-cosmology/answers.html", "image_paths": [ "../answer-assets/9702_s23_ms_43-p16.png" ] }, { "id": "9702-2023-mj-51-q01", "question_id": "9702-2023-mj-51-q01", "subject": "9702", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "51", "question_number": 1, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "1 Defining the problem\n is the independent variable and t is the dependent variable or vary  and measure t 1\nkeep d constant 1\nMethods of data collection\nlabelled diagram of workable experiment including: 1\n plane supported by stand / support\n stand/support on bench/floor/horizontal surface\n minimum of two labels from cube, cylinder, (inclined) plane, method of support, bench/floor/horizontal surface, pulley,\nstring\ndiagram showing method to measure d, e.g. 1\nclamped vertical rule near cylinder\nor\ndrawn rule on plane used to measure d\nor\ndistance d marked on plane and rule used to determine d\nuse a protractor to measure θ 1\nor\nuse a rule(r) to measure appropriate lengths for a trigonometric calculation\nuse a timer/stop-watch to measure t or light gates connected to a timer to measure t 1\n© UCLES 2023 Page 6 of 12\n\n1 Method of Analysis\n\n1 1 1\nplot a graph of against sin  or equivalent (e.g. sin  against )\nt2 t2\nDo not accept logarithms.\n2dABgradient 1\nH \nA\n\n1\n2dAB\n(for sin  against : H  )\nt2 Agradient\n2dABy-intercept 1\nK \nA\n\n1\n(for sin  against : K Hy-intercept)\nt2\n© UCLES 2023 Page 7 of 12\n\n1 Additional detail including safety considerations 6\nD1 Safety precaution linked to falling cylinder, e.g. use of cushion/sand box to collect cylinder/prevent damage to\ncylinder/floor/bench/injury\nD2 protractor correctly positioned on diagram\nor\nappropriate trigonometric relationship for marked lengths\nD3 keep A and B constant\nD4 use a (top-pan) balance to measure A and B\nD5 correct positioning of light gates to determine t, e.g. two light gates either end of distance d, connected to a timer\nor\ncorrect position of video camera with timer in frame of the video to determine t\nD6 method to release cylinder/cube, e.g. cube held by set square, set square moved to release cube.\nD7 reasoned method to keep d constant as θ changes, e.g.\n(when measuring d by position of cylinder) adjust the length of the string or adjust the position of vertical marks or\nadjust the position of the vertical rule or initial position of the cube\nor\n(when measuring d by position of cube) use fixed marks on the plane or ruler placed on the plane with d measured\nbetween the marks\nD8 method to increase t for cylinder to fall, e.g. use large d to increase t\nD9 repeat measurements of t for the same θ and average t\nD10 relationship valid if a straight line is produced (not passing through the origin)\nDo not accept straight line passing through the origin.\n© UCLES 2023 Page 8 of 12", "source_pages": [ 6, 7, 8 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9702_s23_ms_51.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-May-June/9702_s23_ms_51.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_s23_ms_51-p06.png", "../answer-assets/9702_s23_ms_51-p07.png", "../answer-assets/9702_s23_ms_51-p08.png" ] }, { "id": "9702-2023-mj-51-q02", "question_id": "9702-2023-mj-51-q02", "subject": "9702", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "51", "question_number": 2, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "2(a) V  1\ngradient = Rln \nV\n \n0\n\n2(b) 1\nC / 10–4 F T / s\n0.89 or 0.892 13.7  0.8\n1.3 or 1.32 20.4  0.7\n1.6 or 1.58 24.3  0.6\n1.0 or 1.03 16.1  0.8\n1.2 or 1.18 18.3  0.7\n2.1 or 2.08 31.5  0.6\nValues of C and T correct as shown above.\nAbsolute uncertainties in T correct as shown above. 1\n\n2(c)(i) Six points from (b) plotted correctly. 1\nMust be within half a small square. Diameter of points must be less than half a small square.\nError bars in T plotted correctly. 1\nAll error bars to be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.\n© UCLES 2023 Page 9 of 12\n\n2(c)(ii) Straight line of best fit drawn. 1\nDo not accept line from top point to bottom point.\nPoints must be balanced.\nLine must pass between (1.42, 22.0) and (1.45, 22.0) and between (1.95, 30.0) and (2.00, 30.0)\nWorst acceptable line drawn (steepest or shallowest possible line that passes through all the error bars). 1\nAll error bars must be plotted.\n\n2(c)(iii) Gradient determined with clear substitution of data points into y / x. 1\nDistance between data points must be greater than half the length of the drawn line.\nGradient of worst acceptable line determined with clear substitution of data points into y / x. 1\nuncertainty = (gradient of line of best fit – gradient of worst acceptable line)\nor\nuncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)\n\n2(d) – 0.69 or – 0.693 and  0.06 1\n\n2(e)(i) R determined using gradient and R given to 2 or 3 significant figures. 1\ngradient (c)(iii)\nR  \nV  (d)\nln \nV\n 0 \nR correctly determined using gradient and SI unit with correct power of ten for R (e.g. ). 1\n© UCLES 2023 Page 10 of 12\n\n2(e)(ii) Percentage uncertainty in R with method shown. 1\n  V  \nln  \npercentage uncertainty      V 0     gradient 100\n \nV  gradient\n ln  \n V \n  0  \nor\nCorrect substitution for max/min methods.\n© UCLES 2023 Page 11 of 12\n\n2(f) C determined to a minimum of 2 significant figures from (c)(iii) or (d) and (e)(i) with correct substitution. 1\nT 60.0\nC  \ngradient gradient\nor\nT 60.0\nC  \nV  (e)(i)(d)\nRln \nV\n 0 \nAbsolute uncertainty in C determined with correct method used: 1\nUsing gradient to determine C:\n gradient\nC  C\n gradient \nAllow using R to determine C:\n  V  \nln  \nC      V 0     (e)(ii) C\n \nV  100\n ln  \n V \n  0  \n© UCLES 2023 Page 12 of 12", "source_pages": [ 9, 10, 11, 12 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9702_s23_ms_51.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-May-June/9702_s23_ms_51.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_s23_ms_51-p09.png", "../answer-assets/9702_s23_ms_51-p10.png", "../answer-assets/9702_s23_ms_51-p11.png", "../answer-assets/9702_s23_ms_51-p12.png" ] }, { "id": "9702-2023-mj-52-q01", "question_id": "9702-2023-mj-52-q01", "subject": "9702", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "52", "question_number": 1, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "1 Defining the problem\nR is the independent variable and E is the dependent variable or vary R and measure E 1\nkeep V constant 1\nMethods of data collection\nlabelled diagram of workable experiment including: 1\n coil P placed close to coil Q\n separate workable circuit for coil Q\n (a.c.) voltmeter or oscilloscope connected across coil Q\n(Do not accept a power supply connected to coil Q.)\na.c. power supply/signal generator connected to resistor and coil P in series 1\nworkable circuit with power supply and (a.c.) voltmeter/oscilloscope in parallel with resistor and coil P or across terminals of 1\npower supply/signal generator\nmethod to determine R, e.g. measure current in R and p.d. across R and use R = V /I or measure R using an ohmmeter 1\nR\n© UCLES 2023 Page 6 of 12\n\n1 Method of Analysis\n\n1 1 1\nplot a graph of against R or equivalent (e.g. R against )\nE E\nDo not accept logarithms.\n\n1 1\nM \n2fV gradient\n\n1 gradient\n(for R against : M  )\nE 2fV\nk 2fVMy-intercept 1\nor\ny-intercept\nk \ngradient\n\n1\n(for R against : k =  y-intercept)\nE\n© UCLES 2023 Page 7 of 12\n\n1 Additional detail including safety considerations 6\nD1 precaution linked to hot coil (P) / hot resistor, e.g. use of (heat-proof) gloves, wait until circuit cools down\nor\nprecaution linked to shocks from high voltages e.g. use of (insulating) gloves or switch off supply before touching the\ncircuit (to change R)\nD2 keep the number of turns on (both) coils constant\nD3 keep f constant\nD4 keep distance between the coils constant\nD5 method to keep distance between the coils constant, e.g. fix/clamp coils to bench\nD6 method to measure f, e.g. read from signal generator or use of oscilloscope\nD7 method to determine f from oscilloscope, e.g. period from oscilloscope T = time-base  horizontal distance and\nf = 1/T\nD8 method to determine V or E from oscilloscope, e.g. V = y-gain  vertical distance\nD9 method to increase E e.g. use iron core/more turns on coil Q/high frequency/high p.d. (across R and coil P)\nD10 relationship valid if a straight line is produced (not passing through the origin)\nDo not accept straight line passing through the origin.\n© UCLES 2023 Page 8 of 12", "source_pages": [ 6, 7, 8 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9702_s23_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-May-June/9702_s23_ms_52.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_s23_ms_52-p06.png", "../answer-assets/9702_s23_ms_52-p07.png", "../answer-assets/9702_s23_ms_52-p08.png" ] }, { "id": "9702-2023-mj-52-q02", "question_id": "9702-2023-mj-52-q02", "subject": "9702", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "52", "question_number": 2, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "2(a) Yk 1\ngradient =\np\nYkZ\ny-intercept =\np\n\n2(b) 1\nV / 10–5 m3 absolute uncertainty\n3.81 or 3.815  0.03\n3.99 or 3.986  0.03\n4.16 or 4.163  0.04\n4.33 or 4.335  0.04\n4.48 or 4.481  0.04\n4.65 or 4.652  0.04\nValues of V correct as shown above.\nAbsolute uncertainties in V correct as shown above. 1\n\n2(c)(i) Six points from (b) plotted correctly. 1\nMust be within half a small square. Diameter of points must be less than half a small square.\nError bars in V plotted correctly. 1\nAll error bars to be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.\n© UCLES 2023 Page 9 of 12\n\n2(c)(ii) Straight line of best fit drawn. 1\nDo not accept line from top point to bottom point.\nPoints must be balanced.\nLine must pass between (27.5, 3.90) and (29.5, 3.90) and between (82.0, 4.60) and (84.0, 4.60).\nWorst acceptable line drawn (steepest or shallowest possible line that passes through all the error bars). 1\nAll error bars must be plotted.\n\n2(c)(iii) Gradient determined with clear substitution of data points into y / x. 1\nDistance between data points must be greater than half the length of the drawn line.\nGradient of worst acceptable line determined with clear substitution of data points into y / x. 1\nuncertainty = (gradient of line of best fit – gradient of worst acceptable line)\nor\nuncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)\n\n2(c)(iv) y-intercept determined by substitution of correct point with consistent power of ten in m and y into y = mx + c. 1\ny-intercept of worst acceptable line determined by substitution into y = mx + c. 1\nuncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line\nor\nuncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)\nDo not accept ECF from false origin method.\n© UCLES 2023 Page 10 of 12\n\n2(d)(i) Y determined using gradient and Y and Z given to 2 or 3 significant figures. 1\npgradient\nY  7.31881027 gradient\nk\nZ determined using y-intercept and Y and Z given with SI units. 1\npy-intercept y-intercept\nZ  or Z \nYk gradient\nUnits:\nY: no unit\nZ: °C\n\n2(d)(ii) Percentage uncertainty in Y with method shown. 1\np gradient\npercentage uncertainty   100\n p gradient \nor\nCorrect substitution for max/min methods.\n© UCLES 2023 Page 11 of 12\n\n2(e)  determined to a minimum of 2 significant figures from (c)(iii) and (c)(iv) or (d)(i) with correct substitution and correct 1\npowers of ten.\n0.02792\n0.0600\nV  3.67105\n4\nand\npV V V y-intercept\n Z or  Z or \nYk gradient gradient\nor using h directly:\nd2h\ny-intercept\npd2h d2h\n4\n Z or  Z or \n4Yk 4gradient gradient\n© UCLES 2023 Page 12 of 12", "source_pages": [ 9, 10, 11, 12 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9702_s23_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-May-June/9702_s23_ms_52.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_s23_ms_52-p09.png", "../answer-assets/9702_s23_ms_52-p10.png", "../answer-assets/9702_s23_ms_52-p11.png", "../answer-assets/9702_s23_ms_52-p12.png" ] }, { "id": "9702-2023-mj-53-q01", "question_id": "9702-2023-mj-53-q01", "subject": "9702", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "53", "question_number": 1, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "1 Defining the problem\n is the independent variable and t is the dependent variable or vary  and measure t 1\nkeep d constant 1\nMethods of data collection\nlabelled diagram of workable experiment including: 1\n plane supported by stand / support\n stand/support on bench/floor/horizontal surface\n minimum of two labels from cube, cylinder, (inclined) plane, method of support, bench/floor/horizontal surface, pulley,\nstring\ndiagram showing method to measure d, e.g. 1\nclamped vertical rule near cylinder\nor\ndrawn rule on plane used to measure d\nor\ndistance d marked on plane and rule used to determine d\nuse a protractor to measure θ 1\nor\nuse a rule(r) to measure appropriate lengths for a trigonometric calculation\nuse a timer/stop-watch to measure t or light gates connected to a timer to measure t 1\n© UCLES 2023 Page 6 of 12\n\n1 Method of Analysis\n\n1 1 1\nplot a graph of against sin  or equivalent (e.g. sin  against )\nt2 t2\nDo not accept logarithms.\n2dABgradient 1\nH \nA\n\n1\n2dAB\n(for sin  against : H  )\nt2 Agradient\n2dABy-intercept 1\nK \nA\n\n1\n(for sin  against : K Hy-intercept)\nt2\n© UCLES 2023 Page 7 of 12\n\n1 Additional detail including safety considerations 6\nD1 Safety precaution linked to falling cylinder, e.g. use of cushion/sand box to collect cylinder/prevent damage to\ncylinder/floor/bench/injury\nD2 protractor correctly positioned on diagram\nor\nappropriate trigonometric relationship for marked lengths\nD3 keep A and B constant\nD4 use a (top-pan) balance to measure A and B\nD5 correct positioning of light gates to determine t, e.g. two light gates either end of distance d, connected to a timer\nor\ncorrect position of video camera with timer in frame of the video to determine t\nD6 method to release cylinder/cube, e.g. cube held by set square, set square moved to release cube.\nD7 reasoned method to keep d constant as θ changes, e.g.\n(when measuring d by position of cylinder) adjust the length of the string or adjust the position of vertical marks or\nadjust the position of the vertical rule or initial position of the cube\nor\n(when measuring d by position of cube) use fixed marks on the plane or ruler placed on the plane with d measured\nbetween the marks\nD8 method to increase t for cylinder to fall, e.g. use large d to increase t\nD9 repeat measurements of t for the same θ and average t\nD10 relationship valid if a straight line is produced (not passing through the origin)\nDo not accept straight line passing through the origin.\n© UCLES 2023 Page 8 of 12", "source_pages": [ 6, 7, 8 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9702_s23_ms_53.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-May-June/9702_s23_ms_53.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_s23_ms_53-p06.png", "../answer-assets/9702_s23_ms_53-p07.png", "../answer-assets/9702_s23_ms_53-p08.png" ] }, { "id": "9702-2023-mj-53-q02", "question_id": "9702-2023-mj-53-q02", "subject": "9702", "year": 2023, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "53", "question_number": 2, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "2(a) V  1\ngradient = Rln \nV\n \n0\n\n2(b) 1\nC / 10–4 F T / s\n0.89 or 0.892 13.7  0.8\n1.3 or 1.32 20.4  0.7\n1.6 or 1.58 24.3  0.6\n1.0 or 1.03 16.1  0.8\n1.2 or 1.18 18.3  0.7\n2.1 or 2.08 31.5  0.6\nValues of C and T correct as shown above.\nAbsolute uncertainties in T correct as shown above. 1\n\n2(c)(i) Six points from (b) plotted correctly. 1\nMust be within half a small square. Diameter of points must be less than half a small square.\nError bars in T plotted correctly. 1\nAll error bars to be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.\n© UCLES 2023 Page 9 of 12\n\n2(c)(ii) Straight line of best fit drawn. 1\nDo not accept line from top point to bottom point.\nPoints must be balanced.\nLine must pass between (1.42, 22.0) and (1.45, 22.0) and between (1.95, 30.0) and (2.00, 30.0)\nWorst acceptable line drawn (steepest or shallowest possible line that passes through all the error bars). 1\nAll error bars must be plotted.\n\n2(c)(iii) Gradient determined with clear substitution of data points into y / x. 1\nDistance between data points must be greater than half the length of the drawn line.\nGradient of worst acceptable line determined with clear substitution of data points into y / x. 1\nuncertainty = (gradient of line of best fit – gradient of worst acceptable line)\nor\nuncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)\n\n2(d) – 0.69 or – 0.693 and  0.06 1\n\n2(e)(i) R determined using gradient and R given to 2 or 3 significant figures. 1\ngradient (c)(iii)\nR  \nV  (d)\nln \nV\n 0 \nR correctly determined using gradient and SI unit with correct power of ten for R (e.g. ). 1\n© UCLES 2023 Page 10 of 12\n\n2(e)(ii) Percentage uncertainty in R with method shown. 1\n  V  \nln  \npercentage uncertainty      V 0     gradient 100\n \nV  gradient\n ln  \n V \n  0  \nor\nCorrect substitution for max/min methods.\n© UCLES 2023 Page 11 of 12\n\n2(f) C determined to a minimum of 2 significant figures from (c)(iii) or (d) and (e)(i) with correct substitution. 1\nT 60.0\nC  \ngradient gradient\nor\nT 60.0\nC  \nV  (e)(i)(d)\nRln \nV\n 0 \nAbsolute uncertainty in C determined with correct method used: 1\nUsing gradient to determine C:\n gradient\nC  C\n gradient \nAllow using R to determine C:\n  V  \nln  \nC      V 0     (e)(ii) C\n \nV  100\n ln  \n V \n  0  \n© UCLES 2023 Page 12 of 12", "source_pages": [ 9, 10, 11, 12 ], "source_pdf": "_source-pdfs/2023-May-June/ms/9702_s23_ms_53.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-May-June/9702_s23_ms_53.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_s23_ms_53-p09.png", "../answer-assets/9702_s23_ms_53-p10.png", "../answer-assets/9702_s23_ms_53-p11.png", "../answer-assets/9702_s23_ms_53-p12.png" ] }, { "id": "9702-2023-on-41-q01", "question_id": "9702-2023-on-41-q01", "subject": "9702", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 1, "topic": "Electric fields", "topic_slug": "9702-topic-18-electric-fields", "marks": 10, "status": "available", "reason": null, "text": "1(a)(i) direction of the force acting on a (test) mass placed at the point B1\n\n1(a)(ii) change in height negligible compared with radius (of Earth) B1\n(so) field lines are (effectively) parallel B1\n\n1(b)(i) Y = GM / R2 M1\nG is the gravitational constant A1\n\n1(b)(ii) gravitational force is (always) attractive B1\nor\ngravitational force (always) acts towards the centre of the sphere\nforce is in opposite direction to displacement B1\nor\nat a point to the right of the centre, force acts to the left\nor\nat a point to the left of the centre, force acts to the right\n\n1(b)(iii) sketch: smooth curve with decreasing positive gradient, starting at (R, –Y) and reaching 3R with g still negative B1\nor\nsmooth curve with increasing positive gradient, ending at (–R, Y) and reaching –3R with g still positive\nboth of the above curves, in correct quadrants B1\ncurve passing through (2R, 0.25Y) and (3R, 0.11Y) B1\n© UCLES 2023 Page 7 of 16", "source_pages": [ 7 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9702_w23_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-Oct-Nov/9702_w23_ms_41.pdf?download=true", "html": "9702-topic-18-electric-fields/answers.html", "image_paths": [ "../answer-assets/9702_w23_ms_41-p07.png" ] }, { "id": "9702-2023-on-41-q02", "question_id": "9702-2023-on-41-q02", "subject": "9702", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 2, "topic": "Thermodynamics", "topic_slug": "9702-topic-16-thermodynamics", "marks": 12, "status": "available", "reason": null, "text": "2(a) (thermal) energy per unit mass (to change temperature) B1\n(thermal) energy per unit change in temperature B1\n\n2(b)(i) work done = pV A1\n= (2.0  105)  (0.063 – 0.038) = 5000 J\n\n2(b)(ii) gas is expanding (against external pressure) B1\ngas does work / work is done by gas, so (work done on gas is) negative B1\n\n2(b)(iii) U = q + W C1\n7600 = q + (–5000) A1\nq = 12 600 J\n\n2(b)(iv) specific heat capacity = q / mT C1\n= 12600 / (0.35  56)\n= 640 J kg–1 K–1 A1\n\n2(c) same gain in internal energy so same temperature rise B1\nno change in volume so no work done B1\nor\nno work done so less thermal energy needed (for same change in internal energy)\nless thermal energy needed (for same temperature change) so lower specific heat capacity B1\n© UCLES 2023 Page 8 of 16", "source_pages": [ 8 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9702_w23_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-Oct-Nov/9702_w23_ms_41.pdf?download=true", "html": "9702-topic-16-thermodynamics/answers.html", "image_paths": [ "../answer-assets/9702_w23_ms_41-p08.png" ] }, { "id": "9702-2023-on-41-q03", "question_id": "9702-2023-on-41-q03", "subject": "9702", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 3, "topic": "Ideal gases", "topic_slug": "9702-topic-15-ideal-gases", "marks": 13, "status": "available", "reason": null, "text": "3(a)(i) N: number of molecules (of the gas) B1\nm: mass of one molecule (of the gas) B1\n: mean square speed (of molecules) B1\n\n3(a)(ii) pV = NkT M1\nNkT = ⅓Nm and E = ½m leading to E = (3/2) kT A1\nK K\n\n3(b) ½  3.34  10–27  93002 = (3/2)  1.38  10–23  T C1\nT = 6980 K A1\n\n3(c)(i) L = F  4d2 C1\nL = 2.52  10–8  4  (4.16  1016)2 A1\n= 5.48  1026 W\n\n3(c)(ii) L = 4r2T4 C1\n5.48  1026 = 4  5.67  10–8  r2  69804\nr = 5.69  108 m A1\n\n3(d) (very high pressure so) molecules are (very) close together (not just ‘nearer’) B1\nforces between molecules are not negligible B1\nor\nvolume of molecules not negligible compared with gas volume\n© UCLES 2023 Page 9 of 16", "source_pages": [ 9 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9702_w23_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-Oct-Nov/9702_w23_ms_41.pdf?download=true", "html": "9702-topic-15-ideal-gases/answers.html", "image_paths": [ "../answer-assets/9702_w23_ms_41-p09.png" ] }, { "id": "9702-2023-on-41-q04", "question_id": "9702-2023-on-41-q04", "subject": "9702", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 4, "topic": "Oscillations", "topic_slug": "9702-topic-17-oscillations", "marks": 9, "status": "available", "reason": null, "text": "4(a) (motion in which) acceleration is (directly) proportional to displacement B1\n(motion in which): B1\nacceleration is (always) in the opposite direction to displacement\nor\nacceleration is (always) directed towards a fixed point\n\n4(b)(i)  = 2 / T C1\n = 2 / 3.0 A1\n= 2.1 rad s–1\n\n4(b)(ii) E = ½m2x 2 C1\n0\n= ½  0.81  2.12  0.0362 A1\n= 2.3  10–3 J\n\n4(c) sketch: line starting at (0, 0.036) and not reaching x =  0.036 m at any other time B1\nsmooth curve, with no sudden changes in gradient, showing continuously decreasing magnitude of x from B1\nmaximum displacement at t = 0 to final displacement of zero where the gradient is also zero\ndisplacement reaches final value of zero between t = 0.75 s and t = 3.0 s at the latest B1\n© UCLES 2023 Page 10 of 16", "source_pages": [ 10 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9702_w23_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-Oct-Nov/9702_w23_ms_41.pdf?download=true", "html": "9702-topic-17-oscillations/answers.html", "image_paths": [ "../answer-assets/9702_w23_ms_41-p10.png" ] }, { "id": "9702-2023-on-41-q05", "question_id": "9702-2023-on-41-q05", "subject": "9702", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 5, "topic": "Electric fields", "topic_slug": "9702-topic-18-electric-fields", "marks": 8, "status": "available", "reason": null, "text": "5(a) work done per unit charge B1\nwork (done on charge) moving positive charge from infinity (to the point) B1\n\n5(b) Any three points from: B3\nUp to 2 points from:\n• radius of sphere X is 2.0 m\n• radius of sphere Y is 4.0 m\n• radius of Y is double the radius of X\nUp to 2 points from:\n• charge on X is negative\n• charge on Y is positive\n• spheres carry opposite charges\nUp to 1 point from:\n• magnitudes of charges on the spheres are equal\n\n5(c) particle is attracted to X or repelled from Y B1\nor\nresultant force on particle is towards X / away from Y / to the left\nparticle accelerates towards X / away from Y / to the left B1\n(magnitude of) acceleration of particle increases B1\n© UCLES 2023 Page 11 of 16", "source_pages": [ 11 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9702_w23_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-Oct-Nov/9702_w23_ms_41.pdf?download=true", "html": "9702-topic-18-electric-fields/answers.html", "image_paths": [ "../answer-assets/9702_w23_ms_41-p11.png" ] }, { "id": "9702-2023-on-41-q06", "question_id": "9702-2023-on-41-q06", "subject": "9702", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 6, "topic": "Magnetic fields", "topic_slug": "9702-topic-20-magnetic-fields", "marks": 12, "status": "available", "reason": null, "text": "6(a) • force per unit length B2\n• force per unit current\n• length / current perpendicular to field\n1 mark for any two points, 2 marks for all three points\n\n6(b)(i) into the page B1\n\n6(b)(ii) F = Bqv C1\n= 4.8  10–3  1.6  10–19  1.7  107 = 1.3  10–14 N A1\n\n6(b)(iii) arrow at point X pointing down the page B1\n\n6(b)(iv) F = mv2 / r C1\n1.3  10–14 = (9.11  10–31)  (1.7  107)2 / r C1\n(r = 0.020 m) A1\nd = 2r\nd = 0.040 m\n\n6(c) path shows upwards deflection such that the curvature is always anticlockwise within the field B1\ncircular path with larger radius B1\nline enters field at X and leaves field at distance 2d vertically from X B1\n© UCLES 2023 Page 12 of 16", "source_pages": [ 12 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9702_w23_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-Oct-Nov/9702_w23_ms_41.pdf?download=true", "html": "9702-topic-20-magnetic-fields/answers.html", "image_paths": [ "../answer-assets/9702_w23_ms_41-p12.png" ] }, { "id": "9702-2023-on-41-q07", "question_id": "9702-2023-on-41-q07", "subject": "9702", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 7, "topic": "Ideal gases", "topic_slug": "9702-topic-15-ideal-gases", "marks": 8, "status": "available", "reason": null, "text": "7(a)(i) P = I 2R A1\n0\n\n7(a)(ii) P = 4I 2R A1\n0\n\n7(b) sketch: square wave of period T, with P always non-zero B1\nhorizontal lines, from 0 to 0.5T and from 1.0T to 1.5T, all at the same level that the scale indicates to be I 2R B1\n0\nhorizontal lines, from 0.5T to 1.0T and from 1.5T to 2.0T, at a level that is four times higher than the lower lines B1\n\n7(c)(i)

= (5/2)I 2R A1\n0\n\n7(c)(ii)

= I 2R C1\nr.m.s.\nI 2R = (5/2)I 2R A1\nr.m.s. 0\nI = √(5/2) I\nr.m.s. 0\n© UCLES 2023 Page 13 of 16", "source_pages": [ 13 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9702_w23_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-Oct-Nov/9702_w23_ms_41.pdf?download=true", "html": "9702-topic-15-ideal-gases/answers.html", "image_paths": [ "../answer-assets/9702_w23_ms_41-p13.png" ] }, { "id": "9702-2023-on-41-q08", "question_id": "9702-2023-on-41-q08", "subject": "9702", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 8, "topic": "Quantum physics", "topic_slug": "9702-topic-22-quantum-physics", "marks": 10, "status": "available", "reason": null, "text": "8(a)(i) p = E / c M1\nE = hc /  and completion of algebra leading to p = h /  A1\n\n8(a)(ii) wavelength = (6.63  10–34) / (9.5  10–28) = 700  10–9 m so red B1\n\n8(b)(i) power = intensity  area C1\nnumber per unit time = (160  2.5  10–6) / (9.5  10–28  3.00  108) = 1.4  1015 s–1 A1\n\n8(b)(ii) pressure = force / area C1\nforce = rate of change of momentum C1\n= 2  9.5  10–28  1.4  1015\npressure = (2  9.5  10–28  1.4  1015) / (2.5  10–6) A1\n= 1.1  10–6 Pa\n\n8(c) photons have greater momentum B1\nor\nfewer photons per unit time\ngreater photon momentum but smaller number of photons (per unit time) so pressure is the same B1\n© UCLES 2023 Page 14 of 16", "source_pages": [ 14 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9702_w23_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-Oct-Nov/9702_w23_ms_41.pdf?download=true", "html": "9702-topic-22-quantum-physics/answers.html", "image_paths": [ "../answer-assets/9702_w23_ms_41-p14.png" ] }, { "id": "9702-2023-on-41-q09", "question_id": "9702-2023-on-41-q09", "subject": "9702", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 9, "topic": "Nuclear physics", "topic_slug": "9702-topic-23-nuclear-physics", "marks": 10, "status": "available", "reason": null, "text": "9(a) (two small) nuclei join together M1\nto form one larger nucleus A1\n\n9(b) line with a peak at A  56 B1\nline with steep initial positive gradient on the left of peak and shallower negative gradient at all points to the right of peak B1\nand line does not return to 0 binding energy\n\n9(c)(i) X shown at value of A to the right of the peak B1\n\n9(c)(ii) Y shown at value of A close to 1 B1\n\n9(d) energy from 1 nucleus = (1.77  1013) / (6.02  1023) C1\n( = 2.94  10–11 J)\nbinding energy of Z = [(1.25 + 1.81)  10–10] – 2.94  10–11 C1\n( = 2.77  10–10 J)\nnucleon number of Z = 93 + 139 + 2 – 1 C1\n( = 233)\nbinding energy per nucleon = (2.77  10–10) / (233  1.60  10–13) A1\n= 7.43 MeV\n© UCLES 2023 Page 15 of 16", "source_pages": [ 15 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9702_w23_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-Oct-Nov/9702_w23_ms_41.pdf?download=true", "html": "9702-topic-23-nuclear-physics/answers.html", "image_paths": [ "../answer-assets/9702_w23_ms_41-p15.png" ] }, { "id": "9702-2023-on-41-q10", "question_id": "9702-2023-on-41-q10", "subject": "9702", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 10, "topic": "Medical physics", "topic_slug": "9702-topic-24-medical-physics", "marks": 8, "status": "available", "reason": null, "text": "10(a) ultrasound production: vibrating quartz crystal B1\nX-ray production: electrons hitting metal target B1\nultrasound detected wave: reflected B1\nX-ray detected wave: transmitted B1\n\n10(b)(i) I = I exp (–x) C1\n0\nln (0.72) = –6.2 A1\n = 0.053 cm–1\n\n10(b)(ii) I / I = exp (–9.3  0.053) C1\n0\n( = 0.61)\npercentage attenuated = 100  (1.00 – 0.61) A1\n= 39%\n© UCLES 2023 Page 16 of 16", "source_pages": [ 16 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9702_w23_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-Oct-Nov/9702_w23_ms_41.pdf?download=true", "html": "9702-topic-24-medical-physics/answers.html", "image_paths": [ "../answer-assets/9702_w23_ms_41-p16.png" ] }, { "id": "9702-2023-on-42-q01", "question_id": "9702-2023-on-42-q01", "subject": "9702", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 1, "topic": "Motion in a circle", "topic_slug": "9702-topic-12-motion-in-a-circle", "marks": 10, "status": "available", "reason": null, "text": "1(a) angle (subtended at centre of circle) when arc length = radius B1\n\n1(b)  = 2 / T C1\n= 2 / (1.0  60  60) A1\n= 1.7  10–3 rad s–1\n\n1(c)(i) angle = 1.7  10–3  1400 A1\n= 2.4 rad\n\n1(c)(ii) L = arc length / angle C1\n= 0.44 / 2.4\nor\nL = 0.44  (3600 / 1400) / 2\nL = 0.18 m A1\n\n1(c)(iii) a = r2 C1\n= 0.18  (1.745  10–3)2 A1\n= 5.5  10–7 m s–2\n\n1(d) centripetal acceleration is negligible compared with acceleration of free fall B1\nor\nnumerical comparison establishing answer to (c)(iii) ≪ 9.81\nresultant force is negligible compared with weight (of modelling clay) (so variation is negligible) B1\nor\nforce exerted by minute hand (approximately) equal (and opposite) to weight of modelling clay\n© UCLES 2023 Page 6 of 15", "source_pages": [ 6 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9702_w23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-Oct-Nov/9702_w23_ms_42.pdf?download=true", "html": "9702-topic-12-motion-in-a-circle/answers.html", "image_paths": [ "../answer-assets/9702_w23_ms_42-p06.png" ] }, { "id": "9702-2023-on-42-q02", "question_id": "9702-2023-on-42-q02", "subject": "9702", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 2, "topic": "Ideal gases", "topic_slug": "9702-topic-15-ideal-gases", "marks": 12, "status": "available", "reason": null, "text": "2(a)(i) work done per unit mass B1\nwork (done) moving mass from infinity (to the point) B1\n\n2(a)(ii)  = –GM / r C1\n= – (6.67  10–11  7.3  1022) / (1.7  106)\n= – 2.9  106 J kg–1 A1\n\n2(b)(i) E = m B1\nP\n\n2(b)(ii) ½mv2 + m = 0 M1\ncorrect algebra leading to v = √(–2) A1\n\n2(c) speed = √(2  2.9  106) A1\n= 2400 m s–1\n\n2(d) ½m = (3/2)kT C1\n3.34  10–27  = 3  1.38  10–23  400 C1\nc = 2200 m s–1 A1\nr.m.s.\n\n2(e) r.m.s. speed is an average so many molecules have speeds greater than the escape speed B1\nor\nthere is a distribution of molecular speeds (around the r.m.s. value) so many molecules have speeds greater than the\nescape speed\n© UCLES 2023 Page 7 of 15", "source_pages": [ 7 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9702_w23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-Oct-Nov/9702_w23_ms_42.pdf?download=true", "html": "9702-topic-15-ideal-gases/answers.html", "image_paths": [ "../answer-assets/9702_w23_ms_42-p07.png" ] }, { "id": "9702-2023-on-42-q03", "question_id": "9702-2023-on-42-q03", "subject": "9702", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 3, "topic": "Thermodynamics", "topic_slug": "9702-topic-16-thermodynamics", "marks": 8, "status": "available", "reason": null, "text": "3(a) sum of potential energy and kinetic energy (of particles) B1\n(total) energy of random motion of particles B1\n\n3(b)(i) no thermal energy transferred B1\nwork is done on the spring (increasing the potential energy of particles) M1\nso internal energy increases A1\n\n3(b)(ii) thermal energy transferred to water B1\nwork is done by water (expanding against atmosphere as it vaporises) B1\nmore thermal energy transferred than work done so internal energy increases B1\n© UCLES 2023 Page 8 of 15", "source_pages": [ 8 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9702_w23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-Oct-Nov/9702_w23_ms_42.pdf?download=true", "html": "9702-topic-16-thermodynamics/answers.html", "image_paths": [ "../answer-assets/9702_w23_ms_42-p08.png" ] }, { "id": "9702-2023-on-42-q04", "question_id": "9702-2023-on-42-q04", "subject": "9702", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 4, "topic": "Oscillations", "topic_slug": "9702-topic-17-oscillations", "marks": 11, "status": "available", "reason": null, "text": "4(a)(i) amplitude = ½  7.2  10–15 A1\n= 3.6  10–15 m\n\n4(a)(ii)  = 2 / (0.20  10–6) A1\n= 3.1  107 rad s–1\n\n4(a)(iii) v = x C1\n0 0\nv = 3.1  107  3.6  10–15 = 1.1  10–7 m s–1 A1\n0\n\n4(b)(i) I = nAv e C1\n0 0\n= 8.5  1028  4.3  10–4  1.1  10–7  1.60  10–19\n= 0.64 A A1\n\n4(b)(ii) sketch: two cycles of sinusoidal curve of amplitude I and period 0.20 s B1\n0\ncorrect phase, with I = +I at t = 0 B1\n0\n\n4(b)(iii) equation of form I = I cos t M1\n0\nvalue of I used matches answer to (b)(i) and value of  used matches answer to (a)(ii) A1\n0\n[if (a)(ii) and (b)(i) correct then I = 0.64 cos (3.1  107 t)]\n\n4(b)(iv) I = I / √2 A1\nr.m.s. 0\n= 0.64 / √2\n= 0.45 A\n© UCLES 2023 Page 9 of 15", "source_pages": [ 9 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9702_w23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-Oct-Nov/9702_w23_ms_42.pdf?download=true", "html": "9702-topic-17-oscillations/answers.html", "image_paths": [ "../answer-assets/9702_w23_ms_42-p09.png" ] }, { "id": "9702-2023-on-42-q05", "question_id": "9702-2023-on-42-q05", "subject": "9702", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 5, "topic": "Electric fields", "topic_slug": "9702-topic-18-electric-fields", "marks": 11, "status": "available", "reason": null, "text": "5(a) (electric) force is (directly) proportional to product of charges B1\n(electric) force (between point charges) is inversely proportional to the square of their separation B1\n\n5(b) F = Q2 / 4x2 C1\n0\n6.3  10–17 = Q2 / [4  8.85  10–12  (3.8  10–6)2]\ncharge = 3.2  10–19 C A1\n\n5(c)(i) negative B1\n\n5(c)(ii) four straight lines perpendicular to the plates, starting on one plate and finishing on the other B1\nlines equally spaced B1\narrows indicating direction downwards B1\n\n5(c)(iii) E = V / d C1\nmg = EQ C1\nmass = (1200  3.2  10–19) / (9.81  0.052) A1\n= 7.5  10–16 kg\n© UCLES 2023 Page 10 of 15", "source_pages": [ 10 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9702_w23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-Oct-Nov/9702_w23_ms_42.pdf?download=true", "html": "9702-topic-18-electric-fields/answers.html", "image_paths": [ "../answer-assets/9702_w23_ms_42-p10.png" ] }, { "id": "9702-2023-on-42-q06", "question_id": "9702-2023-on-42-q06", "subject": "9702", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 6, "topic": "Capacitance", "topic_slug": "9702-topic-19-capacitance", "marks": 9, "status": "available", "reason": null, "text": "6(a)(i) energy stored = area under graph C1\n= ½  450  10–6  8.0 = 1.8  10–3 J or 1.8 mJ A1\n\n6(a)(ii) C = Q / V or E = ½CV2 C1\nC = (450  10–6) / 8.0 or (2  1.8  10–3) / 8.02 A1\n= 5.6  10–5 F\n\n6(b)(i) V = V exp (– t / RC) and  = RC C1\n0\nV = V exp (– t / ) A1\n0\nV = 8.0 V, and at one time constant, t = \n0\nV / 8.0 = exp (– / ), so ln (V / 8.0) = –1.0 or –ln (V / 8.0) = 1.0\n\n6(b)(ii) [t read from graph at –ln (V / 8.0) = 1.0]:  = 3.2 s A1\n\n6(b)(iii)  = RC C1\nR = 3.2 / (5.6  10–5) A1\n= 5.7  104 \n© UCLES 2023 Page 11 of 15", "source_pages": [ 11 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9702_w23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-Oct-Nov/9702_w23_ms_42.pdf?download=true", "html": "9702-topic-19-capacitance/answers.html", "image_paths": [ "../answer-assets/9702_w23_ms_42-p11.png" ] }, { "id": "9702-2023-on-42-q07", "question_id": "9702-2023-on-42-q07", "subject": "9702", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 7, "topic": "Magnetic fields", "topic_slug": "9702-topic-20-magnetic-fields", "marks": 11, "status": "available", "reason": null, "text": "7(a)(i) V = BI / ntq A1\nH\n= (4.0  10–6  5.4) / (1.5  1016  1.8  10–3  1.60  10–19) = 5.0 V\n\n7(a)(ii) sketch: straight diagonal line from (0, 0) to t = 0.020 s B1\nand\nstraight diagonal line between two non-zero V values of same sign from t = 0.040 to 0.050 s\nH\nhorizontal straight line at V = 5.0 V from t = 0.020 to 0.040 s B1\nH\nhorizontal straight line at V = 2.5 V from t = 0.050 to 0.080 s B1\nH\n\n7(b)(i) e.m.f. = rate of change of (magnetic) flux (linkage) C1\nE = NA ΔB / Δt or E = NA  gradient (at t = 0.010 s) C1\nE = 3000  3.4  10–4  (4.0  10–6) / (0.020) = 2.0  10–4 V A1\n\n7(b)(ii) sketch: line showing non-zero E from t = 0 to t = 0.020 s and from t = 0.040 s to t = 0.050 s, and E = 0 at all other times B1\n‘top hats’ showing constant non-zero E from t = 0 to t = 0.020 s and from t = 0.040 s to t = 0.050 s B1\nmagnitude of E shown as 2.0  10–4 V in both non-zero sections B1\nsign of E in the t = 0 to t = 0.020 s region opposite to the sign of E in the t = 0.040 s to t = 0.050 s region B1\n© UCLES 2023 Page 12 of 15", "source_pages": [ 12 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9702_w23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-Oct-Nov/9702_w23_ms_42.pdf?download=true", "html": "9702-topic-20-magnetic-fields/answers.html", "image_paths": [ "../answer-assets/9702_w23_ms_42-p12.png" ] }, { "id": "9702-2023-on-42-q08", "question_id": "9702-2023-on-42-q08", "subject": "9702", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 8, "topic": "Quantum physics", "topic_slug": "9702-topic-22-quantum-physics", "marks": 8, "status": "available", "reason": null, "text": "8(a) packet / quantum of energy M1\nof electromagnetic radiation A1\n\n8(b)(i) photoelectric effect B1\n\n8(b)(ii) • electron needs a minimum energy to escape B3\nor\nelectron emitted if energy in packet is enough\n• energy must be absorbed in packets that are related to frequency\n• intensity relates to number of packets (not to energy in packet)\n• electron absorbs only a single whole packet\nAny three points, 1 mark each\n\n8(c)(i) Planck constant B1\n\n8(c)(ii) – work function (energy) B1\n© UCLES 2023 Page 13 of 15", "source_pages": [ 13 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9702_w23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-Oct-Nov/9702_w23_ms_42.pdf?download=true", "html": "9702-topic-22-quantum-physics/answers.html", "image_paths": [ "../answer-assets/9702_w23_ms_42-p13.png" ] }, { "id": "9702-2023-on-42-q09", "question_id": "9702-2023-on-42-q09", "subject": "9702", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 9, "topic": "Medical physics", "topic_slug": "9702-topic-24-medical-physics", "marks": 12, "status": "available", "reason": null, "text": "9(a)(i) material introduced into the body B1\nand\n(position in body) can be detected or absorbed by the tissue (being studied)\n\n9(a)(ii) X = + or e+ and P = 1 B1\nQ = 0 and R = 18 B1\n\n9(b)(i) positrons (emitted in the decay) and electrons annihilate B1\nmass of particles becomes energy of gamma photons B1\n\n9(b)(ii) arrival times of photons are processed B1\nimage built up of tracer concentration in the tissue B1\n\n9(c)(i) A = N and  = ln 2 / T C1\nN = n  N C1\nA\n2 photons produced from each decay, so R = 2    n  N A1\n0 A\nR = (2 ln 2) nN / T (allow 0.693 for ln 2)\n0 A\n\n9(c)(ii) sketch: exponential decay curve from t = 0 to t = 2T, starting at (0, R ) and with a negative gradient of continuously B1\n0\ndecreasing magnitude\nline with negative gradient passing through (T, R / 2) and (2T, R / 4) B1\n0 0\n© UCLES 2023 Page 14 of 15", "source_pages": [ 14 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9702_w23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-Oct-Nov/9702_w23_ms_42.pdf?download=true", "html": "9702-topic-24-medical-physics/answers.html", "image_paths": [ "../answer-assets/9702_w23_ms_42-p14.png" ] }, { "id": "9702-2023-on-42-q10", "question_id": "9702-2023-on-42-q10", "subject": "9702", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 10, "topic": "Astronomy and cosmology", "topic_slug": "9702-topic-25-astronomy-and-cosmology", "marks": 8, "status": "available", "reason": null, "text": "10(a) temperature inversely proportional to wavelength M1\ntemperature is thermodynamic temperature of surface, and wavelength is the wavelength at which maximum emission rate A1\noccurs\n\n10(b)(i) (astronomical) object of known luminosity B1\n\n10(b)(ii) star / galaxy is moving away from the student B1\n\n10(b)(iii) one tick placed in correct column in each row: B1\nwavelength: too high\nsurface temperature: too low B1\ndistance: unchanged B1\nradius: too high B1\n© UCLES 2023 Page 15 of 15", "source_pages": [ 15 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9702_w23_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-Oct-Nov/9702_w23_ms_42.pdf?download=true", "html": "9702-topic-25-astronomy-and-cosmology/answers.html", "image_paths": [ "../answer-assets/9702_w23_ms_42-p15.png" ] }, { "id": "9702-2023-on-43-q01", "question_id": "9702-2023-on-43-q01", "subject": "9702", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 1, "topic": "Electric fields", "topic_slug": "9702-topic-18-electric-fields", "marks": 10, "status": "available", "reason": null, "text": "1(a)(i) direction of the force acting on a (test) mass placed at the point B1\n\n1(a)(ii) change in height negligible compared with radius (of Earth) B1\n(so) field lines are (effectively) parallel B1\n\n1(b)(i) Y = GM / R2 M1\nG is the gravitational constant A1\n\n1(b)(ii) gravitational force is (always) attractive B1\nor\ngravitational force (always) acts towards the centre of the sphere\nforce is in opposite direction to displacement B1\nor\nat a point to the right of the centre, force acts to the left\nor\nat a point to the left of the centre, force acts to the right\n\n1(b)(iii) sketch: smooth curve with decreasing positive gradient, starting at (R, –Y) and reaching 3R with g still negative B1\nor\nsmooth curve with increasing positive gradient, ending at (–R, Y) and reaching –3R with g still positive\nboth of the above curves, in correct quadrants B1\ncurve passing through (2R, 0.25Y) and (3R, 0.11Y) B1\n© UCLES 2023 Page 7 of 16", "source_pages": [ 7 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9702_w23_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-Oct-Nov/9702_w23_ms_43.pdf?download=true", "html": "9702-topic-18-electric-fields/answers.html", "image_paths": [ "../answer-assets/9702_w23_ms_43-p07.png" ] }, { "id": "9702-2023-on-43-q02", "question_id": "9702-2023-on-43-q02", "subject": "9702", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 2, "topic": "Thermodynamics", "topic_slug": "9702-topic-16-thermodynamics", "marks": 12, "status": "available", "reason": null, "text": "2(a) (thermal) energy per unit mass (to change temperature) B1\n(thermal) energy per unit change in temperature B1\n\n2(b)(i) work done = pV A1\n= (2.0  105)  (0.063 – 0.038) = 5000 J\n\n2(b)(ii) gas is expanding (against external pressure) B1\ngas does work / work is done by gas, so (work done on gas is) negative B1\n\n2(b)(iii) U = q + W C1\n7600 = q + (–5000) A1\nq = 12 600 J\n\n2(b)(iv) specific heat capacity = q / mT C1\n= 12600 / (0.35  56)\n= 640 J kg–1 K–1 A1\n\n2(c) same gain in internal energy so same temperature rise B1\nno change in volume so no work done B1\nor\nno work done so less thermal energy needed (for same change in internal energy)\nless thermal energy needed (for same temperature change) so lower specific heat capacity B1\n© UCLES 2023 Page 8 of 16", "source_pages": [ 8 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9702_w23_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-Oct-Nov/9702_w23_ms_43.pdf?download=true", "html": "9702-topic-16-thermodynamics/answers.html", "image_paths": [ "../answer-assets/9702_w23_ms_43-p08.png" ] }, { "id": "9702-2023-on-43-q03", "question_id": "9702-2023-on-43-q03", "subject": "9702", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 3, "topic": "Ideal gases", "topic_slug": "9702-topic-15-ideal-gases", "marks": 13, "status": "available", "reason": null, "text": "3(a)(i) N: number of molecules (of the gas) B1\nm: mass of one molecule (of the gas) B1\n: mean square speed (of molecules) B1\n\n3(a)(ii) pV = NkT M1\nNkT = ⅓Nm and E = ½m leading to E = (3/2) kT A1\nK K\n\n3(b) ½  3.34  10–27  93002 = (3/2)  1.38  10–23  T C1\nT = 6980 K A1\n\n3(c)(i) L = F  4d2 C1\nL = 2.52  10–8  4  (4.16  1016)2 A1\n= 5.48  1026 W\n\n3(c)(ii) L = 4r2T4 C1\n5.48  1026 = 4  5.67  10–8  r2  69804\nr = 5.69  108 m A1\n\n3(d) (very high pressure so) molecules are (very) close together (not just ‘nearer’) B1\nforces between molecules are not negligible B1\nor\nvolume of molecules not negligible compared with gas volume\n© UCLES 2023 Page 9 of 16", "source_pages": [ 9 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9702_w23_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-Oct-Nov/9702_w23_ms_43.pdf?download=true", "html": "9702-topic-15-ideal-gases/answers.html", "image_paths": [ "../answer-assets/9702_w23_ms_43-p09.png" ] }, { "id": "9702-2023-on-43-q04", "question_id": "9702-2023-on-43-q04", "subject": "9702", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 4, "topic": "Oscillations", "topic_slug": "9702-topic-17-oscillations", "marks": 9, "status": "available", "reason": null, "text": "4(a) (motion in which) acceleration is (directly) proportional to displacement B1\n(motion in which): B1\nacceleration is (always) in the opposite direction to displacement\nor\nacceleration is (always) directed towards a fixed point\n\n4(b)(i)  = 2 / T C1\n = 2 / 3.0 A1\n= 2.1 rad s–1\n\n4(b)(ii) E = ½m2x 2 C1\n0\n= ½  0.81  2.12  0.0362 A1\n= 2.3  10–3 J\n\n4(c) sketch: line starting at (0, 0.036) and not reaching x =  0.036 m at any other time B1\nsmooth curve, with no sudden changes in gradient, showing continuously decreasing magnitude of x from B1\nmaximum displacement at t = 0 to final displacement of zero where the gradient is also zero\ndisplacement reaches final value of zero between t = 0.75 s and t = 3.0 s at the latest B1\n© UCLES 2023 Page 10 of 16", "source_pages": [ 10 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9702_w23_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-Oct-Nov/9702_w23_ms_43.pdf?download=true", "html": "9702-topic-17-oscillations/answers.html", "image_paths": [ "../answer-assets/9702_w23_ms_43-p10.png" ] }, { "id": "9702-2023-on-43-q05", "question_id": "9702-2023-on-43-q05", "subject": "9702", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 5, "topic": "Electric fields", "topic_slug": "9702-topic-18-electric-fields", "marks": 8, "status": "available", "reason": null, "text": "5(a) work done per unit charge B1\nwork (done on charge) moving positive charge from infinity (to the point) B1\n\n5(b) Any three points from: B3\nUp to 2 points from:\n• radius of sphere X is 2.0 m\n• radius of sphere Y is 4.0 m\n• radius of Y is double the radius of X\nUp to 2 points from:\n• charge on X is negative\n• charge on Y is positive\n• spheres carry opposite charges\nUp to 1 point from:\n• magnitudes of charges on the spheres are equal\n\n5(c) particle is attracted to X or repelled from Y B1\nor\nresultant force on particle is towards X / away from Y / to the left\nparticle accelerates towards X / away from Y / to the left B1\n(magnitude of) acceleration of particle increases B1\n© UCLES 2023 Page 11 of 16", "source_pages": [ 11 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9702_w23_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-Oct-Nov/9702_w23_ms_43.pdf?download=true", "html": "9702-topic-18-electric-fields/answers.html", "image_paths": [ "../answer-assets/9702_w23_ms_43-p11.png" ] }, { "id": "9702-2023-on-43-q06", "question_id": "9702-2023-on-43-q06", "subject": "9702", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 6, "topic": "Magnetic fields", "topic_slug": "9702-topic-20-magnetic-fields", "marks": 12, "status": "available", "reason": null, "text": "6(a) • force per unit length B2\n• force per unit current\n• length / current perpendicular to field\n1 mark for any two points, 2 marks for all three points\n\n6(b)(i) into the page B1\n\n6(b)(ii) F = Bqv C1\n= 4.8  10–3  1.6  10–19  1.7  107 = 1.3  10–14 N A1\n\n6(b)(iii) arrow at point X pointing down the page B1\n\n6(b)(iv) F = mv2 / r C1\n1.3  10–14 = (9.11  10–31)  (1.7  107)2 / r C1\n(r = 0.020 m) A1\nd = 2r\nd = 0.040 m\n\n6(c) path shows upwards deflection such that the curvature is always anticlockwise within the field B1\ncircular path with larger radius B1\nline enters field at X and leaves field at distance 2d vertically from X B1\n© UCLES 2023 Page 12 of 16", "source_pages": [ 12 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9702_w23_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-Oct-Nov/9702_w23_ms_43.pdf?download=true", "html": "9702-topic-20-magnetic-fields/answers.html", "image_paths": [ "../answer-assets/9702_w23_ms_43-p12.png" ] }, { "id": "9702-2023-on-43-q07", "question_id": "9702-2023-on-43-q07", "subject": "9702", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 7, "topic": "Ideal gases", "topic_slug": "9702-topic-15-ideal-gases", "marks": 8, "status": "available", "reason": null, "text": "7(a)(i) P = I 2R A1\n0\n\n7(a)(ii) P = 4I 2R A1\n0\n\n7(b) sketch: square wave of period T, with P always non-zero B1\nhorizontal lines, from 0 to 0.5T and from 1.0T to 1.5T, all at the same level that the scale indicates to be I 2R B1\n0\nhorizontal lines, from 0.5T to 1.0T and from 1.5T to 2.0T, at a level that is four times higher than the lower lines B1\n\n7(c)(i)

= (5/2)I 2R A1\n0\n\n7(c)(ii)

= I 2R C1\nr.m.s.\nI 2R = (5/2)I 2R A1\nr.m.s. 0\nI = √(5/2) I\nr.m.s. 0\n© UCLES 2023 Page 13 of 16", "source_pages": [ 13 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9702_w23_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-Oct-Nov/9702_w23_ms_43.pdf?download=true", "html": "9702-topic-15-ideal-gases/answers.html", "image_paths": [ "../answer-assets/9702_w23_ms_43-p13.png" ] }, { "id": "9702-2023-on-43-q08", "question_id": "9702-2023-on-43-q08", "subject": "9702", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 8, "topic": "Quantum physics", "topic_slug": "9702-topic-22-quantum-physics", "marks": 10, "status": "available", "reason": null, "text": "8(a)(i) p = E / c M1\nE = hc /  and completion of algebra leading to p = h /  A1\n\n8(a)(ii) wavelength = (6.63  10–34) / (9.5  10–28) = 700  10–9 m so red B1\n\n8(b)(i) power = intensity  area C1\nnumber per unit time = (160  2.5  10–6) / (9.5  10–28  3.00  108) = 1.4  1015 s–1 A1\n\n8(b)(ii) pressure = force / area C1\nforce = rate of change of momentum C1\n= 2  9.5  10–28  1.4  1015\npressure = (2  9.5  10–28  1.4  1015) / (2.5  10–6) A1\n= 1.1  10–6 Pa\n\n8(c) photons have greater momentum B1\nor\nfewer photons per unit time\ngreater photon momentum but smaller number of photons (per unit time) so pressure is the same B1\n© UCLES 2023 Page 14 of 16", "source_pages": [ 14 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9702_w23_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-Oct-Nov/9702_w23_ms_43.pdf?download=true", "html": "9702-topic-22-quantum-physics/answers.html", "image_paths": [ "../answer-assets/9702_w23_ms_43-p14.png" ] }, { "id": "9702-2023-on-43-q09", "question_id": "9702-2023-on-43-q09", "subject": "9702", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 9, "topic": "Nuclear physics", "topic_slug": "9702-topic-23-nuclear-physics", "marks": 10, "status": "available", "reason": null, "text": "9(a) (two small) nuclei join together M1\nto form one larger nucleus A1\n\n9(b) line with a peak at A  56 B1\nline with steep initial positive gradient on the left of peak and shallower negative gradient at all points to the right of peak B1\nand line does not return to 0 binding energy\n\n9(c)(i) X shown at value of A to the right of the peak B1\n\n9(c)(ii) Y shown at value of A close to 1 B1\n\n9(d) energy from 1 nucleus = (1.77  1013) / (6.02  1023) C1\n( = 2.94  10–11 J)\nbinding energy of Z = [(1.25 + 1.81)  10–10] – 2.94  10–11 C1\n( = 2.77  10–10 J)\nnucleon number of Z = 93 + 139 + 2 – 1 C1\n( = 233)\nbinding energy per nucleon = (2.77  10–10) / (233  1.60  10–13) A1\n= 7.43 MeV\n© UCLES 2023 Page 15 of 16", "source_pages": [ 15 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9702_w23_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-Oct-Nov/9702_w23_ms_43.pdf?download=true", "html": "9702-topic-23-nuclear-physics/answers.html", "image_paths": [ "../answer-assets/9702_w23_ms_43-p15.png" ] }, { "id": "9702-2023-on-43-q10", "question_id": "9702-2023-on-43-q10", "subject": "9702", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 10, "topic": "Medical physics", "topic_slug": "9702-topic-24-medical-physics", "marks": 8, "status": "available", "reason": null, "text": "10(a) ultrasound production: vibrating quartz crystal B1\nX-ray production: electrons hitting metal target B1\nultrasound detected wave: reflected B1\nX-ray detected wave: transmitted B1\n\n10(b)(i) I = I exp (–x) C1\n0\nln (0.72) = –6.2 A1\n = 0.053 cm–1\n\n10(b)(ii) I / I = exp (–9.3  0.053) C1\n0\n( = 0.61)\npercentage attenuated = 100  (1.00 – 0.61) A1\n= 39%\n© UCLES 2023 Page 16 of 16", "source_pages": [ 16 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9702_w23_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-Oct-Nov/9702_w23_ms_43.pdf?download=true", "html": "9702-topic-24-medical-physics/answers.html", "image_paths": [ "../answer-assets/9702_w23_ms_43-p16.png" ] }, { "id": "9702-2023-on-51-q01", "question_id": "9702-2023-on-51-q01", "subject": "9702", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "51", "question_number": 1, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "1 Defining the problem.\nf is the independent variable and E is the dependent variable or vary f and measure E 1\nkeep V and R constant 1\nMethods of data collection\nlabelled diagram of workable experiment including: 1\n• coils C and D placed with their axes on a straight line\n• separate workable circuit for coil D\n• (a.c.) voltmeter or oscilloscope connected across coil D\n(Do not accept a power supply connected to coil D.)\na.c. power supply/signal generator connected to coil C 1\nworkable circuit for coil C with power supply and (a.c.) voltmeter/oscilloscope in parallel with resistor and coil C 1\nmethod to determine f, e.g. read from signal generator or use of oscilloscope 1\nMethod of Analysis\nplot a graph of lg E against lg f or equivalent (e.g. ln E against ln f) 1\nq = gradient 1\nR 1\np= 10y-intercept\nV\nR\n(for ln E against ln f: p= ey-intercept)\nV\n© UCLES 2023 Page 5 of 10\n\n1 Additional detail including safety considerations 6\nD1 precaution (to prevent burns) from hot coils/hot resistor, e.g. use gloves to handle hot coil/resistor, switch off circuit\nand wait for hot coil/resistor to cool\nD2 keep the number of turns on each coil constant\nD3 keep distance between the coils constant\nD4 workable circuit diagram to determine R. e.g. circuit with ammeter connected in series and voltmeter in parallel with\nresistor\nor\nresistor connected to ohmmeter only\nD5 determination of resistance R:\npotential difference across R ÷ current in R\nor\nuse ohmmeter to measure R\nD6 method to keep distance between the coils constant, e.g. fix/clamp coils to bench\nD7 method to determine f from oscilloscope, e.g. period T = time-base  horizontal distance and f = 1 / T\nD8 method to determine V or E from oscilloscope, e.g. V or E = y-gain  vertical distance\nD9 method to increase E e.g. use iron core, place coils closer, increase V, decrease R\npV \nD10 relationship valid if a straight line is produced (passing through log )\n \n R \nDo not accept line passing through the origin.\n© UCLES 2023 Page 6 of 10", "source_pages": [ 5, 6 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9702_w23_ms_51.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-Oct-Nov/9702_w23_ms_51.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_w23_ms_51-p05.png", "../answer-assets/9702_w23_ms_51-p06.png" ] }, { "id": "9702-2023-on-51-q02", "question_id": "9702-2023-on-51-q02", "subject": "9702", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "51", "question_number": 2, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "2(a) 2g 1\ngradient =\nuZ\n2g\ny-intercept =\nu\n\n2(b) 1\n1\n1 −\n/cm 2\nh\n0.218 or 0.2182\n0.237 or 0.2370\n0.248 or 0.2485\n0.262 or 0.2617\n0.282 or 0.2817\n0.313 or 0.3131\nValues correct as shown above.\n1 1\nUncertainties in from ± 0.001 to ± 0.003.\nh\n\n2(c)(i) Six points from (b) plotted correctly. 1\nMust be within half a small square. Diameter of points must be less than half a small square.\n1 1\nError bars in plotted correctly.\nh\nAll error bars to be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.\n© UCLES 2023 Page 7 of 10\n\n2(c)(ii) Straight line of best fit drawn. 1\nDo not accept line from top point to bottom point.\nPoints must be balanced.\nLine must pass between (605, 0.230) and (615, 0.230) and between (845, 0.300) and (855, 0.300)\nWorst acceptable line drawn (steepest or shallowest possible line that passes through all the error bars). 1\nAll error bars must be plotted.\n\n2(c)(iii) Gradient determined with clear substitution of data points into y / x. 1\nDistance between data points must be greater than half the length of the drawn line.\nGradient determined of worst acceptable line. 1\nuncertainty = (gradient of line of best fit – gradient of worst acceptable line)\nor\nuncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)\n\n2(c)(iv) y-intercept determined by substitution of correct point with consistent power of ten into y = mx + c. 1\ny-intercept of worst acceptable line determined by substitution into y = mx + c. 1\nuncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line\nor\nuncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)\nDo not accept ECF from false origin method.\n© UCLES 2023 Page 8 of 10\n\n2(d)(i) u determined using y-intercept and u and Z given to 2, 3 or 4 significant figures. 1\n2981 44.29\nu = =\ny-intercept (c)(iv)\nZ determined using gradient with method shown and u and Z given with SI units with appropriate powers of ten. 1\n2981 44.29 y-intercept (c)(iv)\nZ = = or Z = =\nu gradient u(c)(iii) gradient (c)(iii)\n\n2(d)(ii) Percentage uncertainty in Z with method shown. 1\ny-intercept gradient\npercentage uncertainty in Z = + 100\n y-intercept gradient \nor\nCorrect substitution for u and\nu gradient\npercentage uncertainty in Z = + 100\n u gradient \nor\nCorrect substitution for max/min methods.\n© UCLES 2023 Page 9 of 10\n\n2(e) M determined to a minimum of 2 significant figures from (c)(iii) and (c)(iv) or (d)(i) with correct substitution. 1\n 1 \n−y-intercept\n \n 25 \nM =\ngradient\nor\nuZ uZ\nM = −Z = −Z\n2gh 221.5\n© UCLES 2023 Page 10 of 10", "source_pages": [ 7, 8, 9, 10 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9702_w23_ms_51.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-Oct-Nov/9702_w23_ms_51.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_w23_ms_51-p07.png", "../answer-assets/9702_w23_ms_51-p08.png", "../answer-assets/9702_w23_ms_51-p09.png", "../answer-assets/9702_w23_ms_51-p10.png" ] }, { "id": "9702-2023-on-52-q01", "question_id": "9702-2023-on-52-q01", "subject": "9702", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "52", "question_number": 1, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "1 Defining the problem\nV is the independent variable and z is the dependent variable or vary V and measure z 1\nkeep h constant 1\nMethods of data collection\nlabelled diagram of workable experiment including: 1\n• pulley supported by stand\n• stand placed on surface/bench/floor\n• minimum of two labels from stand, beaker, oil, surface/bench/floor, pulley, string\nuse (metre) rule to measure h or (metre) rule correctly positioned with h marked on diagram 1\nuse measuring cylinder to measure V 1\ntiming method to measure time t of fall of beaker to determine z 1\ne.g. use timer/stopwatch or use light gate(s) connected to a timer/data logger\n© UCLES 2023 Page 5 of 11\n\n1 Method of Analysis\n\n1 1 1 1 1\nplot a graph of against or equivalent (e.g. against )\nz2 V V z2\nDo not accept logarithms.\n\n1 1\nb=\n2hy-intercept\n\n1 1 Mgradient gradient\n(for against :b= or b=− )\nV z2 ah 2hy-intercept\nM 2My-intercept 1\na= or a=\nbhgradient gradient\n\n1 1\n(for against : a=−2My-intercept)\nV z2\n© UCLES 2023 Page 6 of 11\n\n1 Additional detail including safety considerations 6\nD1 precaution linked to oil spillage, e.g. use of cushion/sand box/tray for falling beaker to land or use of bungs/lids on\nbeakers or use foam on bench/floor or use foam to prevent rising beaker hitting pulley\nD2 precaution linked to oil contact with skin e.g. use gloves to avoid contact with oil\nD3 keep M constant\nD4 use a (top-pan) balance to measure M\nD5 method to keep h constant e.g. use a fiducial mark to release the beaker from the same position or release from the\nsame position on the clamped rule each time\nD6 equation to determine z for method used, e.g.\nfor timing h, z = 2h / t\nor\nfor one light gate, z = L / t where L is the length of the interrupted beam\nor\nfor two light gates, z = distance between light gates / t\nDo not accept h / t.\nD7 additional detail on diagram to measure h, e.g. clamp (metre) rule with stand on surface or use of set squares\npositioned on the surface to side of rule or spirit level positioned to side of rule\nD8 use large value of h to increase time of fall of beaker\nD9 repeat measurements of z for the same V and average z\n 1 \nD10 relationship valid if a straight line is produced (passing through )\n \n2bh\nDo not accept line passing through the origin.\n© UCLES 2023 Page 7 of 11", "source_pages": [ 5, 6, 7 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9702_w23_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-Oct-Nov/9702_w23_ms_52.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_w23_ms_52-p05.png", "../answer-assets/9702_w23_ms_52-p06.png", "../answer-assets/9702_w23_ms_52-p07.png" ] }, { "id": "9702-2023-on-52-q02", "question_id": "9702-2023-on-52-q02", "subject": "9702", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "52", "question_number": 2, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "2(a) t 1\ngradient = −\nR\ny-intercept = ln I R\n0\n\n2(b) 1\n1 / C / 104 F–1 ln (V / V)\n0.91 or 0.909 0.896 or 0.8961\n0.76 or 0.758 1.012 or 1.0116\n0.63 or 0.633 1.115 or 1.1151\n0.61 or 0.606 1.131 or 1.1314\n0.48 or 0.482 1.253 or 1.2528\n0.36 or 0.357 1.348 or 1.3481\nValues correct as shown above.\nUncertainties in ln (V / V) from ± 0.021 or ± 0.020 to ± 0.010 or ± 0.013 1\n\n2(c)(i) Six points from (b) plotted correctly. 1\nMust be within half a small square. Diameter of points must be less than half a small square.\nError bars in ln (V / V) plotted correctly. 1\nAll error bars to be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.\n© UCLES 2023 Page 8 of 11\n\n2(c)(ii) Straight line of best fit drawn. 1\nDo not accept line from top point to bottom point.\nPoints must be balanced.\nLine must pass between (0.820, 0.95) and (0.845, 0.95) and between (0.400, 1.30) and (0.425, 1.30)\nWorst acceptable line drawn (steepest or shallowest possible line that passes through all the error bars). 1\nAll error bars must be plotted.\n\n2(c)(iii) Gradient determined with clear substitution of data points into y / x. 1\nDistance between data points must be greater than half the length of the drawn line.\nGradient must be negative.\nGradient determined of worst acceptable line. 1\nuncertainty = (gradient of line of best fit – gradient of worst acceptable line)\nor\nuncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)\n\n2(c)(iv) y-intercept determined by substitution of correct point with consistent power of ten in m and x into y = mx + c. 1\n© UCLES 2023 Page 9 of 11\n\n2(d)(i) R determined using gradient. 1\n30.0 30.0\nR =− =\ngradient (c)(iii)\nI determined using y-intercept with method shown. 1\n0\ney-intercept e(c)(iv)\nI = =\n0 R (d)(i)\nR and I determined correctly using gradient and y-intercept 1\n0\nand\nR and I given to 2 or 3 significant figures\n0\nand\nR and I given with SI units with appropriate powers of ten.\n0\nUnits:\nR:  or s F-1\nI : A or V F s–1 or V  –1\n0\n\n2(d)(ii) Percentage uncertainty in R with method shown. 1\nt gradient\npercentage uncertainty in R= + 100\n t gradient \nor\nCorrect substitution for max/min methods.\n© UCLES 2023 Page 10 of 11\n\n2(e) C determined to a minimum of 2 significant figures from (c)(iii) and (c)(iv) or (d)(i) with correct substitutions. 1\ngradient gradient\nC = or C =−\nlnV −y-intercept y-intercept−lnV\nor\nt t\nC =− or C =\nR(lnV −lnI R) R(lnI R −lnV)\n0 0\n© UCLES 2023 Page 11 of 11", "source_pages": [ 8, 9, 10, 11 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9702_w23_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-Oct-Nov/9702_w23_ms_52.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_w23_ms_52-p08.png", "../answer-assets/9702_w23_ms_52-p09.png", "../answer-assets/9702_w23_ms_52-p10.png", "../answer-assets/9702_w23_ms_52-p11.png" ] }, { "id": "9702-2023-on-53-q01", "question_id": "9702-2023-on-53-q01", "subject": "9702", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "53", "question_number": 1, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "1 Defining the problem.\nf is the independent variable and E is the dependent variable or vary f and measure E 1\nkeep V and R constant 1\nMethods of data collection\nlabelled diagram of workable experiment including: 1\n• coils C and D placed with their axes on a straight line\n• separate workable circuit for coil D\n• (a.c.) voltmeter or oscilloscope connected across coil D\n(Do not accept a power supply connected to coil D.)\na.c. power supply/signal generator connected to coil C 1\nworkable circuit for coil C with power supply and (a.c.) voltmeter/oscilloscope in parallel with resistor and coil C 1\nmethod to determine f, e.g. read from signal generator or use of oscilloscope 1\nMethod of Analysis\nplot a graph of lg E against lg f or equivalent (e.g. ln E against ln f) 1\nq = gradient 1\nR 1\np= 10y-intercept\nV\nR\n(for ln E against ln f: p= ey-intercept)\nV\n© UCLES 2023 Page 5 of 10\n\n1 Additional detail including safety considerations 6\nD1 precaution (to prevent burns) from hot coils/hot resistor, e.g. use gloves to handle hot coil/resistor, switch off circuit\nand wait for hot coil/resistor to cool\nD2 keep the number of turns on each coil constant\nD3 keep distance between the coils constant\nD4 workable circuit diagram to determine R. e.g. circuit with ammeter connected in series and voltmeter in parallel with\nresistor\nor\nresistor connected to ohmmeter only\nD5 determination of resistance R:\npotential difference across R ÷ current in R\nor\nuse ohmmeter to measure R\nD6 method to keep distance between the coils constant, e.g. fix/clamp coils to bench\nD7 method to determine f from oscilloscope, e.g. period T = time-base  horizontal distance and f = 1 / T\nD8 method to determine V or E from oscilloscope, e.g. V or E = y-gain  vertical distance\nD9 method to increase E e.g. use iron core, place coils closer, increase V, decrease R\npV \nD10 relationship valid if a straight line is produced (passing through log )\n \n R \nDo not accept line passing through the origin.\n© UCLES 2023 Page 6 of 10", "source_pages": [ 5, 6 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9702_w23_ms_53.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-Oct-Nov/9702_w23_ms_53.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_w23_ms_53-p05.png", "../answer-assets/9702_w23_ms_53-p06.png" ] }, { "id": "9702-2023-on-53-q02", "question_id": "9702-2023-on-53-q02", "subject": "9702", "year": 2023, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "53", "question_number": 2, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "2(a) 2g 1\ngradient =\nuZ\n2g\ny-intercept =\nu\n\n2(b) 1\n1\n1 −\n/cm 2\nh\n0.218 or 0.2182\n0.237 or 0.2370\n0.248 or 0.2485\n0.262 or 0.2617\n0.282 or 0.2817\n0.313 or 0.3131\nValues correct as shown above.\n1 1\nUncertainties in from ± 0.001 to ± 0.003.\nh\n\n2(c)(i) Six points from (b) plotted correctly. 1\nMust be within half a small square. Diameter of points must be less than half a small square.\n1 1\nError bars in plotted correctly.\nh\nAll error bars to be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.\n© UCLES 2023 Page 7 of 10\n\n2(c)(ii) Straight line of best fit drawn. 1\nDo not accept line from top point to bottom point.\nPoints must be balanced.\nLine must pass between (605, 0.230) and (615, 0.230) and between (845, 0.300) and (855, 0.300)\nWorst acceptable line drawn (steepest or shallowest possible line that passes through all the error bars). 1\nAll error bars must be plotted.\n\n2(c)(iii) Gradient determined with clear substitution of data points into y / x. 1\nDistance between data points must be greater than half the length of the drawn line.\nGradient determined of worst acceptable line. 1\nuncertainty = (gradient of line of best fit – gradient of worst acceptable line)\nor\nuncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)\n\n2(c)(iv) y-intercept determined by substitution of correct point with consistent power of ten into y = mx + c. 1\ny-intercept of worst acceptable line determined by substitution into y = mx + c. 1\nuncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line\nor\nuncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)\nDo not accept ECF from false origin method.\n© UCLES 2023 Page 8 of 10\n\n2(d)(i) u determined using y-intercept and u and Z given to 2, 3 or 4 significant figures. 1\n2981 44.29\nu = =\ny-intercept (c)(iv)\nZ determined using gradient with method shown and u and Z given with SI units with appropriate powers of ten. 1\n2981 44.29 y-intercept (c)(iv)\nZ = = or Z = =\nu gradient u(c)(iii) gradient (c)(iii)\n\n2(d)(ii) Percentage uncertainty in Z with method shown. 1\ny-intercept gradient\npercentage uncertainty in Z = + 100\n y-intercept gradient \nor\nCorrect substitution for u and\nu gradient\npercentage uncertainty in Z = + 100\n u gradient \nor\nCorrect substitution for max/min methods.\n© UCLES 2023 Page 9 of 10\n\n2(e) M determined to a minimum of 2 significant figures from (c)(iii) and (c)(iv) or (d)(i) with correct substitution. 1\n 1 \n−y-intercept\n \n 25 \nM =\ngradient\nor\nuZ uZ\nM = −Z = −Z\n2gh 221.5\n© UCLES 2023 Page 10 of 10", "source_pages": [ 7, 8, 9, 10 ], "source_pdf": "_source-pdfs/2023-Oct-Nov/ms/9702_w23_ms_53.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2023-Oct-Nov/9702_w23_ms_53.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_w23_ms_53-p07.png", "../answer-assets/9702_w23_ms_53-p08.png", "../answer-assets/9702_w23_ms_53-p09.png", "../answer-assets/9702_w23_ms_53-p10.png" ] }, { "id": "9702-2024-m-42-q01", "question_id": "9702-2024-m-42-q01", "subject": "9702", "year": 2024, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 1, "topic": "Gravitational fields", "topic_slug": "9702-topic-13-gravitational-fields", "marks": 10, "status": "available", "reason": null, "text": "1(a) (gravitational) potential is zero at infinity B1\n(gravitational force between two masses is attractive so)\neither work is done on a mass to move it away from another mass B1\nor work is done on a mass to move it to infinity\n\n1(b)(i) M = (–) gradient / G C1\ne.g. M = (1.76  108) / (3.0  10–8  6.67  10–11) = 8.8  1025 kg A1\n\n1(b)(ii) either GMm / r2 = mr2 and  = 2 / T C1\nor GMm / r2 = mv2 / r and v = 2r / T\nor GMm / r2 = 42mr / T2\nR3 = 6.67  10–11  8.8  1025  (0.72  24  60  60)2 / 42 C1\nR = 8.3  107 m A1\n\n1(b)(iii) E = (GMm / r) – ½mv2 C1\nkinetic energy = (½  1200  84002)\npotential energy = (–)[(6.67  10–11  8.8  1025  1200) / (8.3  107)] C1\nE = [(6.67  10–11  8.8  1025  1200) / (8.3  107)] A1\n– (½  1200  84002)\n= 4.3  1010 J\n© Cambridge University Press & Assessment 2024 Page 6 of 14", "source_pages": [ 6 ], "source_pdf": "_source-pdfs/2024-March/ms/9702_m24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-March/9702_m24_ms_42.pdf?download=true", "html": "9702-topic-13-gravitational-fields/answers.html", "image_paths": [ "../answer-assets/9702_m24_ms_42-p06.png" ] }, { "id": "9702-2024-m-42-q02", "question_id": "9702-2024-m-42-q02", "subject": "9702", "year": 2024, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 2, "topic": "Thermodynamics", "topic_slug": "9702-topic-16-thermodynamics", "marks": 11, "status": "available", "reason": null, "text": "2(a) total kinetic energy associated with random motion of molecules M1\nplus total potential energy (of molecules) but potential energy is zero A1\n\n2(b)(i) W = pV C1\n= 1.01  105  5.20  10–5 A1\n= (+)5.25 J\n\n2(b)(ii) V  T or V / T = constant C1\n1.24 / (273 + 20) = (1.24 + 0.520) / T A1\nT = 416 K\n\n2(b)(iii) c = Q / mT C1\n= 960 / (0.016  (416 – 293)) A1\n= 490 J kg–1 K–1\n\n2(c) no change in volume so no work is done (by the gas) B1\n(same temperature change so) same change in internal energy B1\nless thermal energy needs to be supplied so c is less B1\nQuestion Answer Marks", "source_pages": [ 7 ], "source_pdf": "_source-pdfs/2024-March/ms/9702_m24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-March/9702_m24_ms_42.pdf?download=true", "html": "9702-topic-16-thermodynamics/answers.html", "image_paths": [ "../answer-assets/9702_m24_ms_42-p07.png" ] }, { "id": "9702-2024-m-42-q03", "question_id": "9702-2024-m-42-q03", "subject": "9702", "year": 2024, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 3, "topic": "Oscillations", "topic_slug": "9702-topic-17-oscillations", "marks": 7, "status": "available", "reason": null, "text": "3(a) E = ½ m2x 2 C1\no\n2.2  10–4 = ½  24  10–3  (14  10–3 / 4)2  2 C1\n = 39 rads–1 A1\n© Cambridge University Press & Assessment 2024 Page 7 of 14\n\n3(b)(i) use of acceleration = 9.81 m s–2 C1\nx = 9.81 / 392\no\n= 6.4  10–3 m A1\n\n3(b)(ii) at top of oscillation B1\nany one point from: B1\nwhere the downward acceleration first exceeds free-fall acceleration\nwhere the greatest downwards acceleration occurs\nwhere the resultant force is the maximum downwards\nwhere the contact force is a minimum\nQuestion Answer Marks", "source_pages": [ 7, 8 ], "source_pdf": "_source-pdfs/2024-March/ms/9702_m24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-March/9702_m24_ms_42.pdf?download=true", "html": "9702-topic-17-oscillations/answers.html", "image_paths": [ "../answer-assets/9702_m24_ms_42-p07.png", "../answer-assets/9702_m24_ms_42-p08.png" ] }, { "id": "9702-2024-m-42-q04", "question_id": "9702-2024-m-42-q04", "subject": "9702", "year": 2024, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 4, "topic": "Alternating currents", "topic_slug": "9702-topic-21-alternating-currents", "marks": 10, "status": "available", "reason": null, "text": "4(a) combined capacitance of parallel capacitors = 30 (F) C1\ntotal capacitance = (1 / 45 + 1 / 30)–1 A1\n= 18 F\n\n4(b) E = ½CV2 C1\nE = ½  45  10–6 (9.62 – 8.02) A1\n= 6.3  10–4 J\n\n4(c)(i) gaps in circuit closed and correct symbol for capacitor shown in parallel with load resistor B1\n\n4(c)(ii) two correct pairs of values of t and V read off from within same discharge cycle, e.g. (5.0, 4.0) and (13.0, 3.2) C1\ncorrect substitution of values of V, V and t into V = V exp (–t / ) C1\n0 0\ne.g. 3.2 = 4.0 exp (–8.0 / )\n = 36 ms A1\n© Cambridge University Press & Assessment 2024 Page 8 of 14\n\n4(d)(i) 8.0 W A1\n\n4(d)(ii) 4.0 W A1\nQuestion Answer Marks", "source_pages": [ 8, 9 ], "source_pdf": "_source-pdfs/2024-March/ms/9702_m24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-March/9702_m24_ms_42.pdf?download=true", "html": "9702-topic-21-alternating-currents/answers.html", "image_paths": [ "../answer-assets/9702_m24_ms_42-p08.png", "../answer-assets/9702_m24_ms_42-p09.png" ] }, { "id": "9702-2024-m-42-q05", "question_id": "9702-2024-m-42-q05", "subject": "9702", "year": 2024, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 5, "topic": "Magnetic fields", "topic_slug": "9702-topic-20-magnetic-fields", "marks": 10, "status": "available", "reason": null, "text": "5(a) any 2 points from: B2\n• (angular) displacement\n• velocity\n• momentum\n• (centripetal) acceleration\n• (resultant) force\n\n5(b)(i) Bqv = mv2 / r M1\nv = 2r / T M1\ncompletion of algebra leading to B = 2m / qT A1\n\n5(b)(ii) B = (2  4  1.66  10–27) / (2  1.60  10–19  2.5  10–6) C1\n= 0.052 T A1\n\n5(b)(iii) either the same because T is independent of r B1\nor the same because B, q and m are unchanged\nor the same because both radius and speed have doubled\n\n5(b)(iv) qE = Bqv C1\nE = Bv = 0.052  1.1  106 A1\n= 5.7  104 N C–1\n© Cambridge University Press & Assessment 2024 Page 9 of 14", "source_pages": [ 9 ], "source_pdf": "_source-pdfs/2024-March/ms/9702_m24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-March/9702_m24_ms_42.pdf?download=true", "html": "9702-topic-20-magnetic-fields/answers.html", "image_paths": [ "../answer-assets/9702_m24_ms_42-p09.png" ] }, { "id": "9702-2024-m-42-q06", "question_id": "9702-2024-m-42-q06", "subject": "9702", "year": 2024, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 6, "topic": "Magnetic fields", "topic_slug": "9702-topic-20-magnetic-fields", "marks": 11, "status": "available", "reason": null, "text": "6(a)(i) non-zero horizontal straight line from X to Y B1\n\n6(a)(ii) constant flux density (inside coil) B1\neither (magnetic) flux linkage proportional to flux density B1\nor  = BAN and B, A and N are all constant\n\n6(a)(iii)  = BAN C1\n= 0.080  0.71  10–4  64 A1\n= 3.6  10–4 Wb\n\n6(a)(iv) sketch showing: B1\nE is zero from time 0 to time t and non-zero after time t\nE has constant non-zero magnitude between time t and time 4t B1\nE has non-zero value of one sign between time t and time 2t, and non-zero value of the opposite sign between time 2t and B1\ntime 4t\n\n6(b) current in spring creates a magnetic field around the spring B1\neither (magnetic) fields around adjacent turns interact to cause a force to be exerted (between the turns) B1\nor current in one turn interacts with (magnetic) field due to adjacent turns to cause force to be exerted (between the\nturns)\n(magnetic force) is attractive so distance (between turns) decreases B1\n© Cambridge University Press & Assessment 2024 Page 10 of 14", "source_pages": [ 10 ], "source_pdf": "_source-pdfs/2024-March/ms/9702_m24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-March/9702_m24_ms_42.pdf?download=true", "html": "9702-topic-20-magnetic-fields/answers.html", "image_paths": [ "../answer-assets/9702_m24_ms_42-p10.png" ] }, { "id": "9702-2024-m-42-q07", "question_id": "9702-2024-m-42-q07", "subject": "9702", "year": 2024, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 7, "topic": "Quantum physics", "topic_slug": "9702-topic-22-quantum-physics", "marks": 9, "status": "available", "reason": null, "text": "7(a) p = E / c C1\n= (3.11  10–19) / (3.00  108) A1\n= 1.04  10–27 N s\n\n7(b)(i) E = hf and c = f so C1\nenergy of one photon = hc / \n350  10–3 = N  (6.63  10–34  3.00  108) / (640  10–9)\nN = 1.1  1018 A1\n\n7(b)(ii) F = (change in) momentum / time M1\nClear use of p = E / c and t = E / P to complete the algebra and arrive at the final equation: A1\ne.g. F = [E / c] / [E / P] = P / c\n\n7(c)(i) maximum wavelength (of electromagnetic radiation) that causes electrons to be emitted (from surface of metal) B1\n\n7(c)(ii) work function = 2.26  1.60  10–19 (J) C1\nE = hc /  so\n(2.26  1.60  10–19) = (6.63  10–34  3.00  108) / \n0\n = 5.50  10–7 m A1\n0\nQuestion Answer Marks", "source_pages": [ 11 ], "source_pdf": "_source-pdfs/2024-March/ms/9702_m24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-March/9702_m24_ms_42.pdf?download=true", "html": "9702-topic-22-quantum-physics/answers.html", "image_paths": [ "../answer-assets/9702_m24_ms_42-p11.png" ] }, { "id": "9702-2024-m-42-q08", "question_id": "9702-2024-m-42-q08", "subject": "9702", "year": 2024, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 8, "topic": "Nuclear physics", "topic_slug": "9702-topic-23-nuclear-physics", "marks": 9, "status": "available", "reason": null, "text": "8(a) (minimum) energy required to separate the nucleons (of a nucleus) M1\nto infinity A1\n\n8(b)(i) 4 A1\n© Cambridge University Press & Assessment 2024 Page 11 of 14\n\n8(b)(ii) energy = (142  8.37) + (90  8.72) – (235  7.59) C1\n= 190 MeV A1\n\n8(b)(iii) either it has too many neutrons (for the number of protons) B1\nor its neutron to proton ratio is too high\n\n8(b)(iv) (when t = 6.0 s), N / N = 1 / 32 C1\no\neither C1\n(1 / 32) = exp (– ln2  6.0 / t )\n½\nt = 1.2 s A1\n½\nor (C1)\n32 / 2n = 1 so n = 5 (half-lives)\nt = 6.0 / 5 (A1)\n1/2\n= 1.2 s\nQuestion Answer Marks", "source_pages": [ 11, 12 ], "source_pdf": "_source-pdfs/2024-March/ms/9702_m24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-March/9702_m24_ms_42.pdf?download=true", "html": "9702-topic-23-nuclear-physics/answers.html", "image_paths": [ "../answer-assets/9702_m24_ms_42-p11.png", "../answer-assets/9702_m24_ms_42-p12.png" ] }, { "id": "9702-2024-m-42-q09", "question_id": "9702-2024-m-42-q09", "subject": "9702", "year": 2024, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 9, "topic": "Medical physics", "topic_slug": "9702-topic-24-medical-physics", "marks": 13, "status": "available", "reason": null, "text": "9(a)(i) eV = hc /  C1\n = (6.63  10–34  3.00  108) / (84  103  1.60  10–19) A1\n= 1.5  10–11 m\n\n9(a)(ii) either (some) kinetic energy (of electrons) is converted to thermal energy at target B1\nor some X-rays are absorbed by the target so its temperature increases\n(tungsten) has higher melting point so does not melt quickly / easily B1\n© Cambridge University Press & Assessment 2024 Page 12 of 14\n\n9(b) I = I exp (–t) C1\n0\n0.13 = [exp (–3.0x)]  [exp (–0.22x)] C1\n= exp (–3.22x)\nx = 0.63 cm A1\n\n9(c)(i) product of density and speed M1\nspeed of ultrasound in medium A1\n\n9(c)(ii) I / I = (7.8 – 1.7)2 / (7.8 + 1.7)2 C1\nR 0\n= 0.41\nfraction transmitted = 1.00 – 0.41 = 0.59\npercentage transmitted = 59% A1\n\n9(c)(iii) more than one boundary so more reflections B1\nsome ultrasound is attenuated in matter B1\nQuestion Answer Marks", "source_pages": [ 12, 13 ], "source_pdf": "_source-pdfs/2024-March/ms/9702_m24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-March/9702_m24_ms_42.pdf?download=true", "html": "9702-topic-24-medical-physics/answers.html", "image_paths": [ "../answer-assets/9702_m24_ms_42-p12.png", "../answer-assets/9702_m24_ms_42-p13.png" ] }, { "id": "9702-2024-m-42-q10", "question_id": "9702-2024-m-42-q10", "subject": "9702", "year": 2024, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 10, "topic": "Astronomy and cosmology", "topic_slug": "9702-topic-25-astronomy-and-cosmology", "marks": 10, "status": "available", "reason": null, "text": "10(a)(i) L = 4σr2T4 C1\n3.85  1026 = 4  5.67  10–8  r2  57804\nr = 6.96  108 m A1\n© Cambridge University Press & Assessment 2024 Page 13 of 14\n\n10(a)(ii) F = L / 4d2 C1\n= (3.85  1026) / (4  (1.50  1011)2)\n= 1.36  103 W m–2 A1\n\n10(a)(iii) line of same shape showing peak intensity at greater wavelength B1\nline of same shape showing lower peak intensity B1\n\n10(b)(i) 5 lines in same pattern shifted to longer wavelengths B1\n\n10(b)(ii)  /  = v / c C1\n = (21400 / 300000)  656\n= 46.8 nm\nwavelength = 656 + 46.8 A1\n= 703 nm\n\n10(b)(iii) (peak) wavelength too high so temperature too low B1\n© Cambridge University Press & Assessment 2024 Page 14 of 14", "source_pages": [ 13, 14 ], "source_pdf": "_source-pdfs/2024-March/ms/9702_m24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-March/9702_m24_ms_42.pdf?download=true", "html": "9702-topic-25-astronomy-and-cosmology/answers.html", "image_paths": [ "../answer-assets/9702_m24_ms_42-p13.png", "../answer-assets/9702_m24_ms_42-p14.png" ] }, { "id": "9702-2024-m-52-q01", "question_id": "9702-2024-m-52-q01", "subject": "9702", "year": 2024, "session": "March", "session_code": "m", "paper": 5, "variant": "52", "question_number": 1, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "1 Defining the problem\nL is the independent variable and f is the dependent variable, or vary L and measure f. 1\nKeep  constant 1\nMethods of data collection\nLabelled diagram of workable experiment including: 1\n• rod supported by string / elastic bands from a clamp\n• clamp attached to stand, with stand on bench\n• two labels from stand, clamp, hammer, microphone, rod, string.\nDiagram showing labelled microphone connected to labelled oscilloscope. 1\nMethod to measure L, e.g. use a metre rule 1\nMethod to measure mass (m) (of metal rod), e.g. use a (top-pan) balance 1\nMethod of Analysis\n\n1 1\nPlots a graph of log f against log L or equivalent e.g. log f against log\nL\nn = − gradient 1\n\n1\n(for log f against log : n = gradient)\nL\nE =4102y-intercept 1\n\n1\n(for lg f vs lg : E =4102y-intercept)\nL\n(for ln f against ln L; E =4e2y-intercept)\n© Cambridge University Press & Assessment 2024 Page 5 of 9\n\n1 Additional detail including safety considerations 6\nAny six from:\nD1 Precaution linked to falling rod, e.g. sand tray / cushion (in case rod falls) OR gently hit rod prevent rod falling\nD2 Method to determine area of rod (A) e.g. measure diameter (d) of rod using a micrometer / calipers\nD3 Repeat measurements of diameter along the length of rod / around the rod and average diameter\nm d2\nD4 Method to determine ρ from experimental method, e.g. = and A =\nAL 4\n4m\nor =\nd2L\nd m\nor r = and =", "source_pages": [ 5, 6 ], "source_pdf": "_source-pdfs/2024-March/ms/9702_m24_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-March/9702_m24_ms_52.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_m24_ms_52-p05.png", "../answer-assets/9702_m24_ms_52-p06.png" ] }, { "id": "9702-2024-m-52-q02", "question_id": "9702-2024-m-52-q02", "subject": "9702", "year": 2024, "session": "March", "session_code": "m", "paper": 5, "variant": "52", "question_number": 2, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "2 r2L\nD5 Perform experiment in a quiet room\nD6 Reasoned method to prevent rod hitting microphone, e.g. have a gap between rod and microphone / gently hit rod or\nmethod to obtain measurable signal from the microphone, e.g. use a cone to increase the sound detected by the\nmicrophone\nD7 Method to determine frequency from oscilloscope, e.g. T = time-base  (horizontal) length (of one wave)\nand f = 1/T\nD8 Method to reduce uncertainties e.g.\nuse large values of L to reduce (percentage) uncertainty in L\nor\nadjust time-base to display as few waves as possible or Z waves on oscilloscope and divide time by Z\nor\nwait for the wave(form) / frequency to stabilise (and reach resonance)\nD9 Repeat measurements of f for each value of L and average f\n© Cambridge University Press & Assessment 2024 Page 6 of 9\n\n2(a) 3 1\nGradient =\nE\n4Z\ny-intercept =\nE\n\n2(b) 1\n1\n/ A−1\nI\n4440 or 4444\n5410 or 5405\n6250 or 6250\n7140 or 7143\n8000 or 8000\n8700 or 8696\n1 1\nUncertainties in\nI\nFrom  90–110 to  360–400\n© Cambridge University Press & Assessment 2024 Page 7 of 9\n\n2(c)(i) Six points from (b) plotted correctly. 1\nMust be within half a small square. Diameter of points must be less than half a small square.\n1 1\nError bars in plotted correctly.\nI\nAll error bars to be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.\n\n2(c)(ii) Straight line of best fit drawn. 1\nDo not accept line from top plot to bottom plot.\nPoints must be balanced.\nLine must pass between (1.8, 5000) and (2.1, 5000) and between (7.2, 8500) and (7.5, 8500)\nWorst acceptable line drawn. 1\nSteepest or shallowest possible line that passes through all the error bars.\nAll error bars must be plotted.\n\n2(c)(iii) Gradient determined with clear substitution of data points into y / x; distance between data points must be greater than 1\nhalf the length of the drawn line.\nGradient determined of worst acceptable line with clear substitution of data points into y / x; 1\nuncertainty = (gradient of line of best fit – gradient of worst acceptable line)\nor\nuncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)\n\n2(c)(iv) y-intercept determined by substitution of correct point with consistent power of ten in m and x into y = mx + c 1\ny-intercept of worst acceptable line determined by substitution into y = mx + c 1\nuncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line, or\nuncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)\n© Cambridge University Press & Assessment 2024 Page 8 of 9\n\n2(d)(i) E determined using gradient and 1\nE and Z given to 2, 3 or 4 sf.\n3\nE =\ngradient\nZ determined using y-intercept and 1\nE and Z given with SI units with correct powers of ten\nEy-intercept 3  y-intercept\nZ = or Z =\n4 4  gradient\nUnit of E: V or A \nUnit of Z: \n\n2(d)(ii) Percentage uncertainty in Z with method shown. 1\n gradient y-intercept\n%uncertainty= + \n gradient y-intercept \nor\nCorrect substitution for max/min methods.\n\n2(e) R determined to a minimum of 2sf from (c)(iii) and (c)(iv) or (d)(i) with correct substitution and correct powers of ten. 1\n0.1 mA = 0.1  10–3 A and\n1\n− y-intercept\n0.1010−3\nR = or\ngradient\nE 4Z\nR = −\n30.1010−3 3\n© Cambridge University Press & Assessment 2024 Page 9 of 9", "source_pages": [ 6, 7, 8, 9 ], "source_pdf": "_source-pdfs/2024-March/ms/9702_m24_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-March/9702_m24_ms_52.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_m24_ms_52-p06.png", "../answer-assets/9702_m24_ms_52-p07.png", "../answer-assets/9702_m24_ms_52-p08.png", "../answer-assets/9702_m24_ms_52-p09.png" ] }, { "id": "9702-2024-mj-41-q01", "question_id": "9702-2024-mj-41-q01", "subject": "9702", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 1, "topic": "Gravitational fields", "topic_slug": "9702-topic-13-gravitational-fields", "marks": 10, "status": "available", "reason": null, "text": "1(a) work done per unit mass B1\nwork done moving mass from infinity (to the point) B1\n\n1(b)(i) potential is zero at infinity B1\nwork is done by (two) masses in moving them closer together B1\nor\nwork is done on (two) masses in moving them apart\n\n1(b)(ii) magnitude of potential shown as 4 B1\npotential negative and shown as a multiple of – [potential = –4 if fully correct] B1\n\n1(b)(iii) field strength at X:  / 4R A1\nfield strength at Y: 4 / R A1\npotential energy at X: –M A1\npotential energy at Y: –8M A1\n© Cambridge University Press & Assessment 2024 Page 6 of 15", "source_pages": [ 6 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9702_s24_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-May-June/9702_s24_ms_41.pdf?download=true", "html": "9702-topic-13-gravitational-fields/answers.html", "image_paths": [ "../answer-assets/9702_s24_ms_41-p06.png" ] }, { "id": "9702-2024-mj-41-q02", "question_id": "9702-2024-mj-41-q02", "subject": "9702", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 2, "topic": "Temperature", "topic_slug": "9702-topic-14-temperature", "marks": 7, "status": "available", "reason": null, "text": "2(a)(i) 0 K B1\n\n2(a)(ii) (measurement) depends on properties of the liquid B1\n\n2(b)(i)  resistivity varies with temperature B2\n variation with temperature is linear\n unique value of resistivity for each (different value of) temperature\nAny two points, 1 mark each\n\n2(b)(ii) thermometer has high heat capacity/specific heat capacity B1\nor\nenergy transfer needed for thermometer to reach correct temperature\nor\nthermometer takes time to reach the correct temperature\n\n2(b)(iii) thermocouple B1\n\n2(c) (variation is) inverse B1\nor\n(variation is) non-linear\n© Cambridge University Press & Assessment 2024 Page 7 of 15", "source_pages": [ 7 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9702_s24_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-May-June/9702_s24_ms_41.pdf?download=true", "html": "9702-topic-14-temperature/answers.html", "image_paths": [ "../answer-assets/9702_s24_ms_41-p07.png" ] }, { "id": "9702-2024-mj-41-q03", "question_id": "9702-2024-mj-41-q03", "subject": "9702", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 3, "topic": "Thermodynamics", "topic_slug": "9702-topic-16-thermodynamics", "marks": 12, "status": "available", "reason": null, "text": "3(a)(i) gas for which pV  T M1\nwhere T is thermodynamic temperature A1\n\n3(a)(ii) no intermolecular forces B1\n(so) potential energy is zero B1\n\n3(b)(i) pV = NkT C1\nN = (2.0  105  0.26) / (1.38  10–23  290) A1\n= 1.3  1025\n\n3(b)(ii) E = (3/2) kT C1\nK\nE = (3/2)  1.38  10–23  290 A1\nK\n= 6.0  10–21 J\n\n3(b)(iii) internal energy = total KE + PE of molecules B1\nor\nPE = 0 so internal energy = total KE of molecules\ninternal energy = 1.3  1025  6.0  10–21 A1\n= 7.8  104 J\n\n3(c) straight line with positive gradient B1\nline passing through the origin B1\n© Cambridge University Press & Assessment 2024 Page 8 of 15", "source_pages": [ 8 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9702_s24_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-May-June/9702_s24_ms_41.pdf?download=true", "html": "9702-topic-16-thermodynamics/answers.html", "image_paths": [ "../answer-assets/9702_s24_ms_41-p08.png" ] }, { "id": "9702-2024-mj-41-q04", "question_id": "9702-2024-mj-41-q04", "subject": "9702", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 4, "topic": "Oscillations", "topic_slug": "9702-topic-17-oscillations", "marks": 8, "status": "available", "reason": null, "text": "4(a) oscillation (of object) at maximum amplitude B1\nwhen driving frequency = natural frequency (of system) B1\n\n4(b)(i) light damping B1\n\n4(b)(ii) oscillations (of ball) lose energy B1\n(due to) resistive forces (acting on ball) B1\n\n4(b)(iii) frequency = 1 / 0.25 A1\n= 4.0 Hz\n\n4(c) curve showing a maximum amplitude at a single non-zero frequency B1\nsingle maximum amplitude shown at 4.0 Hz B1\n© Cambridge University Press & Assessment 2024 Page 9 of 15", "source_pages": [ 9 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9702_s24_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-May-June/9702_s24_ms_41.pdf?download=true", "html": "9702-topic-17-oscillations/answers.html", "image_paths": [ "../answer-assets/9702_s24_ms_41-p09.png" ] }, { "id": "9702-2024-mj-41-q05", "question_id": "9702-2024-mj-41-q05", "subject": "9702", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 5, "topic": "Magnetic fields", "topic_slug": "9702-topic-20-magnetic-fields", "marks": 13, "status": "available", "reason": null, "text": "5(a) force per unit charge B1\nforce on positive charge B1\n\n5(b)(i) four straight vertical parallel lines, approximately evenly spaced B1\narrows downwards B1\n\n5(b)(ii) E = V / d C1\nE = 430 / 0.067 A1\n= 6.4  103 N C–1\n\n5(b)(iii) smooth curve within plates and straight lines outside plates B1\ndirection of deflection shown as upwards B1\n\n5(c)(i) into the page B1\n\n5(c)(ii) forces are in opposite directions B1\n(undeviated) when (magnitudes of) forces are equal B1\n\n5(c)(iii) Eq = Bqv C1\nB = E / v = (6.4  103) / (2.6  107) A1\n= 2.5  10–4 T\n© Cambridge University Press & Assessment 2024 Page 10 of 15", "source_pages": [ 10 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9702_s24_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-May-June/9702_s24_ms_41.pdf?download=true", "html": "9702-topic-20-magnetic-fields/answers.html", "image_paths": [ "../answer-assets/9702_s24_ms_41-p10.png" ] }, { "id": "9702-2024-mj-41-q06", "question_id": "9702-2024-mj-41-q06", "subject": "9702", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 6, "topic": "Capacitance", "topic_slug": "9702-topic-19-capacitance", "marks": 10, "status": "available", "reason": null, "text": "6(a)  p.d. across capacitor proportional to charge on capacitor B2\n p.d. across capacitor = p.d. across resistor\n current in resistor proportional to p.d. across resistor\n current in resistor = rate of decrease of charge on capacitor\nAny two points, 1 mark each\ncharge proportional to current so rate of decrease of current decreases as current decreases (therefore exponential shape) B1\n\n6(b)(i) R = V / I C1\n= 12 / (0.13  10–3)\n= 9.2  104  A1\n\n6(b)(ii) correct read-off of at least one pair of values for I and t C1\nattempted read-off of t when I = 0.048 mA C1\nor\nsubstitution of a correct pair of values of I and t into I = 0.13 exp (– t / )\n = 4.3 s A1\n\n6(c)  = RC C1\nC =  / R = 4.3 / (9.2  104) A1\n= 4.7  10–5 F\n© Cambridge University Press & Assessment 2024 Page 11 of 15", "source_pages": [ 11 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9702_s24_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-May-June/9702_s24_ms_41.pdf?download=true", "html": "9702-topic-19-capacitance/answers.html", "image_paths": [ "../answer-assets/9702_s24_ms_41-p11.png" ] }, { "id": "9702-2024-mj-41-q07", "question_id": "9702-2024-mj-41-q07", "subject": "9702", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 7, "topic": "Temperature", "topic_slug": "9702-topic-14-temperature", "marks": 10, "status": "available", "reason": null, "text": "7(a) rectification (of the input voltage) M1\nfull-wave A1\n\n7(b)(i) P = V2 / R C1\nor\nmaximum V = 9.0 V\nP = 9.02 / 370 = 0.22 W A1\nMAX\n\n7(b)(ii) sinusoidal shape with minima sitting on the time axis B1\ncorrect frequency and phase, with minima at 0, 0.02, 0.04, 0.06 and 0.08 s and maxima at 0.01, 0.03, 0.05 and 0.07 s B1\nall maxima shown at 0.22 W B1\n\n7(b)(iii) mean power = peak power / 2 = 0.22 / 2 A1\n= 0.11 W\n\n7(c) power–time graph is identical B1\n(so) mean powers are equal B1\n© Cambridge University Press & Assessment 2024 Page 12 of 15", "source_pages": [ 12 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9702_s24_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-May-June/9702_s24_ms_41.pdf?download=true", "html": "9702-topic-14-temperature/answers.html", "image_paths": [ "../answer-assets/9702_s24_ms_41-p12.png" ] }, { "id": "9702-2024-mj-41-q08", "question_id": "9702-2024-mj-41-q08", "subject": "9702", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 8, "topic": "Medical physics", "topic_slug": "9702-topic-24-medical-physics", "marks": 13, "status": "available", "reason": null, "text": "8(a) packet / quantum of energy M1\nof electromagnetic radiation A1\n\n8(b)(i) electron(s) B1\n\n8(b)(ii) X labelled – and Y labelled + B1\n\n8(c)(i) 0.032 MeV A1\n\n8(c)(ii) momentum = E / c C1\nmomentum = (0.032 × 1.60  10–13) / (3.00  108) A1\n= 1.7  10–23 N s\n\n8(c)(iii) E = hf and  = c / f C1\n = hc / E C1\n= (6.63  10–34 × 3.00  108) / (0.032  1.60 × 10–13)\n = 3.9  10–11 m A1\n\n8(d) discussion of bone and soft tissue B1\ndiscussion of different attenuation (coefficients) B1\nor\ndiscussion differences in penetration / transmission / absorption\ntransmitted intensities (by bone and tissue) are very different (leading to good contrast images) B1\n© Cambridge University Press & Assessment 2024 Page 13 of 15", "source_pages": [ 13 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9702_s24_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-May-June/9702_s24_ms_41.pdf?download=true", "html": "9702-topic-24-medical-physics/answers.html", "image_paths": [ "../answer-assets/9702_s24_ms_41-p13.png" ] }, { "id": "9702-2024-mj-41-q09", "question_id": "9702-2024-mj-41-q09", "subject": "9702", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 9, "topic": "Nuclear physics", "topic_slug": "9702-topic-23-nuclear-physics", "marks": 8, "status": "available", "reason": null, "text": "9(a) time for activity (of sample) to halve B1\n\n9(b)(i) activity (of X at time t) B1\n\n9(b)(ii)  Y is a stable isotope B3\n total number of nuclei is constant\n half-life (of X) is 13.6 s\n decay constant (of X) is 0.051 s–1\n amount (of X) at t = 0 is 0.066 mol\n activity (of X) at t = 0 is 2.0  1021 Bq\nAny three points, 1 mark each\n\n9(c) mass of 1 nucleus = (7.3  10–4) / (4.0  1022) C1\nnucleon number = mass of nucleus / (1.66  10–27) C1\n= (7.3  10–4) / (4.0 × 1022  1.66 × 10–27) A1\n= 11 and given as an integer\n© Cambridge University Press & Assessment 2024 Page 14 of 15", "source_pages": [ 14 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9702_s24_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-May-June/9702_s24_ms_41.pdf?download=true", "html": "9702-topic-23-nuclear-physics/answers.html", "image_paths": [ "../answer-assets/9702_s24_ms_41-p14.png" ] }, { "id": "9702-2024-mj-41-q10", "question_id": "9702-2024-mj-41-q10", "subject": "9702", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 10, "topic": "Astronomy and cosmology", "topic_slug": "9702-topic-25-astronomy-and-cosmology", "marks": 9, "status": "available", "reason": null, "text": "10(a)(i) total power B1\npower radiated (by the star) B1\n\n10(a)(ii) standard candle has known luminosity B1\nradiant flux intensity measured by observer B1\n(distance calculated using) F = L / 4d2 B1\n\n10(b)(i) luminosity = 4 r2T4 C1\n= 4  5.67  10–8  (6.96  108)2 × 57804\n= 3.85 × 1026 W A1\n\n10(b)(ii)  T = constant C1\nMAX\ntemperature = (5780  501) / 624 A1\n= 4640 K\n© Cambridge University Press & Assessment 2024 Page 15 of 15", "source_pages": [ 15 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9702_s24_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-May-June/9702_s24_ms_41.pdf?download=true", "html": "9702-topic-25-astronomy-and-cosmology/answers.html", "image_paths": [ "../answer-assets/9702_s24_ms_41-p15.png" ] }, { "id": "9702-2024-mj-42-q01", "question_id": "9702-2024-mj-42-q01", "subject": "9702", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 1, "topic": "Motion in a circle", "topic_slug": "9702-topic-12-motion-in-a-circle", "marks": 10, "status": "available", "reason": null, "text": "1(a) angle (subtended at the centre of a circle) when arc (length) = radius B1\n\n1(b)(i) arrow, labelled V, pointing in NE direction B1\n\n1(b)(ii) arrow, labelled A, pointing in NW direction B1\n\n1(c)(i) v = r C1\n = 0.68 / (0.093 – 0.012) A1\n= 8.4 rad s–1\n\n1(c)(ii) a = v2 / r or a = r2 C1\na = 0.682 / (0.093 – 0.012) or (0.093 – 0.012)  8.42 A1\n= 5.7 m s–2\n\n1(d) angular speed: same for both pieces B1\nlinear speed: less for second piece than first piece B1\nacceleration: less for second piece than first piece B1\n© Cambridge University Press & Assessment 2024 Page 6 of 16", "source_pages": [ 6 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9702_s24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-May-June/9702_s24_ms_42.pdf?download=true", "html": "9702-topic-12-motion-in-a-circle/answers.html", "image_paths": [ "../answer-assets/9702_s24_ms_42-p06.png" ] }, { "id": "9702-2024-mj-42-q02", "question_id": "9702-2024-mj-42-q02", "subject": "9702", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 2, "topic": "Ideal gases", "topic_slug": "9702-topic-15-ideal-gases", "marks": 10, "status": "available", "reason": null, "text": "2(a) (if in thermal contact) no net transfer of (thermal) energy (between them) B1\n\n2(b)(i) pV = nRT C1\nT = (1.20  105  0.0260) / (0.740  8.31) M1\n( = 507 K)\ntemperature = 507 – 273 = 234 °C A1\n\n2(b)(ii) thermal equilibrium so temperatures (of X and Y) are equal B1\npV = NkT C1\nN = (2.90  105  0.0430) / (1.38  10–23  507) A1\n= 1.78  1024\n\n2(b)(iii)  (molecular) kinetic energy is proportional to temperature B2\nor\nkinetic energy (of molecules) is same in both cylinders\n kinetic energy proportional to mass  mean-square speed\nor\ntemperature proportional to mass  mean-square speed\nor\nr.m.s. speed proportional to √(temperature / mass)\n mean-square speed inversely proportional to mass\nor\nr.m.s. speed inversely proportional to √(mass)\nAny two bulleted points, 1 mark each\nr.m.s. speed (of molecules) in X is half r.m.s. speed (of molecules) in Y B1\n© Cambridge University Press & Assessment 2024 Page 7 of 16", "source_pages": [ 7 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9702_s24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-May-June/9702_s24_ms_42.pdf?download=true", "html": "9702-topic-15-ideal-gases/answers.html", "image_paths": [ "../answer-assets/9702_s24_ms_42-p07.png" ] }, { "id": "9702-2024-mj-42-q03", "question_id": "9702-2024-mj-42-q03", "subject": "9702", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 3, "topic": "Thermodynamics", "topic_slug": "9702-topic-16-thermodynamics", "marks": 8, "status": "available", "reason": null, "text": "3(a) sum of potential energy and kinetic energy B1\n(total) energy of random motion of particles B1\n\n3(b)(i) no change in separation so no change in (molecular) potential energy B1\ntemperature increases so kinetic energy (of molecules) increases B1\nkinetic energy increases and potential energy unchanged, so internal energy increases B1\n\n3(b)(ii) temperature constant so no change in (molecular) kinetic energy B1\nseparation increases so potential energy (of molecules) increases B1\npotential energy increases and kinetic energy unchanged, so internal energy increases B1\n© Cambridge University Press & Assessment 2024 Page 8 of 16", "source_pages": [ 8 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9702_s24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-May-June/9702_s24_ms_42.pdf?download=true", "html": "9702-topic-16-thermodynamics/answers.html", "image_paths": [ "../answer-assets/9702_s24_ms_42-p08.png" ] }, { "id": "9702-2024-mj-42-q04", "question_id": "9702-2024-mj-42-q04", "subject": "9702", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 4, "topic": "Oscillations", "topic_slug": "9702-topic-17-oscillations", "marks": 9, "status": "available", "reason": null, "text": "4(a) straight line through the origin shows that a is proportional to x B1\nnegative gradient shows that a and x are (always) in opposite directions B1\n\n4(b)(i) a = –2x A1\n = √(2A / 3Y)\n\n4(b)(ii) v = x C1\n0 0\n= 3Y  √(2A / 3Y) A1\n= √(6AY)\n\n4(b)(iii) E = ½ m2x 2 C1\n0\n= ½ m  (2A / 3Y)  (3Y)2 A1\n= 3mAY\n\n4(c)  = 2 / T C1\n( = 2 / 0.75)\nx = 1.8 sin (8.4 t) A1\n© Cambridge University Press & Assessment 2024 Page 9 of 16", "source_pages": [ 9 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9702_s24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-May-June/9702_s24_ms_42.pdf?download=true", "html": "9702-topic-17-oscillations/answers.html", "image_paths": [ "../answer-assets/9702_s24_ms_42-p09.png" ] }, { "id": "9702-2024-mj-42-q05", "question_id": "9702-2024-mj-42-q05", "subject": "9702", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 5, "topic": "Electric fields", "topic_slug": "9702-topic-18-electric-fields", "marks": 7, "status": "available", "reason": null, "text": "5(a) work done per unit charge B1\nwork (done) moving positive charge from infinity (to the point) B1\n\n5(b) Any three points from: B3\nUp to 2 points from:\n radius of sphere X is 0.30 m\n radius of sphere Y is 0.10 m\n radius of X is treble the radius of Y\nUp to 2 points from:\n charge on X is positive\n charge on Y is positive\n spheres X and Y carry charges of the same sign\nUp to 1 point from:\n (magnitudes of) charges on the spheres are equal\n charges on the spheres have the same magnitude\n\n5(c) proton remains at rest (in the position of release) M1\npotential energy of proton is (already) at its minimum A1\nor\n(electric) forces (from spheres) on proton are equal and opposite\nor\nno resultant (electric) force on proton\nor\nresultant electric field strength (at proton) is zero\n© Cambridge University Press & Assessment 2024 Page 10 of 16", "source_pages": [ 10 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9702_s24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-May-June/9702_s24_ms_42.pdf?download=true", "html": "9702-topic-18-electric-fields/answers.html", "image_paths": [ "../answer-assets/9702_s24_ms_42-p10.png" ] }, { "id": "9702-2024-mj-42-q06", "question_id": "9702-2024-mj-42-q06", "subject": "9702", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 6, "topic": "Capacitance", "topic_slug": "9702-topic-19-capacitance", "marks": 10, "status": "available", "reason": null, "text": "6(a) equal charge on both capacitors B1\nV + V = V M1\nX Y\n(Q / C ) + (Q / C ) = (Q / C ) leading to (1 / C ) + (1 / C ) = (1 / C ) A1\nX Y T X Y T\nor\n(V / Q) + (V / Q) = (V / Q) leading to (1 / C ) + (1 / C ) = (1 / C )\nX Y X Y T\n\n6(b)(i) E = ½CV 2 C1\nV = √[(2  2.5  10–3) / (200  10–6)] = 5.0 V A1\n\n6(b)(ii) total capacitance = 600 F C1\nE = ½  600  10–6  5.02 C1\n( = 7.5  10–3 J)\n= 7.5 mJ A1\n\n6(b)(iii) line with positive gradient starting at (0, 2.5) B1\nstraight line passing through (400, 7.5) B1\n© Cambridge University Press & Assessment 2024 Page 11 of 16", "source_pages": [ 11 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9702_s24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-May-June/9702_s24_ms_42.pdf?download=true", "html": "9702-topic-19-capacitance/answers.html", "image_paths": [ "../answer-assets/9702_s24_ms_42-p11.png" ] }, { "id": "9702-2024-mj-42-q07", "question_id": "9702-2024-mj-42-q07", "subject": "9702", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 7, "topic": "Magnetic fields", "topic_slug": "9702-topic-20-magnetic-fields", "marks": 14, "status": "available", "reason": null, "text": "7(a) (induced) e.m.f. is (directly) proportional to rate M1\nof change of (magnetic) flux (linkage) A1\n\n7(b)(i)  = BA C1\n= 7.2  10–3  3.2  10–4 A1\n= 2.3  10–6 Wb\n\n7(b)(ii) tangent drawn at steepest point on Fig. 7.2 C1\nevidence of multiplication by 340 C1\nmaximum rate of change of flux = 0.82 Wb s–1 A1\n\n7(b)(iii) V = 0.82 V A1\n0\nor\nV given as identical numerical answer to the answer in (b)(ii)\n0\n\n7(b)(iv) sinusoidal curve of period 2.0 ms from t = 0 to t = 6.0 ms B1\nall peaks at +V and all troughs at –V B1\n0 0\nline showing V = 0 at (and only at) t = 0, 1.0, 2.0, 3.0, 4.0, 5.0 and 6.0 ms B1\n\n7(b)(v) A = 0.82 V A1\nor\nA has same numerical value as answer in (b)(iii), with unit V\nB = 2 / (2.0  10–3) C1\n= 3100 rad s–1 A1\n© Cambridge University Press & Assessment 2024 Page 12 of 16", "source_pages": [ 12 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9702_s24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-May-June/9702_s24_ms_42.pdf?download=true", "html": "9702-topic-20-magnetic-fields/answers.html", "image_paths": [ "../answer-assets/9702_s24_ms_42-p12.png" ] }, { "id": "9702-2024-mj-42-q08", "question_id": "9702-2024-mj-42-q08", "subject": "9702", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 8, "topic": "Astronomy and cosmology", "topic_slug": "9702-topic-25-astronomy-and-cosmology", "marks": 12, "status": "available", "reason": null, "text": "8(a)(i) movement of star causes change in (observed) frequency B1\nor\nmovement of star causes redshift\nobserved frequency is lower (than emitted frequency) B1\n\n8(a)(ii) all three lines shown to left of corresponding printed lines B1\ndistance between drawn line and corresponding printed line approximately the same for all three lines B1\n\n8(b)(i) E = hf and  = c / f C1\nE = (6.63  10–34  3.00  108) / (488  10–9) A1\n= 4.08  10–19 J\n\n8(b)(ii) photon energy = (4.08  10–19) / (1.60  10–19) C1\n= 2.55 eV\nenergy level = –3.40 + 2.55 A1\n= –0.85 eV\n\n8(b)(iii)  =   (v / c) C1\n= (488  6.2  106) / (3.00  108)\n( = 10 nm)\nobserved wavelength = 488 +  = 488 + 10 A1\n= 498 nm\n© Cambridge University Press & Assessment 2024 Page 13 of 16\n\n8(c) v = H d C1\n0\nd = (6.2  106) / (2.3  10–18) A1\n= 2.7  1024 m\n© Cambridge University Press & Assessment 2024 Page 14 of 16", "source_pages": [ 13, 14 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9702_s24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-May-June/9702_s24_ms_42.pdf?download=true", "html": "9702-topic-25-astronomy-and-cosmology/answers.html", "image_paths": [ "../answer-assets/9702_s24_ms_42-p13.png", "../answer-assets/9702_s24_ms_42-p14.png" ] }, { "id": "9702-2024-mj-42-q09", "question_id": "9702-2024-mj-42-q09", "subject": "9702", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 9, "topic": "Nuclear physics", "topic_slug": "9702-topic-23-nuclear-physics", "marks": 13, "status": "available", "reason": null, "text": "9(a) energy required to separate (all) the nucleons (in the nucleus) M1\nto infinity A1\n\n9(b)(i) m = {[(84  1.007276) + (128  1.008665)] – 211.942749} (u) C1\n( = 1.778 u)\n= 1.778  1.66  10–27 (kg) C1\n= 2.95  10–27 kg A1\n\n9(b)(ii) E = ()mc2 C1\nbinding energy = 2.95  10–27  (3.00  108)2 A1\n= 2.66  10–10 J\n\n9(b)(iii) binding energy per nucleon = (2.66  10–10) / 212 A1\n= 1.25  10–12 J\n\n9(c)(i) line rising to a single peak that is to the left of the ‘9’ in the Fig. 9.1 label and then continually decreasing B1\nsteep positive gradient on the left of the peak and shallow negative gradient on the right B1\n\n9(c)(ii) X shown on the line at a value of A that is to the right of the left-hand edge of the ‘A’ in the axis label, and to the left of ‘2’ in B1\nthe 250 label\n\n9(c)(iii) nucleus formed (as a result of the decay) has a lower nucleon number B1\n(nucleus formed has a) greater binding energy per nucleon B1\n© Cambridge University Press & Assessment 2024 Page 15 of 16", "source_pages": [ 15 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9702_s24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-May-June/9702_s24_ms_42.pdf?download=true", "html": "9702-topic-23-nuclear-physics/answers.html", "image_paths": [ "../answer-assets/9702_s24_ms_42-p15.png" ] }, { "id": "9702-2024-mj-42-q10", "question_id": "9702-2024-mj-42-q10", "subject": "9702", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 10, "topic": "Medical physics", "topic_slug": "9702-topic-24-medical-physics", "marks": 7, "status": "available", "reason": null, "text": "10(a) time gives information about depth (of boundary) B1\nintensity gives information about nature of boundary B1\n\n10(b)(i) product of density and speed M1\nspeed of ultrasound in medium (and density of medium) A1\n\n10(b)(ii) Z = 1000  1420 (= 1.42  106 kg m–2 s–1) C1\nwater\nand\nZ = 2500  4560 (= 11.4  106 kg m–2 s–1)\nglass\nintensity reflection coefficient = (11.4 – 1.42)2 / (11.4 + 1.42)2 C1\n= 0.61 A1\n© Cambridge University Press & Assessment 2024 Page 16 of 16", "source_pages": [ 16 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9702_s24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-May-June/9702_s24_ms_42.pdf?download=true", "html": "9702-topic-24-medical-physics/answers.html", "image_paths": [ "../answer-assets/9702_s24_ms_42-p16.png" ] }, { "id": "9702-2024-mj-43-q01", "question_id": "9702-2024-mj-43-q01", "subject": "9702", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 1, "topic": "Gravitational fields", "topic_slug": "9702-topic-13-gravitational-fields", "marks": 10, "status": "available", "reason": null, "text": "1(a) work done per unit mass B1\nwork done moving mass from infinity (to the point) B1\n\n1(b)(i) potential is zero at infinity B1\nwork is done by (two) masses in moving them closer together B1\nor\nwork is done on (two) masses in moving them apart\n\n1(b)(ii) magnitude of potential shown as 4 B1\npotential negative and shown as a multiple of – [potential = –4 if fully correct] B1\n\n1(b)(iii) field strength at X:  / 4R A1\nfield strength at Y: 4 / R A1\npotential energy at X: –M A1\npotential energy at Y: –8M A1\n© Cambridge University Press & Assessment 2024 Page 6 of 15", "source_pages": [ 6 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9702_s24_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-May-June/9702_s24_ms_43.pdf?download=true", "html": "9702-topic-13-gravitational-fields/answers.html", "image_paths": [ "../answer-assets/9702_s24_ms_43-p06.png" ] }, { "id": "9702-2024-mj-43-q02", "question_id": "9702-2024-mj-43-q02", "subject": "9702", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 2, "topic": "Temperature", "topic_slug": "9702-topic-14-temperature", "marks": 7, "status": "available", "reason": null, "text": "2(a)(i) 0 K B1\n\n2(a)(ii) (measurement) depends on properties of the liquid B1\n\n2(b)(i)  resistivity varies with temperature B2\n variation with temperature is linear\n unique value of resistivity for each (different value of) temperature\nAny two points, 1 mark each\n\n2(b)(ii) thermometer has high heat capacity/specific heat capacity B1\nor\nenergy transfer needed for thermometer to reach correct temperature\nor\nthermometer takes time to reach the correct temperature\n\n2(b)(iii) thermocouple B1\n\n2(c) (variation is) inverse B1\nor\n(variation is) non-linear\n© Cambridge University Press & Assessment 2024 Page 7 of 15", "source_pages": [ 7 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9702_s24_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-May-June/9702_s24_ms_43.pdf?download=true", "html": "9702-topic-14-temperature/answers.html", "image_paths": [ "../answer-assets/9702_s24_ms_43-p07.png" ] }, { "id": "9702-2024-mj-43-q03", "question_id": "9702-2024-mj-43-q03", "subject": "9702", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 3, "topic": "Thermodynamics", "topic_slug": "9702-topic-16-thermodynamics", "marks": 12, "status": "available", "reason": null, "text": "3(a)(i) gas for which pV  T M1\nwhere T is thermodynamic temperature A1\n\n3(a)(ii) no intermolecular forces B1\n(so) potential energy is zero B1\n\n3(b)(i) pV = NkT C1\nN = (2.0  105  0.26) / (1.38  10–23  290) A1\n= 1.3  1025\n\n3(b)(ii) E = (3/2) kT C1\nK\nE = (3/2)  1.38  10–23  290 A1\nK\n= 6.0  10–21 J\n\n3(b)(iii) internal energy = total KE + PE of molecules B1\nor\nPE = 0 so internal energy = total KE of molecules\ninternal energy = 1.3  1025  6.0  10–21 A1\n= 7.8  104 J\n\n3(c) straight line with positive gradient B1\nline passing through the origin B1\n© Cambridge University Press & Assessment 2024 Page 8 of 15", "source_pages": [ 8 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9702_s24_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-May-June/9702_s24_ms_43.pdf?download=true", "html": "9702-topic-16-thermodynamics/answers.html", "image_paths": [ "../answer-assets/9702_s24_ms_43-p08.png" ] }, { "id": "9702-2024-mj-43-q04", "question_id": "9702-2024-mj-43-q04", "subject": "9702", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 4, "topic": "Oscillations", "topic_slug": "9702-topic-17-oscillations", "marks": 8, "status": "available", "reason": null, "text": "4(a) oscillation (of object) at maximum amplitude B1\nwhen driving frequency = natural frequency (of system) B1\n\n4(b)(i) light damping B1\n\n4(b)(ii) oscillations (of ball) lose energy B1\n(due to) resistive forces (acting on ball) B1\n\n4(b)(iii) frequency = 1 / 0.25 A1\n= 4.0 Hz\n\n4(c) curve showing a maximum amplitude at a single non-zero frequency B1\nsingle maximum amplitude shown at 4.0 Hz B1\n© Cambridge University Press & Assessment 2024 Page 9 of 15", "source_pages": [ 9 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9702_s24_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-May-June/9702_s24_ms_43.pdf?download=true", "html": "9702-topic-17-oscillations/answers.html", "image_paths": [ "../answer-assets/9702_s24_ms_43-p09.png" ] }, { "id": "9702-2024-mj-43-q05", "question_id": "9702-2024-mj-43-q05", "subject": "9702", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 5, "topic": "Magnetic fields", "topic_slug": "9702-topic-20-magnetic-fields", "marks": 13, "status": "available", "reason": null, "text": "5(a) force per unit charge B1\nforce on positive charge B1\n\n5(b)(i) four straight vertical parallel lines, approximately evenly spaced B1\narrows downwards B1\n\n5(b)(ii) E = V / d C1\nE = 430 / 0.067 A1\n= 6.4  103 N C–1\n\n5(b)(iii) smooth curve within plates and straight lines outside plates B1\ndirection of deflection shown as upwards B1\n\n5(c)(i) into the page B1\n\n5(c)(ii) forces are in opposite directions B1\n(undeviated) when (magnitudes of) forces are equal B1\n\n5(c)(iii) Eq = Bqv C1\nB = E / v = (6.4  103) / (2.6  107) A1\n= 2.5  10–4 T\n© Cambridge University Press & Assessment 2024 Page 10 of 15", "source_pages": [ 10 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9702_s24_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-May-June/9702_s24_ms_43.pdf?download=true", "html": "9702-topic-20-magnetic-fields/answers.html", "image_paths": [ "../answer-assets/9702_s24_ms_43-p10.png" ] }, { "id": "9702-2024-mj-43-q06", "question_id": "9702-2024-mj-43-q06", "subject": "9702", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 6, "topic": "Capacitance", "topic_slug": "9702-topic-19-capacitance", "marks": 10, "status": "available", "reason": null, "text": "6(a)  p.d. across capacitor proportional to charge on capacitor B2\n p.d. across capacitor = p.d. across resistor\n current in resistor proportional to p.d. across resistor\n current in resistor = rate of decrease of charge on capacitor\nAny two points, 1 mark each\ncharge proportional to current so rate of decrease of current decreases as current decreases (therefore exponential shape) B1\n\n6(b)(i) R = V / I C1\n= 12 / (0.13  10–3)\n= 9.2  104  A1\n\n6(b)(ii) correct read-off of at least one pair of values for I and t C1\nattempted read-off of t when I = 0.048 mA C1\nor\nsubstitution of a correct pair of values of I and t into I = 0.13 exp (– t / )\n = 4.3 s A1\n\n6(c)  = RC C1\nC =  / R = 4.3 / (9.2  104) A1\n= 4.7  10–5 F\n© Cambridge University Press & Assessment 2024 Page 11 of 15", "source_pages": [ 11 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9702_s24_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-May-June/9702_s24_ms_43.pdf?download=true", "html": "9702-topic-19-capacitance/answers.html", "image_paths": [ "../answer-assets/9702_s24_ms_43-p11.png" ] }, { "id": "9702-2024-mj-43-q07", "question_id": "9702-2024-mj-43-q07", "subject": "9702", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 7, "topic": "Temperature", "topic_slug": "9702-topic-14-temperature", "marks": 10, "status": "available", "reason": null, "text": "7(a) rectification (of the input voltage) M1\nfull-wave A1\n\n7(b)(i) P = V2 / R C1\nor\nmaximum V = 9.0 V\nP = 9.02 / 370 = 0.22 W A1\nMAX\n\n7(b)(ii) sinusoidal shape with minima sitting on the time axis B1\ncorrect frequency and phase, with minima at 0, 0.02, 0.04, 0.06 and 0.08 s and maxima at 0.01, 0.03, 0.05 and 0.07 s B1\nall maxima shown at 0.22 W B1\n\n7(b)(iii) mean power = peak power / 2 = 0.22 / 2 A1\n= 0.11 W\n\n7(c) power–time graph is identical B1\n(so) mean powers are equal B1\n© Cambridge University Press & Assessment 2024 Page 12 of 15", "source_pages": [ 12 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9702_s24_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-May-June/9702_s24_ms_43.pdf?download=true", "html": "9702-topic-14-temperature/answers.html", "image_paths": [ "../answer-assets/9702_s24_ms_43-p12.png" ] }, { "id": "9702-2024-mj-43-q08", "question_id": "9702-2024-mj-43-q08", "subject": "9702", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 8, "topic": "Medical physics", "topic_slug": "9702-topic-24-medical-physics", "marks": 13, "status": "available", "reason": null, "text": "8(a) packet / quantum of energy M1\nof electromagnetic radiation A1\n\n8(b)(i) electron(s) B1\n\n8(b)(ii) X labelled – and Y labelled + B1\n\n8(c)(i) 0.032 MeV A1\n\n8(c)(ii) momentum = E / c C1\nmomentum = (0.032 × 1.60  10–13) / (3.00  108) A1\n= 1.7  10–23 N s\n\n8(c)(iii) E = hf and  = c / f C1\n = hc / E C1\n= (6.63  10–34 × 3.00  108) / (0.032  1.60 × 10–13)\n = 3.9  10–11 m A1\n\n8(d) discussion of bone and soft tissue B1\ndiscussion of different attenuation (coefficients) B1\nor\ndiscussion differences in penetration / transmission / absorption\ntransmitted intensities (by bone and tissue) are very different (leading to good contrast images) B1\n© Cambridge University Press & Assessment 2024 Page 13 of 15", "source_pages": [ 13 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9702_s24_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-May-June/9702_s24_ms_43.pdf?download=true", "html": "9702-topic-24-medical-physics/answers.html", "image_paths": [ "../answer-assets/9702_s24_ms_43-p13.png" ] }, { "id": "9702-2024-mj-43-q09", "question_id": "9702-2024-mj-43-q09", "subject": "9702", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 9, "topic": "Nuclear physics", "topic_slug": "9702-topic-23-nuclear-physics", "marks": 8, "status": "available", "reason": null, "text": "9(a) time for activity (of sample) to halve B1\n\n9(b)(i) activity (of X at time t) B1\n\n9(b)(ii)  Y is a stable isotope B3\n total number of nuclei is constant\n half-life (of X) is 13.6 s\n decay constant (of X) is 0.051 s–1\n amount (of X) at t = 0 is 0.066 mol\n activity (of X) at t = 0 is 2.0  1021 Bq\nAny three points, 1 mark each\n\n9(c) mass of 1 nucleus = (7.3  10–4) / (4.0  1022) C1\nnucleon number = mass of nucleus / (1.66  10–27) C1\n= (7.3  10–4) / (4.0 × 1022  1.66 × 10–27) A1\n= 11 and given as an integer\n© Cambridge University Press & Assessment 2024 Page 14 of 15", "source_pages": [ 14 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9702_s24_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-May-June/9702_s24_ms_43.pdf?download=true", "html": "9702-topic-23-nuclear-physics/answers.html", "image_paths": [ "../answer-assets/9702_s24_ms_43-p14.png" ] }, { "id": "9702-2024-mj-43-q10", "question_id": "9702-2024-mj-43-q10", "subject": "9702", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 10, "topic": "Astronomy and cosmology", "topic_slug": "9702-topic-25-astronomy-and-cosmology", "marks": 9, "status": "available", "reason": null, "text": "10(a)(i) total power B1\npower radiated (by the star) B1\n\n10(a)(ii) standard candle has known luminosity B1\nradiant flux intensity measured by observer B1\n(distance calculated using) F = L / 4d2 B1\n\n10(b)(i) luminosity = 4 r2T4 C1\n= 4  5.67  10–8  (6.96  108)2 × 57804\n= 3.85 × 1026 W A1\n\n10(b)(ii)  T = constant C1\nMAX\ntemperature = (5780  501) / 624 A1\n= 4640 K\n© Cambridge University Press & Assessment 2024 Page 15 of 15", "source_pages": [ 15 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9702_s24_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-May-June/9702_s24_ms_43.pdf?download=true", "html": "9702-topic-25-astronomy-and-cosmology/answers.html", "image_paths": [ "../answer-assets/9702_s24_ms_43-p15.png" ] }, { "id": "9702-2024-mj-51-q01", "question_id": "9702-2024-mj-51-q01", "subject": "9702", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "51", "question_number": 1, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "1 Defining the problem\nt is the independent variable and T is the dependent variable or vary t and measure T 1\nC C\nkeep T constant 1\nR\nMethods of data collection\nlabelled diagram of workable experiment including: 1\n solid cylinder cooling\n insulation surrounding all of the cylinder\n thermometer touching cylinder inside insulation\n insulation and thermometer labelled\nmethod to heat the cylinder uniformly, e.g. place in oven/immerse in hot water or diagram showing cylinder in oven or hot 1\nwater\nmethod to determine time t, e.g. stopwatch or temperature sensor connected to a data logger 1\nmethod to measure L e.g. use a ruler/calipers/micrometer 1\nand\nmethod to measure d e.g. use calipers/micrometer\nMethod of Analysis\nplot a graph of ln (T – T ) against t or equivalent 1\nC R\nmcgradient 1\nU \nA\nZ = ey-intercept 1\n© Cambridge University Press & Assessment 2024 Page 5 of 9\n\n1 Additional detail including safety considerations 6\nD1 precaution to prevent burns or use of hot cylinder / oven / hot water e.g. use of gloves, use of tongs\nD2 keep thickness of the insulating material constant (for each T )\nC\nD3 method to measure m, e.g. use a (top-pan) balance\nD4 for water bath/oven methods, wait for initial temperature of the cylinder to become uniform or constant throughout\nthe cylinder\nd2 d2 \nD5 (surface) AdL or dL2 ", "source_pages": [ 5, 6 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9702_s24_ms_51.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-May-June/9702_s24_ms_51.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_s24_ms_51-p05.png", "../answer-assets/9702_s24_ms_51-p06.png" ] }, { "id": "9702-2024-mj-51-q02", "question_id": "9702-2024-mj-51-q02", "subject": "9702", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "51", "question_number": 2, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "2 4\n \nD6 repeat measurements of d along the length of the cylinder / in different directions and determine the average value\nof d\nD7 description of how c is determined from a separate experiment by heating the cylinder using electrical heater and\nE\nc \nm\nD8 method of determining energy supplied to electrical heater to determine c, e.g. use of joulemeter for E or electrical\nmethod using ammeter and voltmeter to determine IVt\nD9 use several temperature sensors and determine the average T\nC\nD10 relationship valid if a straight line is produced (with y-intercept = ln Z)\nDo not accept line passing through the origin.\n© Cambridge University Press & Assessment 2024 Page 6 of 9\n\n2(a) 1 1\ngradient = \nkf\ns\n1\ny-intercept =\nf\ns\n\n2(b) 1\n1\nv / ms–1 / 10–3 Hz–1\nf\n3.5  0.4 1.118 or 1.1183\n6.3  0.4 1.110 or 1.1096\n8.7  0.5 1.101 or 1.1013\n11.4  0.5 1.092 or 1.0919\n13.9  0.6 1.083 or 1.0827\n16.2  0.6 1.074 or 1.0739\n1\nValues of v and correct as shown above.\nf\nUncertainties in v correct as shown above. 1\n\n2(c)(i) Six points from (b) plotted correctly. 1\nMust be within half a small square. Diameter of points must be less than half a small square.\nError bars in v plotted correctly. 1\nAll error bars to be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.\n© Cambridge University Press & Assessment 2024 Page 7 of 9\n\n2(c)(ii) Straight line of best fit drawn. 1\nDo not accept line from top point to bottom point.\nLine must pass between (14.5, 1.080) and (14.9, 1.080) and between (4.5, 1.115) and (4.8, 1.115).\nWorst acceptable line drawn (steepest or shallowest possible line that passes through all the error bars). 1\nAll error bars must be plotted.\n\n2(c)(iii) Gradient determined with clear substitution of data points into y / x. 1\nGradient must be negative.\nDistance between data points must be greater than half the length of the drawn line.\nGradient determined of worst acceptable line with clear substitution of data points into y / x. 1\nuncertainty = (gradient of line of best fit – gradient of worst acceptable line)\nor\nuncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)\n\n2(c)(iv) y-intercept determined by substitution of correct point with consistent power of ten in m and y into y = mx + c. 1\ny-intercept of worst acceptable line determined by substitution into y = mx + c. 1\nuncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line\nor\nuncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)\nDo not accept ECF from false origin method.\n© Cambridge University Press & Assessment 2024 Page 8 of 9\n\n2(d)(i) f determined using y-intercept and f given to 2, 3 or 4 significant figures and k given to 2 or 3 significant figures. 1\ns s\n1\nf \ns y-intercept\nk determined using gradient with method shown and f and k given with SI units with appropriate powers of ten. 1\ns\ny-intercept 1\nk  or k \ngradient gradientf\ns\nUnits of f : Hz\ns\nUnits of k: m s–1\n\n2(d)(ii) Percentage uncertainty in k with method shown. 1\ny-intercept gradient\npercentage uncertainty  100\n y-intercept gradient \nor\ncorrect substitution for max/min methods.\n\n2(e) v determined (non-zero) to a minimum of 2 significant figures from (c)(iii) and (c)(iv) or (d)(i) with correct substitution. 1\n1\ny-intercept\nf\nv \ngradient\nor\nkf\nv k  s\nf\n© Cambridge University Press & Assessment 2024 Page 9 of 9", "source_pages": [ 6, 7, 8, 9 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9702_s24_ms_51.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-May-June/9702_s24_ms_51.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_s24_ms_51-p06.png", "../answer-assets/9702_s24_ms_51-p07.png", "../answer-assets/9702_s24_ms_51-p08.png", "../answer-assets/9702_s24_ms_51-p09.png" ] }, { "id": "9702-2024-mj-52-q01", "question_id": "9702-2024-mj-52-q01", "subject": "9702", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "52", "question_number": 1, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "1 Defining the problem\nt is the independent variable and s is the dependent variable or vary t and measure s 1\nkeep k constant 1\nMethods of data collection\nlabelled diagram of workable experiment including: 1\n spring connected to magnet\n vertical rule parallel to spring to determine s\n rule held in position by a stand\n stand resting on the bench\n rule labelled and at least one other label from stand, clamp, card, (magnetic) sheet, (cylindrical) magnet, spring\ns = (new) length/position of spring – original length/position of spring 1\nuse a micrometer to measure t 1\nmeasure B using a (calibrated) Hall probe and rotate probe until maximum value 1\nor\nmeasure B using Hall probe first in one direction, then in the opposite direction and average\n© Cambridge University Press & Assessment 2024 Page 5 of 10\n\n1 Method of Analysis\n\n1 1\nplot a graph of s against or equivalent\nt\n(allow lg s against lg t)\nrelationship valid if a straight line that passes through the origin is produced 1\n(for lg s against lg t: relationship valid if a straight line with gradient 1)\nk 1\nZ  gradient\nALB\nk10y-intercept\n(for lg s against lg t: Z  )\nALB\n© Cambridge University Press & Assessment 2024 Page 6 of 10\n\n1 Additional detail including safety considerations 6\nD1 precaution related to spring and/or magnet hitting eyes, e.g. use of goggles/use of safety screen around experiment\nD2 keep A, L and B constant\nD3 use a rule to measure L\nD4 micrometer/calipers to measure diameter d of the magnet and A = d2 / 4\nD5 description of method to determine k, e.g. add mass to spring and k = mg / extension\nor\nuse newton meter to measure force applied to spring and k = force / extension\nor\ntake several readings of force and extension, plot a force–extension graph and k = gradient\nD6 (magnetic) sheet clamped to bench\nD7 use pointer(s)/marker(s) on the spring to read off values from the rule\nD8 method to use video recorder and replay to determine maximum length of the spring\nor\nincrease s or force gradually/slowly until magnet (just) leaves the card\nD9 repeat measurements of t in different positions on the card and average t\nor\nrepeat measurements of s for each value of t and average s\nD10 method to check that the spring has not exceeded the elastic limit\nD11 use of non-magnetic stand or named non-magnetic material for stand, e.g. wood\n© Cambridge University Press & Assessment 2024 Page 7 of 10", "source_pages": [ 5, 6, 7 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9702_s24_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-May-June/9702_s24_ms_52.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_s24_ms_52-p05.png", "../answer-assets/9702_s24_ms_52-p06.png", "../answer-assets/9702_s24_ms_52-p07.png" ] }, { "id": "9702-2024-mj-52-q02", "question_id": "9702-2024-mj-52-q02", "subject": "9702", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "52", "question_number": 2, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "2(a) gradient = n 1\n\n2\ny-intercept = lg\nC\n\n2(b) 1\nlg (L / cm) lg (T / 10–5 s)\n1.73 or 1.732 1.38 or 1.380  0.02\n1.85 or 1.845 1.51 or 1.505  0.01\n1.93 or 1.934 1.59 or 1.591  0.01\n2.033 or 2.0334 1.69 or 1.690  0.02\n2.146 or 2.1461 1.81 or 1.806  0.01\n2.223 or 2.2227 1.87 or 1.869  0.01\nValues of lg (L/ cm) and lg (T/ 10–5 s) correct as shown above.\nUncertainties in lg (T/ 10–5 s) correct as shown above. 1\n\n2(c)(i) Six points from (b) plotted correctly. 1\nMust be within half a small square. Diameter of points must be less than half a small square.\nError bars in lg T plotted correctly. 1\nAll error bars to be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.\n© Cambridge University Press & Assessment 2024 Page 8 of 10\n\n2(c)(ii) Straight line of best fit drawn. 1\nDo not accept line from top point to bottom point.\nLine must pass between (1.780, 1.45) and (1.800, 1.45) and between (2.085, 1.75) and (2.100, 1.75)\nWorst acceptable line drawn (steepest or shallowest possible line that passes through all the error bars). 1\nAll error bars must be plotted.\n\n2(c)(iii) Gradient determined with clear substitution of data points into y / x. 1\nDistance between data points must be greater than half the length of the drawn line.\nGradient determined of worst acceptable line with clear substitution of data points into y / x. 1\nuncertainty = (gradient of line of best fit – gradient of worst acceptable line)\nor\nuncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)\n\n2(c)(iv) y-intercept determined by substitution of correct point with consistent power of ten in m and x into y = mx + c. 1\ny-intercept of worst acceptable line determined by substitution into y = mx + c. 1\nuncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line\nor\nuncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)\nDo not accept ECF from false origin method.\n© Cambridge University Press & Assessment 2024 Page 9 of 10\n\n2(d) Value of n determined using gradient (n = gradient) and C given to 2 or 3 significant figures. 1\nValue of C determined using y-intercept with method shown. 1\n\n2\nC \n10y-intercept\nAbsolute uncertainties in n and C. 1\nuncertainty in n = uncertainty in gradient\nand\n\n2 2\n\n\n2 10minworst y-intercept 10maxworst y-intercept\nC=C or C =\n10worst y-intercept 2\nClear method must be shown with C correctly evaluated.\n\n2(e) Value of L determined (non-zero) to a minimum of 2 significant figures from (c)(iii) and (c)(iv) or (d) with correct substitution 1\nand correct power of ten.\nUnits of T and either C or y-intercept must be consistent.\n\n2\nlogT log\nC log10y-intercept\nlogL \nn n\nlog10y-intercept\nL10 n\nor\nTC\nL n\n\n2\n© Cambridge University Press & Assessment 2024 Page 10 of 10", "source_pages": [ 8, 9, 10 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9702_s24_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-May-June/9702_s24_ms_52.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_s24_ms_52-p08.png", "../answer-assets/9702_s24_ms_52-p09.png", "../answer-assets/9702_s24_ms_52-p10.png" ] }, { "id": "9702-2024-mj-53-q01", "question_id": "9702-2024-mj-53-q01", "subject": "9702", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "53", "question_number": 1, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "1 Defining the problem\nt is the independent variable and T is the dependent variable or vary t and measure T 1\nC C\nkeep T constant 1\nR\nMethods of data collection\nlabelled diagram of workable experiment including: 1\n solid cylinder cooling\n insulation surrounding all of the cylinder\n thermometer touching cylinder inside insulation\n insulation and thermometer labelled\nmethod to heat the cylinder uniformly, e.g. place in oven/immerse in hot water or diagram showing cylinder in oven or hot 1\nwater\nmethod to determine time t, e.g. stopwatch or temperature sensor connected to a data logger 1\nmethod to measure L e.g. use a ruler/calipers/micrometer 1\nand\nmethod to measure d e.g. use calipers/micrometer\nMethod of Analysis\nplot a graph of ln (T – T ) against t or equivalent 1\nC R\nmcgradient 1\nU \nA\nZ = ey-intercept 1\n© Cambridge University Press & Assessment 2024 Page 5 of 9\n\n1 Additional detail including safety considerations 6\nD1 precaution to prevent burns or use of hot cylinder / oven / hot water e.g. use of gloves, use of tongs\nD2 keep thickness of the insulating material constant (for each T )\nC\nD3 method to measure m, e.g. use a (top-pan) balance\nD4 for water bath/oven methods, wait for initial temperature of the cylinder to become uniform or constant throughout\nthe cylinder\nd2 d2 \nD5 (surface) AdL or dL2 ", "source_pages": [ 5, 6 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9702_s24_ms_53.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-May-June/9702_s24_ms_53.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_s24_ms_53-p05.png", "../answer-assets/9702_s24_ms_53-p06.png" ] }, { "id": "9702-2024-mj-53-q02", "question_id": "9702-2024-mj-53-q02", "subject": "9702", "year": 2024, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "53", "question_number": 2, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "2 4\n \nD6 repeat measurements of d along the length of the cylinder / in different directions and determine the average value\nof d\nD7 description of how c is determined from a separate experiment by heating the cylinder using electrical heater and\nE\nc \nm\nD8 method of determining energy supplied to electrical heater to determine c, e.g. use of joulemeter for E or electrical\nmethod using ammeter and voltmeter to determine IVt\nD9 use several temperature sensors and determine the average T\nC\nD10 relationship valid if a straight line is produced (with y-intercept = ln Z)\nDo not accept line passing through the origin.\n© Cambridge University Press & Assessment 2024 Page 6 of 9\n\n2(a) 1 1\ngradient = \nkf\ns\n1\ny-intercept =\nf\ns\n\n2(b) 1\n1\nv / ms–1 / 10–3 Hz–1\nf\n3.5  0.4 1.118 or 1.1183\n6.3  0.4 1.110 or 1.1096\n8.7  0.5 1.101 or 1.1013\n11.4  0.5 1.092 or 1.0919\n13.9  0.6 1.083 or 1.0827\n16.2  0.6 1.074 or 1.0739\n1\nValues of v and correct as shown above.\nf\nUncertainties in v correct as shown above. 1\n\n2(c)(i) Six points from (b) plotted correctly. 1\nMust be within half a small square. Diameter of points must be less than half a small square.\nError bars in v plotted correctly. 1\nAll error bars to be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.\n© Cambridge University Press & Assessment 2024 Page 7 of 9\n\n2(c)(ii) Straight line of best fit drawn. 1\nDo not accept line from top point to bottom point.\nLine must pass between (14.5, 1.080) and (14.9, 1.080) and between (4.5, 1.115) and (4.8, 1.115).\nWorst acceptable line drawn (steepest or shallowest possible line that passes through all the error bars). 1\nAll error bars must be plotted.\n\n2(c)(iii) Gradient determined with clear substitution of data points into y / x. 1\nGradient must be negative.\nDistance between data points must be greater than half the length of the drawn line.\nGradient determined of worst acceptable line with clear substitution of data points into y / x. 1\nuncertainty = (gradient of line of best fit – gradient of worst acceptable line)\nor\nuncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)\n\n2(c)(iv) y-intercept determined by substitution of correct point with consistent power of ten in m and y into y = mx + c. 1\ny-intercept of worst acceptable line determined by substitution into y = mx + c. 1\nuncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line\nor\nuncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)\nDo not accept ECF from false origin method.\n© Cambridge University Press & Assessment 2024 Page 8 of 9\n\n2(d)(i) f determined using y-intercept and f given to 2, 3 or 4 significant figures and k given to 2 or 3 significant figures. 1\ns s\n1\nf \ns y-intercept\nk determined using gradient with method shown and f and k given with SI units with appropriate powers of ten. 1\ns\ny-intercept 1\nk  or k \ngradient gradientf\ns\nUnits of f : Hz\ns\nUnits of k: m s–1\n\n2(d)(ii) Percentage uncertainty in k with method shown. 1\ny-intercept gradient\npercentage uncertainty  100\n y-intercept gradient \nor\ncorrect substitution for max/min methods.\n\n2(e) v determined (non-zero) to a minimum of 2 significant figures from (c)(iii) and (c)(iv) or (d)(i) with correct substitution. 1\n1\ny-intercept\nf\nv \ngradient\nor\nkf\nv k  s\nf\n© Cambridge University Press & Assessment 2024 Page 9 of 9", "source_pages": [ 6, 7, 8, 9 ], "source_pdf": "_source-pdfs/2024-May-June/ms/9702_s24_ms_53.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-May-June/9702_s24_ms_53.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_s24_ms_53-p06.png", "../answer-assets/9702_s24_ms_53-p07.png", "../answer-assets/9702_s24_ms_53-p08.png", "../answer-assets/9702_s24_ms_53-p09.png" ] }, { "id": "9702-2024-on-41-q01", "question_id": "9702-2024-on-41-q01", "subject": "9702", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 1, "topic": "Gravitational fields", "topic_slug": "9702-topic-13-gravitational-fields", "marks": 12, "status": "available", "reason": null, "text": "1(a) (gravitational) force is (directly) proportional to product of masses B1\nforce (between point masses) is inversely proportional to the square of their separation B1\n\n1(b)(i) (gravitational) force acts perpendicular to direction of motion B1\ngravitational force provides centripetal acceleration B1\n\n1(b)(ii) (F =) GMm / x2 = mx2 and  = 2 / T M1\nor\nGMm / x2 = 42mx / T2\ncompletion of algebra leading to x3 = GMT2 / 42 A1\nclear indication that B = radius of planet and that A = mass (of planet) B1\n\n1(b)(iii) gradient = 3√(42 / GA) C1\ne.g. (1280 – 360) / (12  106) = 3√(42 / [6.67  10–11  A]) C1\nA = 1.3  1024 kg A1\nintercept = gradient  B C1\ne.g. 360 = ((1280 – 360)  B) / (12  106) A1\nB = 4.7  106 m\n© Cambridge University Press & Assessment 2024 Page 6 of 15", "source_pages": [ 6 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9702_w24_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-Oct-Nov/9702_w24_ms_41.pdf?download=true", "html": "9702-topic-13-gravitational-fields/answers.html", "image_paths": [ "../answer-assets/9702_w24_ms_41-p06.png" ] }, { "id": "9702-2024-on-41-q02", "question_id": "9702-2024-on-41-q02", "subject": "9702", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 2, "topic": "Temperature", "topic_slug": "9702-topic-14-temperature", "marks": 8, "status": "available", "reason": null, "text": "2(a) (thermal) energy per unit mass (to change temperature) B1\n(thermal) energy per unit change in temperature B1\n\n2(b)(i) Any three bulleted points from: B3\n• the blocks end up in thermal equilibrium\n• heat capacity of Y is larger than heat capacity of X\n• no heat loss to the surroundings\nUp to 2 points from these six:\n• initial temperature of X = 85 °C\n• initial temperature of Y = 25 °C\n• the temperature change of X = 45 °C\n• the temperature change of Y = 15 °C\n• the temperature change in X is three times that in Y\n• final temperature of both = 40 °C\n\n2(b)(ii)  = 45 °C for X and 15 °C for Y C1\nmc  45 = 1.3  m  901  15 C1\nc = 390 J kg K–1 A1\n© Cambridge University Press & Assessment 2024 Page 7 of 15", "source_pages": [ 7 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9702_w24_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-Oct-Nov/9702_w24_ms_41.pdf?download=true", "html": "9702-topic-14-temperature/answers.html", "image_paths": [ "../answer-assets/9702_w24_ms_41-p07.png" ] }, { "id": "9702-2024-on-41-q03", "question_id": "9702-2024-on-41-q03", "subject": "9702", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 3, "topic": "Thermodynamics", "topic_slug": "9702-topic-16-thermodynamics", "marks": 8, "status": "available", "reason": null, "text": "3(a)(i) number of particles per unit amount of substance B1\n\n3(a)(ii) N = R / k B1\nA\n\n3(b)(i) X pressure and Y pressure both = NkT / V B1\nX amount = N / N and Y amount = 2N / N B1\nA A\nX mean-square speed = 3kT / m and Y mean-square speed = 3kT / 2m B1\nX internal energy = 3NkT / 2 and Y internal energy = 3NkT B1\n\n3(b)(ii) line passing through the origin and not returning to either axis B1\ncurve with positive decreasing gradient B1\n© Cambridge University Press & Assessment 2024 Page 8 of 15", "source_pages": [ 8 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9702_w24_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-Oct-Nov/9702_w24_ms_41.pdf?download=true", "html": "9702-topic-16-thermodynamics/answers.html", "image_paths": [ "../answer-assets/9702_w24_ms_41-p08.png" ] }, { "id": "9702-2024-on-41-q04", "question_id": "9702-2024-on-41-q04", "subject": "9702", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 4, "topic": "Oscillations", "topic_slug": "9702-topic-17-oscillations", "marks": 11, "status": "available", "reason": null, "text": "4(a) (motion in which) acceleration is (directly) proportional to displacement B1\n(motion in which): B1\nacceleration is (always) in the opposite direction to displacement\nor\nacceleration is (always) directed towards a fixed point\n\n4(b)(i) amplitude = (9.5 – 3.5) / 2 A1\n= 3.0 cm\n\n4(b)(ii)  = v / x C1\n0 0\n= 9.5 / 3.0 = 3.2 rad s–1 A1\n\n4(b)(iii) T = 2 /  C1\n= 2 / 3.2 A1\n= 2.0 s\n\n4(b)(iv) attempted sinusoidal curve starting with a minimum at t = 0 B1\nsinusoidal curve of period 2.0 s from t = 0 to t = 6.0 s B1\nall peaks shown at h = 9.5 cm B1\nall troughs shown at h = 3.5 cm B1\n© Cambridge University Press & Assessment 2024 Page 9 of 15", "source_pages": [ 9 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9702_w24_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-Oct-Nov/9702_w24_ms_41.pdf?download=true", "html": "9702-topic-17-oscillations/answers.html", "image_paths": [ "../answer-assets/9702_w24_ms_41-p09.png" ] }, { "id": "9702-2024-on-41-q05", "question_id": "9702-2024-on-41-q05", "subject": "9702", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 5, "topic": "Electric fields", "topic_slug": "9702-topic-18-electric-fields", "marks": 10, "status": "available", "reason": null, "text": "5(a) (electric) field equals (electric) potential gradient M1\nreference to minus sign A1\n\n5(b) • for potential to be zero, one potential must be positive and the other potential must be negative B3\n• for potential to be zero, the charges must have opposite sign\n• for field to be zero, the fields (due to X and Y) must be in opposite directions\n• for field to be zero, the charges must have the same sign\n• the signs of the charges cannot (simultaneously) be both the same and opposite (so not possible)\nAny three points, 1 mark each\n\n5(c)(i) V = (–) Q / 4ε x and V = (–) 2Q / 4ε y C1\nX 0 Y 0\n(V + V = 0 so) Q / 4ε x = 2Q / 4ε y leading to y = 2x A1\nX Y 0 0\n\n5(c)(ii) E = Q / 4ε x2 A1\nX 0\n\n5(c)(iii) E = 2Q / 4ε (2x)2 C1\nY 0\n( = Q / 8ε x2)\n0\n(opposite charges so fields in same direction so magnitudes add): A1\nE = (Q / 4ε x2) + (Q / 8ε x2)\n0 0\n= 3Q / 8ε x2\n0\n© Cambridge University Press & Assessment 2024 Page 10 of 15", "source_pages": [ 10 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9702_w24_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-Oct-Nov/9702_w24_ms_41.pdf?download=true", "html": "9702-topic-18-electric-fields/answers.html", "image_paths": [ "../answer-assets/9702_w24_ms_41-p10.png" ] }, { "id": "9702-2024-on-41-q06", "question_id": "9702-2024-on-41-q06", "subject": "9702", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 6, "topic": "Alternating currents", "topic_slug": "9702-topic-21-alternating-currents", "marks": 11, "status": "available", "reason": null, "text": "6(a)(i) conversion (from a.c.) to d.c. B1\n\n6(a)(ii) half-wave: voltage in one direction is removed B1\nfull-wave: voltage in one direction is reversed B1\n\n6(b)(i) one gap connected by a single diode and other gap connected directly B1\ndiode drawn (in a circuit) with correct circuit symbol B1\n\n6(b)(ii) smoothing B1\n\n6(c)(i) E = ½CV2 C1\nC = 2  0.041 / 122 = 5.7  10–4 F = 570 F A1\n\n6(c)(ii) 8.0 = 12.0 exp (– 0.010 / RC) C1\nln (8.0 / 12.0) = – 0.010 / (R  5.7  10–4) C1\nR = 43  A1\n© Cambridge University Press & Assessment 2024 Page 11 of 15", "source_pages": [ 11 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9702_w24_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-Oct-Nov/9702_w24_ms_41.pdf?download=true", "html": "9702-topic-21-alternating-currents/answers.html", "image_paths": [ "../answer-assets/9702_w24_ms_41-p11.png" ] }, { "id": "9702-2024-on-41-q07", "question_id": "9702-2024-on-41-q07", "subject": "9702", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 7, "topic": "Magnetic fields", "topic_slug": "9702-topic-20-magnetic-fields", "marks": 11, "status": "available", "reason": null, "text": "7(a) • force per unit length B2\n• force per unit current\n• length / current perpendicular to field\n1 mark for any two points, 2 marks for all three points\n\n7(b) concentric circles around the wire (at least two circles needed) B1\nspacing between circles increases with distance from wire (at least four circles needed) B1\narrows showing direction of field is clockwise B1\n\n7(c)(i) (each) wire sits in the (magnetic) field created by the other B1\ncurrent (in one wire) is perpendicular to (magnetic) field (due to other wire) so (magnetic) force acts (on wire) B1\n\n7(c)(ii) arrow drawn, starting from X and pointing towards Y, labelled F B1\n\n7(c)(iii) (forces have) equal magnitudes B1\n(forces are in) opposite directions B1\n\n7(c)(iv) no change (in the direction of the force) since both the current in X and the field due to Y have reversed B1\n© Cambridge University Press & Assessment 2024 Page 12 of 15", "source_pages": [ 12 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9702_w24_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-Oct-Nov/9702_w24_ms_41.pdf?download=true", "html": "9702-topic-20-magnetic-fields/answers.html", "image_paths": [ "../answer-assets/9702_w24_ms_41-p12.png" ] }, { "id": "9702-2024-on-41-q08", "question_id": "9702-2024-on-41-q08", "subject": "9702", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 8, "topic": "Quantum physics", "topic_slug": "9702-topic-22-quantum-physics", "marks": 9, "status": "available", "reason": null, "text": "8(a) photoelectric effect B1\n\n8(b)(i) E = hf C1\nwork function = 6.63  10–34  8.8  1014 A1\n= 5.8  10–19 J\n\n8(b)(ii) hf =  + ½ mv 2 C1\nMAX\n6.63  10–34  11  1014 = (5.8  10–19) + (½  9.11  10–31  v 2) C1\nMAX\nv = 5.7  105 m s–1 A1\nMAX\n\n8(c) E shown as zero from f = 8.0 to 8.8 and non-zero from f = 8.8 to 11 B1\nMAX\nall non-zero E shown as a single straight line with a positive gradient B1\nMAX\nline passing through (11, 1.45) B1\n© Cambridge University Press & Assessment 2024 Page 13 of 15", "source_pages": [ 13 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9702_w24_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-Oct-Nov/9702_w24_ms_41.pdf?download=true", "html": "9702-topic-22-quantum-physics/answers.html", "image_paths": [ "../answer-assets/9702_w24_ms_41-p13.png" ] }, { "id": "9702-2024-on-41-q09", "question_id": "9702-2024-on-41-q09", "subject": "9702", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 9, "topic": "Medical physics", "topic_slug": "9702-topic-24-medical-physics", "marks": 10, "status": "available", "reason": null, "text": "9(a)(i) positron B1\n\n9(a)(ii)  = ln 2 / (110  60) = 1.05  10–4 s–1 A1\n\n9(a)(iii) N = M / (18 u) or (M in grams  N / 18) C1\nA\nN = (2.1  10–12) / (18  1.66  10–27) or (2.1  10–9  6.02  1023) / 18\n( = 7.0  1013)\nA = N C1\n= 1.05  10–4  7.0  1013 A1\n= 7.4  109 Bq\n\n9(b)(i) • (pair) annihilation occurs B3\n• the mass of the two particles is converted into energy\n• two gamma photons are formed and travel in opposite directions\nor\ntwo gamma photons are formed and leave the body\n• difference in arrival times of photons (at detector) is processed\nAny three points, 1 mark each\n\n9(b)(ii) with a shorter half-life: sample would (almost) fully decay before the test is complete B1\na longer half-life: exposes patient to harmful/ionising radiation unnecessarily B1\nor\nwith a longer half-life: a larger dose (of tracer) needed to produce detectable activity\n© Cambridge University Press & Assessment 2024 Page 14 of 15", "source_pages": [ 14 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9702_w24_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-Oct-Nov/9702_w24_ms_41.pdf?download=true", "html": "9702-topic-24-medical-physics/answers.html", "image_paths": [ "../answer-assets/9702_w24_ms_41-p14.png" ] }, { "id": "9702-2024-on-41-q10", "question_id": "9702-2024-on-41-q10", "subject": "9702", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 10, "topic": "Astronomy and cosmology", "topic_slug": "9702-topic-25-astronomy-and-cosmology", "marks": 10, "status": "available", "reason": null, "text": "10(a) • redshift is the increase in observed wavelength / decrease in observed frequency (caused by Doppler effect) B3\n• radiation from distant galaxies is observed to be redshifted\n• redshift provides evidence that galaxies are moving apart\n• galaxies moving apart means Universe must be expanding\nAny three points, 1 mark each\n\n10(b)(i) F = L / 4d2 C1\nd = √(1.90  1036 / [4  8.42  10–16]) A1\n= 1.34  1025 m\n\n10(b)(ii)  /  = v / c C1\n(726 – 658) / 658 = v / (3.00  108)\nv = 3.1  107 m s–1 A1\n\n10(c)(i) line with positive gradient passing through the origin B1\nstraight line with positive gradient B1\n\n10(c)(ii) Hubble constant B1\n© Cambridge University Press & Assessment 2024 Page 15 of 15", "source_pages": [ 15 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9702_w24_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-Oct-Nov/9702_w24_ms_41.pdf?download=true", "html": "9702-topic-25-astronomy-and-cosmology/answers.html", "image_paths": [ "../answer-assets/9702_w24_ms_41-p15.png" ] }, { "id": "9702-2024-on-42-q01", "question_id": "9702-2024-on-42-q01", "subject": "9702", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 1, "topic": "Magnetic fields", "topic_slug": "9702-topic-20-magnetic-fields", "marks": 13, "status": "available", "reason": null, "text": "1(a)(i) v = r C1\n= 0.85  140 A1\n= 120 m s–1\n\n1(a)(ii) a = r2 or a = v2 / r C1\na = 0.85  1402 or 1202 / 0.85 A1\n= 1.7  104 m s–2\n\n1(b)(i) direction of (induced) e.m.f. M1\nis such as to (produce effects that) oppose the change that caused it A1\n\n1(b)(ii) T = 2 /  A1\n= 2 / 140 = 0.045 s = 45 ms\n\n1(b)(iii)  = BA C1\n= 0.18    0.852 C1\n= 0.41 Wb A1\n\n1(b)(iv) E =  / t C1\n= 0.41 / 0.045 A1\n= 9.1 V\n\n1(b)(v) force (on spoke) must be anticlockwise, so current is from A to X (by Fleming’s left hand rule), so X is at the higher potential B1\n© Cambridge University Press & Assessment 2024 Page 5 of 14", "source_pages": [ 5 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9702_w24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-Oct-Nov/9702_w24_ms_42.pdf?download=true", "html": "9702-topic-20-magnetic-fields/answers.html", "image_paths": [ "../answer-assets/9702_w24_ms_42-p05.png" ] }, { "id": "9702-2024-on-42-q02", "question_id": "9702-2024-on-42-q02", "subject": "9702", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 2, "topic": "Astronomy and cosmology", "topic_slug": "9702-topic-25-astronomy-and-cosmology", "marks": 12, "status": "available", "reason": null, "text": "2(a) force per unit mass B1\n\n2(b)(i) g = GM / x2 C1\n= (6.67  10–11  1.99  1030) / (1.47  1011)2 A1\n= 6.14  10–3 N kg– 1\n\n2(b)(ii) E = – GMm / x C1\nP\n= – (6.67  10–11  1.99  1030  2.63) / (1.47  1011)\n= – 2.37  109 J A1\n\n2(c)(i) F = L / 4x2 C1\n(g = GM / x2 and so) x2 = GM / g M1\nand\nx2 = L / 4F\nelimination of x and subsequent algebra shown leading to g = 4GMF / L A1\n\n2(c)(ii) correct read-off of pair of values of g and F and full substitution of values of g, G, M and F into equation C1\ne.g. L = (4  6.67  10–11  1.99  1030  1.83  103) / (8.0  10–3)\nL = 3.8  1026 W A1\n\n2(c)(iii) L = 4 r2T4 C1\n3.8  1026 = (4  5.67  10–8  57804)  r2\nr = 6.9  108 m A1\n© Cambridge University Press & Assessment 2024 Page 6 of 14", "source_pages": [ 6 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9702_w24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-Oct-Nov/9702_w24_ms_42.pdf?download=true", "html": "9702-topic-25-astronomy-and-cosmology/answers.html", "image_paths": [ "../answer-assets/9702_w24_ms_42-p06.png" ] }, { "id": "9702-2024-on-42-q03", "question_id": "9702-2024-on-42-q03", "subject": "9702", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 3, "topic": "Thermodynamics", "topic_slug": "9702-topic-16-thermodynamics", "marks": 11, "status": "available", "reason": null, "text": "3(a) (thermal) energy per unit mass (to cause change of state) B1\n(thermal) energy to change state at constant temperature B1\n\n3(b)(i) W = pV C1\n= 1.0  105  0.017 = 1700 J = 1.7 kJ A1\n\n3(b)(ii) U = Q + W C1\nQ = 17.6 + 1.7 A1\n= 19.3 kJ\n\n3(b)(iii) mass = 710  7.2  10–5 C1\n( = 0.051 kg)\nL = 19.3 / 0.051 A1\n= 380 kJ kg–1\n\n3(c) fusion involves (much) smaller volume change (than vaporisation) B1\nsmaller change in intermolecular spacing so smaller change in internal energy B1\nnegligible work done (by substance during fusion) so L is less (than L ) B1\nF V\n© Cambridge University Press & Assessment 2024 Page 7 of 14", "source_pages": [ 7 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9702_w24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-Oct-Nov/9702_w24_ms_42.pdf?download=true", "html": "9702-topic-16-thermodynamics/answers.html", "image_paths": [ "../answer-assets/9702_w24_ms_42-p07.png" ] }, { "id": "9702-2024-on-42-q04", "question_id": "9702-2024-on-42-q04", "subject": "9702", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 4, "topic": "Ideal gases", "topic_slug": "9702-topic-15-ideal-gases", "marks": 9, "status": "available", "reason": null, "text": "4(a) • molecules are in (constant) random motion B3\n• (all) collisions between molecules are (perfectly) elastic\n• no forces between molecules (except during collisions)\n• volume of molecules is negligible (compared with volume of gas)\n• collisions involving molecules are instantaneous\nAny three points, 1 mark each\n\n4(b) • molecules collide with (walls of) container B3\n• momentum of molecule changes during collision (with walls)\n• change in momentum is caused by force on molecule by wall\n• molecule experiences force from wall so molecule exerts force on wall\n• many molecules exerting force across the area of the wall leads to pressure (on the wall)\nAny three points, 1 mark each\n\n4(c) Any three bulleted points from: B3\n• both gases are ideal\nUp to 2 points from:\n• mass of one molecule of gas X is 3.3  10–27 kg\n• mass of one molecule of gas Y is 6.6  10–27 kg\n• mass of one molecule of gas Y is double mass of one molecule of gas X\nUp to 2 points from:\n• sample of X contains 0.27 mol / 1.6  1023 molecules\n• sample of Y contains 0.81 mol / 4.9  1023 molecules\n• sample of Y contains treble the amount of gas / number of molecules as sample of X\nUp to 2 points from:\n• mass of gas X is 5.4  10–4 kg\n• mass of gas Y is 3.2  10–3 kg\n• mass of gas Y is six times mass of gas X\n© Cambridge University Press & Assessment 2024 Page 8 of 14", "source_pages": [ 8 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9702_w24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-Oct-Nov/9702_w24_ms_42.pdf?download=true", "html": "9702-topic-15-ideal-gases/answers.html", "image_paths": [ "../answer-assets/9702_w24_ms_42-p08.png" ] }, { "id": "9702-2024-on-42-q05", "question_id": "9702-2024-on-42-q05", "subject": "9702", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 5, "topic": "Oscillations", "topic_slug": "9702-topic-17-oscillations", "marks": 9, "status": "available", "reason": null, "text": "5(a) arrow from sphere, perpendicular to string, pointing left and down B1\n\n5(b)(i) amplitude = 0.016 m A1\n\n5(b)(ii) angular frequency = 2 / T C1\n= 2 / 0.40 A1\n= 16 rad s–1\n\n5(b)(iii) total energy = ½m2x 2 C1\n0\n= ½  0.15  15.72  0.0162 A1\n= 4.7  10–3 J\n\n5(c) dome-shaped curve starting and ending on the x-axis, with peak at x = 0 B1\nmaximum E shown as 4.7  10–3 J B1\nK\nminimum x shown as –0.016 m and maximum x shown as +0.016 m at the ends of the line B1\n© Cambridge University Press & Assessment 2024 Page 9 of 14", "source_pages": [ 9 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9702_w24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-Oct-Nov/9702_w24_ms_42.pdf?download=true", "html": "9702-topic-17-oscillations/answers.html", "image_paths": [ "../answer-assets/9702_w24_ms_42-p09.png" ] }, { "id": "9702-2024-on-42-q06", "question_id": "9702-2024-on-42-q06", "subject": "9702", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 6, "topic": "Electric fields", "topic_slug": "9702-topic-18-electric-fields", "marks": 10, "status": "available", "reason": null, "text": "6(a) (electric) force is (directly) proportional to product of charges B1\nforce (between point charges) is inversely proportional to the square of their separation B1\n\n6(b) at least four straight, radial lines to/from surface of sphere B1\nat least four straight radial lines drawn, approximately equally spaced B1\narrows pointing away from the surface of the sphere B1\n\n6(c)(i) radius = 3.2 cm A1\n\n6(c)(ii) E = Q / (4x2) C1\n0\nQ = e.g. 2.2  105  4  8.85  10–12  0.0322 C1\n= 2.5  10–8 C A1\n\n6(c)(iii) • the (positive) charge is all the way around the surface B1\n• a charge placed inside the sphere is pulled equally in all directions\n• if the field was not zero, the charges would move (until field is zero)\n• electric field lines go from positive charge to negative charge, and there are no negative charges inside the sphere\nAny point, 1 mark\n© Cambridge University Press & Assessment 2024 Page 10 of 14", "source_pages": [ 10 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9702_w24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-Oct-Nov/9702_w24_ms_42.pdf?download=true", "html": "9702-topic-18-electric-fields/answers.html", "image_paths": [ "../answer-assets/9702_w24_ms_42-p10.png" ] }, { "id": "9702-2024-on-42-q07", "question_id": "9702-2024-on-42-q07", "subject": "9702", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 7, "topic": "Capacitance", "topic_slug": "9702-topic-19-capacitance", "marks": 10, "status": "available", "reason": null, "text": "7(a) charge / potential (difference) M1\ncharge is charge on one plate, and potential is p.d. between the plates A1\n\n7(b)(i) straight line starting at the origin B1\nline with positive gradient ending at (V, Q) B1\n\n7(b)(ii) work done is the area under the graph B1\nW = ½QV A1\n\n7(c)(i) final p.d. shown as V / 4 for both capacitors B1\nfinal charges add together to give Q B1\ncharge on Y = 3  charge on X (and both charges shown as a multiple of Q) B1\nFully correct answer:\nX Y\nfinal p.d. V / 4 V / 4\nfinal charge Q / 4 3Q / 4\n\n7(c)(ii) less than B1\n© Cambridge University Press & Assessment 2024 Page 11 of 14", "source_pages": [ 11 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9702_w24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-Oct-Nov/9702_w24_ms_42.pdf?download=true", "html": "9702-topic-19-capacitance/answers.html", "image_paths": [ "../answer-assets/9702_w24_ms_42-p11.png" ] }, { "id": "9702-2024-on-42-q08", "question_id": "9702-2024-on-42-q08", "subject": "9702", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 8, "topic": "Alternating currents", "topic_slug": "9702-topic-21-alternating-currents", "marks": 9, "status": "available", "reason": null, "text": "8(a) number of cycles per unit time B1\n\n8(b)(i) period = 2 / 40 = 0.050 s = 50 ms A1\n\n8(b)(ii) sinusoidal curve, starting at (0, 0) and initially increasing from there B1\nperiodic line showing 2 cycles with period 50 ms from t = 0 to t = 100 ms B1\nall peaks shown at I = +3.5 A and all troughs shown at I = –3.5 A B1\n\n8(b)(iii) I = 3.5 / √2 A1\nr.m.s\n= 2.5 A\n\n8(c) P = I2R C1\npeak power = 3.52  680 (= 8330 W) M1\nor\nmean power = 2.472  680 (= 4170 W)\npeak and mean powers both calculated correctly, with supporting working, and compared leading to conclusion that mean A1\npower is half the peak power\n© Cambridge University Press & Assessment 2024 Page 12 of 14", "source_pages": [ 12 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9702_w24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-Oct-Nov/9702_w24_ms_42.pdf?download=true", "html": "9702-topic-21-alternating-currents/answers.html", "image_paths": [ "../answer-assets/9702_w24_ms_42-p12.png" ] }, { "id": "9702-2024-on-42-q09", "question_id": "9702-2024-on-42-q09", "subject": "9702", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 9, "topic": "Quantum physics", "topic_slug": "9702-topic-22-quantum-physics", "marks": 9, "status": "available", "reason": null, "text": "9(a) diffraction is characteristic of wave behaviour so shows that electrons can behave like waves B1\n\n9(b) qV = ½mv2 C1\np = mv C1\np = m  √(2qV / m) A1\n= √(2qVm)\n\n9(c) (electrons have) greater momentum so smaller (de Broglie) wavelength B1\nfringes become closer together B1\n\n9(d)(i) straight line with positive gradient B1\nline with positive gradient passing through the origin B1\n\n9(d)(ii) Planck constant B1\n© Cambridge University Press & Assessment 2024 Page 13 of 14", "source_pages": [ 13 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9702_w24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-Oct-Nov/9702_w24_ms_42.pdf?download=true", "html": "9702-topic-22-quantum-physics/answers.html", "image_paths": [ "../answer-assets/9702_w24_ms_42-p13.png" ] }, { "id": "9702-2024-on-42-q10", "question_id": "9702-2024-on-42-q10", "subject": "9702", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 10, "topic": "Nuclear physics", "topic_slug": "9702-topic-23-nuclear-physics", "marks": 8, "status": "available", "reason": null, "text": "10(a)(i) cannot predict when a particular nucleus will decay B1\nor\ncannot predict which nucleus will decay next\n\n10(a)(ii) (decay is) not affected by external (environmental) factors B1\n\n10(a)(iii) fluctuations in (measured) count rate B1\n\n10(b)(i) • large nuclei undergo fission whereas small nuclei undergo fusion B3\n• fission involves one nucleus splitting into two (or more) (smaller) nuclei\n• fusion involves two nuclei joining together to form one (larger) nucleus\n• fission is (usually) initiated by neutron bombardment\n• fusion is (usually) initiated by (very) high temperatures\nAny three points, 1 mark each\n\n10(b)(ii) binding energy per nucleon is greatest for intermediate nucleon numbers B1\n(may be shown on sketch graph with axes labelled ‘binding energy per nucleon’ and ‘nucleon number’)\nboth fusion and fission involve an increase in binding energy (per nucleon) B1\n© Cambridge University Press & Assessment 2024 Page 14 of 14", "source_pages": [ 14 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9702_w24_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-Oct-Nov/9702_w24_ms_42.pdf?download=true", "html": "9702-topic-23-nuclear-physics/answers.html", "image_paths": [ "../answer-assets/9702_w24_ms_42-p14.png" ] }, { "id": "9702-2024-on-43-q01", "question_id": "9702-2024-on-43-q01", "subject": "9702", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 1, "topic": "Gravitational fields", "topic_slug": "9702-topic-13-gravitational-fields", "marks": 12, "status": "available", "reason": null, "text": "1(a) (gravitational) force is (directly) proportional to product of masses B1\nforce (between point masses) is inversely proportional to the square of their separation B1\n\n1(b)(i) (gravitational) force acts perpendicular to direction of motion B1\ngravitational force provides centripetal acceleration B1\n\n1(b)(ii) (F =) GMm / x2 = mx2 and  = 2 / T M1\nor\nGMm / x2 = 42mx / T2\ncompletion of algebra leading to x3 = GMT2 / 42 A1\nclear indication that B = radius of planet and that A = mass (of planet) B1\n\n1(b)(iii) gradient = 3√(42 / GA) C1\ne.g. (1280 – 360) / (12  106) = 3√(42 / [6.67  10–11  A]) C1\nA = 1.3  1024 kg A1\nintercept = gradient  B C1\ne.g. 360 = ((1280 – 360)  B) / (12  106) A1\nB = 4.7  106 m\n© Cambridge University Press & Assessment 2024 Page 6 of 15", "source_pages": [ 6 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9702_w24_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-Oct-Nov/9702_w24_ms_43.pdf?download=true", "html": "9702-topic-13-gravitational-fields/answers.html", "image_paths": [ "../answer-assets/9702_w24_ms_43-p06.png" ] }, { "id": "9702-2024-on-43-q02", "question_id": "9702-2024-on-43-q02", "subject": "9702", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 2, "topic": "Temperature", "topic_slug": "9702-topic-14-temperature", "marks": 8, "status": "available", "reason": null, "text": "2(a) (thermal) energy per unit mass (to change temperature) B1\n(thermal) energy per unit change in temperature B1\n\n2(b)(i) Any three bulleted points from: B3\n• the blocks end up in thermal equilibrium\n• heat capacity of Y is larger than heat capacity of X\n• no heat loss to the surroundings\nUp to 2 points from these six:\n• initial temperature of X = 85 °C\n• initial temperature of Y = 25 °C\n• the temperature change of X = 45 °C\n• the temperature change of Y = 15 °C\n• the temperature change in X is three times that in Y\n• final temperature of both = 40 °C\n\n2(b)(ii)  = 45 °C for X and 15 °C for Y C1\nmc  45 = 1.3  m  901  15 C1\nc = 390 J kg K–1 A1\n© Cambridge University Press & Assessment 2024 Page 7 of 15", "source_pages": [ 7 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9702_w24_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-Oct-Nov/9702_w24_ms_43.pdf?download=true", "html": "9702-topic-14-temperature/answers.html", "image_paths": [ "../answer-assets/9702_w24_ms_43-p07.png" ] }, { "id": "9702-2024-on-43-q03", "question_id": "9702-2024-on-43-q03", "subject": "9702", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 3, "topic": "Thermodynamics", "topic_slug": "9702-topic-16-thermodynamics", "marks": 8, "status": "available", "reason": null, "text": "3(a)(i) number of particles per unit amount of substance B1\n\n3(a)(ii) N = R / k B1\nA\n\n3(b)(i) X pressure and Y pressure both = NkT / V B1\nX amount = N / N and Y amount = 2N / N B1\nA A\nX mean-square speed = 3kT / m and Y mean-square speed = 3kT / 2m B1\nX internal energy = 3NkT / 2 and Y internal energy = 3NkT B1\n\n3(b)(ii) line passing through the origin and not returning to either axis B1\ncurve with positive decreasing gradient B1\n© Cambridge University Press & Assessment 2024 Page 8 of 15", "source_pages": [ 8 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9702_w24_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-Oct-Nov/9702_w24_ms_43.pdf?download=true", "html": "9702-topic-16-thermodynamics/answers.html", "image_paths": [ "../answer-assets/9702_w24_ms_43-p08.png" ] }, { "id": "9702-2024-on-43-q04", "question_id": "9702-2024-on-43-q04", "subject": "9702", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 4, "topic": "Oscillations", "topic_slug": "9702-topic-17-oscillations", "marks": 11, "status": "available", "reason": null, "text": "4(a) (motion in which) acceleration is (directly) proportional to displacement B1\n(motion in which): B1\nacceleration is (always) in the opposite direction to displacement\nor\nacceleration is (always) directed towards a fixed point\n\n4(b)(i) amplitude = (9.5 – 3.5) / 2 A1\n= 3.0 cm\n\n4(b)(ii)  = v / x C1\n0 0\n= 9.5 / 3.0 = 3.2 rad s–1 A1\n\n4(b)(iii) T = 2 /  C1\n= 2 / 3.2 A1\n= 2.0 s\n\n4(b)(iv) attempted sinusoidal curve starting with a minimum at t = 0 B1\nsinusoidal curve of period 2.0 s from t = 0 to t = 6.0 s B1\nall peaks shown at h = 9.5 cm B1\nall troughs shown at h = 3.5 cm B1\n© Cambridge University Press & Assessment 2024 Page 9 of 15", "source_pages": [ 9 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9702_w24_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-Oct-Nov/9702_w24_ms_43.pdf?download=true", "html": "9702-topic-17-oscillations/answers.html", "image_paths": [ "../answer-assets/9702_w24_ms_43-p09.png" ] }, { "id": "9702-2024-on-43-q05", "question_id": "9702-2024-on-43-q05", "subject": "9702", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 5, "topic": "Electric fields", "topic_slug": "9702-topic-18-electric-fields", "marks": 10, "status": "available", "reason": null, "text": "5(a) (electric) field equals (electric) potential gradient M1\nreference to minus sign A1\n\n5(b) • for potential to be zero, one potential must be positive and the other potential must be negative B3\n• for potential to be zero, the charges must have opposite sign\n• for field to be zero, the fields (due to X and Y) must be in opposite directions\n• for field to be zero, the charges must have the same sign\n• the signs of the charges cannot (simultaneously) be both the same and opposite (so not possible)\nAny three points, 1 mark each\n\n5(c)(i) V = (–) Q / 4ε x and V = (–) 2Q / 4ε y C1\nX 0 Y 0\n(V + V = 0 so) Q / 4ε x = 2Q / 4ε y leading to y = 2x A1\nX Y 0 0\n\n5(c)(ii) E = Q / 4ε x2 A1\nX 0\n\n5(c)(iii) E = 2Q / 4ε (2x)2 C1\nY 0\n( = Q / 8ε x2)\n0\n(opposite charges so fields in same direction so magnitudes add): A1\nE = (Q / 4ε x2) + (Q / 8ε x2)\n0 0\n= 3Q / 8ε x2\n0\n© Cambridge University Press & Assessment 2024 Page 10 of 15", "source_pages": [ 10 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9702_w24_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-Oct-Nov/9702_w24_ms_43.pdf?download=true", "html": "9702-topic-18-electric-fields/answers.html", "image_paths": [ "../answer-assets/9702_w24_ms_43-p10.png" ] }, { "id": "9702-2024-on-43-q06", "question_id": "9702-2024-on-43-q06", "subject": "9702", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 6, "topic": "Alternating currents", "topic_slug": "9702-topic-21-alternating-currents", "marks": 11, "status": "available", "reason": null, "text": "6(a)(i) conversion (from a.c.) to d.c. B1\n\n6(a)(ii) half-wave: voltage in one direction is removed B1\nfull-wave: voltage in one direction is reversed B1\n\n6(b)(i) one gap connected by a single diode and other gap connected directly B1\ndiode drawn (in a circuit) with correct circuit symbol B1\n\n6(b)(ii) smoothing B1\n\n6(c)(i) E = ½CV2 C1\nC = 2  0.041 / 122 = 5.7  10–4 F = 570 F A1\n\n6(c)(ii) 8.0 = 12.0 exp (– 0.010 / RC) C1\nln (8.0 / 12.0) = – 0.010 / (R  5.7  10–4) C1\nR = 43  A1\n© Cambridge University Press & Assessment 2024 Page 11 of 15", "source_pages": [ 11 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9702_w24_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-Oct-Nov/9702_w24_ms_43.pdf?download=true", "html": "9702-topic-21-alternating-currents/answers.html", "image_paths": [ "../answer-assets/9702_w24_ms_43-p11.png" ] }, { "id": "9702-2024-on-43-q07", "question_id": "9702-2024-on-43-q07", "subject": "9702", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 7, "topic": "Magnetic fields", "topic_slug": "9702-topic-20-magnetic-fields", "marks": 11, "status": "available", "reason": null, "text": "7(a) • force per unit length B2\n• force per unit current\n• length / current perpendicular to field\n1 mark for any two points, 2 marks for all three points\n\n7(b) concentric circles around the wire (at least two circles needed) B1\nspacing between circles increases with distance from wire (at least four circles needed) B1\narrows showing direction of field is clockwise B1\n\n7(c)(i) (each) wire sits in the (magnetic) field created by the other B1\ncurrent (in one wire) is perpendicular to (magnetic) field (due to other wire) so (magnetic) force acts (on wire) B1\n\n7(c)(ii) arrow drawn, starting from X and pointing towards Y, labelled F B1\n\n7(c)(iii) (forces have) equal magnitudes B1\n(forces are in) opposite directions B1\n\n7(c)(iv) no change (in the direction of the force) since both the current in X and the field due to Y have reversed B1\n© Cambridge University Press & Assessment 2024 Page 12 of 15", "source_pages": [ 12 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9702_w24_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-Oct-Nov/9702_w24_ms_43.pdf?download=true", "html": "9702-topic-20-magnetic-fields/answers.html", "image_paths": [ "../answer-assets/9702_w24_ms_43-p12.png" ] }, { "id": "9702-2024-on-43-q08", "question_id": "9702-2024-on-43-q08", "subject": "9702", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 8, "topic": "Quantum physics", "topic_slug": "9702-topic-22-quantum-physics", "marks": 9, "status": "available", "reason": null, "text": "8(a) photoelectric effect B1\n\n8(b)(i) E = hf C1\nwork function = 6.63  10–34  8.8  1014 A1\n= 5.8  10–19 J\n\n8(b)(ii) hf =  + ½ mv 2 C1\nMAX\n6.63  10–34  11  1014 = (5.8  10–19) + (½  9.11  10–31  v 2) C1\nMAX\nv = 5.7  105 m s–1 A1\nMAX\n\n8(c) E shown as zero from f = 8.0 to 8.8 and non-zero from f = 8.8 to 11 B1\nMAX\nall non-zero E shown as a single straight line with a positive gradient B1\nMAX\nline passing through (11, 1.45) B1\n© Cambridge University Press & Assessment 2024 Page 13 of 15", "source_pages": [ 13 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9702_w24_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-Oct-Nov/9702_w24_ms_43.pdf?download=true", "html": "9702-topic-22-quantum-physics/answers.html", "image_paths": [ "../answer-assets/9702_w24_ms_43-p13.png" ] }, { "id": "9702-2024-on-43-q09", "question_id": "9702-2024-on-43-q09", "subject": "9702", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 9, "topic": "Medical physics", "topic_slug": "9702-topic-24-medical-physics", "marks": 10, "status": "available", "reason": null, "text": "9(a)(i) positron B1\n\n9(a)(ii)  = ln 2 / (110  60) = 1.05  10–4 s–1 A1\n\n9(a)(iii) N = M / (18 u) or (M in grams  N / 18) C1\nA\nN = (2.1  10–12) / (18  1.66  10–27) or (2.1  10–9  6.02  1023) / 18\n( = 7.0  1013)\nA = N C1\n= 1.05  10–4  7.0  1013 A1\n= 7.4  109 Bq\n\n9(b)(i) • (pair) annihilation occurs B3\n• the mass of the two particles is converted into energy\n• two gamma photons are formed and travel in opposite directions\nor\ntwo gamma photons are formed and leave the body\n• difference in arrival times of photons (at detector) is processed\nAny three points, 1 mark each\n\n9(b)(ii) with a shorter half-life: sample would (almost) fully decay before the test is complete B1\na longer half-life: exposes patient to harmful/ionising radiation unnecessarily B1\nor\nwith a longer half-life: a larger dose (of tracer) needed to produce detectable activity\n© Cambridge University Press & Assessment 2024 Page 14 of 15", "source_pages": [ 14 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9702_w24_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-Oct-Nov/9702_w24_ms_43.pdf?download=true", "html": "9702-topic-24-medical-physics/answers.html", "image_paths": [ "../answer-assets/9702_w24_ms_43-p14.png" ] }, { "id": "9702-2024-on-43-q10", "question_id": "9702-2024-on-43-q10", "subject": "9702", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 10, "topic": "Astronomy and cosmology", "topic_slug": "9702-topic-25-astronomy-and-cosmology", "marks": 10, "status": "available", "reason": null, "text": "10(a) • redshift is the increase in observed wavelength / decrease in observed frequency (caused by Doppler effect) B3\n• radiation from distant galaxies is observed to be redshifted\n• redshift provides evidence that galaxies are moving apart\n• galaxies moving apart means Universe must be expanding\nAny three points, 1 mark each\n\n10(b)(i) F = L / 4d2 C1\nd = √(1.90  1036 / [4  8.42  10–16]) A1\n= 1.34  1025 m\n\n10(b)(ii)  /  = v / c C1\n(726 – 658) / 658 = v / (3.00  108)\nv = 3.1  107 m s–1 A1\n\n10(c)(i) line with positive gradient passing through the origin B1\nstraight line with positive gradient B1\n\n10(c)(ii) Hubble constant B1\n© Cambridge University Press & Assessment 2024 Page 15 of 15", "source_pages": [ 15 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9702_w24_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-Oct-Nov/9702_w24_ms_43.pdf?download=true", "html": "9702-topic-25-astronomy-and-cosmology/answers.html", "image_paths": [ "../answer-assets/9702_w24_ms_43-p15.png" ] }, { "id": "9702-2024-on-51-q01", "question_id": "9702-2024-on-51-q01", "subject": "9702", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "51", "question_number": 1, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "1 Defining the problem\ns is the independent variable and v is the dependent variable or vary s and measure v 1\nkeep D constant 1\nMethods of data collection\nlabelled diagram of workable experiment including: 1\n• light gate positioned at P\n• light gate connected to timer / data logger\n• labels for light gate and P and data logger / timer and at least one other label from block, magnet(s), trolley, s and D\nmeasure D with a rule(r) and measure L with a rule(r) or calipers 1\ndescription to determine v at P, e.g. (measure length of) card to interrupt beam 1\nmethod to measure s, e.g. use calipers 1\n© Cambridge University Press & Assessment 2024 Page 5 of 11\n\n1 Method of Analysis\n\n1 1 1\nplot a graph of v2 against or equivalent (e.g. against v2)\ns4 s4\nDo not accept logarithms.\nmgradient 1\nK =\n2DA2B2L2\nm 1\n(or K = for against v2)\n2DA2B2L2gradient s4\nmy-intercept 1\nQ=−\n2D\nmy-intercept 1\n(or Q =KA2B2L2y-intercept or Q = for against v2)\n2Dgradient s4\n© Cambridge University Press & Assessment 2024 Page 6 of 11\n\n1 Additional detail including safety considerations 6\nD1 method to stop the trolley (after passing point P), e.g. labelled block / buffer / cushion drawn after P\nor\nplace a block / buffer / cushion after P to stop the trolley\nD2 keep L, A, m and B constant\nd2\nD3 use micrometer / calipers to measure diameter (d) of the magnet and A =\n4\nD4 method to secure block to bench, e.g. clamp block to bench or (heavy) mass on top of block\nor\nmethod to secure magnets, e.g. use glue to stick magnets to trolley / block\nD5 method to increase the accuracy of measuring s or D, e.g. use a marker to left of the trolley\nD6 measure B using a (calibrated) Hall probe and adjust / rotate probe until maximum value\nor\nmeasure B using Hall probe first in one direction, then in the opposite direction and average\nD7 use a (top-pan) balance to measure m\nD8 use of strong magnets to increase v\nD9 repeat measurements of v for each value of s and average v\n 2DQ\nD10 relationship valid if a straight line is produced (passing through −  )\n m \nDo not accept line passing through the origin.\n© Cambridge University Press & Assessment 2024 Page 7 of 11", "source_pages": [ 5, 6, 7 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9702_w24_ms_51.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-Oct-Nov/9702_w24_ms_51.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_w24_ms_51-p05.png", "../answer-assets/9702_w24_ms_51-p06.png", "../answer-assets/9702_w24_ms_51-p07.png" ] }, { "id": "9702-2024-on-51-q02", "question_id": "9702-2024-on-51-q02", "subject": "9702", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "51", "question_number": 2, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "2(a) 3 1\ngradient =\n3E −E\ns\n2Z\ny-intercept =\n3E −E\ns\n\n2(b) 1\n1\n/ A−1\nI\n5150 or 5155\n5560 or 5556\n5810 or 5814\n6250\n6670 or 6667\n6940 or 6944\nValues correct as shown above.\n1 1\nUncertainties in from 50 or 60 to 90 or 100.\nI\n\n2(c)(i) Six points from (b) plotted correctly. 1\nMust be within half a small square. Diameter of points must be less than half a small square.\n1 1\nError bars in plotted correctly.\nI\nAll error bars to be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.\n© Cambridge University Press & Assessment 2024 Page 8 of 11\n\n2(c)(ii) Straight line of best fit drawn. 1\nDo not accept line from top point to bottom point.\nPoints must be balanced.\nLine must pass between (1.63, 5400) and (1.67, 5400) and between (2.58, 6800) and (2.62, 6800).\nWorst acceptable line drawn (steepest or shallowest possible line that passes through all the error bars). 1\nAll error bars must be plotted.\n\n2(c)(iii) Gradient determined with clear substitution of data points into y / x. 1\nDistance between data points must be greater than half the length of the drawn line.\nGradient determined of worst acceptable line with clear substitution of data points into y / x. 1\nuncertainty = (gradient of line of best fit – gradient of worst acceptable line)\nor\nuncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)\n\n2(c)(iv) y-intercept determined by substitution of correct point with consistent power of ten in m and x into y = mx + c. 1\ny-intercept of worst acceptable line determined by substitution into y = mx + c. 1\nuncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line\nor\nuncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)\nDo not accept ECF from false origin method.\n© Cambridge University Press & Assessment 2024 Page 9 of 11\n\n2(d)(i) E determined using gradient 1\nand\nE and Z given to 2 or 3 or 4 significant figures.\n1 3  3+gradientE 1 E\nE =  +E = s = + s\n3gradient s  3gradient gradient 3\n1\nE = +0.733\ngradient\nZ determined using y-intercept 1\nand\nE and Z given with SI units with correct powers of ten.\n(3E −E )y-intercept 3y-intercept\nZ = s or Z =\n\n2 2gradient\nUnit of E: V\nUnit of Z: \n\n2(d)(ii) Absolute uncertainty in E with method shown. 1\n gradient 1  0.05\nuncertainty=  +\n gradient gradient 3\nor\ncorrect substitution for max/min methods.\n© Cambridge University Press & Assessment 2024 Page 10 of 11\n\n2(e) Value of R determined to a minimum of two significant figures from (c)(iii) and (c)(iv) or (d)(i) with correct substitution and 1\ncorrect use of power of ten.\n1\n−y-intercept\n25010−6\nR =\ngradient\nor\n1 2Z\nR = −\ngradient25010−6 3\nor\n3E −2.2 2Z\nR = −\n325010−6 3\n© Cambridge University Press & Assessment 2024 Page 11 of 11", "source_pages": [ 8, 9, 10, 11 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9702_w24_ms_51.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-Oct-Nov/9702_w24_ms_51.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_w24_ms_51-p08.png", "../answer-assets/9702_w24_ms_51-p09.png", "../answer-assets/9702_w24_ms_51-p10.png", "../answer-assets/9702_w24_ms_51-p11.png" ] }, { "id": "9702-2024-on-52-q01", "question_id": "9702-2024-on-52-q01", "subject": "9702", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "52", "question_number": 1, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "1 Defining the problem\nL is the independent variable and  or temperature change/increase is the dependent variable 1\nor\nvary L and measure  or temperature change/increase\nkeep t constant 1\nMethods of data collection\nlabelled diagram of workable experiment including: 1\n• oil in a beaker/container (on a bench)\n• coil fully submerged in oil\n• (bulb of) thermometer in the oil\n• at least three labels from thermometer, coil or resistance wire, oil, beaker/container, clamp/stand, bench\nDo not accept other heating sources.\nmethod to determine V – diagram of workable circuit including: 1\n• power supply connected to wire\n• voltmeter positioned to measure V across the coil\nmeasure the initial and final temperature and find the difference  1\nmethod to determine t, e.g. use stopwatch/timer 1\nand\nmethod to determine L e.g. use a rule(r) to measure L / length of wire or e.g. using number of turns and measure the\ndiameter of the coil with rule(r) / calipers\n© Cambridge University Press & Assessment 2024 Page 5 of 10\n\n1 Method of Analysis\n\n1 1 1\nplot a graph of  against or equivalent, e.g. against \nL L\nDo not accept logarithms.\n\n1\n\n1 1\nfor  against for against \nL L\nAtV2 AtV2gradient\nK = K =\nmgradient m\n\n1\n\n1 1\nfor  against for against \nL L\nZ =−mKy-intercept Z = AtV2y-intercept\nor\nAtV2y-intercept\nZ =−\ngradient\n© Cambridge University Press & Assessment 2024 Page 6 of 10\n\n1 Additional detail including safety considerations 6\nD1 precaution linked to hot oil / beaker / wire, e.g. use of gloves to prevent burns from oil\nor\nprecaution linked to spillage of oil, e.g. perform experiment in a tray\nD2 keep A and m and V constant\nd2\nD3 use a micrometer to measure the diameter (d) of the wire and A =\n4\nD4 repeat measurements of d along the wire and average\nD5 method to reduce heat loss e.g. add insulation around the container / add a lid to the container\nD6 method to keep V constant, e.g. adjust / change a variable resistor / power supply to keep V or voltmeter reading\nconstant\nD7 use a balance to determine the mass of the oil\nand\nmass of oil = mass of (beaker + oil) − mass of beaker\nor\nplace beaker on balance and zero balance, then add oil and read balance\nD8 stir the oil for uniform temperature\nor\nkeep the initial temperature (of oil) constant\nD9 repeat the experiment for the same value of L and average  / average temperature change\n Z \nD10 relationship valid if a straight line is produced (passing through  − )\n mK \nDo not accept line passing through the origin.\nD11 method to determine L accurately, e.g. measure length of unwound coil\n© Cambridge University Press & Assessment 2024 Page 7 of 10", "source_pages": [ 5, 6, 7 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9702_w24_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-Oct-Nov/9702_w24_ms_52.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_w24_ms_52-p05.png", "../answer-assets/9702_w24_ms_52-p06.png", "../answer-assets/9702_w24_ms_52-p07.png" ] }, { "id": "9702-2024-on-52-q02", "question_id": "9702-2024-on-52-q02", "subject": "9702", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "52", "question_number": 2, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "2(a) gradient = Bn2 1\ny-intercept = −B\n\n2(b) 1\nd2 / cm2\n615 or 615.0\n458 or 458.0\n292 or 292.4\n216 or 216.1\n166 or 166.4\n139 or 139.2\nValues correct as shown above.\nUncertainties in d2 decreasing from 10 to 4 or 5. 1\n\n2(c)(i) Six points from (b) plotted correctly. 1\nMust be within half a small square. Diameter of points must be less than half a small square.\nError bars in d2 plotted correctly. 1\nAll error bars to be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.\n\n2(c)(ii) Straight line of best fit drawn. 1\nDo not accept line from top point to bottom point.\nPoints must be balanced.\nLine must pass between (1.90, 250) and (2.00, 250) and between (3.55, 500) and (3.65, 500).\nWorst acceptable line drawn (steepest or shallowest possible line that passes through all the error bars). 1\nAll error bars must be plotted.\n© Cambridge University Press & Assessment 2024 Page 8 of 10\n\n2(c)(iii) Gradient determined with clear substitution of data points into y / x. 1\nDistance between data points must be greater than half the length of the drawn line.\nGradient determined of worst acceptable line with clear substitution of data points into y / x. 1\nuncertainty = (gradient of line of best fit – gradient of worst acceptable line)\nor\nuncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)\n\n2(c)(iv) y-intercept determined by substitution of correct point with consistent power of ten in m and x into y = mx + c. 1\ny-intercept of worst acceptable line determined by substitution into y = mx + c. 1\nuncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line\nor\nuncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)\nDo not accept ECF from false origin method.\n\n2(d)(i) B determined using y-intercept (B = – y-intercept) and B and n given to 2 or 3 or 4 significant figures. 1\nn determined using gradient 1\nand\nB and n given with SI units with correct powers of ten.\ngradient gradient\nn= or n=\nB −y-intercept\nUnit for B: cm2\nNo unit for n.\n© Cambridge University Press & Assessment 2024 Page 9 of 10\n\n2(d)(ii) Percentage uncertainty in n determined with method shown. 1\n1y-intercept gradient\npercentage uncertainty=  + 100\n2 y-intercept gradient \nor\ncorrect substitution for max/min methods.\n\n2(e)  determined to a minimum of two significant figures from (c)(iii) and (c)(iv) or (d)(i) with correct substitution and correct 1\npower of ten.\ngradient\n=sin−1\n−y-intercept+900\nor\nn2B\n=sin−1\nB+900\n© Cambridge University Press & Assessment 2024 Page 10 of 10", "source_pages": [ 8, 9, 10 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9702_w24_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-Oct-Nov/9702_w24_ms_52.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_w24_ms_52-p08.png", "../answer-assets/9702_w24_ms_52-p09.png", "../answer-assets/9702_w24_ms_52-p10.png" ] }, { "id": "9702-2024-on-53-q01", "question_id": "9702-2024-on-53-q01", "subject": "9702", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "53", "question_number": 1, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "1 Defining the problem\ns is the independent variable and v is the dependent variable or vary s and measure v 1\nkeep D constant 1\nMethods of data collection\nlabelled diagram of workable experiment including: 1\n• light gate positioned at P\n• light gate connected to timer / data logger\n• labels for light gate and P and data logger / timer and at least one other label from block, magnet(s), trolley, s and D\nmeasure D with a rule(r) and measure L with a rule(r) or calipers 1\ndescription to determine v at P, e.g. (measure length of) card to interrupt beam 1\nmethod to measure s, e.g. use calipers 1\n© Cambridge University Press & Assessment 2024 Page 5 of 11\n\n1 Method of Analysis\n\n1 1 1\nplot a graph of v2 against or equivalent (e.g. against v2)\ns4 s4\nDo not accept logarithms.\nmgradient 1\nK =\n2DA2B2L2\nm 1\n(or K = for against v2)\n2DA2B2L2gradient s4\nmy-intercept 1\nQ=−\n2D\nmy-intercept 1\n(or Q =KA2B2L2y-intercept or Q = for against v2)\n2Dgradient s4\n© Cambridge University Press & Assessment 2024 Page 6 of 11\n\n1 Additional detail including safety considerations 6\nD1 method to stop the trolley (after passing point P), e.g. labelled block / buffer / cushion drawn after P\nor\nplace a block / buffer / cushion after P to stop the trolley\nD2 keep L, A, m and B constant\nd2\nD3 use micrometer / calipers to measure diameter (d) of the magnet and A =\n4\nD4 method to secure block to bench, e.g. clamp block to bench or (heavy) mass on top of block\nor\nmethod to secure magnets, e.g. use glue to stick magnets to trolley / block\nD5 method to increase the accuracy of measuring s or D, e.g. use a marker to left of the trolley\nD6 measure B using a (calibrated) Hall probe and adjust / rotate probe until maximum value\nor\nmeasure B using Hall probe first in one direction, then in the opposite direction and average\nD7 use a (top-pan) balance to measure m\nD8 use of strong magnets to increase v\nD9 repeat measurements of v for each value of s and average v\n 2DQ\nD10 relationship valid if a straight line is produced (passing through −  )\n m \nDo not accept line passing through the origin.\n© Cambridge University Press & Assessment 2024 Page 7 of 11", "source_pages": [ 5, 6, 7 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9702_w24_ms_53.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-Oct-Nov/9702_w24_ms_53.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_w24_ms_53-p05.png", "../answer-assets/9702_w24_ms_53-p06.png", "../answer-assets/9702_w24_ms_53-p07.png" ] }, { "id": "9702-2024-on-53-q02", "question_id": "9702-2024-on-53-q02", "subject": "9702", "year": 2024, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "53", "question_number": 2, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "2(a) 3 1\ngradient =\n3E −E\ns\n2Z\ny-intercept =\n3E −E\ns\n\n2(b) 1\n1\n/ A−1\nI\n5150 or 5155\n5560 or 5556\n5810 or 5814\n6250\n6670 or 6667\n6940 or 6944\nValues correct as shown above.\n1 1\nUncertainties in from 50 or 60 to 90 or 100.\nI\n\n2(c)(i) Six points from (b) plotted correctly. 1\nMust be within half a small square. Diameter of points must be less than half a small square.\n1 1\nError bars in plotted correctly.\nI\nAll error bars to be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.\n© Cambridge University Press & Assessment 2024 Page 8 of 11\n\n2(c)(ii) Straight line of best fit drawn. 1\nDo not accept line from top point to bottom point.\nPoints must be balanced.\nLine must pass between (1.63, 5400) and (1.67, 5400) and between (2.58, 6800) and (2.62, 6800).\nWorst acceptable line drawn (steepest or shallowest possible line that passes through all the error bars). 1\nAll error bars must be plotted.\n\n2(c)(iii) Gradient determined with clear substitution of data points into y / x. 1\nDistance between data points must be greater than half the length of the drawn line.\nGradient determined of worst acceptable line with clear substitution of data points into y / x. 1\nuncertainty = (gradient of line of best fit – gradient of worst acceptable line)\nor\nuncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)\n\n2(c)(iv) y-intercept determined by substitution of correct point with consistent power of ten in m and x into y = mx + c. 1\ny-intercept of worst acceptable line determined by substitution into y = mx + c. 1\nuncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line\nor\nuncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)\nDo not accept ECF from false origin method.\n© Cambridge University Press & Assessment 2024 Page 9 of 11\n\n2(d)(i) E determined using gradient 1\nand\nE and Z given to 2 or 3 or 4 significant figures.\n1 3  3+gradientE 1 E\nE =  +E = s = + s\n3gradient s  3gradient gradient 3\n1\nE = +0.733\ngradient\nZ determined using y-intercept 1\nand\nE and Z given with SI units with correct powers of ten.\n(3E −E )y-intercept 3y-intercept\nZ = s or Z =\n\n2 2gradient\nUnit of E: V\nUnit of Z: \n\n2(d)(ii) Absolute uncertainty in E with method shown. 1\n gradient 1  0.05\nuncertainty=  +\n gradient gradient 3\nor\ncorrect substitution for max/min methods.\n© Cambridge University Press & Assessment 2024 Page 10 of 11\n\n2(e) Value of R determined to a minimum of two significant figures from (c)(iii) and (c)(iv) or (d)(i) with correct substitution and 1\ncorrect use of power of ten.\n1\n−y-intercept\n25010−6\nR =\ngradient\nor\n1 2Z\nR = −\ngradient25010−6 3\nor\n3E −2.2 2Z\nR = −\n325010−6 3\n© Cambridge University Press & Assessment 2024 Page 11 of 11", "source_pages": [ 8, 9, 10, 11 ], "source_pdf": "_source-pdfs/2024-Oct-Nov/ms/9702_w24_ms_53.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2024-Oct-Nov/9702_w24_ms_53.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_w24_ms_53-p08.png", "../answer-assets/9702_w24_ms_53-p09.png", "../answer-assets/9702_w24_ms_53-p10.png", "../answer-assets/9702_w24_ms_53-p11.png" ] }, { "id": "9702-2025-m-42-q01", "question_id": "9702-2025-m-42-q01", "subject": "9702", "year": 2025, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 1, "topic": "Motion in a circle", "topic_slug": "9702-topic-12-motion-in-a-circle", "marks": 9, "status": "available", "reason": null, "text": "1(a) arrow vertically downwards labelled ‘weight’ and arrow perpendicular to cone, inwards and upwards, labelled ‘normal B1\ncontact force’\n\n1(b) vertical component of contact force = weight (so no resultant force vertically) B1\nhorizontal component of contact force is resultant force towards centre (of circle) B1\n\n1(c) a = v2 / r C1\na = g tan 52° C1\n9.81  tan 52° = v2 / 0.15 leading to v = 1.4 m s–1 A1\n\n1(d) v = r or a = r2 C1\n = 1.4 / 0.15 or √(9.81  tan 52° / 0.15) A1\n= 9.3 rads–1\n\n1(e) same resultant force / same acceleration so v2 is proportional to r (so if speed increases radius must also increase) A1\nQuestion Answer Marks", "source_pages": [ 7 ], "source_pdf": "_source-pdfs/2025-March/ms/9702_m25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-March/9702_m25_ms_42.pdf?download=true", "html": "9702-topic-12-motion-in-a-circle/answers.html", "image_paths": [ "../answer-assets/9702_m25_ms_42-p07.png" ] }, { "id": "9702-2025-m-42-q02", "question_id": "9702-2025-m-42-q02", "subject": "9702", "year": 2025, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 2, "topic": "Electric fields", "topic_slug": "9702-topic-18-electric-fields", "marks": 7, "status": "available", "reason": null, "text": "2(a) sketch: B1\nline from x = R to x = 4R entirely in the negative  region\ncurve with continuously decreasing magnitude and with gradient of continuously decreasing magnitude, starting at (R, ) B1\nline passing through (2R, ½) and (4R, ¼) B1\n\n2(b) horizontal straight line from t = 0 to t = 24 hours B1\nline starting at (0, –) B1\n© Cambridge University Press & Assessment 2025 Page 7 of 14\n\n2(c) straight line with non-zero gradient from 0 to d B1\nline with negative gradient from (0, V) to (d, 0) B1\nQuestion Answer Marks", "source_pages": [ 7, 8 ], "source_pdf": "_source-pdfs/2025-March/ms/9702_m25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-March/9702_m25_ms_42.pdf?download=true", "html": "9702-topic-18-electric-fields/answers.html", "image_paths": [ "../answer-assets/9702_m25_ms_42-p07.png", "../answer-assets/9702_m25_ms_42-p08.png" ] }, { "id": "9702-2025-m-42-q03", "question_id": "9702-2025-m-42-q03", "subject": "9702", "year": 2025, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 3, "topic": "Temperature", "topic_slug": "9702-topic-14-temperature", "marks": 12, "status": "available", "reason": null, "text": "3(a)(i) (P and Q are at the) same temperature B1\nno net transfer of thermal energy (between P and Q) B1\n\n3(a)(ii) Q = mcT C1\n24 103 = (0.54  390  T) + (0.37  910  T) C1\nT = 44K A1\n\n3(b)(i) work done = pV C1\n= (1.6  105)  (0.18 – 0.32) C1\n= –2.2  104J A1\n\n3(b)(ii) pV = NkT C1\nN = (1.6  105  0.18) / (1.38  10–23  273) A1\n= 7.6  1024\n\n3(b)(iii) ½m = (3 / 2)kT C1\nr.m.s. speed = √[(3  1.38  10–23  (210 + 273) / (4.7  10–26)] A1\n= 650ms–1\n© Cambridge University Press & Assessment 2025 Page 8 of 14", "source_pages": [ 8 ], "source_pdf": "_source-pdfs/2025-March/ms/9702_m25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-March/9702_m25_ms_42.pdf?download=true", "html": "9702-topic-14-temperature/answers.html", "image_paths": [ "../answer-assets/9702_m25_ms_42-p08.png" ] }, { "id": "9702-2025-m-42-q04", "question_id": "9702-2025-m-42-q04", "subject": "9702", "year": 2025, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 4, "topic": "Oscillations", "topic_slug": "9702-topic-17-oscillations", "marks": 12, "status": "available", "reason": null, "text": "4(a)  = 2 / T C1\n= 2 / (0.15  10–6) = 4.2  107 rads–1 A1\n\n4(b) a = 2x C1\n0 0\n= (4.2  107)2  40  10–6 A1\n= 7.1  1010ms–2\n\n4(c) E = ½m2x 2 C1\no\n= ½  2.4  10–4  (4.2  107)2  (40  10–6)2 C1\n= 340J A1\n\n4(d)(i) apply alternating p.d. (to / across crystal) B1\napplying p.d. to / across crystal causes it to distort B1\n\n4(d)(ii) Z = c C1\nZ = 1100  1600 (= 1.76  106)\nm\nZ = 1900  4100 (= 7.79  106)\nb\nintensity reflection co-efficient= [(7.79 – 1.76) / (7.79 + 1.76)]2 C1\n= 0.40 or 40%\npercentage transmitted = 60% A1\n© Cambridge University Press & Assessment 2025 Page 9 of 14", "source_pages": [ 9 ], "source_pdf": "_source-pdfs/2025-March/ms/9702_m25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-March/9702_m25_ms_42.pdf?download=true", "html": "9702-topic-17-oscillations/answers.html", "image_paths": [ "../answer-assets/9702_m25_ms_42-p09.png" ] }, { "id": "9702-2025-m-42-q05", "question_id": "9702-2025-m-42-q05", "subject": "9702", "year": 2025, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 5, "topic": "Capacitance", "topic_slug": "9702-topic-19-capacitance", "marks": 9, "status": "available", "reason": null, "text": "5(a) Q = Q = Q and V = V + V M1\n1 2 1 2\nV = Q / C so: A1\nQ / C = Q / C + Q / C leading to 1 / C = 1 / C + 1 / C\n1 2 1 2\n\n5(b) total capacitance = C + ½C = (3 / 2)C C1\ntotal capacitance = gradient C1\n= 400  10–6 / 6.0\neither: C = (2  400  10–6) / (3  6.0) = 4.4  10–5F = 44F A1\nor: C = (2  400) / (3  6.0) = 44F\n\n5(c)(i) τ = RC C1\n= 54  103  (3/2)  44  10–6 A1\n= 3.6s\n\n5(c)(ii) 0.15 = exp(–t / 3.6) C1\nt = 6.8s A1\nQuestion Answer Marks", "source_pages": [ 10 ], "source_pdf": "_source-pdfs/2025-March/ms/9702_m25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-March/9702_m25_ms_42.pdf?download=true", "html": "9702-topic-19-capacitance/answers.html", "image_paths": [ "../answer-assets/9702_m25_ms_42-p10.png" ] }, { "id": "9702-2025-m-42-q06", "question_id": "9702-2025-m-42-q06", "subject": "9702", "year": 2025, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 6, "topic": "Magnetic fields", "topic_slug": "9702-topic-20-magnetic-fields", "marks": 11, "status": "available", "reason": null, "text": "6(a) F = F B1\nB E\neither: Bqu = qE and E = V / d leading to u = V / Bd B1\nor: Bqu = qV / d leading to u = V / Bd\n© Cambridge University Press & Assessment 2025 Page 10 of 14\n\n6(b) E = ½mu2 C1\nK\nu = √[(2  4.1  10–17) / (3.2  10–27)] C1\n= 1.6  105ms–1\nB = 980 / (3.6  10–2  1.6  105) A1\n= 0.17T\n\n6(c) expression is independent of mass and charge A1\n\n6(d) either: electric force is downwards so magnetic force is upwards B1\nor: no resultant force so magnetic force is upwards\n(positive ions so) current is from left to right B1\nfrom (Fleming’s) left-hand rule, magnetic field is into the page B1\n\n6(e) curved path inside plates with consistent direction of curvature and with no discontinuity at entry or in curvature B1\ndirection of deflection is upwards B1\nQuestion Answer Marks", "source_pages": [ 10, 11 ], "source_pdf": "_source-pdfs/2025-March/ms/9702_m25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-March/9702_m25_ms_42.pdf?download=true", "html": "9702-topic-20-magnetic-fields/answers.html", "image_paths": [ "../answer-assets/9702_m25_ms_42-p10.png", "../answer-assets/9702_m25_ms_42-p11.png" ] }, { "id": "9702-2025-m-42-q07", "question_id": "9702-2025-m-42-q07", "subject": "9702", "year": 2025, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 7, "topic": "Magnetic fields", "topic_slug": "9702-topic-20-magnetic-fields", "marks": 9, "status": "available", "reason": null, "text": "7(a) (induced) e.m.f. is (directly) proportional to rate M1\nof change of (magnetic) flux (linkage) A1\n\n7(b)(i) (uniform acceleration so) velocity is (directly) proportional to time M1\n(Fig. 7.2 shows) e.m.f. is (directly) proportional to time so E is proportional to v. A1\n© Cambridge University Press & Assessment 2025 Page 11 of 14\n\n7(b)(ii) (v = at so) distance moved in time t = att C1\n = BA C1\nE = ( / t) = B  L  (att) / t = BLat A1\n\n7(b)(iii) B = (0.30  10–3) / (0.45  7.8  2.0) C1\n= 4.3  10–5T A1\nQuestion Answer Marks", "source_pages": [ 11, 12 ], "source_pdf": "_source-pdfs/2025-March/ms/9702_m25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-March/9702_m25_ms_42.pdf?download=true", "html": "9702-topic-20-magnetic-fields/answers.html", "image_paths": [ "../answer-assets/9702_m25_ms_42-p11.png", "../answer-assets/9702_m25_ms_42-p12.png" ] }, { "id": "9702-2025-m-42-q08", "question_id": "9702-2025-m-42-q08", "subject": "9702", "year": 2025, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 8, "topic": "Quantum physics", "topic_slug": "9702-topic-22-quantum-physics", "marks": 10, "status": "available", "reason": null, "text": "8(a) • quantum of energy M1\n• of electromagnetic radiation A1\n\n8(b)(i) ()E = hc /  C1\n = (6.63  10–34  3.00  108) / (1.96  1.60  10–19) C1\n= 6.3  10–7m A1\n\n8(b)(ii) number per unit time = power / energy per photon A1\n= (1.0  10–2) / (1.96  1.60  10–19)\n= 3.2  1016s–1\n© Cambridge University Press & Assessment 2025 Page 12 of 14\n\n8(b)(iii) either: force = rate of change of momentum C1\nor: F = p / t\np = E / c C1\nhalf the photons have change in momentum p, the other half have change in momentum 2p C1\nF = [(1.96  1.60  10–19) / (3.00  108)]  3.2  1016  [(2 + 1) / 2] A1\n= 5.0  10–11N\nQuestion Answer Marks", "source_pages": [ 12, 13 ], "source_pdf": "_source-pdfs/2025-March/ms/9702_m25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-March/9702_m25_ms_42.pdf?download=true", "html": "9702-topic-22-quantum-physics/answers.html", "image_paths": [ "../answer-assets/9702_m25_ms_42-p12.png", "../answer-assets/9702_m25_ms_42-p13.png" ] }, { "id": "9702-2025-m-42-q09", "question_id": "9702-2025-m-42-q09", "subject": "9702", "year": 2025, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 9, "topic": "Medical physics", "topic_slug": "9702-topic-24-medical-physics", "marks": 10, "status": "available", "reason": null, "text": "9(a)(i) either: cannot predict when a (particular) nucleus will decay B1\nor: cannot predict which nucleus will decay next\n\n9(a)(ii) not affected by external / environmental factors B1\n\n9(b) time for activity to halve B1\n\n9(c) energy = (189  7.826) + (4  7.074) – (193  7.774) C1\n= 7.03eV A1\n\n9(d)(i) decay constant A1\n\n9(d)(ii) decay constant / magnitude of gradient = 1.4 / 0.84 C1\nhalf-life = ln2 / (1.4 / 0.84) A1\n= 0.42ms\n\n9(e)(i) positrons collide with electrons and annihilate B1\n\n9(e)(ii) long enough to have time to conduct investigation, not so long as to cause patient unnecessary exposure to radiation B1\n© Cambridge University Press & Assessment 2025 Page 13 of 14", "source_pages": [ 13 ], "source_pdf": "_source-pdfs/2025-March/ms/9702_m25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-March/9702_m25_ms_42.pdf?download=true", "html": "9702-topic-24-medical-physics/answers.html", "image_paths": [ "../answer-assets/9702_m25_ms_42-p13.png" ] }, { "id": "9702-2025-m-42-q10", "question_id": "9702-2025-m-42-q10", "subject": "9702", "year": 2025, "session": "March", "session_code": "m", "paper": 4, "variant": "42", "question_number": 10, "topic": "Astronomy and cosmology", "topic_slug": "9702-topic-25-astronomy-and-cosmology", "marks": 11, "status": "available", "reason": null, "text": "10(a)(i) total power of radiation emitted (by the star) B1\n\n10(a)(ii) standard candle has known luminosity B1\nmeasure the radiant flux intensity B1\nuse F = L / (4d2) to calculate d B1\n\n10(b)(i) v = 2R / T C1\nv = 3.00  108  (656.2877 – 656.2831) / 656.2831 C1\nR = [3.00  108  (656.2877 – 656.2831) / 656.2831]  (2.07  106) / 2 = 6.93  108m A1\n\n10(b)(ii) Z is moving towards Earth M1\nso observed wavelength is less than the emitted wavelength A1\n\n10(b)(iii) L = 4 r2T4 C1\n3.8  1026 = 4  5.67  10–8  (6.93  108)2  T4\nT = 5800K A1\n© Cambridge University Press & Assessment 2025 Page 14 of 14", "source_pages": [ 14 ], "source_pdf": "_source-pdfs/2025-March/ms/9702_m25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-March/9702_m25_ms_42.pdf?download=true", "html": "9702-topic-25-astronomy-and-cosmology/answers.html", "image_paths": [ "../answer-assets/9702_m25_ms_42-p14.png" ] }, { "id": "9702-2025-m-52-q01", "question_id": "9702-2025-m-52-q01", "subject": "9702", "year": 2025, "session": "March", "session_code": "m", "paper": 5, "variant": "52", "question_number": 1, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "1 Defining the problem\nVary p and measure B OR p is the independent variable and B is the dependent variable. 1\nKeep V constant or potential difference between the ends of each conductor constant. 1\nMethods of data collection\nLabelled diagram of workable experiment including: 1\n• conductors in parallel connected in series to power supply and resistor\n• circuit symbols for (variable) resistor and power supply\n• X labelled and one other label from L, P, Q, p and q.\nVoltmeter connected in parallel with conductors (to measure V) and conductors in parallel connected to a power supply. 1\nMethod to measure L and p and q e.g. use a rule / ruler / calipers. 1\nMethod to measure B, e.g. use a (calibrated) Hall probe and adjust / rotate probe until maximum value. 1\nMethod of Analysis\n\n1 1\nPlots a graph of B against or equivalent.\np\nDo not accept logarithms.\n\n1\n\n1 1\nB against against B\np p\nLgradient L\nY = Y =\nAV AV gradient\n© Cambridge University Press & Assessment 2025 Page 7 of 12\n\n1 1\n\n1 1\nB against against B\np p\nqy-intercept Z =−qy-intercept\nZ =\ngradient\nOR\nLqy-intercept\nZ =\nYAV\nAdditional detail including safety considerations 6\nAny six from:\nD1 precaution linked to high current / hot conductors, e.g. use gloves / switch off power supply when not measuring\nB / between measurements / allow conductors to cool\nD2 keep A and L and q constant\nd2\nD3 use calipers / micrometer to measure diameter / d of conductor and A=\n4\nD4 repeat measurements of d in different positions and average d\nD5 method to determine the position of X in relation to the conductors, e.g. divide L by two to find the midpoint of P / Q and\nuse a set square / protractor / plumb line to mark X\nOR\ndivide L by two to find the midpoint of P / Q and use a grid to mark X\nD6 measure B (using Hall probe) first in one direction and then in the opposite direction and average B\nOR\nMeasure B with current / p.d. in one direction and then in the opposite direction and average B\nD7 additional detail on measuring p and / or q, e.g. measure to the conductor and add on the radius\n© Cambridge University Press & Assessment 2025 Page 8 of 12\n\n1 D8 description of method to keep q constant, e.g. tape / adhesive putty to fix conductor Q to the bench OR for vertical\nmethods fix conductor Q in clamp(s) attached to stand(s) to keep q constant\nD9 method to keep P and Q parallel, e.g. measure the separation (between the conductors) at different points\nYZAV\nD10 relationship valid if a straight line is produced (with a y-intercept = ).\nLq\nDo not accept passing through the origin.\nD11 method to keep V constant, e.g. adjust / change variable resistor / power supply to keep voltmeter reading constant.\nQuestion Answer Marks", "source_pages": [ 7, 8, 9 ], "source_pdf": "_source-pdfs/2025-March/ms/9702_m25_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-March/9702_m25_ms_52.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_m25_ms_52-p07.png", "../answer-assets/9702_m25_ms_52-p08.png", "../answer-assets/9702_m25_ms_52-p09.png" ] }, { "id": "9702-2025-m-52-q02", "question_id": "9702-2025-m-52-q02", "subject": "9702", "year": 2025, "session": "March", "session_code": "m", "paper": 5, "variant": "52", "question_number": 2, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "2(a) 1 1\ngradient = −\nK\ny-intercept = ln( − )\n0 R\n© Cambridge University Press & Assessment 2025 Page 9 of 12\n\n2(b)\n( – ) / C ln (( – ) / C)\nR R\n56.5  1.0 4.034 or 4.0342  0.018\n46.0  1.0 3.829 or 3.8286  0.022\n38.5  1.0 3.651 or 3.6507  0.026\n31.5  1.0 3.450 or 3.4500  0.032\n26.0  1.0 3.258 or 3.2581  0.038\n22.5  1.0 3.114 or 3.1135  0.044\nValues of ( –  R) / C and ln (( –  R ) / C) 1\nUncertainties in ( –  R) and ln (( –  R ) / C) 1\n\n2(c)(i) Six points from (b) plotted correctly. 1\nMust be within half a small square. Diameter of points must be less than half a small square.\nError bars in ln (( – ) / C) plotted correctly. 1\nR\nAll error bars to be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.\n© Cambridge University Press & Assessment 2025 Page 10 of 12\n\n2(c)(ii) Straight line of best fit drawn. 1\nDo not accept line from top plot to bottom plot.\nLine must pass between\n(31.5, 3.2) and (33.0, 3.2) and between\n(16.0, 3.7) and (17.0, 3.7)\nWorst acceptable line drawn. 1\nSteepest or shallowest possible line that passes through all the error bars.\nAll error bars must be plotted.\n\n2(c)(iii) Gradient must be negative. 1\nGradient determined with clear substitution of data into y / x; distance between data points must be greater than half the\nlength of the drawn line.\nGradient determined of worst acceptable line with clear substitution of data into y / x; 1\nuncertainty = (gradient of line of best fit – gradient of worst acceptable line)\nor\nuncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)\n\n2(c)(iv) y-intercept determined by substitution of correct point with consistent unit of time into y = mx + c 1\ny-intercept of worst acceptable line determined by substitution into y = mx + c. 1\nuncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line, or\nuncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)\nDo not accept ecf from false origin method.\n© Cambridge University Press & Assessment 2025 Page 11 of 12\n\n2(d)(i) K determined using gradient and 1\nK and  given to 3 or 4 sf.\n0\n1\nK =−\ngradient\n determined using y-intercept and 1\n0\nK and \n0\ngiven with units with appropriate powers of ten\n =ey-intercept +18.5\n0\nUnit of K: min or minute(s)\nunit of : C\n0\n\n2(d)(ii) Absolute uncertainty determined with clear method shown. 1\n = ( emaxy-intercept +19 ) −(ey-intercept +18.5)\n0\nOR\n = (ey-intercept +18.5)− ( eminy-intercept +18 )\n0\nOR\n( emaxy-intercept +19 ) − ( eminy-intercept +18 )\n =\n0 2\n\n2(e) t determined to a minimum of 2sf from (c)(iii) and (c)(iv) OR (d)(i) with correct substitution and correct power of ten. 1\nln(25.0−18.5)−y-intercept\nt =\ngradient\nOR\nt =−K(ln(25.0−18.5)−y-intercept)\nOR\n25.0−18.5\nt =−Kln \n −18.5\n \n0\n© Cambridge University Press & Assessment 2025 Page 12 of 12", "source_pages": [ 9, 10, 11, 12 ], "source_pdf": "_source-pdfs/2025-March/ms/9702_m25_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-March/9702_m25_ms_52.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_m25_ms_52-p09.png", "../answer-assets/9702_m25_ms_52-p10.png", "../answer-assets/9702_m25_ms_52-p11.png", "../answer-assets/9702_m25_ms_52-p12.png" ] }, { "id": "9702-2025-mj-41-q01", "question_id": "9702-2025-mj-41-q01", "subject": "9702", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 1, "topic": "Gravitational fields", "topic_slug": "9702-topic-13-gravitational-fields", "marks": 9, "status": "available", "reason": null, "text": "1(a) work done per unit mass B1\nwork (done in) moving mass from infinity (to the point) B1\n\n1(b)(i) evidence of addition of 3.4 × 106 to 1.7 × 106 or 6.8 × 106 C1\nGM × 122 / (5.1 × 106) or GM × 122 / (10.2 × 106) C1\n6.67 × 10–11 × M × 122 × [(5.1 × 106)–1 – (10.2 × 106)–1] = 5.1 × 108 A1\nleading to M = 6.4 × 1023 kg\n\n1(b)(ii)  = (–) (6.67 × 10–11 × 6.4 × 1023) / (3.4 × 106) C1\n= –1.3 × 107 J kg–1 A1\n\n1(c)(i) Mars takes (just under) 25 hours to rotate once on its axis B1\n\n1(c)(ii) orbit is equatorial B1\nor\norbit is in same direction as direction of rotation of Mars\n© Cambridge University Press & Assessment 2025 Page 8 of 19", "source_pages": [ 8 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9702_s25_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-May-June/9702_s25_ms_41.pdf?download=true", "html": "9702-topic-13-gravitational-fields/answers.html", "image_paths": [ "../answer-assets/9702_s25_ms_41-p08.png" ] }, { "id": "9702-2025-mj-41-q02", "question_id": "9702-2025-mj-41-q02", "subject": "9702", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 2, "topic": "Nuclear physics", "topic_slug": "9702-topic-23-nuclear-physics", "marks": 11, "status": "available", "reason": null, "text": "2(a) (electric) force is (directly) proportional to product of charges B1\nforce (between point charges) is inversely proportional to the square of their separation B1\n\n2(b)(i) charge = (+)2e A1\n\n2(b)(ii) F = 2 × (1.60 × 10–19)2 / [4 × 8.85 × 10–12 × (170 × 10–12)2] = 1.6 × 10–8 N A1\n\n2(c)(i) F = mv2 / r C1\nv = [ (1.6 × 10–8 × 170 × 10–12) / (9.11 × 10–31) ]½ A1\n= 1.7 × 106 m s–1\n\n2(c)(ii) F = mr2 and  = 2 / T C1\nF = 42mr / T2\nT = [ (42 × 9.11 × 10–31 × 170 × 10–12) / (1.6 × 10–8) ]½ A1\n= 6.2 × 10–16 s\nor\nv = 2r / T (C1)\nT = (2 × 170 × 10–12) / (1.73 × 106) (A1)\n= 6.2 × 10–16 s\n© Cambridge University Press & Assessment 2025 Page 9 of 19\n\n2(d)(i) E  Q / r2 C1\nratio = [1.60 × 10–19 × (170 × 10–12)2] / [3.2 × 10–19 × (340 × 10–12)2] A1\n= 0.13\n\n2(d)(ii) resultant force slightly less (than 1.6 × 10–8 N) so speed lower B1\nor\nresultant force slightly less (than 1.6 × 10–8 N) so period greater\n© Cambridge University Press & Assessment 2025 Page 10 of 19", "source_pages": [ 9, 10 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9702_s25_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-May-June/9702_s25_ms_41.pdf?download=true", "html": "9702-topic-23-nuclear-physics/answers.html", "image_paths": [ "../answer-assets/9702_s25_ms_41-p09.png", "../answer-assets/9702_s25_ms_41-p10.png" ] }, { "id": "9702-2025-mj-41-q03", "question_id": "9702-2025-mj-41-q03", "subject": "9702", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 3, "topic": "Temperature", "topic_slug": "9702-topic-14-temperature", "marks": 9, "status": "available", "reason": null, "text": "3(a) (thermal) energy per unit mass (to cause state change) B1\n(thermal) energy to change state at constant temperature B1\n\n3(b) (for vaporisation): B1\ninvolves greater change in volume (of substance)\nor\ninvolves greater increase in separation of molecules\nmore work has to be done by molecules (to separate) M1\nor\ngreater increase in potential energy of molecules\nkinetic energy of molecules unchanged, so more thermal energy needed A1\n\n3(c) Q = mc and Q = mL C1\n for the water = 26.4 – 10.3 C1\n(37.0 × L) + (37.0 × 4.18 × 10.3) = (208 × 4.18 × 16.1) C1\nL = 335 J g–1 A1\n© Cambridge University Press & Assessment 2025 Page 11 of 19", "source_pages": [ 11 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9702_s25_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-May-June/9702_s25_ms_41.pdf?download=true", "html": "9702-topic-14-temperature/answers.html", "image_paths": [ "../answer-assets/9702_s25_ms_41-p11.png" ] }, { "id": "9702-2025-mj-41-q04", "question_id": "9702-2025-mj-41-q04", "subject": "9702", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 4, "topic": "Thermodynamics", "topic_slug": "9702-topic-16-thermodynamics", "marks": 8, "status": "available", "reason": null, "text": "4(a)(i) sum of potential energy and kinetic energy B1\n(total) energy of random motion of particles B1\n\n4(a)(ii) potential energy (of molecules) (in an ideal gas) is zero, so the internal energy of the gas is equal to the total kinetic energy B1\n(of molecules)\nkinetic energy of molecules is proportional to (thermodynamic) temperature (so internal energy is proportional to B1\n(thermodynamic) temperature))\n\n4(b) cooling work done = 0 B1\ncompression increase in internal energy = +2U B1\ncooling change in internal energy = –U B1\nboth rows: thermal energy adds to work to give increase in internal energy in terms of U and/or W B1\n(if fully correct, thermal energy for compression = 2U – W and thermal energy for cooling = –U:\ncompression +W 2U – W +2U\ncooling 0 –U –U\n)\n© Cambridge University Press & Assessment 2025 Page 12 of 19", "source_pages": [ 12 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9702_s25_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-May-June/9702_s25_ms_41.pdf?download=true", "html": "9702-topic-16-thermodynamics/answers.html", "image_paths": [ "../answer-assets/9702_s25_ms_41-p12.png" ] }, { "id": "9702-2025-mj-41-q05", "question_id": "9702-2025-mj-41-q05", "subject": "9702", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 5, "topic": "Oscillations", "topic_slug": "9702-topic-17-oscillations", "marks": 8, "status": "available", "reason": null, "text": "5(a)(i) amplitude = 0.60 m A1\n\n5(a)(ii) oscillations are simple harmonic B1\n\n5(b) Any three points from: B3\n• mean / equilibrium position is at h = 1.4 m\n• total energy of oscillations = 9.0 J\n• angular frequency of oscillations = 1.2 rad s–1\nor\nperiod of oscillations = 5.1 s\nor\nfrequency of oscillation = 0.19 Hz\n• maximum speed of block = 0.73 m s–1\n• mass of block = 33 kg\n\n5(c) U-shaped curve resting on h axis (with minimum at E = 0) B1\nP\ncurve from h = 0.8 m to h = 2.0 m, with minimum E shown at h = 1.4 m B1\nP\nboth end-points of curve shown at E = 9.0 J B1\nP\n© Cambridge University Press & Assessment 2025 Page 13 of 19", "source_pages": [ 13 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9702_s25_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-May-June/9702_s25_ms_41.pdf?download=true", "html": "9702-topic-17-oscillations/answers.html", "image_paths": [ "../answer-assets/9702_s25_ms_41-p13.png" ] }, { "id": "9702-2025-mj-41-q06", "question_id": "9702-2025-mj-41-q06", "subject": "9702", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 6, "topic": "Alternating currents", "topic_slug": "9702-topic-21-alternating-currents", "marks": 12, "status": "available", "reason": null, "text": "6(a)(i) conversion from a.c. to d.c. B1\n\n6(a)(ii) smoothing B1\n\n6(b)(i) A = 12 V A1\nB = 2 / (20 × 10–3) A1\n= 310 rad s–1\n\n6(b)(ii) full-wave (rectification) B1\n\n6(b)(iii) four diodes shown, with correct circuit symbols B1\nfour diodes correctly connected to form a bridge rectifier B1\n\n6(b)(iv) V = V exp (–t / ) C1\n0\nor\nV = V exp (–t / RC) and  = RC\n0\n8.0 = 12 exp (– 7.3 × 10–3 / ) C1\n = 0.018 s A1\n\n6(c) time constant = RC C1\nR = (0.018 / 570 × 10–6) A1\n= 32 \n© Cambridge University Press & Assessment 2025 Page 14 of 19", "source_pages": [ 14 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9702_s25_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-May-June/9702_s25_ms_41.pdf?download=true", "html": "9702-topic-21-alternating-currents/answers.html", "image_paths": [ "../answer-assets/9702_s25_ms_41-p14.png" ] }, { "id": "9702-2025-mj-41-q07", "question_id": "9702-2025-mj-41-q07", "subject": "9702", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 7, "topic": "Magnetic fields", "topic_slug": "9702-topic-20-magnetic-fields", "marks": 10, "status": "available", "reason": null, "text": "7(a) • force per unit length B2\n• force per unit current\n• length / current perpendicular to field\n1 mark for any two points, 2 marks for all three points.\n\n7(b)(i) F = BQv B1\n\n7(b)(ii) arrow at Y pointing vertically upwards B1\n\n7(b)(iii) upwards deflection showing circular path B1\n\n7(c)(i) electric field applied vertically downwards (may be shown on a labelled diagram) B1\nelectric force on particle in opposite direction to magnetic force (may be shown on a labelled diagram) B1\nparticle undeflected when magnitudes of electric and magnetic forces are equal B1\n\n7(c)(ii) EQ = BQv B1\nv = E / B A1\n© Cambridge University Press & Assessment 2025 Page 15 of 19", "source_pages": [ 15 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9702_s25_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-May-June/9702_s25_ms_41.pdf?download=true", "html": "9702-topic-20-magnetic-fields/answers.html", "image_paths": [ "../answer-assets/9702_s25_ms_41-p15.png" ] }, { "id": "9702-2025-mj-41-q08", "question_id": "9702-2025-mj-41-q08", "subject": "9702", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 8, "topic": "Quantum physics", "topic_slug": "9702-topic-22-quantum-physics", "marks": 13, "status": "available", "reason": null, "text": "8(a) wavelength associated with a moving particle B1\n\n8(b)  = h / p C1\n= (6.63 × 10–34) / (9.11 × 10–31 × 4.9 × 107) A1\n= 1.5 × 10–11 m\n\n8(c) similarity: any one point from: B1\n• same mass\n• same magnitude of charge\n• both leptons\ndifference: any one point from: B1\n• electron has negative charge, positron has positive charge\n• positron is anti-particle of electron\n• electron is a particle, positron is an anti-particle\n\n8(d)(i) (pair) annihilation B1\n\n8(d)(ii) their mass gets converted into energy B1\n(their mass–energy) becomes the energy of the gamma photons B1\n\n8(d)(iii) they travel in opposite directions to conserve momentum B1\n\n8(d)(iv) kinetic energy = ½ × 9.11 × 10–31 × (4.9 × 107)2 = 1.1 × 10–15 J A1\n© Cambridge University Press & Assessment 2025 Page 16 of 19\n\n8(d)(v) E = mc2 C1\nE = hc /  C1\nor\nE = hf and c = f\n(1.1 × 10–15) + (9.11 × 10–31 × (3.00 × 108)2) = (6.63 × 10–34 × 3.00 × 108) /  A1\n = 2.39 × 10–12 m\n© Cambridge University Press & Assessment 2025 Page 17 of 19", "source_pages": [ 16, 17 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9702_s25_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-May-June/9702_s25_ms_41.pdf?download=true", "html": "9702-topic-22-quantum-physics/answers.html", "image_paths": [ "../answer-assets/9702_s25_ms_41-p16.png", "../answer-assets/9702_s25_ms_41-p17.png" ] }, { "id": "9702-2025-mj-41-q09", "question_id": "9702-2025-mj-41-q09", "subject": "9702", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 9, "topic": "Nuclear physics", "topic_slug": "9702-topic-23-nuclear-physics", "marks": 10, "status": "available", "reason": null, "text": "9(a) number of nuclear disintegrations per unit time B1\n\n9(b) activity is proportional to the number of undecayed nuclei B1\nactivity = (–) rate of change of number of undecayed nuclei B1\nN is proportional to the rate of change of N (so exponential variation) B1\n\n9(c)(i) 120 = 180 exp (–  × 8.4) C1\n = 0.048 min–1 A1\n\n9(c)(ii) half-life = ln 2 / 0.048 A1\n= 14 min\n\n9(c)(iii) line with negative gradient throughout, starting at (0, 180) B1\ncurve with negative gradient passing through (8.4, 120) B1\ncurve with decreasing negative gradient, from t = 0 to t = 24 min, passing through (14, 90) B1\n© Cambridge University Press & Assessment 2025 Page 18 of 19", "source_pages": [ 18 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9702_s25_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-May-June/9702_s25_ms_41.pdf?download=true", "html": "9702-topic-23-nuclear-physics/answers.html", "image_paths": [ "../answer-assets/9702_s25_ms_41-p18.png" ] }, { "id": "9702-2025-mj-41-q10", "question_id": "9702-2025-mj-41-q10", "subject": "9702", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "41", "question_number": 10, "topic": "Astronomy and cosmology", "topic_slug": "9702-topic-25-astronomy-and-cosmology", "marks": 10, "status": "available", "reason": null, "text": "10(a) speed is (directly) proportional to distance M1\nspeed is speed of recession of galaxy from an observer, and distance is the distance of the galaxy from the observer A1\n\n10(b)(i) galaxy is receding from the Earth B1\nobserved wavelength is redshifted from emitted wavelength B1\n\n10(b)(ii)  /  = v / c C1\n(4.91 – 4.62) / 4.62 = v / (3.00 × 108)\nv = 1.9 × 107 m s–1 A1\n\n10(b)(iii) wavelength (of maximum intensity) is inversely proportional to temperature B1\nobserved wavelength too high, so determined temperature too low B1\n\n10(c) v = H d C1\n0\nd = (1.9 × 107) / (2.3 × 10–18) A1\n= 8.3 × 1024 m\n© Cambridge University Press & Assessment 2025 Page 19 of 19", "source_pages": [ 19 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9702_s25_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-May-June/9702_s25_ms_41.pdf?download=true", "html": "9702-topic-25-astronomy-and-cosmology/answers.html", "image_paths": [ "../answer-assets/9702_s25_ms_41-p19.png" ] }, { "id": "9702-2025-mj-42-q01", "question_id": "9702-2025-mj-42-q01", "subject": "9702", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 1, "topic": "Motion in a circle", "topic_slug": "9702-topic-12-motion-in-a-circle", "marks": 10, "status": "available", "reason": null, "text": "1(a) angle (subtended at centre of a circle) when arc (length) = radius B1\n\n1(b)(i) v = r C1\n = 17 / 0.46 A1\n= 37 rad s–1\n\n1(b)(ii) T = 2r / v or T = 2 /  C1\n= 2 × 0.46 / 17 or 2 / 37 A1\n= 0.17 s\n\n1(b)(iii) distance = 2 × 0.038 = 0.24 m A1\n\n1(b)(iv) angle = arc length / radius C1\n= 0.24 / 0.15 A1\n= 1.6 rad\n\n1(c) point X moves through a smaller distance in the same time B1\nor\n(linear) speed of movement of point X / chain decreases\nor\n(linear) speed of (circumference of) both cogs decreases\nangular speed (of pedals) decreases B1\n© Cambridge University Press & Assessment 2025 Page 8 of 18", "source_pages": [ 8 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9702_s25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-May-June/9702_s25_ms_42.pdf?download=true", "html": "9702-topic-12-motion-in-a-circle/answers.html", "image_paths": [ "../answer-assets/9702_s25_ms_42-p08.png" ] }, { "id": "9702-2025-mj-42-q02", "question_id": "9702-2025-mj-42-q02", "subject": "9702", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 2, "topic": "Magnetic fields", "topic_slug": "9702-topic-20-magnetic-fields", "marks": 12, "status": "available", "reason": null, "text": "2(a)(i) direction of force B1\nforce acting on a (test) mass B1\n\n2(a)(ii) at least four radial lines from the Earth’s surface, equally spaced around the surface B1\narrows indicating direction towards Earth B1\n\n2(b)(i) top pole labelled S and bottom pole labelled N B1\n\n2(b)(ii) solenoid field pattern at the poles: B1\nat least two field lines either side of both poles, close to the poles, clustered closely together, leaving the surface\napproximately perpendicularly to the surface and curving away from the axis of the poles as their distance from the surface\nincreases\nsolenoid field pattern above the equator: B1\nat least one field line either side of the Earth connecting two points on the surface that are on the same side of the poles,\none north of the equator and one south of it, passing above the surface near the magnetic equator approximately parallel to\nthe surface\n\n2(c)(i) (around the surface) lines are evenly spaced B1\nall lines perpendicular to surface B1\nor\npointing down towards surface (at all points around the surface)\n© Cambridge University Press & Assessment 2025 Page 9 of 18\n\n2(c)(ii) Any three bulleted points from: B3\n• strongest at the poles\n• weakest near the Equator\nUp to two points from:\n• perpendicular to surface at the poles\n• parallel to the surface near the Equator\n• angle to surface increases from Equator to poles\nQuestion Answer Marks", "source_pages": [ 9, 10 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9702_s25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-May-June/9702_s25_ms_42.pdf?download=true", "html": "9702-topic-20-magnetic-fields/answers.html", "image_paths": [ "../answer-assets/9702_s25_ms_42-p09.png", "../answer-assets/9702_s25_ms_42-p10.png" ] }, { "id": "9702-2025-mj-42-q03", "question_id": "9702-2025-mj-42-q03", "subject": "9702", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 3, "topic": "Thermodynamics", "topic_slug": "9702-topic-16-thermodynamics", "marks": 13, "status": "available", "reason": null, "text": "3(a) (thermal) energy per unit mass (to cause temperature change) B1\n(thermal) energy per unit change in temperature B1\n\n3(b)(i) density = mass / volume C1\nmass = 2.700 × 103 × 3.612 × 10–3 A1\n= 9.752 kg\n\n3(b)(ii) volume = 3.612 × 10–3 × (2.700 / 2.620) = 3.722 × 10–3 m3 A1\nor\nvolume = 9.752 / (2.620 × 103) = 3.722 × 10–3 m3\n© Cambridge University Press & Assessment 2025 Page 10 of 18\n\n3(b)(iii) W = pV C1\n= 1.01 × 105 × (3.722 – 3.612) × 10–3 A1\n= 11.1 J\n\n3(b)(iv) volume (of block) increases B1\nwork is done against the atmosphere so work done (on block) is negative B1\n\n3(b)(v) thermal energy = (4.38 × 106) + 11.1 B1\nspecific heat capacity = (4.38 × 106) / (9.75 × 500) C1\n= 898 J kg–1 °C–1 A1\n\n3(c) work done is negligible compared with (change in) internal energy so (answer in (b)(v) would be) unchanged B1\n© Cambridge University Press & Assessment 2025 Page 11 of 18", "source_pages": [ 10, 11 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9702_s25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-May-June/9702_s25_ms_42.pdf?download=true", "html": "9702-topic-16-thermodynamics/answers.html", "image_paths": [ "../answer-assets/9702_s25_ms_42-p10.png", "../answer-assets/9702_s25_ms_42-p11.png" ] }, { "id": "9702-2025-mj-42-q04", "question_id": "9702-2025-mj-42-q04", "subject": "9702", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 4, "topic": "Ideal gases", "topic_slug": "9702-topic-15-ideal-gases", "marks": 8, "status": "available", "reason": null, "text": "4(a)(i) thermodynamic temperature B1\n\n4(a)(ii) molar gas constant B1\n\n4(b)(i) m: mass of one molecule (of the gas) B1\n〈c2〉: mean-square speed (of molecules) B1\n\n4(b)(ii) 1 M1\nNBT / A = Nm〈c2〉\n3\n1 A1\nclear use of E = m〈c2〉 leading to E = 3BT / 2A\nK K\n2\n\n4(c) line with positive gradient passing through the origin B1\nsmooth curve with decreasing positive gradient B1\n© Cambridge University Press & Assessment 2025 Page 12 of 18", "source_pages": [ 12 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9702_s25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-May-June/9702_s25_ms_42.pdf?download=true", "html": "9702-topic-15-ideal-gases/answers.html", "image_paths": [ "../answer-assets/9702_s25_ms_42-p12.png" ] }, { "id": "9702-2025-mj-42-q05", "question_id": "9702-2025-mj-42-q05", "subject": "9702", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 5, "topic": "Oscillations", "topic_slug": "9702-topic-17-oscillations", "marks": 9, "status": "available", "reason": null, "text": "5(a) (motion in which) acceleration is (directly) proportional to displacement B1\n(motion in which) B1\nacceleration is (always) in the opposite direction to displacement\nor\nacceleration is (always) directed towards a fixed point\n\n5(b)(i) period = 2 / 16 A1\n= 0.39 s\n\n5(b)(ii) v = x or v =  ( x 2 −02) C1\n0 0 0 0\nx = 0.56 / 16 A1\n0\n= 0.035 m\n\n5(b)(iii) v = ±16 √(0.0352 – x2) A1\n\n5(b)(iv) closed loop surrounding the origin B1\nloop crosses v = 0 at maximum values of x at x = ± 3.5 cm B1\nloop crosses x = 0 at maximum values of v at v = ± 0.56 m s–1 B1\n© Cambridge University Press & Assessment 2025 Page 13 of 18", "source_pages": [ 13 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9702_s25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-May-June/9702_s25_ms_42.pdf?download=true", "html": "9702-topic-17-oscillations/answers.html", "image_paths": [ "../answer-assets/9702_s25_ms_42-p13.png" ] }, { "id": "9702-2025-mj-42-q06", "question_id": "9702-2025-mj-42-q06", "subject": "9702", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 6, "topic": "Electric fields", "topic_slug": "9702-topic-18-electric-fields", "marks": 11, "status": "available", "reason": null, "text": "6(a) plate X marked as negative and plate Y marked as positive B1\n\n6(b)(i) E = V / x C1\n= (58 × 103) / 0.041 A1\n= 1.4 × 106 N C–1\n\n6(b)(ii) ma = eE C1\na = (1.60 × 10–19 × 1.41 × 106) / (9.11 × 10–31) A1\n= 2.5 × 1017 m s–2\n\n6(c)(i) eV = hc /  C1\nor\neV = hf and f = c / \n(1.60 × 10–19 × 58 × 103) = (6.63 × 10–34 × 3.00 × 108) /  M1\nclear conversion from m to pm leading to  = 21 pm A1\n\n6(c)(ii) X-rays B1\n\n6(c)(iii) Any two points from: B2\n• waves are passed into structure and transmitted waves detected\n• different parts of the structure absorb different fractions of energy\n• difference in detected / transmitted intensities used (to form image)\n© Cambridge University Press & Assessment 2025 Page 14 of 18", "source_pages": [ 14 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9702_s25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-May-June/9702_s25_ms_42.pdf?download=true", "html": "9702-topic-18-electric-fields/answers.html", "image_paths": [ "../answer-assets/9702_s25_ms_42-p14.png" ] }, { "id": "9702-2025-mj-42-q07", "question_id": "9702-2025-mj-42-q07", "subject": "9702", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 7, "topic": "Capacitance", "topic_slug": "9702-topic-19-capacitance", "marks": 10, "status": "available", "reason": null, "text": "7(a) V = V B1\nC R\n\n7(b)(i) C = Q / V C1\n= (7.2 × 10–3) / 12 (= 6.0 × 10–4 F) A1\n= 600 F\n\n7(b)(ii) R = V / I C1\n= 12 / (1.5 × 10–3) (= 8000 ) A1\n= 8.0 k\n\n7(b)(iii)  = RC C1\n= 8000 × 6.0 × 10–4 A1\n= 4.8 s\n\n7(c) Any two points from: B2\n• charge and current are both (directly) proportional to voltage\n• charge is (directly) proportional to current\n• current is the rate of change of charge\nQ is proportional to the rate of change of Q (so exponential variation) B1\n© Cambridge University Press & Assessment 2025 Page 15 of 18", "source_pages": [ 15 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9702_s25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-May-June/9702_s25_ms_42.pdf?download=true", "html": "9702-topic-19-capacitance/answers.html", "image_paths": [ "../answer-assets/9702_s25_ms_42-p15.png" ] }, { "id": "9702-2025-mj-42-q08", "question_id": "9702-2025-mj-42-q08", "subject": "9702", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 8, "topic": "Alternating currents", "topic_slug": "9702-topic-21-alternating-currents", "marks": 10, "status": "available", "reason": null, "text": "8(a)(i) half-wave (rectification) B1\n\n8(a)(ii) A1\nV = 6.0 × 2\n0\n= 8.5 V\n\n8(b)(i) P = V2 / R C1\nP = 8.52 / 45 A1\n0\n= 1.6 W\n\n8(b)(ii) two humps of width 0.5T and two sections of zero power of width 0.5T B1\nall humps drawn have width 0.5T, minima at P = 0 and peaks at P = P B1\n0\ncorrect sinusoidal shape, with smooth troughs sitting on t-axis at P = 0 B1\n\n8(b)(iii) 1 B1\nmean power within each hump is P from the symmetry of the curve\n2 0\nadditional half factor from removal of half of the power in each cycle B1\n\n8(b)(iv) 〈P〉 = V 2 / R A1\nr.m.s.\nV = (1.6/4)45\nr.m.s.  \n= 4.2 V\n© Cambridge University Press & Assessment 2025 Page 16 of 18", "source_pages": [ 16 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9702_s25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-May-June/9702_s25_ms_42.pdf?download=true", "html": "9702-topic-21-alternating-currents/answers.html", "image_paths": [ "../answer-assets/9702_s25_ms_42-p16.png" ] }, { "id": "9702-2025-mj-42-q09", "question_id": "9702-2025-mj-42-q09", "subject": "9702", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 9, "topic": "Quantum physics", "topic_slug": "9702-topic-22-quantum-physics", "marks": 8, "status": "available", "reason": null, "text": "9(a) emission of electrons (from a metal surface) B1\nwhen electromagnetic radiation is incident (on surface / electrons) B1\n\n9(b)(i) current falls to zero when applied voltage equals energy per unit charge of emitted electrons B1\nenergy of photon depends on frequency B1\nmaximum energy of electron depends on energy of photon B1\n\n9(b)(ii) Any three points from: B3\n• threshold frequency = 1.5 × 1015 Hz\n• threshold wavelength = 2.0 × 10–7 m\n• work function = 6.2 eV (or 9.9 × 10–19 J)\n• Planck constant = 6.6 × 10–34 J s (not 6.63 × 10–34 J s)\n• number per unit time (of photons / electrons) = 1.7 × 1016 s–1 (in stage 1)\n• power of incident radiation = 0.028 W (in stage 1)\n© Cambridge University Press & Assessment 2025 Page 17 of 18", "source_pages": [ 17 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9702_s25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-May-June/9702_s25_ms_42.pdf?download=true", "html": "9702-topic-22-quantum-physics/answers.html", "image_paths": [ "../answer-assets/9702_s25_ms_42-p17.png" ] }, { "id": "9702-2025-mj-42-q10", "question_id": "9702-2025-mj-42-q10", "subject": "9702", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "42", "question_number": 10, "topic": "Nuclear physics", "topic_slug": "9702-topic-23-nuclear-physics", "marks": 9, "status": "available", "reason": null, "text": "10(a) (decay is) not affected by external / environmental factors B1\n\n10(b)(i) half-life of X = 2T and half-life of Y = 3T B1\nboth samples show decay constant, in terms of 1 / T, equal to ln 2 / half-life B1\n(decay constant of X = ln 2 / 2T and decay constant of Y = ln 2 / 3T if both half-lives correct)\nboth samples show N , in terms of AT, equal to initial activity / decay constant B1\n0\n(N for X = 8AT / ln 2 and N for Y = 3AT / ln 2 if both decay constants correct)\n0 0\n(Fully correct table:\nhalf-life decay constant A N\n0 0\nX 2T ln 2 / 2T 4A 8AT / ln 2\nY 3T ln 2 / 3T A 3AT / ln 2\n)\n\n10(b)(ii) correct substitution of A and  into A exp (–t) for sample X or sample Y C1\n0 0\n4A exp (–t ln 2 / 2T) = A exp (–t ln 2 / 3T) C1\nt = 12T A1\n\n10(c) Any two points from: B2\n• (radiation) emitted in all directions, not just in direction of detector\n• some radiation absorbed by air / sample / window of detector\n• some radiation may not register even though it reaches detector\n© Cambridge University Press & Assessment 2025 Page 18 of 18", "source_pages": [ 18 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9702_s25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-May-June/9702_s25_ms_42.pdf?download=true", "html": "9702-topic-23-nuclear-physics/answers.html", "image_paths": [ "../answer-assets/9702_s25_ms_42-p18.png" ] }, { "id": "9702-2025-mj-43-q01", "question_id": "9702-2025-mj-43-q01", "subject": "9702", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 1, "topic": "Gravitational fields", "topic_slug": "9702-topic-13-gravitational-fields", "marks": 9, "status": "available", "reason": null, "text": "1(a) work done per unit mass B1\nwork (done in) moving mass from infinity (to the point) B1\n\n1(b)(i) evidence of addition of 3.4 × 106 to 1.7 × 106 or 6.8 × 106 C1\nGM × 122 / (5.1 × 106) or GM × 122 / (10.2 × 106) C1\n6.67 × 10–11 × M × 122 × [(5.1 × 106)–1 – (10.2 × 106)–1] = 5.1 × 108 A1\nleading to M = 6.4 × 1023 kg\n\n1(b)(ii)  = (–) (6.67 × 10–11 × 6.4 × 1023) / (3.4 × 106) C1\n= –1.3 × 107 J kg–1 A1\n\n1(c)(i) Mars takes (just under) 25 hours to rotate once on its axis B1\n\n1(c)(ii) orbit is equatorial B1\nor\norbit is in same direction as direction of rotation of Mars\n© Cambridge University Press & Assessment 2025 Page 8 of 19", "source_pages": [ 8 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9702_s25_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-May-June/9702_s25_ms_43.pdf?download=true", "html": "9702-topic-13-gravitational-fields/answers.html", "image_paths": [ "../answer-assets/9702_s25_ms_43-p08.png" ] }, { "id": "9702-2025-mj-43-q02", "question_id": "9702-2025-mj-43-q02", "subject": "9702", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 2, "topic": "Nuclear physics", "topic_slug": "9702-topic-23-nuclear-physics", "marks": 11, "status": "available", "reason": null, "text": "2(a) (electric) force is (directly) proportional to product of charges B1\nforce (between point charges) is inversely proportional to the square of their separation B1\n\n2(b)(i) charge = (+)2e A1\n\n2(b)(ii) F = 2 × (1.60 × 10–19)2 / [4 × 8.85 × 10–12 × (170 × 10–12)2] = 1.6 × 10–8 N A1\n\n2(c)(i) F = mv2 / r C1\nv = [ (1.6 × 10–8 × 170 × 10–12) / (9.11 × 10–31) ]½ A1\n= 1.7 × 106 m s–1\n\n2(c)(ii) F = mr2 and  = 2 / T C1\nF = 42mr / T2\nT = [ (42 × 9.11 × 10–31 × 170 × 10–12) / (1.6 × 10–8) ]½ A1\n= 6.2 × 10–16 s\nor\nv = 2r / T (C1)\nT = (2 × 170 × 10–12) / (1.73 × 106) (A1)\n= 6.2 × 10–16 s\n© Cambridge University Press & Assessment 2025 Page 9 of 19\n\n2(d)(i) E  Q / r2 C1\nratio = [1.60 × 10–19 × (170 × 10–12)2] / [3.2 × 10–19 × (340 × 10–12)2] A1\n= 0.13\n\n2(d)(ii) resultant force slightly less (than 1.6 × 10–8 N) so speed lower B1\nor\nresultant force slightly less (than 1.6 × 10–8 N) so period greater\n© Cambridge University Press & Assessment 2025 Page 10 of 19", "source_pages": [ 9, 10 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9702_s25_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-May-June/9702_s25_ms_43.pdf?download=true", "html": "9702-topic-23-nuclear-physics/answers.html", "image_paths": [ "../answer-assets/9702_s25_ms_43-p09.png", "../answer-assets/9702_s25_ms_43-p10.png" ] }, { "id": "9702-2025-mj-43-q03", "question_id": "9702-2025-mj-43-q03", "subject": "9702", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 3, "topic": "Temperature", "topic_slug": "9702-topic-14-temperature", "marks": 9, "status": "available", "reason": null, "text": "3(a) (thermal) energy per unit mass (to cause state change) B1\n(thermal) energy to change state at constant temperature B1\n\n3(b) (for vaporisation): B1\ninvolves greater change in volume (of substance)\nor\ninvolves greater increase in separation of molecules\nmore work has to be done by molecules (to separate) M1\nor\ngreater increase in potential energy of molecules\nkinetic energy of molecules unchanged, so more thermal energy needed A1\n\n3(c) Q = mc and Q = mL C1\n for the water = 26.4 – 10.3 C1\n(37.0 × L) + (37.0 × 4.18 × 10.3) = (208 × 4.18 × 16.1) C1\nL = 335 J g–1 A1\n© Cambridge University Press & Assessment 2025 Page 11 of 19", "source_pages": [ 11 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9702_s25_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-May-June/9702_s25_ms_43.pdf?download=true", "html": "9702-topic-14-temperature/answers.html", "image_paths": [ "../answer-assets/9702_s25_ms_43-p11.png" ] }, { "id": "9702-2025-mj-43-q04", "question_id": "9702-2025-mj-43-q04", "subject": "9702", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 4, "topic": "Thermodynamics", "topic_slug": "9702-topic-16-thermodynamics", "marks": 8, "status": "available", "reason": null, "text": "4(a)(i) sum of potential energy and kinetic energy B1\n(total) energy of random motion of particles B1\n\n4(a)(ii) potential energy (of molecules) (in an ideal gas) is zero, so the internal energy of the gas is equal to the total kinetic energy B1\n(of molecules)\nkinetic energy of molecules is proportional to (thermodynamic) temperature (so internal energy is proportional to B1\n(thermodynamic) temperature))\n\n4(b) cooling work done = 0 B1\ncompression increase in internal energy = +2U B1\ncooling change in internal energy = –U B1\nboth rows: thermal energy adds to work to give increase in internal energy in terms of U and/or W B1\n(if fully correct, thermal energy for compression = 2U – W and thermal energy for cooling = –U:\ncompression +W 2U – W +2U\ncooling 0 –U –U\n)\n© Cambridge University Press & Assessment 2025 Page 12 of 19", "source_pages": [ 12 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9702_s25_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-May-June/9702_s25_ms_43.pdf?download=true", "html": "9702-topic-16-thermodynamics/answers.html", "image_paths": [ "../answer-assets/9702_s25_ms_43-p12.png" ] }, { "id": "9702-2025-mj-43-q05", "question_id": "9702-2025-mj-43-q05", "subject": "9702", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 5, "topic": "Oscillations", "topic_slug": "9702-topic-17-oscillations", "marks": 8, "status": "available", "reason": null, "text": "5(a)(i) amplitude = 0.60 m A1\n\n5(a)(ii) oscillations are simple harmonic B1\n\n5(b) Any three points from: B3\n• mean / equilibrium position is at h = 1.4 m\n• total energy of oscillations = 9.0 J\n• angular frequency of oscillations = 1.2 rad s–1\nor\nperiod of oscillations = 5.1 s\nor\nfrequency of oscillation = 0.19 Hz\n• maximum speed of block = 0.73 m s–1\n• mass of block = 33 kg\n\n5(c) U-shaped curve resting on h axis (with minimum at E = 0) B1\nP\ncurve from h = 0.8 m to h = 2.0 m, with minimum E shown at h = 1.4 m B1\nP\nboth end-points of curve shown at E = 9.0 J B1\nP\n© Cambridge University Press & Assessment 2025 Page 13 of 19", "source_pages": [ 13 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9702_s25_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-May-June/9702_s25_ms_43.pdf?download=true", "html": "9702-topic-17-oscillations/answers.html", "image_paths": [ "../answer-assets/9702_s25_ms_43-p13.png" ] }, { "id": "9702-2025-mj-43-q06", "question_id": "9702-2025-mj-43-q06", "subject": "9702", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 6, "topic": "Alternating currents", "topic_slug": "9702-topic-21-alternating-currents", "marks": 12, "status": "available", "reason": null, "text": "6(a)(i) conversion from a.c. to d.c. B1\n\n6(a)(ii) smoothing B1\n\n6(b)(i) A = 12 V A1\nB = 2 / (20 × 10–3) A1\n= 310 rad s–1\n\n6(b)(ii) full-wave (rectification) B1\n\n6(b)(iii) four diodes shown, with correct circuit symbols B1\nfour diodes correctly connected to form a bridge rectifier B1\n\n6(b)(iv) V = V exp (–t / ) C1\n0\nor\nV = V exp (–t / RC) and  = RC\n0\n8.0 = 12 exp (– 7.3 × 10–3 / ) C1\n = 0.018 s A1\n\n6(c) time constant = RC C1\nR = (0.018 / 570 × 10–6) A1\n= 32 \n© Cambridge University Press & Assessment 2025 Page 14 of 19", "source_pages": [ 14 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9702_s25_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-May-June/9702_s25_ms_43.pdf?download=true", "html": "9702-topic-21-alternating-currents/answers.html", "image_paths": [ "../answer-assets/9702_s25_ms_43-p14.png" ] }, { "id": "9702-2025-mj-43-q07", "question_id": "9702-2025-mj-43-q07", "subject": "9702", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 7, "topic": "Magnetic fields", "topic_slug": "9702-topic-20-magnetic-fields", "marks": 10, "status": "available", "reason": null, "text": "7(a) • force per unit length B2\n• force per unit current\n• length / current perpendicular to field\n1 mark for any two points, 2 marks for all three points.\n\n7(b)(i) F = BQv B1\n\n7(b)(ii) arrow at Y pointing vertically upwards B1\n\n7(b)(iii) upwards deflection showing circular path B1\n\n7(c)(i) electric field applied vertically downwards (may be shown on a labelled diagram) B1\nelectric force on particle in opposite direction to magnetic force (may be shown on a labelled diagram) B1\nparticle undeflected when magnitudes of electric and magnetic forces are equal B1\n\n7(c)(ii) EQ = BQv B1\nv = E / B A1\n© Cambridge University Press & Assessment 2025 Page 15 of 19", "source_pages": [ 15 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9702_s25_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-May-June/9702_s25_ms_43.pdf?download=true", "html": "9702-topic-20-magnetic-fields/answers.html", "image_paths": [ "../answer-assets/9702_s25_ms_43-p15.png" ] }, { "id": "9702-2025-mj-43-q08", "question_id": "9702-2025-mj-43-q08", "subject": "9702", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 8, "topic": "Quantum physics", "topic_slug": "9702-topic-22-quantum-physics", "marks": 13, "status": "available", "reason": null, "text": "8(a) wavelength associated with a moving particle B1\n\n8(b)  = h / p C1\n= (6.63 × 10–34) / (9.11 × 10–31 × 4.9 × 107) A1\n= 1.5 × 10–11 m\n\n8(c) similarity: any one point from: B1\n• same mass\n• same magnitude of charge\n• both leptons\ndifference: any one point from: B1\n• electron has negative charge, positron has positive charge\n• positron is anti-particle of electron\n• electron is a particle, positron is an anti-particle\n\n8(d)(i) (pair) annihilation B1\n\n8(d)(ii) their mass gets converted into energy B1\n(their mass–energy) becomes the energy of the gamma photons B1\n\n8(d)(iii) they travel in opposite directions to conserve momentum B1\n\n8(d)(iv) kinetic energy = ½ × 9.11 × 10–31 × (4.9 × 107)2 = 1.1 × 10–15 J A1\n© Cambridge University Press & Assessment 2025 Page 16 of 19\n\n8(d)(v) E = mc2 C1\nE = hc /  C1\nor\nE = hf and c = f\n(1.1 × 10–15) + (9.11 × 10–31 × (3.00 × 108)2) = (6.63 × 10–34 × 3.00 × 108) /  A1\n = 2.39 × 10–12 m\n© Cambridge University Press & Assessment 2025 Page 17 of 19", "source_pages": [ 16, 17 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9702_s25_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-May-June/9702_s25_ms_43.pdf?download=true", "html": "9702-topic-22-quantum-physics/answers.html", "image_paths": [ "../answer-assets/9702_s25_ms_43-p16.png", "../answer-assets/9702_s25_ms_43-p17.png" ] }, { "id": "9702-2025-mj-43-q09", "question_id": "9702-2025-mj-43-q09", "subject": "9702", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 9, "topic": "Nuclear physics", "topic_slug": "9702-topic-23-nuclear-physics", "marks": 10, "status": "available", "reason": null, "text": "9(a) number of nuclear disintegrations per unit time B1\n\n9(b) activity is proportional to the number of undecayed nuclei B1\nactivity = (–) rate of change of number of undecayed nuclei B1\nN is proportional to the rate of change of N (so exponential variation) B1\n\n9(c)(i) 120 = 180 exp (–  × 8.4) C1\n = 0.048 min–1 A1\n\n9(c)(ii) half-life = ln 2 / 0.048 A1\n= 14 min\n\n9(c)(iii) line with negative gradient throughout, starting at (0, 180) B1\ncurve with negative gradient passing through (8.4, 120) B1\ncurve with decreasing negative gradient, from t = 0 to t = 24 min, passing through (14, 90) B1\n© Cambridge University Press & Assessment 2025 Page 18 of 19", "source_pages": [ 18 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9702_s25_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-May-June/9702_s25_ms_43.pdf?download=true", "html": "9702-topic-23-nuclear-physics/answers.html", "image_paths": [ "../answer-assets/9702_s25_ms_43-p18.png" ] }, { "id": "9702-2025-mj-43-q10", "question_id": "9702-2025-mj-43-q10", "subject": "9702", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 4, "variant": "43", "question_number": 10, "topic": "Astronomy and cosmology", "topic_slug": "9702-topic-25-astronomy-and-cosmology", "marks": 10, "status": "available", "reason": null, "text": "10(a) speed is (directly) proportional to distance M1\nspeed is speed of recession of galaxy from an observer, and distance is the distance of the galaxy from the observer A1\n\n10(b)(i) galaxy is receding from the Earth B1\nobserved wavelength is redshifted from emitted wavelength B1\n\n10(b)(ii)  /  = v / c C1\n(4.91 – 4.62) / 4.62 = v / (3.00 × 108)\nv = 1.9 × 107 m s–1 A1\n\n10(b)(iii) wavelength (of maximum intensity) is inversely proportional to temperature B1\nobserved wavelength too high, so determined temperature too low B1\n\n10(c) v = H d C1\n0\nd = (1.9 × 107) / (2.3 × 10–18) A1\n= 8.3 × 1024 m\n© Cambridge University Press & Assessment 2025 Page 19 of 19", "source_pages": [ 19 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9702_s25_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-May-June/9702_s25_ms_43.pdf?download=true", "html": "9702-topic-25-astronomy-and-cosmology/answers.html", "image_paths": [ "../answer-assets/9702_s25_ms_43-p19.png" ] }, { "id": "9702-2025-mj-51-q01", "question_id": "9702-2025-mj-51-q01", "subject": "9702", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "51", "question_number": 1, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "1 Defining the problem\nvary f and measure V or f is the independent variable and V is the dependent variable 1\nkeep E constant 1\nMethods of data collection\nlabelled diagram of workable experiment including: 1\n• circuit with a.c. supply\n• oscilloscope connected in parallel with the resistor\n• workable circuit\n• oscilloscope and a.c. supply labelled\nlabelled signal generator or variable frequency power supply connected across the terminals 1\nmethod to determine V or E from oscilloscope, e.g. multiply amplitude / height of wave by y-gain on oscilloscope 1\nmethod to determine f from oscilloscope, e.g. determine period T by multiplying number of divisions in 1 cycle or horizontal 1\ndistance in 1 cycle by the time base and f = 1/T\nMethod of Analysis\n\n1 1 1\nplot a graph of against f or equivalent, e.g. f against\nV V\nAllow logarithms e.g. lg V against lg f.\nrelationship valid if a straight line is produced passing through the origin 1\n(for lg V against lg f: relationship valid if a straight line is produced with gradient = −1)\n© Cambridge University Press & Assessment 2025 Page 7 of 12\n\n1 1\n\n1 1\nagainst f f against\nV V\nlES lES 1\nK = gradient K = \nAN2 AN2 gradient\nElS\n(for lg V against lg f: K = 10−y-intercept).\nAN2\nAdditional detail including safety considerations 6\nD1 precaution linked to hot coil or hot resistor or prevention of burns from coil or resistor, e.g. use gloves / switch off\npower supply when not measuring V to prevent burns from coil / resistor\nD2 keep N and A and l and S constant\nD3 method to keep S constant, e.g. switch off power supply between readings to prevent heating of resistor or to allow\nresistor to cool\nd2\nD4 method to determine A, e.g. use calipers / micrometer to measure diameter (of coil) / d and A=\n4\nD5 repeat measurements of diameter d along the length of the coil / in different directions and determine the average\nvalue of d\nD6 method to determine the value of S, e.g. separate circuit diagram showing resistor connected to ohmmeter, or\ncircuit diagram showing resistor connected to a power supply with an ammeter and voltmeter and S = V / I\nD7 measure l with a ruler / calipers\nD8 oscilloscope drawn connected across terminals / across signal generator and description to determine E\nD9 adjust y-gain for maximum amplitude\nor\nadjust time base for length of one wave or measure n waves and divide measured time by n\n© Cambridge University Press & Assessment 2025 Page 8 of 12\n\n1 D10 method to keep E constant, e.g. check p.d. and alter supply\nor\nmethod to keep l constant, e.g. tape coil\nor\nmethod to keep A constant, e.g. wind wire on a cylinder\nQuestion Answer Marks", "source_pages": [ 7, 8, 9 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9702_s25_ms_51.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-May-June/9702_s25_ms_51.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_s25_ms_51-p07.png", "../answer-assets/9702_s25_ms_51-p08.png", "../answer-assets/9702_s25_ms_51-p09.png" ] }, { "id": "9702-2025-mj-51-q02", "question_id": "9702-2025-mj-51-q02", "subject": "9702", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "51", "question_number": 2, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "2(a) R 1\ngradient =\nE\nZ\ny-intercept =\nE\n\n2(b) 1\n1\n/ 103 A–1\nI\n2.20 or 2.198\n1.90 or 1.905\n1.72 or 1.724\n1.57 or 1.575\n1.46 or 1.460\n1.31 or 1.307\n1\nValues of / 103 A–1 correct as shown above.\nI\n© Cambridge University Press & Assessment 2025 Page 9 of 12\n\n2(b) 1 1\nUncertainties in / 103 A–1 from  0.02 or  0.03 decreasing to  0.01.\nI\n\n2(c)(i) Six points from (b) plotted correctly. 1\nMust be within half a small square. Diameter of points must be less than half a small square.\n1 1\nError bars in plotted correctly.\nI\nAll error bars to be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.\n\n2(c)(ii) Straight line of best fit drawn. 1\nThickness of the line must be less than half a small square.\nDo not accept line from top point to bottom point.\nLine must pass between (0.101, 1.40) and (0.104, 1.40) and between (0.189, 2.10) and (0.194, 2.10)\nWorst acceptable straight line drawn (steepest or shallowest possible line that passes through all the error bars). 1\nThickness of the line must be less than half a small square.\nAll error bars must be plotted.\n\n2(c)(iii) Gradient determined with clear substitution of data points into y / x. 1\nDistance between data points must be greater than half the length of the drawn line.\nGradient determined of worst acceptable line with clear substitution of data points into y / x. 1\nuncertainty = (gradient of line of best fit – gradient of worst acceptable line)\nor\nuncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)\n© Cambridge University Press & Assessment 2025 Page 10 of 12\n\n2(c)(iv) y-intercept determined by substitution of correct point with consistent power of ten in m and y into y = mx + c. 1\ny-intercept of worst acceptable line determined by substitution into y = mx + c. 1\nuncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line\nor\nuncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)\nDo not accept ECF from false origin method.\n\n2(d)(i) R determined using gradient and R and Z given to 2 or 3 significant figures. 1\nR = gradient  5.8\nZ determined using y-intercept and R and Z given with units with appropriate powers of ten. 1\nZ = y-intercept  5.8\nunit of R:  or V A–1\nunit of Z:  or V A–1\n\n2(d)(ii) Percentage uncertainty determined using E = 0.2 (V) with method shown. 1\nE gradient\nR%= + 100\n E gradient \nor\n0.2 gradient\nR%= + 100\n5.8 gradient \n© Cambridge University Press & Assessment 2025 Page 11 of 12\n\n2(e) I determined to a minimum of 2 significant figures from (c)(iii) and (c)(iv) or (d)(i) with correct substitution. 1\n1\nI =\ngradient\n+y-intercept\n20\nor\nE\nI =\n R \n+Z\n \n20 \n© Cambridge University Press & Assessment 2025 Page 12 of 12", "source_pages": [ 9, 10, 11, 12 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9702_s25_ms_51.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-May-June/9702_s25_ms_51.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_s25_ms_51-p09.png", "../answer-assets/9702_s25_ms_51-p10.png", "../answer-assets/9702_s25_ms_51-p11.png", "../answer-assets/9702_s25_ms_51-p12.png" ] }, { "id": "9702-2025-mj-52-q01", "question_id": "9702-2025-mj-52-q01", "subject": "9702", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "52", "question_number": 1, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "1 Defining the problem\nvary m and measure v or m is the independent variable and v is the dependent variable 1\nkeep h constant 1\nMethods of data collection\nlabelled diagram of workable experiment including: 1\n• axle resting on support(s) (on stands)\n• supports placed on bench\n• light gate (connected to timer) positioned at a distance h\n• light gate labelled and h indicated\nvertical metre rule clamped to a stand in a position close to block to measure h 1\nmethod to determine v using an interrupt length, e.g. v = length of block / time recorded by the timer 1\nmethod to measure m, e.g. use a (top-pan) balance 1\nMethod of analysis\n\n1 1 1\nplot a graph of against or equivalent\nv2 m\nDo not accept logarithms.\n© Cambridge University Press & Assessment 2025 Page 7 of 12\n\n1 1\n\n1 1 1 1\nagainst against\nv2 m m v2\n\n1 gradient\nP = P =−\nhy-intercept y-intercepth\nor\nr2zgradient\nP =\n2Qh\n\n1\n\n1 1 1 1\nagainst against\nv2 m m v2\nr2z r2zy-intercept\nQ= Q=−\n2Phgradient 2\nor\nr2zy-intercept\nQ=\n2gradient\nAdditional detail including safety considerations 6\nD1 precaution linked to falling block resulting in damage to block / bench, e.g. use a cushion / sand tray to prevent\ndamage to bench\nor\nprecaution linked to stands falling, e.g. clamp stand(s) to the bench prevent stand falling\nD2 keep r and z constant\nD3 method to determine r e.g. use calipers / ruler to measure diameter d and r = d / 2\nD4 measure z with a micrometer / calipers\n© Cambridge University Press & Assessment 2025 Page 8 of 12\n\n1 D5 set square correctly positioned between rule and bench to ensure that rule to measure h is vertical\nD6 method to keep h constant by identifying constant initial position of the bottom of the block, e.g. clamped pin / rod to\nindicate the starting point each time or (fiducial) marker on rule\nD7 description of method to ensure that axle can rotate, e.g. axle is lubricated at the supports to enable axle to rotate,\naxle is not fixed at the supports so axle can rotate\nD8 use a large length of block to reduce (percentage) uncertainty in interrupt time or increase the time light gate is\ninterrupted\nD9 repeat experiment for the same value of m and determine the average v\n\n1\nD10 relationship valid if a straight line is produced (with y-intercept = ).\nhP\nDo not accept line passing through the origin.\n© Cambridge University Press & Assessment 2025 Page 9 of 12", "source_pages": [ 7, 8, 9 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9702_s25_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-May-June/9702_s25_ms_52.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_s25_ms_52-p07.png", "../answer-assets/9702_s25_ms_52-p08.png", "../answer-assets/9702_s25_ms_52-p09.png" ] }, { "id": "9702-2025-mj-52-q02", "question_id": "9702-2025-mj-52-q02", "subject": "9702", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "52", "question_number": 2, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "2(a) C 1\ngradient =\nEA\n1\ny-intercept =\nE\n\n2(b) 1\n1\nV/ V / V–1\nV\n4.25 0.235 or 0.2353\n3.70 0.270 or 0.2703\n3.25 0.308 or 0.3077\n2.90 0.345 or 0.3448\n2.60 0.385 or 0.3846\n2.35 0.426 or 0.4255\n1\nValues of V / V and / V–1 correct as shown above.\nV\nUncertainties in V all  0.05 1\nand\n1\nuncertainties in from  0.002 or  0.003 increasing to  0.009.\nV\n© Cambridge University Press & Assessment 2025 Page 10 of 12\n\n2(c)(i) Six points from (b) plotted correctly. 1\nMust be within half a small square. Diameter of points must be less than half a small square.\n1 1\nError bars in plotted correctly.\nV\nAll error bars must be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.\n\n2(c)(ii) Straight line of best fit drawn. 1\nThickness of the line must be less than half a small square.\nDo not accept line from top point to bottom point.\nLine must pass between (2.65, 0.26) and (2.80, 0.26) and between (6.30, 0.40) and (6.50, 0.40).\nWorst acceptable straight line drawn (steepest or shallowest possible line that passes through all the error bars). 1\nThickness of the line must be less than half a small square.\nAll error bars must be plotted.\n\n2(c)(iii) Gradient determined with clear substitution of data points into y / x. 1\nDistance between data points must be greater than half the length of the drawn line.\nGradient determined of worst acceptable line with clear substitution of data points into y / x. 1\nuncertainty = (gradient of line of best fit – gradient of worst acceptable line)\nor\nuncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)\n\n2(c)(iv) y-intercept determined by substitution of correct point into y = mx + c. 1\ny-intercept of worst acceptable line determined by substitution into y = mx + c. 1\nuncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line\nor\nuncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)\nDo not accept ECF from false origin method.\n© Cambridge University Press & Assessment 2025 Page 11 of 12\n\n2(d)(i) E determined using y-intercept and E and C given to 2 or 3 significant figures. 1\n1\nE =\ny-intercept\nC determined using gradient and E and C given with correct units with appropriate powers of ten. 1\nAgradient\nC = or C=AEgradient\ny-intercept\nunit of E: V\nunit of C: F\n\n2(d)(ii) Percentage uncertainty determined with method shown. 1\nA gradient y-intercept\nC%= + + 100\n A gradient y-intercept \nor\nA gradient E\nC%= + + 100 with method to determine E shown\n A gradient E \n\n2(e) V determined to a minimum of 2 significant figures from (c)(iii) and (c)(iv) or (d)(i) with correct substitution. 1\n1\nV =\n10gradient+y-intercept\nor\nEA\nV =\n10C +A\n© Cambridge University Press & Assessment 2025 Page 12 of 12", "source_pages": [ 10, 11, 12 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9702_s25_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-May-June/9702_s25_ms_52.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_s25_ms_52-p10.png", "../answer-assets/9702_s25_ms_52-p11.png", "../answer-assets/9702_s25_ms_52-p12.png" ] }, { "id": "9702-2025-mj-53-q01", "question_id": "9702-2025-mj-53-q01", "subject": "9702", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "53", "question_number": 1, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "1 Defining the problem\nvary f and measure V or f is the independent variable and V is the dependent variable 1\nkeep E constant 1\nMethods of data collection\nlabelled diagram of workable experiment including: 1\n• circuit with a.c. supply\n• oscilloscope connected in parallel with the resistor\n• workable circuit\n• oscilloscope and a.c. supply labelled\nlabelled signal generator or variable frequency power supply connected across the terminals 1\nmethod to determine V or E from oscilloscope, e.g. multiply amplitude / height of wave by y-gain on oscilloscope 1\nmethod to determine f from oscilloscope, e.g. determine period T by multiplying number of divisions in 1 cycle or horizontal 1\ndistance in 1 cycle by the time base and f = 1/T\nMethod of Analysis\n\n1 1 1\nplot a graph of against f or equivalent, e.g. f against\nV V\nAllow logarithms e.g. lg V against lg f.\nrelationship valid if a straight line is produced passing through the origin 1\n(for lg V against lg f: relationship valid if a straight line is produced with gradient = −1)\n© Cambridge University Press & Assessment 2025 Page 7 of 12\n\n1 1\n\n1 1\nagainst f f against\nV V\nlES lES 1\nK = gradient K = \nAN2 AN2 gradient\nElS\n(for lg V against lg f: K = 10−y-intercept).\nAN2\nAdditional detail including safety considerations 6\nD1 precaution linked to hot coil or hot resistor or prevention of burns from coil or resistor, e.g. use gloves / switch off\npower supply when not measuring V to prevent burns from coil / resistor\nD2 keep N and A and l and S constant\nD3 method to keep S constant, e.g. switch off power supply between readings to prevent heating of resistor or to allow\nresistor to cool\nd2\nD4 method to determine A, e.g. use calipers / micrometer to measure diameter (of coil) / d and A=\n4\nD5 repeat measurements of diameter d along the length of the coil / in different directions and determine the average\nvalue of d\nD6 method to determine the value of S, e.g. separate circuit diagram showing resistor connected to ohmmeter, or\ncircuit diagram showing resistor connected to a power supply with an ammeter and voltmeter and S = V / I\nD7 measure l with a ruler / calipers\nD8 oscilloscope drawn connected across terminals / across signal generator and description to determine E\nD9 adjust y-gain for maximum amplitude\nor\nadjust time base for length of one wave or measure n waves and divide measured time by n\n© Cambridge University Press & Assessment 2025 Page 8 of 12\n\n1 D10 method to keep E constant, e.g. check p.d. and alter supply\nor\nmethod to keep l constant, e.g. tape coil\nor\nmethod to keep A constant, e.g. wind wire on a cylinder\nQuestion Answer Marks", "source_pages": [ 7, 8, 9 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9702_s25_ms_53.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-May-June/9702_s25_ms_53.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_s25_ms_53-p07.png", "../answer-assets/9702_s25_ms_53-p08.png", "../answer-assets/9702_s25_ms_53-p09.png" ] }, { "id": "9702-2025-mj-53-q02", "question_id": "9702-2025-mj-53-q02", "subject": "9702", "year": 2025, "session": "May/June", "session_code": "mj", "paper": 5, "variant": "53", "question_number": 2, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "2(a) R 1\ngradient =\nE\nZ\ny-intercept =\nE\n\n2(b) 1\n1\n/ 103 A–1\nI\n2.20 or 2.198\n1.90 or 1.905\n1.72 or 1.724\n1.57 or 1.575\n1.46 or 1.460\n1.31 or 1.307\n1\nValues of / 103 A–1 correct as shown above.\nI\n© Cambridge University Press & Assessment 2025 Page 9 of 12\n\n2(b) 1 1\nUncertainties in / 103 A–1 from  0.02 or  0.03 decreasing to  0.01.\nI\n\n2(c)(i) Six points from (b) plotted correctly. 1\nMust be within half a small square. Diameter of points must be less than half a small square.\n1 1\nError bars in plotted correctly.\nI\nAll error bars to be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.\n\n2(c)(ii) Straight line of best fit drawn. 1\nThickness of the line must be less than half a small square.\nDo not accept line from top point to bottom point.\nLine must pass between (0.101, 1.40) and (0.104, 1.40) and between (0.189, 2.10) and (0.194, 2.10)\nWorst acceptable straight line drawn (steepest or shallowest possible line that passes through all the error bars). 1\nThickness of the line must be less than half a small square.\nAll error bars must be plotted.\n\n2(c)(iii) Gradient determined with clear substitution of data points into y / x. 1\nDistance between data points must be greater than half the length of the drawn line.\nGradient determined of worst acceptable line with clear substitution of data points into y / x. 1\nuncertainty = (gradient of line of best fit – gradient of worst acceptable line)\nor\nuncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)\n© Cambridge University Press & Assessment 2025 Page 10 of 12\n\n2(c)(iv) y-intercept determined by substitution of correct point with consistent power of ten in m and y into y = mx + c. 1\ny-intercept of worst acceptable line determined by substitution into y = mx + c. 1\nuncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line\nor\nuncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)\nDo not accept ECF from false origin method.\n\n2(d)(i) R determined using gradient and R and Z given to 2 or 3 significant figures. 1\nR = gradient  5.8\nZ determined using y-intercept and R and Z given with units with appropriate powers of ten. 1\nZ = y-intercept  5.8\nunit of R:  or V A–1\nunit of Z:  or V A–1\n\n2(d)(ii) Percentage uncertainty determined using E = 0.2 (V) with method shown. 1\nE gradient\nR%= + 100\n E gradient \nor\n0.2 gradient\nR%= + 100\n5.8 gradient \n© Cambridge University Press & Assessment 2025 Page 11 of 12\n\n2(e) I determined to a minimum of 2 significant figures from (c)(iii) and (c)(iv) or (d)(i) with correct substitution. 1\n1\nI =\ngradient\n+y-intercept\n20\nor\nE\nI =\n R \n+Z\n \n20 \n© Cambridge University Press & Assessment 2025 Page 12 of 12", "source_pages": [ 9, 10, 11, 12 ], "source_pdf": "_source-pdfs/2025-May-June/ms/9702_s25_ms_53.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-May-June/9702_s25_ms_53.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_s25_ms_53-p09.png", "../answer-assets/9702_s25_ms_53-p10.png", "../answer-assets/9702_s25_ms_53-p11.png", "../answer-assets/9702_s25_ms_53-p12.png" ] }, { "id": "9702-2025-on-41-q01", "question_id": "9702-2025-on-41-q01", "subject": "9702", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 1, "topic": "Oscillations", "topic_slug": "9702-topic-17-oscillations", "marks": 15, "status": "available", "reason": null, "text": "1(a) velocity and acceleration both have constant magnitude B1\nvelocity is (always) perpendicular to acceleration B1\n\n1(b)(i) v = R A1\n\n1(b)(ii) a = R2 or a = v2 / R C1\na = v A1\n\n1(c)(i) x = R sin  A1\n\n1(c)(ii)  = t A1\n\n1(c)(iii) clear substitution of  = t into x = R sin  leading to x = R sin t A1\n\n1(c)(iv) equation is of the form x = x sin t (so simple harmonic motion) B1\n0\n\n1(d)(i) amplitude = 0.46 / 2 A1\n= 0.23 m\n\n1(d)(ii)  = 2 / T C1\nperiod = 2 / 1.9 A1\n= 3.3 s\n© Cambridge University Press & Assessment 2025 Page 8 of 17\n\n1(d)(iii) a = 2x C1\n0 0\n= 1.92  0.23 A1\n= 0.83 m s–2\n\n1(e) shadow on screen, labelled A, above left-hand edge of the circular path B1\nQuestion Answer Marks", "source_pages": [ 8, 9 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9702_w25_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-Oct-Nov/9702_w25_ms_41.pdf?download=true", "html": "9702-topic-17-oscillations/answers.html", "image_paths": [ "../answer-assets/9702_w25_ms_41-p08.png", "../answer-assets/9702_w25_ms_41-p09.png" ] }, { "id": "9702-2025-on-41-q02", "question_id": "9702-2025-on-41-q02", "subject": "9702", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 2, "topic": "Thermodynamics", "topic_slug": "9702-topic-16-thermodynamics", "marks": 8, "status": "available", "reason": null, "text": "2(a) work done on / by system B1\nthermal energy supplied to / removed from system B1\n\n2(b)(i) no thermal energy transferred to / from system (due to lack of time) B1\nwork is done on the gas to compress it / to decrease its volume B1\ninternal energy increases so temperature increases B1\n\n2(b)(ii) (during vaporisation) molecular separation increases B1\n(heating causes) potential energy of molecules to increase B1\nkinetic energy of molecules unchanged so temperature unchanged B1\n© Cambridge University Press & Assessment 2025 Page 9 of 17", "source_pages": [ 9 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9702_w25_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-Oct-Nov/9702_w25_ms_41.pdf?download=true", "html": "9702-topic-16-thermodynamics/answers.html", "image_paths": [ "../answer-assets/9702_w25_ms_41-p09.png" ] }, { "id": "9702-2025-on-41-q03", "question_id": "9702-2025-on-41-q03", "subject": "9702", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 3, "topic": "Gravitational fields", "topic_slug": "9702-topic-13-gravitational-fields", "marks": 9, "status": "available", "reason": null, "text": "3(a) force per unit mass B1\n\n3(b)(i) F = GMm / x2 C1\ng = F / m A1\ng = [GMm / x2] / m = GM / x2 and G = gravitational constant\n\n3(b)(ii) arrow drawn at P pointing directly towards the point mass B1\n\n3(b)(iii) fields are in opposite directions B1\nfield strength at Q is four times the field strength at P B1\n\n3(c) line starting at (R, –g ) and ending at (L – R, +g ) B1\n0 0\nline passing through (L / 2, 0) B1\ncurve becoming shallower from R to (L / 2) and then steeper from (L / 2) to (L – R) B1\n© Cambridge University Press & Assessment 2025 Page 10 of 17", "source_pages": [ 10 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9702_w25_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-Oct-Nov/9702_w25_ms_41.pdf?download=true", "html": "9702-topic-13-gravitational-fields/answers.html", "image_paths": [ "../answer-assets/9702_w25_ms_41-p10.png" ] }, { "id": "9702-2025-on-41-q04", "question_id": "9702-2025-on-41-q04", "subject": "9702", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 4, "topic": "Temperature", "topic_slug": "9702-topic-14-temperature", "marks": 10, "status": "available", "reason": null, "text": "4(a)(i) temperature = –273.15 °C A1\n\n4(a)(ii) temperature = 0 K A1\n\n4(b)(i) gas is ideal B1\n\n4(b)(ii) pV = NkT C1\nN = 270 / (8.0  10–21) A1\n= 3.4  1022\n\n4(b)(iii) n = (3.4  1022) / (6.02  1023) A1\n= 0.056 mol\n\n4(c) ½ m = (3 / 2) kT C1\n½  m  19002 = 1.5  8.0  10–21 C1\n(m = 6.65  10–27 kg)\nm = (6.65  10–27) / (1.66  10–27) C1\n= 4.0 u A1\n© Cambridge University Press & Assessment 2025 Page 11 of 17", "source_pages": [ 11 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9702_w25_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-Oct-Nov/9702_w25_ms_41.pdf?download=true", "html": "9702-topic-14-temperature/answers.html", "image_paths": [ "../answer-assets/9702_w25_ms_41-p11.png" ] }, { "id": "9702-2025-on-41-q05", "question_id": "9702-2025-on-41-q05", "subject": "9702", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 5, "topic": "Electric fields", "topic_slug": "9702-topic-18-electric-fields", "marks": 10, "status": "available", "reason": null, "text": "5(a) work done per unit charge B1\nwork (done) moving positive charge from infinity (to the point) B1\n\n5(b)(i) potential (due to proton) = (1.60  10–19) / (4  8.85  10–12  10  10–12) C1\nor\npotential (due to electron) = (–1.60  10–19) / (4  8.85  10–12  110  10–12)\nV = [(1.60  10–19) / (4  8.85  10–12)]  [(10–1 – 110–1)  1012] = 130 V A1\n\n5(b)(ii) V = [(1.60  10–19) / (4  8.85  10–12)]  [(30–1 – 90–1)  1012] C1\n= (+) 32 V A1\n\n5(b)(iii) cross drawn midway between the electron and the proton B1\n\n5(b)(iv) line from (10, +130) to (110, –130) B1\ncurve getting shallower until x = 60 pm, crossing V = 0 at (60, 0) and then getting steeper after x = 60 pm B1\ncurve passing through (30, ±32) and (90, ±32) B1\n© Cambridge University Press & Assessment 2025 Page 12 of 17", "source_pages": [ 12 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9702_w25_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-Oct-Nov/9702_w25_ms_41.pdf?download=true", "html": "9702-topic-18-electric-fields/answers.html", "image_paths": [ "../answer-assets/9702_w25_ms_41-p12.png" ] }, { "id": "9702-2025-on-41-q06", "question_id": "9702-2025-on-41-q06", "subject": "9702", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 6, "topic": "Capacitance", "topic_slug": "9702-topic-19-capacitance", "marks": 11, "status": "available", "reason": null, "text": "6(a) series charges: Q = Q = Q B1\nS 1 2\nseries p.d.s: V = V + V B1\nS 1 2\nparallel charges: Q = Q + Q B1\nS 1 2\nparallel p.d.s: V = V = V B1\nS 1 2\n\n6(b)(i) E = ½ CV2 C1\np.d. = [(2  19  10–3) / (470  10–6)]½ A1\n= 9.0 V\n\n6(b)(ii) E = Q2 / 2C or C = Q / V C1\nQ = (19 × 10–3 × 2 × 470 × 10–6)½ A1\nor\nQ = 470 × 10–6 × 9.0\nQ = 4.2  10–3 C\n\n6(b)(iii) total charge unchanged C1\ntotal capacitance = (470 + 180)  10–6 (F) C1\nE = Q2 / 2C = (4.23  10–3)2 / (2  650  10–6) (= 0.014 J) A1\nE = 14 mJ\n© Cambridge University Press & Assessment 2025 Page 13 of 17", "source_pages": [ 13 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9702_w25_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-Oct-Nov/9702_w25_ms_41.pdf?download=true", "html": "9702-topic-19-capacitance/answers.html", "image_paths": [ "../answer-assets/9702_w25_ms_41-p13.png" ] }, { "id": "9702-2025-on-41-q07", "question_id": "9702-2025-on-41-q07", "subject": "9702", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 7, "topic": "Magnetic fields", "topic_slug": "9702-topic-20-magnetic-fields", "marks": 11, "status": "available", "reason": null, "text": "7(a) (induced) e.m.f. is (directly) proportional to rate M1\nof change of (magnetic) flux (linkage) A1\n\n7(b)(i) flux = e.m.f.  time C1\nflux = 0.54  15 A1\n= 8.1 Wb\n\n7(b)(ii)  = BA C1\narea = 8.1 / (38  10–6) A1\n= 2.1  105 m2\n\n7(b)(iii) area = speed  time  width C1\nv = (2.1  105) / (15  68) A1\n= 210 m s–1\n\n7(b)(iv) opposing force (due to current in wings) must be backwards B1\nfrom Fleming’s left-hand rule, current (in wings) must be from Q to P B1\ncurrent is from – to + inside an e.m.f. source so P is at higher potential B1\n© Cambridge University Press & Assessment 2025 Page 14 of 17", "source_pages": [ 14 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9702_w25_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-Oct-Nov/9702_w25_ms_41.pdf?download=true", "html": "9702-topic-20-magnetic-fields/answers.html", "image_paths": [ "../answer-assets/9702_w25_ms_41-p14.png" ] }, { "id": "9702-2025-on-41-q08", "question_id": "9702-2025-on-41-q08", "subject": "9702", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 8, "topic": "Nuclear physics", "topic_slug": "9702-topic-23-nuclear-physics", "marks": 11, "status": "available", "reason": null, "text": "8(a) packet / quantum of energy M1\nof electromagnetic radiation A1\n\n8(b)(i) E = c2m C1\nm = (4.274  106  1.60  10–19) / (1.66  10–27  (3.00  108)2) C1\n( = 0.00458 u)\nm = 233.915174 + 4.000407 + 0.00458 A1\n= 237.92016 u\n\n8(b)(ii) E = hc /  C1\nor\nE = hf and c = f\n(4.274 – 4.200)  1.60  10–13 = (6.63  10–34  3.00  108) /  C1\n = 1.7  10–11 m A1\n\n8(b)(iii) (true) energy of gamma photon is smaller so (true) wavelength is larger B1\n\n8(c) (anti)neutrinos are emitted during beta decay B1\nparticles emitted during beta decay carry varying amounts of energy, so energy of gamma photon is also variable (between B1\ndecays)\n© Cambridge University Press & Assessment 2025 Page 15 of 17", "source_pages": [ 15 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9702_w25_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-Oct-Nov/9702_w25_ms_41.pdf?download=true", "html": "9702-topic-23-nuclear-physics/answers.html", "image_paths": [ "../answer-assets/9702_w25_ms_41-p15.png" ] }, { "id": "9702-2025-on-41-q09", "question_id": "9702-2025-on-41-q09", "subject": "9702", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 9, "topic": "Astronomy and cosmology", "topic_slug": "9702-topic-25-astronomy-and-cosmology", "marks": 7, "status": "available", "reason": null, "text": "9(a) temperature inversely proportional to wavelength M1\ntemperature is thermodynamic temperature of surface of star and wavelength is the wavelength at which maximum A1\nemission rate from star occurs\n\n9(b) Any three points from: B3\n• (surface) temperature of star X = 7000 K\nor\nstar X has a higher temperature than the Sun\n• star X has a higher luminosity than the Sun\n• luminosity of star X = 2.7  1027 W\n• radius of star X = 1.3  109 m\n\n9(c) light (from star X) is redshifted B1\nwavelength of peak emission rate would be greater (using observed data) B1\n© Cambridge University Press & Assessment 2025 Page 16 of 17", "source_pages": [ 16 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9702_w25_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-Oct-Nov/9702_w25_ms_41.pdf?download=true", "html": "9702-topic-25-astronomy-and-cosmology/answers.html", "image_paths": [ "../answer-assets/9702_w25_ms_41-p16.png" ] }, { "id": "9702-2025-on-41-q10", "question_id": "9702-2025-on-41-q10", "subject": "9702", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "41", "question_number": 10, "topic": "Medical physics", "topic_slug": "9702-topic-24-medical-physics", "marks": 8, "status": "available", "reason": null, "text": "10(a) product of density and speed M1\nspeed of sound in medium (and density of the medium) A1\n\n10(b) ultrasound waves cause crystal to vibrate B1\nvibrations (of crystal) cause induced e.m.f. (across crystal) B1\n\n10(c)(i) intensity reflection coefficient= (40.4 – 1.48)2 / (40.4 + 1.48)2 C1\n= 0.86 A1\n\n10(c)(ii) Z values are very similar B1\n(almost) all the ultrasound will be transmitted B1\nor\n(almost) none of the ultrasound will be reflected\n© Cambridge University Press & Assessment 2025 Page 17 of 17", "source_pages": [ 17 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9702_w25_ms_41.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-Oct-Nov/9702_w25_ms_41.pdf?download=true", "html": "9702-topic-24-medical-physics/answers.html", "image_paths": [ "../answer-assets/9702_w25_ms_41-p17.png" ] }, { "id": "9702-2025-on-42-q01", "question_id": "9702-2025-on-42-q01", "subject": "9702", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 1, "topic": "Motion in a circle", "topic_slug": "9702-topic-12-motion-in-a-circle", "marks": 9, "status": "available", "reason": null, "text": "1(a)(i) radius = 6.37  106  cos 52.2° = 3.90  106 m A1\n\n1(a)(ii) period = 24 hours C1\nv = 2r / T C1\nor\nv = r and  = 2 / T\nv = (2  3.90  106) / (24  60  60) A1\n= 280 m s–1\n\n1(b)(i) F = mv2 / r C1\n= (58.6  2802) / (3.90  106) A1\n= 1.2 N\n\n1(b)(ii) arrow pointing horizontally to the left B1\n\n1(b)(iii) arrow from student pointing along the dotted line, labelled ‘weight’ B1\nupwards arrow from student pointing in a direction to the left of normal and above the tangent to the Earth, labelled ‘contact B1\nforce’\n© Cambridge University Press & Assessment 2025 Page 8 of 17", "source_pages": [ 8 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9702_w25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-Oct-Nov/9702_w25_ms_42.pdf?download=true", "html": "9702-topic-12-motion-in-a-circle/answers.html", "image_paths": [ "../answer-assets/9702_w25_ms_42-p08.png" ] }, { "id": "9702-2025-on-42-q02", "question_id": "9702-2025-on-42-q02", "subject": "9702", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 2, "topic": "Ideal gases", "topic_slug": "9702-topic-15-ideal-gases", "marks": 11, "status": "available", "reason": null, "text": "2(a) (gravitational) force is (directly) proportional to product of masses B1\nforce (between point masses) is inversely proportional to the square of their separation B1\n\n2(b) Any two points from: B2\n• molecules are in continuous random motion\n• molecules have negligible volume compared with volume of gas\n• collisions (involving molecules) are (perfectly) elastic\n• collisions (of molecules) are instantaneous\n\n2(c)(i) pV = nRT C1\np = (0.0160  8.31  282) / (1.87  10–4) A1\n= 2.01  105 Pa\n\n2(c)(ii) number of molecules = 0.0160  6.02  1023 C1\nseparation = 3√[(1.87  10–4) / (0.0160  6.02  1023)] A1\n= 2.7  10–9 m (allow any answer that is 3  10–9 m to one significant figure)\n\n2(d)(i) F = 6.67  10–11  (3.34  10–27)2 / (2.7  10–9)2 C1\n= 1.0  10–46 N A1\n\n2(d)(ii) numerical comparison between 10–46 N (F) and 10–26 N (the weight of molecule) leading to a conclusion that the assumption B1\nis supported\n© Cambridge University Press & Assessment 2025 Page 9 of 17", "source_pages": [ 9 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9702_w25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-Oct-Nov/9702_w25_ms_42.pdf?download=true", "html": "9702-topic-15-ideal-gases/answers.html", "image_paths": [ "../answer-assets/9702_w25_ms_42-p09.png" ] }, { "id": "9702-2025-on-42-q03", "question_id": "9702-2025-on-42-q03", "subject": "9702", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 3, "topic": "Temperature", "topic_slug": "9702-topic-14-temperature", "marks": 9, "status": "available", "reason": null, "text": "3(a) same temperature B1\nno net transfer of thermal energy (between them) B1\n\n3(b)(i) density B1\n\n3(b)(ii) Any two points from: B2\n• large response time / large time to reach equilibrium\nor\ncannot measure rapidly changing temperatures\n• reaching equilibrium requires (significant) transfer of energy\nor\nchanges temperature of environment being measured\nor\ncannot measure temperature of small objects\n• bulky / difficult to set up\nor\ndifficult to take readings / scale not calibrated to read temperature\nor\ncannot measure temperature of solid objects\n\n3(b)(iii) substance with large mass B1\nor\ntemperature that is constant (over time)\nor\nto calibrate other thermometers (in a laboratory)\n© Cambridge University Press & Assessment 2025 Page 10 of 17\n\n3(b)(iv) 0 °C = 273 K C1\nT = 273  (7.83 – 2.31) / (8.69 – 2.31) C1\n( = 236 K)\n = 236 – 273 A1\n= – 37 °C\nQuestion Answer Marks", "source_pages": [ 10, 11 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9702_w25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-Oct-Nov/9702_w25_ms_42.pdf?download=true", "html": "9702-topic-14-temperature/answers.html", "image_paths": [ "../answer-assets/9702_w25_ms_42-p10.png", "../answer-assets/9702_w25_ms_42-p11.png" ] }, { "id": "9702-2025-on-42-q04", "question_id": "9702-2025-on-42-q04", "subject": "9702", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 4, "topic": "Thermodynamics", "topic_slug": "9702-topic-16-thermodynamics", "marks": 10, "status": "available", "reason": null, "text": "4(a) change in internal energy = work done + energy transfer by heating C1\nincrease in internal energy = work done on system + energy transferred to the system by heating A1\n\n4(b)(i) U = (3 / 2) pV A1\n\n4(b)(ii) pV = NkT and k identified as Boltzmann constant B1\nU = (3 / 2) NkT A1\n\n4(c)(i) W = (+)8XY A1\n\n4(c)(ii) W = –20XY A1\n\n4(d) work done during stages BC and DA = 0 B1\nchange in internal energy (over complete cycle) = 0 C1\nthermal energy supplied = 20XY – 8XY A1\n= (+)12XY\n© Cambridge University Press & Assessment 2025 Page 11 of 17", "source_pages": [ 11 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9702_w25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-Oct-Nov/9702_w25_ms_42.pdf?download=true", "html": "9702-topic-16-thermodynamics/answers.html", "image_paths": [ "../answer-assets/9702_w25_ms_42-p11.png" ] }, { "id": "9702-2025-on-42-q05", "question_id": "9702-2025-on-42-q05", "subject": "9702", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 5, "topic": "Oscillations", "topic_slug": "9702-topic-17-oscillations", "marks": 10, "status": "available", "reason": null, "text": "5(a)(i) straight line through the origin shows that a is proportional to x B1\nnegative gradient shows that a is always in the opposite direction to x B1\n\n5(a)(ii) a = 2x C1\n0 0\n = 2 / T C1\nT = 2 √(x / a ) A1\n0 0\n= 2 √ (1.2 / 13)\n= 1.9 s\n\n5(b)(i) loss of energy of oscillations B1\ndue to resistive force(s) B1\n\n5(b)(ii) line starting from x = 1.2 cm at t = 0 B1\nline starting from non-zero value of x from t = 0 to t = 2T that is entirely either above or below the t-axis B1\ncurve from t = 0 starting from non-zero x value, with both magnitude of x value and magnitude of gradient continuously B1\ndecreasing\n© Cambridge University Press & Assessment 2025 Page 12 of 17", "source_pages": [ 12 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9702_w25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-Oct-Nov/9702_w25_ms_42.pdf?download=true", "html": "9702-topic-17-oscillations/answers.html", "image_paths": [ "../answer-assets/9702_w25_ms_42-p12.png" ] }, { "id": "9702-2025-on-42-q06", "question_id": "9702-2025-on-42-q06", "subject": "9702", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 6, "topic": "Capacitance", "topic_slug": "9702-topic-19-capacitance", "marks": 11, "status": "available", "reason": null, "text": "6(a) force per unit positive charge B1\n\n6(b)(i) radial lines B1\narrows pointing away from the sphere B1\n\n6(b)(ii) C = Q / V C1\nV = 83 / 69 A1\n= (+)1.2 V\n\n6(b)(iii) V = Q / 4ε r C1\n0\nr = (83  10–12) / (4  8.85  10–12  1.2)\n= 0.62 m A1\n\n6(b)(iv) E = Q / 4ε r2 C1\n0\n= (83  10–12) / (4  8.85  10–12  0.622) A1\n= 1.9 N C–1\n\n6(c) 26 = 83 exp [– t / (120  106  69  10–12)] C1\nt = 9.6  10–3 s A1\n© Cambridge University Press & Assessment 2025 Page 13 of 17", "source_pages": [ 13 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9702_w25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-Oct-Nov/9702_w25_ms_42.pdf?download=true", "html": "9702-topic-19-capacitance/answers.html", "image_paths": [ "../answer-assets/9702_w25_ms_42-p13.png" ] }, { "id": "9702-2025-on-42-q07", "question_id": "9702-2025-on-42-q07", "subject": "9702", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 7, "topic": "Alternating currents", "topic_slug": "9702-topic-21-alternating-currents", "marks": 8, "status": "available", "reason": null, "text": "7(a)(i) T = 2 / 40 = 0.050 s A1\n\n7(a)(ii) V = 18 / √2 A1\nr.m.s.\n= 13 V\n\n7(b) sinusoidal curve of period 50 ms from t = 0 to t = 100 ms B1\ncorrect phase (V at t = 0, 50, 100 ms and –V at 25, 75 ms etc.) B1\nMAX MAX\nmaximum and minimum voltages shown as 18 V B1\n\n7(c) Any three points from: B3\n• rectification is full-wave\n• mean power = 14 W\n• resistance of R = 12 \n• peak current in R = 1.6 A\nor\nr.m.s. current in R = 1.1 A\n• period of output voltage / power = 25 ms\nor\nfrequency of output voltage / power = 40 Hz\nor\nangular frequency of output voltage / power = 250 rad s–1\n© Cambridge University Press & Assessment 2025 Page 14 of 17", "source_pages": [ 14 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9702_w25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-Oct-Nov/9702_w25_ms_42.pdf?download=true", "html": "9702-topic-21-alternating-currents/answers.html", "image_paths": [ "../answer-assets/9702_w25_ms_42-p14.png" ] }, { "id": "9702-2025-on-42-q08", "question_id": "9702-2025-on-42-q08", "subject": "9702", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 8, "topic": "Quantum physics", "topic_slug": "9702-topic-22-quantum-physics", "marks": 10, "status": "available", "reason": null, "text": "8(a) Any three points from: B3\n• electrons moving between levels emit a single photon\n• energy of photon = difference between energy levels\n• energy of photon depends on frequency\n• discrete frequencies (in spectrum) so differences between electron energies must be discrete\n• discrete differences between electron energies means energy levels must be discrete\n\n8(b)(i) energy = – (13.6  1.60  10–19) A1\n= – 2.18  10–18 J\n\n8(b)(ii) E = hf C1\n= (6.63  10–34  2.47  1015) / (1.60  10–19) = 10.2 eV A1\n\n8(b)(iii) n = 2 energy level = – 3.4 eV A1\nn = 3 energy difference = 12.1 eV A1\nn = 4 energy difference = 12.8 eV A1\nn = 3 energy level = – 1.5 eV and n = 4 energy level = – 0.8 eV A1\n© Cambridge University Press & Assessment 2025 Page 15 of 17", "source_pages": [ 15 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9702_w25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-Oct-Nov/9702_w25_ms_42.pdf?download=true", "html": "9702-topic-22-quantum-physics/answers.html", "image_paths": [ "../answer-assets/9702_w25_ms_42-p15.png" ] }, { "id": "9702-2025-on-42-q09", "question_id": "9702-2025-on-42-q09", "subject": "9702", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 9, "topic": "Astronomy and cosmology", "topic_slug": "9702-topic-25-astronomy-and-cosmology", "marks": 13, "status": "available", "reason": null, "text": "9(a) difference between mass of nucleus and mass of (constituent) nucleons M1\nwhen nucleons are separated to infinity A1\n\n9(b) m = (2  2.013553) – (4.001505) (u) C1\n( = 0.025601 u)\nE = c2m C1\nenergy from one He-4 nucleus= 0.025601  1.66  10–27  (3.00  108)2 C1\n(= 3.82  10–12 J)\nenergy to form 1.00 mol= 3.82  10–12  6.02  1023 A1\n= 2.30  1012 J\n\n9(c)(i) L = 1.09  1011  (3.00  108)2 C1\n= 9.81  1027 W A1\n\n9(c)(ii) L = 4 r2T4 C1\n9.81  1027 = 4  5.67  10–8  (1.19  109)2  T4\nT = 9930 K A1\n\n9(d) standard candles have known luminosity B1\nradiant flux intensity (from star) measured (on the Earth) B1\ndistance found from F = L / (4d2) B1\n© Cambridge University Press & Assessment 2025 Page 16 of 17", "source_pages": [ 16 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9702_w25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-Oct-Nov/9702_w25_ms_42.pdf?download=true", "html": "9702-topic-25-astronomy-and-cosmology/answers.html", "image_paths": [ "../answer-assets/9702_w25_ms_42-p16.png" ] }, { "id": "9702-2025-on-42-q10", "question_id": "9702-2025-on-42-q10", "subject": "9702", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "42", "question_number": 10, "topic": "Medical physics", "topic_slug": "9702-topic-24-medical-physics", "marks": 9, "status": "available", "reason": null, "text": "10(a) difference in degrees of blackening B1\n\n10(b)(i) I = I exp (–x) A1\n0\n= I exp (– 5.8  0.35) = 0.13 I\n0 0\n\n10(b)(ii) use of exp {–(0.35  3.7)} factor C1\n0.053I = I exp {–[(0.35  3.7) + 2.1]} C1\n0 0\n = 0.78 cm–1 A1\n\n10(b)(iii) factor of only 2.5 between the (detected) intensities (so not good contrast) B1\n\n10(c) (structure) scanned in (thin) sections B1\n(many) scans (of each section) taken from different angles B1\nscanning repeated for all sections and (data) compiled (to form 3D image) B1\n© Cambridge University Press & Assessment 2025 Page 17 of 17", "source_pages": [ 17 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9702_w25_ms_42.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-Oct-Nov/9702_w25_ms_42.pdf?download=true", "html": "9702-topic-24-medical-physics/answers.html", "image_paths": [ "../answer-assets/9702_w25_ms_42-p17.png" ] }, { "id": "9702-2025-on-43-q01", "question_id": "9702-2025-on-43-q01", "subject": "9702", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 1, "topic": "Oscillations", "topic_slug": "9702-topic-17-oscillations", "marks": 15, "status": "available", "reason": null, "text": "1(a) velocity and acceleration both have constant magnitude B1\nvelocity is (always) perpendicular to acceleration B1\n\n1(b)(i) v = R A1\n\n1(b)(ii) a = R2 or a = v2 / R C1\na = v A1\n\n1(c)(i) x = R sin  A1\n\n1(c)(ii)  = t A1\n\n1(c)(iii) clear substitution of  = t into x = R sin  leading to x = R sin t A1\n\n1(c)(iv) equation is of the form x = x sin t (so simple harmonic motion) B1\n0\n\n1(d)(i) amplitude = 0.46 / 2 A1\n= 0.23 m\n\n1(d)(ii)  = 2 / T C1\nperiod = 2 / 1.9 A1\n= 3.3 s\n© Cambridge University Press & Assessment 2025 Page 8 of 17\n\n1(d)(iii) a = 2x C1\n0 0\n= 1.92  0.23 A1\n= 0.83 m s–2\n\n1(e) shadow on screen, labelled A, above left-hand edge of the circular path B1\nQuestion Answer Marks", "source_pages": [ 8, 9 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9702_w25_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-Oct-Nov/9702_w25_ms_43.pdf?download=true", "html": "9702-topic-17-oscillations/answers.html", "image_paths": [ "../answer-assets/9702_w25_ms_43-p08.png", "../answer-assets/9702_w25_ms_43-p09.png" ] }, { "id": "9702-2025-on-43-q02", "question_id": "9702-2025-on-43-q02", "subject": "9702", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 2, "topic": "Thermodynamics", "topic_slug": "9702-topic-16-thermodynamics", "marks": 8, "status": "available", "reason": null, "text": "2(a) work done on / by system B1\nthermal energy supplied to / removed from system B1\n\n2(b)(i) no thermal energy transferred to / from system (due to lack of time) B1\nwork is done on the gas to compress it / to decrease its volume B1\ninternal energy increases so temperature increases B1\n\n2(b)(ii) (during vaporisation) molecular separation increases B1\n(heating causes) potential energy of molecules to increase B1\nkinetic energy of molecules unchanged so temperature unchanged B1\n© Cambridge University Press & Assessment 2025 Page 9 of 17", "source_pages": [ 9 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9702_w25_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-Oct-Nov/9702_w25_ms_43.pdf?download=true", "html": "9702-topic-16-thermodynamics/answers.html", "image_paths": [ "../answer-assets/9702_w25_ms_43-p09.png" ] }, { "id": "9702-2025-on-43-q03", "question_id": "9702-2025-on-43-q03", "subject": "9702", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 3, "topic": "Gravitational fields", "topic_slug": "9702-topic-13-gravitational-fields", "marks": 9, "status": "available", "reason": null, "text": "3(a) force per unit mass B1\n\n3(b)(i) F = GMm / x2 C1\ng = F / m A1\ng = [GMm / x2] / m = GM / x2 and G = gravitational constant\n\n3(b)(ii) arrow drawn at P pointing directly towards the point mass B1\n\n3(b)(iii) fields are in opposite directions B1\nfield strength at Q is four times the field strength at P B1\n\n3(c) line starting at (R, –g ) and ending at (L – R, +g ) B1\n0 0\nline passing through (L / 2, 0) B1\ncurve becoming shallower from R to (L / 2) and then steeper from (L / 2) to (L – R) B1\n© Cambridge University Press & Assessment 2025 Page 10 of 17", "source_pages": [ 10 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9702_w25_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-Oct-Nov/9702_w25_ms_43.pdf?download=true", "html": "9702-topic-13-gravitational-fields/answers.html", "image_paths": [ "../answer-assets/9702_w25_ms_43-p10.png" ] }, { "id": "9702-2025-on-43-q04", "question_id": "9702-2025-on-43-q04", "subject": "9702", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 4, "topic": "Temperature", "topic_slug": "9702-topic-14-temperature", "marks": 10, "status": "available", "reason": null, "text": "4(a)(i) temperature = –273.15 °C A1\n\n4(a)(ii) temperature = 0 K A1\n\n4(b)(i) gas is ideal B1\n\n4(b)(ii) pV = NkT C1\nN = 270 / (8.0  10–21) A1\n= 3.4  1022\n\n4(b)(iii) n = (3.4  1022) / (6.02  1023) A1\n= 0.056 mol\n\n4(c) ½ m = (3 / 2) kT C1\n½  m  19002 = 1.5  8.0  10–21 C1\n(m = 6.65  10–27 kg)\nm = (6.65  10–27) / (1.66  10–27) C1\n= 4.0 u A1\n© Cambridge University Press & Assessment 2025 Page 11 of 17", "source_pages": [ 11 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9702_w25_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-Oct-Nov/9702_w25_ms_43.pdf?download=true", "html": "9702-topic-14-temperature/answers.html", "image_paths": [ "../answer-assets/9702_w25_ms_43-p11.png" ] }, { "id": "9702-2025-on-43-q05", "question_id": "9702-2025-on-43-q05", "subject": "9702", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 5, "topic": "Electric fields", "topic_slug": "9702-topic-18-electric-fields", "marks": 10, "status": "available", "reason": null, "text": "5(a) work done per unit charge B1\nwork (done) moving positive charge from infinity (to the point) B1\n\n5(b)(i) potential (due to proton) = (1.60  10–19) / (4  8.85  10–12  10  10–12) C1\nor\npotential (due to electron) = (–1.60  10–19) / (4  8.85  10–12  110  10–12)\nV = [(1.60  10–19) / (4  8.85  10–12)]  [(10–1 – 110–1)  1012] = 130 V A1\n\n5(b)(ii) V = [(1.60  10–19) / (4  8.85  10–12)]  [(30–1 – 90–1)  1012] C1\n= (+) 32 V A1\n\n5(b)(iii) cross drawn midway between the electron and the proton B1\n\n5(b)(iv) line from (10, +130) to (110, –130) B1\ncurve getting shallower until x = 60 pm, crossing V = 0 at (60, 0) and then getting steeper after x = 60 pm B1\ncurve passing through (30, ±32) and (90, ±32) B1\n© Cambridge University Press & Assessment 2025 Page 12 of 17", "source_pages": [ 12 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9702_w25_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-Oct-Nov/9702_w25_ms_43.pdf?download=true", "html": "9702-topic-18-electric-fields/answers.html", "image_paths": [ "../answer-assets/9702_w25_ms_43-p12.png" ] }, { "id": "9702-2025-on-43-q06", "question_id": "9702-2025-on-43-q06", "subject": "9702", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 6, "topic": "Capacitance", "topic_slug": "9702-topic-19-capacitance", "marks": 11, "status": "available", "reason": null, "text": "6(a) series charges: Q = Q = Q B1\nS 1 2\nseries p.d.s: V = V + V B1\nS 1 2\nparallel charges: Q = Q + Q B1\nS 1 2\nparallel p.d.s: V = V = V B1\nS 1 2\n\n6(b)(i) E = ½ CV2 C1\np.d. = [(2  19  10–3) / (470  10–6)]½ A1\n= 9.0 V\n\n6(b)(ii) E = Q2 / 2C or C = Q / V C1\nQ = (19 × 10–3 × 2 × 470 × 10–6)½ A1\nor\nQ = 470 × 10–6 × 9.0\nQ = 4.2  10–3 C\n\n6(b)(iii) total charge unchanged C1\ntotal capacitance = (470 + 180)  10–6 (F) C1\nE = Q2 / 2C = (4.23  10–3)2 / (2  650  10–6) (= 0.014 J) A1\nE = 14 mJ\n© Cambridge University Press & Assessment 2025 Page 13 of 17", "source_pages": [ 13 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9702_w25_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-Oct-Nov/9702_w25_ms_43.pdf?download=true", "html": "9702-topic-19-capacitance/answers.html", "image_paths": [ "../answer-assets/9702_w25_ms_43-p13.png" ] }, { "id": "9702-2025-on-43-q07", "question_id": "9702-2025-on-43-q07", "subject": "9702", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 7, "topic": "Magnetic fields", "topic_slug": "9702-topic-20-magnetic-fields", "marks": 11, "status": "available", "reason": null, "text": "7(a) (induced) e.m.f. is (directly) proportional to rate M1\nof change of (magnetic) flux (linkage) A1\n\n7(b)(i) flux = e.m.f.  time C1\nflux = 0.54  15 A1\n= 8.1 Wb\n\n7(b)(ii)  = BA C1\narea = 8.1 / (38  10–6) A1\n= 2.1  105 m2\n\n7(b)(iii) area = speed  time  width C1\nv = (2.1  105) / (15  68) A1\n= 210 m s–1\n\n7(b)(iv) opposing force (due to current in wings) must be backwards B1\nfrom Fleming’s left-hand rule, current (in wings) must be from Q to P B1\ncurrent is from – to + inside an e.m.f. source so P is at higher potential B1\n© Cambridge University Press & Assessment 2025 Page 14 of 17", "source_pages": [ 14 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9702_w25_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-Oct-Nov/9702_w25_ms_43.pdf?download=true", "html": "9702-topic-20-magnetic-fields/answers.html", "image_paths": [ "../answer-assets/9702_w25_ms_43-p14.png" ] }, { "id": "9702-2025-on-43-q08", "question_id": "9702-2025-on-43-q08", "subject": "9702", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 8, "topic": "Nuclear physics", "topic_slug": "9702-topic-23-nuclear-physics", "marks": 11, "status": "available", "reason": null, "text": "8(a) packet / quantum of energy M1\nof electromagnetic radiation A1\n\n8(b)(i) E = c2m C1\nm = (4.274  106  1.60  10–19) / (1.66  10–27  (3.00  108)2) C1\n( = 0.00458 u)\nm = 233.915174 + 4.000407 + 0.00458 A1\n= 237.92016 u\n\n8(b)(ii) E = hc /  C1\nor\nE = hf and c = f\n(4.274 – 4.200)  1.60  10–13 = (6.63  10–34  3.00  108) /  C1\n = 1.7  10–11 m A1\n\n8(b)(iii) (true) energy of gamma photon is smaller so (true) wavelength is larger B1\n\n8(c) (anti)neutrinos are emitted during beta decay B1\nparticles emitted during beta decay carry varying amounts of energy, so energy of gamma photon is also variable (between B1\ndecays)\n© Cambridge University Press & Assessment 2025 Page 15 of 17", "source_pages": [ 15 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9702_w25_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-Oct-Nov/9702_w25_ms_43.pdf?download=true", "html": "9702-topic-23-nuclear-physics/answers.html", "image_paths": [ "../answer-assets/9702_w25_ms_43-p15.png" ] }, { "id": "9702-2025-on-43-q09", "question_id": "9702-2025-on-43-q09", "subject": "9702", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 9, "topic": "Astronomy and cosmology", "topic_slug": "9702-topic-25-astronomy-and-cosmology", "marks": 7, "status": "available", "reason": null, "text": "9(a) temperature inversely proportional to wavelength M1\ntemperature is thermodynamic temperature of surface of star and wavelength is the wavelength at which maximum A1\nemission rate from star occurs\n\n9(b) Any three points from: B3\n• (surface) temperature of star X = 7000 K\nor\nstar X has a higher temperature than the Sun\n• star X has a higher luminosity than the Sun\n• luminosity of star X = 2.7  1027 W\n• radius of star X = 1.3  109 m\n\n9(c) light (from star X) is redshifted B1\nwavelength of peak emission rate would be greater (using observed data) B1\n© Cambridge University Press & Assessment 2025 Page 16 of 17", "source_pages": [ 16 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9702_w25_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-Oct-Nov/9702_w25_ms_43.pdf?download=true", "html": "9702-topic-25-astronomy-and-cosmology/answers.html", "image_paths": [ "../answer-assets/9702_w25_ms_43-p16.png" ] }, { "id": "9702-2025-on-43-q10", "question_id": "9702-2025-on-43-q10", "subject": "9702", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 4, "variant": "43", "question_number": 10, "topic": "Medical physics", "topic_slug": "9702-topic-24-medical-physics", "marks": 8, "status": "available", "reason": null, "text": "10(a) product of density and speed M1\nspeed of sound in medium (and density of the medium) A1\n\n10(b) ultrasound waves cause crystal to vibrate B1\nvibrations (of crystal) cause induced e.m.f. (across crystal) B1\n\n10(c)(i) intensity reflection coefficient= (40.4 – 1.48)2 / (40.4 + 1.48)2 C1\n= 0.86 A1\n\n10(c)(ii) Z values are very similar B1\n(almost) all the ultrasound will be transmitted B1\nor\n(almost) none of the ultrasound will be reflected\n© Cambridge University Press & Assessment 2025 Page 17 of 17", "source_pages": [ 17 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9702_w25_ms_43.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-Oct-Nov/9702_w25_ms_43.pdf?download=true", "html": "9702-topic-24-medical-physics/answers.html", "image_paths": [ "../answer-assets/9702_w25_ms_43-p17.png" ] }, { "id": "9702-2025-on-51-q01", "question_id": "9702-2025-on-51-q01", "subject": "9702", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "51", "question_number": 1, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "1 Defining the problem\nvary r and measure v or r is the independent variable and v is the dependent variable 1\nkeep x constant 1\nMethods of data collection\nlabelled diagram of workable experiment including: 1\n• one end of spring resting against block clamped to bench using G-clamp\n• light gate positioned at P\n• light gate connected to timer\n• apparatus shown on bench\n• labels for light gate and P and at least one other label from bench, block, stand, spring, ball, timer\nmethod to determine r, e.g. use calipers or micrometer to measure diameter d and r = d / 2 1\ndescription of method to determine v, use diameter of ball (to interrupt beam) ÷ measured time at light gate positioned at P 1\ninstrument to determine x, e.g. rule(r) or calipers 1\nMethod of Analysis\nplot a graph of (2 lg v) or (lg v2) against lg r 1\nor\nplot a graph of (lg v) against (lg r) or equivalent, e.g. (ln v) against (ln r)\nn=−gradient for (2 lg v) or (lg v2) against lg r 1\nor\nn=−2gradient for (lg v) against lg r\n© Cambridge University Press & Assessment 2025 Page 7 of 11\n\n1 10y-intercept 1\nY = for (2 lg v) or (lg v2) against lg r\nkx2\nor\n102y-intercept\nY = for (lg v) against lg r\nkx2\nAdditional detail including safety considerations 6\nD1 precaution to prevent ball leaving bench, e.g. screens around apparatus / cushions on bench (to stop the ball)\nD2 keep k and  constant\nD3 description of method to determine k, e.g. add mass to spring and k = mg / extension\nor\nuse newton meter to measure force applied to spring and k = force / extension\nor\ntake several readings of force and extension, plot a force–extension graph and k = gradient\nm\nD4 description of experimental method to determine , e.g. measure mass of ball using a balance and =\n4\nr3\n3\nD5 repeat measurements of diameter or d in different directions and determine the average value of d\nD6 method to keep x constant, e.g. use a pin / ruler / card to indicate the starting point each time to keep x constant\nD7 x = original length of spring – compressed length of spring\nD8 adjust (vertical) position of light gate so that the diameter of (each) ball cuts the beam\nD9 repeat experiment for the same value of r and determine the average v\n© Cambridge University Press & Assessment 2025 Page 8 of 11\n\n1 Ykx2  1 Ykx2 \nD10 relationship valid if a straight line is produced (with y-intercept = lg  or lg ).\n 2 \n   \nDo not accept line through the origin.\nQuestion Answer Marks", "source_pages": [ 7, 8, 9 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9702_w25_ms_51.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-Oct-Nov/9702_w25_ms_51.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_w25_ms_51-p07.png", "../answer-assets/9702_w25_ms_51-p08.png", "../answer-assets/9702_w25_ms_51-p09.png" ] }, { "id": "9702-2025-on-51-q02", "question_id": "9702-2025-on-51-q02", "subject": "9702", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "51", "question_number": 2, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "2(a) gradient = H 1\n0\ny-intercept = \n0\n\n2(b) 1\nd\n/ 1015s\nc\n(1.6 or 1.60)  0.40\n(3.47 or 3.467)  0.40\n(4.83 or 4.833)  0.40\n(6.00 or 6.000)  0.40\n(9.50 or 9.500)  0.40\n(12.5 or 12.50)  0.40\nd\nValues of correct as shown above.\nc\nd 1\nUncertainties in correct as shown above.\nc\n© Cambridge University Press & Assessment 2025 Page 9 of 11\n\n2(c)(i) Six points from (b) plotted correctly. 1\nMust be within half a small square. Diameter of points must be less than half a small square.\nd 1\nError bars in plotted correctly.\nc\nAll error bars must be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.\n\n2(c)(ii) Straight line of best fit drawn. 1\nThickness of the line must be less than half a small square.\nDo not accept line from top point to bottom point.\nLine must pass between (2.5, 660.0) and (2.9, 660.0) and between (11.2, 676.0) and (11.6, 676.0).\nWorst acceptable line drawn (steepest or shallowest possible line that passes through all the error bars). 1\nThickness of the line must be less than half a small square.\nAll error bars must be plotted.\n\n2(c)(iii) Gradient determined with clear substitution of data points into y / x. 1\nDistance between data points must be greater than half the length of the drawn line.\nGradient determined of worst acceptable line with clear substitution of data points into y / x. 1\nuncertainty = (gradient of line of best fit – gradient of worst acceptable line)\nor\nuncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)\n\n2(c)(iv) y-intercept determined by substitution of correct point with consistent power of ten in m and x and y into y = mx + c. 1\ny-intercept of worst acceptable line determined by substitution into y = mx + c. 1\nuncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line\nor\nuncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)\nDo not accept ECF from false origin method.\n© Cambridge University Press & Assessment 2025 Page 10 of 11\n\n2(d)  determined using y-intercept and  given to 3 or 4 significant figures and H given to 2, 3 or 4 significant figures. 1\n0 0\n = y-intercept\n0\nH determined using gradient and  and H given with SI units with appropriate powers of ten. 1\n0\ngradient gradient\nH = or H =\ny-intercept \n0\nUnit of : m, nm, m\n0\nUnit of H: s−1\n\n2(e) Value of T determined to a minimum of two significant figures from (d) and correct power of ten. 1\n1\nT =\nH\nAbsolute uncertainty determined with correct substitution. 1\ny-intercept gradient\nT = + T\n y-intercept gradient \nor\n max  maxy-intercept\nT = 0 −T or T = −T\nmin gradient  min gradient \nor\n min  miny-intercept\nT = 0 −T or T = −T\nmax gradient  max gradient \n© Cambridge University Press & Assessment 2025 Page 11 of 11", "source_pages": [ 9, 10, 11 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9702_w25_ms_51.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-Oct-Nov/9702_w25_ms_51.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_w25_ms_51-p09.png", "../answer-assets/9702_w25_ms_51-p10.png", "../answer-assets/9702_w25_ms_51-p11.png" ] }, { "id": "9702-2025-on-52-q01", "question_id": "9702-2025-on-52-q01", "subject": "9702", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "52", "question_number": 1, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "1 Defining the problem\nvary v and measure I or v is the independent variable and I is the dependent variable 1\nkeep A and R constant 1\nMethods of data collection\nlabelled diagram of workable experiment including: 1\n• fan positioned in line with the turbine so that blades of both fan and turbine overlap\n• fan on bench\n• fan labelled and one other label from bench, turbine, blade(s) (of turbine), terminals, L\nworkable circuit diagram showing resistor connected to an ammeter in series with the terminals of the turbine using correct 1\ncircuit symbols\nmethod to vary v, e.g. change speed of fan / change distance between fan and blades / vary current in or p.d. across fan 1\nmethod to determine temperature T, e.g. use a thermometer 1\nMethod of Analysis\nplot a graph of I2 against v3 or equivalent (e.g. lg I against lgv or 2 lnI against lnv) 1\nrelationship valid if a straight line is produced passing through the origin 1\n(For lg I against lg v: relationship valid if a straight line with gradient = 1.5 is produced)\n2TRgradient 1\nQ=\nAP\n2TR102y-intercept\n(For lg I against lg v: Q = )\nAP\n© Cambridge University Press & Assessment 2025 Page 7 of 11\n\n1 Additional detail including safety considerations 6\nD1 precaution with reason linked to (moving) fan blades / turbine blades, e.g. keep away from the fan to avoid (moving)\nblades or use a screen around the fan / turbine to avoid (moving) blades\nor\nprecaution with reason linked to prevent air / dust particles in eye, e.g. use goggles to avoid air stream (into eye)\nD2 clamp turbine / fan to bench\nD3 keep P and T constant\nD4 T = t + 273\nD5 method to determine A: use a rule(r) / calipers to measure L and A = L2\nD6 repeat measurements of L in different positions / different blades and average\nD7 method to measure v, e.g. use an anemometer or air speed meter\nor\nmethod to measure P, e.g. use a manometer or barometer or pressure gauge\nD8 wait for steady / constant air flow / movement of blades / current\nD9 method to determine R, e.g.:\nseparate circuit showing ohmmeter connected to R only\nor\nterminals of turbine connected correctly to resistor and ammeter and voltmeter across resistor and R = V / I\nor\nseparate workable circuit with power supply resistor, ammeter and voltmeter across R and R = V / I\nD10 method to check temperature / pressure is constant, e.g. measure temperature / pressure several times / before and\nafter\n© Cambridge University Press & Assessment 2025 Page 8 of 11", "source_pages": [ 7, 8 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9702_w25_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-Oct-Nov/9702_w25_ms_52.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_w25_ms_52-p07.png", "../answer-assets/9702_w25_ms_52-p08.png" ] }, { "id": "9702-2025-on-52-q02", "question_id": "9702-2025-on-52-q02", "subject": "9702", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "52", "question_number": 2, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "2(a) gradient = n 1\n2\ny-intercept = lg\nk\n\n2(b) 1\nlg (r / 108 m) lg (T / 103 s)\n0.152 or 0.1523 (1.72 or 1.716)  0.04\n0.270 or 0.2695 (1.91 or 1.908)  0.03\n0.377 or 0.3766 (2.08 or 2.079 or 2.0792)  0.04\n0.470 or 0.4698 (2.23 or 2.230 or 2.2304)  0.03\n0.576 or 0.5763 (2.38 or 2.380 or 2.3802)  0.04\n0.723 or 0.7226 (2.59 or 2.591 or 2.5911)  0.03\nValues of lg (r / 108 m) and lg (T / 103 s) correct as shown above.\nUncertainties in lg (T / 103 s) correct as shown above. 1\n\n2(c)(i) Six points from (b) plotted correctly. 1\nMust be within half a small square. Diameter of points must be less than half a small square.\nError bars in lg (T / 103 s) plotted correctly. 1\nAll error bars must be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.\n© Cambridge University Press & Assessment 2025 Page 9 of 11\n\n2(c)(ii) Straight line of best fit drawn. 1\nThickness of the line must be less than half a small square.\nDo not accept line from top point to bottom point.\nLine must pass between (0.19, 1.80) and (0.21, 1.80) and between (0.645, 2.50) and (0.66, 2.50).\nWorst acceptable line drawn (steepest or shallowest possible line that passes through all the error bars). 1\nThickness of the line must be less than half a small square.\nAll error bars must be plotted.\n\n2(c)(iii) Gradient determined with clear substitution of data points into y / x. 1\nDistance between data points must be greater than half the length of the drawn line.\nGradient determined of worst acceptable line with clear substitution of data points into y / x. 1\nuncertainty = (gradient of line of best fit – gradient of worst acceptable line)\nor\nuncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)\n\n2(c)(iv) y-intercept determined by substitution of correct point with consistent powers of ten into y = mx + c. 1\ny-intercept of worst acceptable line determined by substitution into y = mx + c. 1\nuncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line\nor\nuncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)\nDo not accept ECF from false origin method.\n© Cambridge University Press & Assessment 2025 Page 10 of 11\n\n2(d) Value of n determined using gradient and n and k given to 2 or 3 significant figures. 1\nn=gradient=(c)(iii)\nValue of k determined using y-intercept. 1\nCorrect method must be seen.\n2 2\nk = =\n10y-intercept 10(c)(iv)\nAbsolute uncertainties in n and k determined. 1\nAbsolute uncertainty in n = absolute uncertainty in gradient\nand\n2 2\nk = −\n10y-intercept 10WAL y-intercept\nCorrect method must be seen.\n\n2(e) r determined to a minimum of 2 significant figures from (c)(iii) and (c)(iv) or (d) with correct substitution and correct power 1\nof ten.\nlg1380−y-intercept\nr =10 gradient 108\nor\nk1380\nr = n 108\n2π\n© Cambridge University Press & Assessment 2025 Page 11 of 11", "source_pages": [ 9, 10, 11 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9702_w25_ms_52.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-Oct-Nov/9702_w25_ms_52.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_w25_ms_52-p09.png", "../answer-assets/9702_w25_ms_52-p10.png", "../answer-assets/9702_w25_ms_52-p11.png" ] }, { "id": "9702-2025-on-53-q01", "question_id": "9702-2025-on-53-q01", "subject": "9702", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "53", "question_number": 1, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "1 Defining the problem\nvary r and measure v or r is the independent variable and v is the dependent variable 1\nkeep x constant 1\nMethods of data collection\nlabelled diagram of workable experiment including: 1\n• one end of spring resting against block clamped to bench using G-clamp\n• light gate positioned at P\n• light gate connected to timer\n• apparatus shown on bench\n• labels for light gate and P and at least one other label from bench, block, stand, spring, ball, timer\nmethod to determine r, e.g. use calipers or micrometer to measure diameter d and r = d / 2 1\ndescription of method to determine v, use diameter of ball (to interrupt beam) ÷ measured time at light gate positioned at P 1\ninstrument to determine x, e.g. rule(r) or calipers 1\nMethod of Analysis\nplot a graph of (2 lg v) or (lg v2) against lg r 1\nor\nplot a graph of (lg v) against (lg r) or equivalent, e.g. (ln v) against (ln r)\nn=−gradient for (2 lg v) or (lg v2) against lg r 1\nor\nn=−2gradient for (lg v) against lg r\n© Cambridge University Press & Assessment 2025 Page 7 of 11\n\n1 10y-intercept 1\nY = for (2 lg v) or (lg v2) against lg r\nkx2\nor\n102y-intercept\nY = for (lg v) against lg r\nkx2\nAdditional detail including safety considerations 6\nD1 precaution to prevent ball leaving bench, e.g. screens around apparatus / cushions on bench (to stop the ball)\nD2 keep k and  constant\nD3 description of method to determine k, e.g. add mass to spring and k = mg / extension\nor\nuse newton meter to measure force applied to spring and k = force / extension\nor\ntake several readings of force and extension, plot a force–extension graph and k = gradient\nm\nD4 description of experimental method to determine , e.g. measure mass of ball using a balance and =\n4\nr3\n3\nD5 repeat measurements of diameter or d in different directions and determine the average value of d\nD6 method to keep x constant, e.g. use a pin / ruler / card to indicate the starting point each time to keep x constant\nD7 x = original length of spring – compressed length of spring\nD8 adjust (vertical) position of light gate so that the diameter of (each) ball cuts the beam\nD9 repeat experiment for the same value of r and determine the average v\n© Cambridge University Press & Assessment 2025 Page 8 of 11\n\n1 Ykx2  1 Ykx2 \nD10 relationship valid if a straight line is produced (with y-intercept = lg  or lg ).\n 2 \n   \nDo not accept line through the origin.\nQuestion Answer Marks", "source_pages": [ 7, 8, 9 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9702_w25_ms_53.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-Oct-Nov/9702_w25_ms_53.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_w25_ms_53-p07.png", "../answer-assets/9702_w25_ms_53-p08.png", "../answer-assets/9702_w25_ms_53-p09.png" ] }, { "id": "9702-2025-on-53-q02", "question_id": "9702-2025-on-53-q02", "subject": "9702", "year": 2025, "session": "Oct/Nov", "session_code": "on", "paper": 5, "variant": "53", "question_number": 2, "topic": "Practical skills", "topic_slug": "9702-practical-skills", "marks": 15, "status": "available", "reason": null, "text": "2(a) gradient = H 1\n0\ny-intercept = \n0\n\n2(b) 1\nd\n/ 1015s\nc\n(1.6 or 1.60)  0.40\n(3.47 or 3.467)  0.40\n(4.83 or 4.833)  0.40\n(6.00 or 6.000)  0.40\n(9.50 or 9.500)  0.40\n(12.5 or 12.50)  0.40\nd\nValues of correct as shown above.\nc\nd 1\nUncertainties in correct as shown above.\nc\n© Cambridge University Press & Assessment 2025 Page 9 of 11\n\n2(c)(i) Six points from (b) plotted correctly. 1\nMust be within half a small square. Diameter of points must be less than half a small square.\nd 1\nError bars in plotted correctly.\nc\nAll error bars must be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.\n\n2(c)(ii) Straight line of best fit drawn. 1\nThickness of the line must be less than half a small square.\nDo not accept line from top point to bottom point.\nLine must pass between (2.5, 660.0) and (2.9, 660.0) and between (11.2, 676.0) and (11.6, 676.0).\nWorst acceptable line drawn (steepest or shallowest possible line that passes through all the error bars). 1\nThickness of the line must be less than half a small square.\nAll error bars must be plotted.\n\n2(c)(iii) Gradient determined with clear substitution of data points into y / x. 1\nDistance between data points must be greater than half the length of the drawn line.\nGradient determined of worst acceptable line with clear substitution of data points into y / x. 1\nuncertainty = (gradient of line of best fit – gradient of worst acceptable line)\nor\nuncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)\n\n2(c)(iv) y-intercept determined by substitution of correct point with consistent power of ten in m and x and y into y = mx + c. 1\ny-intercept of worst acceptable line determined by substitution into y = mx + c. 1\nuncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line\nor\nuncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept)\nDo not accept ECF from false origin method.\n© Cambridge University Press & Assessment 2025 Page 10 of 11\n\n2(d)  determined using y-intercept and  given to 3 or 4 significant figures and H given to 2, 3 or 4 significant figures. 1\n0 0\n = y-intercept\n0\nH determined using gradient and  and H given with SI units with appropriate powers of ten. 1\n0\ngradient gradient\nH = or H =\ny-intercept \n0\nUnit of : m, nm, m\n0\nUnit of H: s−1\n\n2(e) Value of T determined to a minimum of two significant figures from (d) and correct power of ten. 1\n1\nT =\nH\nAbsolute uncertainty determined with correct substitution. 1\ny-intercept gradient\nT = + T\n y-intercept gradient \nor\n max  maxy-intercept\nT = 0 −T or T = −T\nmin gradient  min gradient \nor\n min  miny-intercept\nT = 0 −T or T = −T\nmax gradient  max gradient \n© Cambridge University Press & Assessment 2025 Page 11 of 11", "source_pages": [ 9, 10, 11 ], "source_pdf": "_source-pdfs/2025-Oct-Nov/ms/9702_w25_ms_53.pdf", "source_pdf_url": "https://pastpapers.co/api/file/caie/A-Level/Physics-9702/2025-Oct-Nov/9702_w25_ms_53.pdf?download=true", "html": "9702-practical-skills/answers.html", "image_paths": [ "../answer-assets/9702_w25_ms_53-p09.png", "../answer-assets/9702_w25_ms_53-p10.png", "../answer-assets/9702_w25_ms_53-p11.png" ] } ] }