Hooke's law

3(a) kmg ( x−a ) 4a B1 AG. Shown convincingly. Use Hooke’s Law: 4mg = leading to k = a x−a 1 3(b) Gain in KE + gain in EPE = loss in GPE B1 One correct EPE term seen. 1 ×6m× ga + 1 kmg     x+ a −a  2 −( x−a )2  =6mg× a M A 1 1 A Al l l l t 3 e r ty m p s e c s o o r f r e e c n t e . rgy included in energy equation. 2 9 2 a   3   3   Simplify and substitute for k from part (a) M1 Obtain linear equation in x and a M1 5 A1 (k = 6) x = a 3 6 © UCLES 2021 Page 8 of 13

3(a) kmg ( x−a ) 4a B1 AG. Shown convincingly. Use Hooke’s Law: 4mg = leading to k = a x−a 1 3(b) Gain in KE + gain in EPE = loss in GPE B1 One correct EPE term seen. 1 ×6m× ga + 1 kmg     x+ a −a  2 −( x−a )2  =6mg× a M A 1 1 A Al l l l t 3 e r ty m p s e c s o o r f r e e c n t e . rgy included in energy equation. 2 9 2 a   3   3   Simplify and substitute for k from part (a) M1 Obtain linear equation in x and a M1 5 A1 (k = 6) x = a 3 6 © UCLES 2021 Page 8 of 13

2 1 × 36x2 (= 22.5x2) B1 EPE correct. 2 0.8 Loss in GPE + loss in KE = gain in EPE *M1 Energy equation with only GPE, EPE and KE terms, allow sign 1 1 36x2 errors, allow missing g for M1 only, weight must be resolved x mgsinα+ m×2= × 2 2 0.8 (allow sin or cos). All terms correct A1   6 x+ 1 = 9 x2   [leading to45x2 −24x−4 = 0] DM1 Simplify to 3-term quadratic and attempt to solve. 5 5 4  ( )( ) 3x−2 15x+2 =0 A1 2 x= only 3 5 © UCLES 2021 Page 7 of 16

3 Loss in EPE = 1 × 12mge2 − 1 × 12mg ×   e− a  2   = 2mg( 6e−a )  B1 E ither term correct. 2 a 2 a  3  3  1 mga B1 Gain in KE = mv2 and Gain in GPE = 2 3 Gain in KE + Gain in GPE = Loss in EPE M1 KE, GPE and at least one EPE term. 1 mv2 + mga = 2mg( 6e−a ) A1 All terms correct. 2 3 3 Simplify to a linear equation in e. M1 1 A1 e= a 2 6 © UCLES 2021 Page 8 of 12

2 At the collision of P and Q: ( m+km ) v=kmu M1 Momentum conserved, allow missing k on RHS. A1 k 4ga So v= ( 1+k ) 2 B1 1 5mg a  mga EPE = × × =     2 a 5  10  2 M1 Energy equation, LHS correct, EPE dimensionally correct. Loss in KE = Gain in EPE: 1 m ( k+1 ) v2 = 1 × 5mg ×   a  2 2 a 5 Substitute for v and rearrange to form quadratic equation in k M1 20k2 =1+k 1 A1 k = 4 6 © UCLES 2021 Page 7 of 12

3 Loss in EPE = 1 × 12mge2 − 1 × 12mg ×   e− a  2   = 2mg( 6e−a )  B1 E ither term correct. 2 a 2 a  3  3  1 mga B1 Gain in KE = mv2 and Gain in GPE = 2 3 Gain in KE + Gain in GPE = Loss in EPE M1 KE, GPE and at least one EPE term. 1 mv2 + mga = 2mg( 6e−a ) A1 All terms correct. 2 3 3 Simplify to a linear equation in e. M1 1 A1 e= a 2 6 © UCLES 2021 Page 8 of 12

1(a) Tcos7.5, Tsin10 B1 1 B1 T   7.52 102 2 = 12.5 N 2 1(b) 50x B1 Hooke’s law: T  , x0.2 0.8 10 M1 x0.8sin1 12.5 4 A1 Vertical distance 0.8 5 3 Question Answer Marks Guidance

1(a) Tcos7.5, Tsin10 B1 1 B1 T   7.52 102 2 = 12.5 N 2 1(b) 50x B1 Hooke’s law: T  , x0.2 0.8 10 M1 x0.8sin1 12.5 4 A1 Vertical distance 0.8 5 3 Question Answer Marks Guidance

2 4 B1 EPE term correct mg 1 . 3 x2 2 a 1 B1 Work term correct mgx 3 Loss in KE = gain in EPE + work done against friction M1 Energy equation with 3 terms, allow sign error. 4 mg 1 mv2  1  3 x2  1 mgx 2 2 a 3 2 M1 g 1  1 ga 3 x2  1 gx 2 4 a 3 Obtain and attempt to solve a 3-term quadratic equation. 16x2 8ax3a2 0 4xa4x3a0 1 A1 x a 4 5 © UCLES 2022 Page 7 of 13

2(a) Loss in KE = Gain in EPE, so B1 EPE terms correct. 1 1 v 2 1 4mg 1  2 1  2 mv2 − m  =   a − a  M1 All 4 terms and no extras. 2 2 2 2 a  2  4     3 4mg 3 M1 Simplify. mv2 =  a2 4 a 16 v2 =ag, v= ag A1 4 2(b) 4mg 1 M1 Hooke’s law: tension =  a (= 2mg) a 2 2mg A1 Accept -2g. Acceleration = =2g m 2 © UCLES 2022 Page 6 of 12

3 16 B1 Mge 3 3 In equilibrium, =4Mg, e= a a 4 In subsequent motion, M1 Energy equation with GPE and KE terms correct and at least one Loss in GPE = gain in EPE + gain in KE EPE term. Dimensionally correct. 6Mga 1 16 Mg  3a 2 1 B1 EPE correct. = . . .a2 −  + .6Mv2 4 2 3 a    4    2 A1 All correct. 3Mga 8Mg 7 M1 Attempt to find v in terms of a and g. = . a2 +3Mv2 etc 2 3a 16 ga 1 A1 =3v2, v= ga 3 3 6 © UCLES 2022 Page 9 of 15

2(a) Loss in KE = Gain in EPE, so B1 EPE terms correct. 1 1 v 2 1 4mg 1  2 1  2 mv2 − m  =   a − a  M1 All 4 terms and no extras. 2 2 2 2 a  2  4     3 4mg 3 M1 Simplify. mv2 =  a2 4 a 16 v2 =ag, v= ag A1 4 2(b) 4mg 1 M1 Hooke’s law: tension =  a (= 2mg) a 2 2mg A1 Accept -2g. Acceleration = =2g m 2 © UCLES 2022 Page 6 of 12

1(a) 3mg B1 Correct EPE term seen 2a2 2a 1 mv2 mg  3a 3 a   3mg 2a2 M1 D on i e m K en E s , i o o n n a e l l E y P c E o r t r e e r c m t e a n n e d r g a y t l e e q a u st a t o i n o e n . G M P u E s . t have 2  4  2a Allow sign errors. 15 A1 AEF v ag 2.74 ag   2 3 1(b) 3mg M1 N2L and Hooke’s law T mg mA and T  2a a Acceleration = 5g [upwards] A1 Allow 50 or 5g 2 © UCLES 2023 Page 6 of 15

1(a) 3mg B1 Correct EPE term seen 2a2 2a 1 mv2 mg  3a 3 a   3mg 2a2 M1 D on i e m K en E s , i o o n n a e l l E y P c E o r t r e e r c m t e a n n e d r g a y t l e e q a u st a t o i n o e n . G M P u E s . t have 2  4  2a Allow sign errors. 15 A1 AEF v ag 2.74 ag   2 3 1(b) 3mg M1 N2L and Hooke’s law T mg mA and T  2a a Acceleration = 5g [upwards] A1 Allow 50 or 5g 2 © UCLES 2023 Page 6 of 15

2 1 M1 Kinetic energy = elastic potential energy, 4 terms, 1 m  v2  v2    2 mg     2 a   2    1 a   2  dimensionally correct, allow sign errors. 2  4  a   3  3    1 A1 With v substituted. 1  1  2 mg  2  2 1  2 m  2ag ag    a    a   2  2  a  3  3     3 3  M1 Solve to find value for  dependent on energy Solve [ v2 g  a  ] equation with 3 or 4 terms 4 9  9 A1 9  SCB2 for  mg if given  not used 2 2 4 © UCLES 2023 Page 5 of 14

4(a) Energy equation: gain in KE = loss in EPE B1 At least one correct EPE. M1 3-term energy equation. 1 m80ag= 1  5mg( (2L−8a)2 −(4a)2) A1 All correct. 2 2 8a 4L2 −8aL−20a2 =0 M1 Simplify to quadratic in L. L=10a A1 Correct single answer. 5 4(b) At P: 2Tcos=macceleration M1 5mg 15 M1 Hooke’s law: T = 12a= mg 8a 2 Solve, acceleration = 12g A1 3 © UCLES 2023 Page 9 of 13

7(a) 2 B1 1 6  Gain in KE = m 2ag and Gain in GPE = mg3asin   2 5  1 B1 kmg Loss in EPE = 2 ( (3a)2 −a2 ) a 1 M1 KE, GPE and at least one EPE present, allow sign errors, 2 kmg Energy equation: 1 m   6 2ag   +mg3asin= 2 ( (3a)2 −a2 ) dimensionally correct. 2 5  a A1 All correct. 36 12 A1 mg+ mg =4kmg 25 5 24 k= 25 5 © UCLES 2023 Page 13 of 15 7(b) 3a 72 M1 Hooke’s law. In lower position, tension T in spring = kmg =3kmg = mg a 25 Perpendicular to rod, T =(F +mg)cos 72 A1 So, (F+mg)cos= mg, 25 19 F = mg 5 Alternative method for question 7(b) 3a 72 M1 In lower position, tension T in spring = kmg =3kmg = mg Hooke’s law. a 25 Eliminate T. Tcos=Tsin Tsin+Tcos =F+mg 19 A1 F = mg 5 2 © UCLES 2023 Page 14 of 15 7(c) Let tension in rod = T' M1 Parallel to rod, T=(F +mg)sin 96 A1 T= mg 25 Alternative method for question 7(c) Tcos=Tsin M1 At least one equation seen with their T and/or F. Tsin+Tcos =F+mg 96 A1 T= mg 25 2 © UCLES 2023 Page 15 of 15

4(a) Energy equation: gain in KE = loss in EPE B1 At least one correct EPE. M1 3-term energy equation. 1 m80ag= 1  5mg( (2L−8a)2 −(4a)2) A1 All correct. 2 2 8a 4L2 −8aL−20a2 =0 M1 Simplify to quadratic in L. L=10a A1 Correct single answer. 5 4(b) At P: 2Tcos=macceleration M1 5mg 15 M1 Hooke’s law: T = 12a= mg 8a 2 Solve, acceleration = 12g A1 3 © UCLES 2023 Page 9 of 13

2(a) In equilibrium: 2Tcosmg M1  5a   B1 Hooke’s law: T    2a   2a  2  4 Equate and use cos 3: 10mg A1 5 3 3 2(b) 1 20  2 B1 80 EPE loss = 2 4a   3 a4a  27 mga 1 8a 80 M1 A1 All 3 terms required, dimensionally correct, their Energy equation: mv2 mg ' mga' . 2 3 27 4 ga A1 Any equivalent form. v 3 3 4 © Cambridge University Press & Assessment 2024 Page 7 of 13

2(a) In equilibrium: 2Tcosmg M1  5a   B1 Hooke’s law: T    2a   2a  2  4 Equate and use cos 3: 10mg A1 5 3 3 2(b) 1 20  2 B1 80 EPE loss = 2 4a   3 a4a  27 mga 1 8a 80 M1 A1 All 3 terms required, dimensionally correct, their Energy equation: mv2 mg ' mga' . 2 3 27 4 ga A1 Any equivalent form. v 3 3 4 © Cambridge University Press & Assessment 2024 Page 7 of 13

4 Consider one situation: B1 kmg  1  2 3 Accept unsimplified. A to B: Loss in EPE = a2   a   = kmga 2a  2   8   1 mga 3  3  M1A1 KE, GPE, EPE terms present. Energy: mV2  sin kmga ( V2  gak1  Must be dimensionally correct. 2 2 8  4  Must include sin or cos in GPE. Consider a second situation: B1 kmg   a 2 15 Accept unsimplified. A to C: Loss in EPE = a2      = kmga 2a   4  32   1 1  2 mg5a 15  15  M1 KE, GPE, EPE terms present. Energy: m  V   sin kmga  V2  agk2  Must be dimensionally correct. 2 2  4 32  4  Must include sin or cos in GPE. Third possible situation: (B1) This may be used in combination with either of kmg  a 2  a 2 3 the first two situations. B to C : Loss in EPE =        = kmga Mark to the candidate’s benefit, 2a  2  4  32   1 1  2 1 1  2 mg3a 3  1  (M1) KE, GPE, EPE terms present. Energy: m V  m V  sin kmga  V2  ag6k  Must be dimensionally correct . 2 2  2 4  4 32  4  Must include sin or cos in GPE. Eliminate V2 from two energy equations to obtain expression involving only k, a M1 At least one of the energy equations must have scored M1. and possibly sin 9 A1 k  4 7 © Cambridge University Press & Assessment 2024 Page 9 of 15

3(a) 2mg M1 Equilibrium position. Hooke’s law: T = extension and T =mg 2 Extension = 1 m A1 1 2mg B1 EPE loss =  (1+d)2 2 2 Gain in GPE = mg(2+1+d) B1 1 M1 Equate: mg(1+d)2 =mg(3+d) 2 d = 5 A1 SC: 3 marks for final answer of 5+1 . SC: 2 marks for final answer of 5+k , k 1 . 6 3(b) 1 1 2mg M1 GPE, KE, EPE terms. Energy equation: mV2 +mg(1+d)=  (1+d)2 2 2 2 V2 =g ( d2 −1 ) V = 40 =2 10 A1 Alternatively: M1 Using KE and GPE from 2 m below O to point O 1 mV2 =2mg 2 V2 =4g V = 40 =2 10 A1 2 © Cambridge University Press & Assessment 2024 Page 7 of 12

2 1 5mg  a 2  B1 Both terms seen. Extension when P comes to rest is x, EPE loss =    −x2 2 a  2    1 5mg  a 2  a  M1 At least one EPE term and a GPE term, Energy:    −x2=mg −x dimensionally correct. 2 a  2  2    Solve: 20x2 −8ax−a2 =0 M1 Obtain homogeneous quadratic equation in x and a Must come from an energy equation involving two EPE terms. Note that the correct case simplifies to a linear 5mga  equation  +x=mg and this scores M1. 2a 2  a 9 A1 x=− , so distance of P below O is a 10 10 © Cambridge University Press & Assessment 2024 Page 7 of 19 2 Alternative method for question 2 Distance of P below O when it comes to rest comes to rest is h B1 EPE loss = = 1  5mg      a  2 −(h−a)2   2 a  2    Energy: 1  5mg      a  2 −(h−a)2  =mg   3 a−h   M1 A di t m le e a n s s t i o o n n a e l l E y P c E o r t r e e r c m t. and a GPE term, 2 a  2  2    20h2 −48ah+27a2 =0 M1 Obtain homogeneous quadratic/linear equation in h and a. Must come from an energy equation involving two EPE terms. 9 A1 (10h−9a)( 2h−3a)=0, h= a 10 4 © Cambridge University Press & Assessment 2024 Page 8 of 19

3(a) 2mg M1 Equilibrium position. Hooke’s law: T = extension and T =mg 2 Extension = 1 m A1 1 2mg B1 EPE loss =  (1+d)2 2 2 Gain in GPE = mg(2+1+d) B1 1 M1 Equate: mg(1+d)2 =mg(3+d) 2 d = 5 A1 SC: 3 marks for final answer of 5+1 . SC: 2 marks for final answer of 5+k , k 1 . 6 3(b) 1 1 2mg M1 GPE, KE, EPE terms. Energy equation: mV2 +mg(1+d)=  (1+d)2 2 2 2 V2 =g ( d2 −1 ) V = 40 =2 10 A1 Alternatively: M1 Using KE and GPE from 2 m below O to point O 1 mV2 =2mg 2 V2 =4g V = 40 =2 10 A1 2 © Cambridge University Press & Assessment 2024 Page 7 of 12

5 At v , T =2gsin   =14g  M1 Allow T =mgsin. max 25 14x 14g A1 Find extension at maximum velocity. = leading to x=1. 0.5 25 5 140.32 B1 FT One correct EPE term seen anywhere. EPE at top: =1.26 FT only on their x found using a correct method 20.5 (dependent on first M mark). 14("their x") 2 EPE at bottom: =0.56 20.5 140.32 14("their x") 2 M1 Dimensionally correct, including KE, GPE, and 2gsin(0.8+0.5+"their x")+ = + 12v2 at least one EPE. 20.5 20.5 2 A1 FT Correct equation. FT only on their x found using a correct method (dependent on first M mark). v2 =0.91g =9.1 leading to v=3.02ms–1. A1 v=3.016620625 6 © Cambridge University Press & Assessment 2025 Page 13 of 17

5 At v , T =2gsin   =14g  M1 Allow T =mgsin. max 25 14x 14g A1 Find extension at maximum velocity. = leading to x=1. 0.5 25 5 140.32 B1 FT One correct EPE term seen anywhere. EPE at top: =1.26 FT only on their x found using a correct method 20.5 (dependent on first M mark). 14("their x") 2 EPE at bottom: =0.56 20.5 140.32 14("their x") 2 M1 Dimensionally correct, including KE, GPE, and 2gsin(0.8+0.5+"their x")+ = + 12v2 at least one EPE. 20.5 20.5 2 A1 FT Correct equation. FT only on their x found using a correct method (dependent on first M mark). v2 =0.91g =9.1 leading to v=3.02ms–1. A1 v=3.016620625 6 © Cambridge University Press & Assessment 2025 Page 13 of 17

2(a) mg M1 Energy equation with all three terms. (x−a)2 =mgxsin30−11mgx 2a 30 Dimensionally correct. B1 Two terms correct (need not be in equation). A1 Correct energy equation. 15x2 −34ax+15a2 =0 M1 Simplify to three-term homogeneous quadratic equation and attempt to solve. (3x−5a)(5x−3a)=0 x= 5a A1 CAO, must reject x= 3a if seen. 3 5 5 2(b) T = 2mg B1 3 F =T −mgsin30 = 2mg−1mg M1 Equation for F. Allow sin/cos mix.   3 2 Allow sign errors. 1mg A1 6 3 © Cambridge University Press & Assessment 2025 Page 10 of 17

1 Energy equation: M1 Energy equation with two EPE terms and at least one GPE term, 20mg(2a)2 20mgx2 dimensionally correct, allow sign errors. 5 +3mga= +mg(a+x) Award equivalent marks if any other x is used. 2a 5 2a B1 One correct EPE term seen anywhere. A1 Correct energy equation. 50x2+5ax−6a2 =0 M1 Dependent on two EPE terms used. Simplify to 3-term quadratic with terms in x2, ax and a2 only. x= 3 a A1 Single correct answer. If seen, x=−2a must be rejected. 10 5 Alternative method for question 1 Energy equation: M1 Energy equation with two EPE terms and at least one GPE term, 20mg(2a)2 20mg(h−a)2 dimensionally correct, allow sign errors. 5 +3mga= +mgh Award equivalent marks if any other h is used. 2a 5 2a B1 One correct EPE term seen anywhere. A1 Correct energy equation. 50h2 −95ah+39a2 =0 M1 Dependent on two EPE terms used. Simplify to 3-term quadratic with terms in h2, ah and a2 only. h=13a , so the maximum extension is 3 a. A1 Single correct answer. 10 10 If seen, h= 3a leading to x=−2a must be rejected. 5 5 5 © Cambridge University Press & Assessment 2025 Page 9 of 17

4 5mge B1 In equilibrium, =5mg, e=a. a B1 At least one correct EPE term seen. Let x be the length of the string when speed of P is 7ag . 5 Energy equation in subsequent motion: M1 Dimensionally correct energy equation with one KE term, 5mga2 5mg(x−a)2 two EPE terms and at least one GPE term. − =mg(e+a−x)+ 1m(7ag) 2a 2a 2 5 A1 Fully correct energy equation. 25x2 −60ax+27a2 =0 M1 Three-term homogeneous quadratic equation in x and a . x= 9a only. A1 5 Alternative method for question 4 5mge B1 In equilibrium, =5mg, e=a. a Let x be the extension when P is released. B1 At least one correct EPE term seen. Energy equation in subsequent motion: 5mga2 5mgx2 M1 Dimensionally correct energy equation with one KE term, − =mg(e−x)+1m (7ag ) two EPE terms and at least one GPE term. 2a 2a 2 5 A1 Fully correct energy equation. 25x2 −10ax−8a2 =0 M1 Three-term homogeneous quadratic equation in x and a . (5x+2a)(5x−4a)=0, x= 4a only. A1 5 Required length of string = 9a. 5 6 © Cambridge University Press & Assessment 2025 Page 12 of 17

2(a) EPE= 2mg (1a )2 B1 Correct EPE seen OR correct work done by 2a 3 friction seen. Work done by friction=mg (1a ) 3 2mg (1a )2 =mg (1a ) M1 Equation or inequality involving EPE and 2a 3 3 work done terms only. = 1 A1 CWO 3 3 © Cambridge University Press & Assessment 2025 Page 9 of 17 2(b) Let x be the extension in the string when P comes to rest. M1 Dimensionally correct equation with two EPE 2mg (1a )2 = 2mg x2 +1mg (1a−x ) terms and one work done term. 2a 3 2a 2 3 Use of 1mgx as work done term can be 2 awarded M1 only. A1 Correct energy equation. May be unsimplified. 18x2 −9ax+a2 =0, (6x−a)(3x−a)=0 x= 1a only. A1 CWO   6 Alternative method for question 2(b) Let x be the distance moved by P before coming to rest. M1 Dimensionally correct equation with two EPE 2mg (1a )2 = 2mg (1a−x )2 +1mgx terms and one work done term. 2a 3 2a 3 2 Use of 1mg (1a−x ) as work done term can 2 3 be awarded M1 only. A1 Correct energy equation. May be unsimplified. 6x2−ax=0, x=1a A1 CWO 6 Extension = 1a−x= 1a 3 6 3 © Cambridge University Press & Assessment 2025 Page 10 of 17

4 5mge B1 In equilibrium, =5mg, e=a. a B1 At least one correct EPE term seen. Let x be the length of the string when speed of P is 7ag . 5 Energy equation in subsequent motion: M1 Dimensionally correct energy equation with one KE term, 5mga2 5mg(x−a)2 two EPE terms and at least one GPE term. − =mg(e+a−x)+ 1m(7ag) 2a 2a 2 5 A1 Fully correct energy equation. 25x2 −60ax+27a2 =0 M1 Three-term homogeneous quadratic equation in x and a . x= 9a only. A1 5 Alternative method for question 4 5mge B1 In equilibrium, =5mg, e=a. a Let x be the extension when P is released. B1 At least one correct EPE term seen. Energy equation in subsequent motion: 5mga2 5mgx2 M1 Dimensionally correct energy equation with one KE term, − =mg(e−x)+1m (7ag ) two EPE terms and at least one GPE term. 2a 2a 2 5 A1 Fully correct energy equation. 25x2 −10ax−8a2 =0 M1 Three-term homogeneous quadratic equation in x and a . (5x+2a)(5x−4a)=0, x= 4a only. A1 5 Required length of string = 9a. 5 6 © Cambridge University Press & Assessment 2025 Page 12 of 17

6(a) N2L: T =T +kmg B1 1 2 5mg(22a−BP−4a) B1 Apply Hooke’s law to string AP. T = 1 4a 6mg(BP−8a) B1 Apply Hooke’s law to string BP. T = 2 8a 5mg(22a−BP−4a) 6mg(BP−8a) M1 Substitute T and T into N2L and obtain linear = +kmg 1 2 4a 8a expression for BP in terms of a and k . 8BP=114a−4ak A1 AG, shown convincingly. 57a−2ak BP= 4 Alternative method for question 6(a) N2L: T =T +kmg B1 1 2 5mgx B1 Apply Hooke’s law to string AP, where x denotes the T = 1 4a extension in string AP. 6mg(18a−x−8a) 6mg(10a−x) B1 Apply Hooke’s law to string BP. T = =  2 8a 8a   5mgx 6mg(10a−x) M1 Substitute T and T into N2L and obtain linear = +kmg 1 2 4a 8a expression for x in terms of a and k AND substitute 15a+2ak their x into BP=18a−x. x= 4 A1 AG, shown convincingly. 15a+2ak 57a−2ak BP=18a−x=18a− = 4 4 © Cambridge University Press & Assessment 2025 Page 15 of 18 6(a) Alternative method for question 6(a) N2L: T =T +kmg B1 1 2 5mg(x−4a) B1 Apply Hooke’s law to string AP, where x denotes the T = length of string AP. 1 4a 6mg(22a−x−8a) 6mg(14a−x) B1 Apply Hooke’s law to string BP. T = =  2 8a 8a   5mg(x−4a) 6mg(14a−x) M1 Substitute T and T into N2L and obtain linear = +kmg 1 2 4a 8a expression for x in terms of a and k AND substitute 31a+2ak 31a+2ak 57a−2ak their x into BP=22a−x. x= , BP=22a−x=22a− = 4 4 4 A1 AG, shown convincingly. Alternative method for question 6(a) N2L: T =T +kmg B1 1 2 5mg(10a−x) B1 Apply Hooke’s law to string AP, where x denotes the T = 1 4a extension in string BP. 6mgx B1 Apply Hooke’s law to string BP. T = 2 8a 5mg(10a−x) 6mgx M1 Substitute T and T into N2L and obtain linear = +kmg 1 2 4a 8a expression for x in terms of a and k AND substitute 25a−2ak 25a−2ak 57a−2ak their x into BP=8a+x. x= , BP=8a+x=8a+ = 4 4 4 A1 AG, shown convincingly. © Cambridge University Press & Assessment 2025 Page 16 of 18 6(a) Alternative method for question 6(a) N2L: T =T +kmg B1 1 2 5mgx 6mgx B1 Apply Hooke’s law to string AP or BP, where x T = A or T = B A 1 4a 2 8a denotes the extension in string AP and x denotes the B extension in string BP. 5mgx 6mgx B1 Apply Hooke’s law to string AP and BP and recognise T = A , T = B , x +x =10a 1 4a 2 8a A B that the sum of the extensions in both strings is 10a. 5mgx 6mgx M1 Substitute T and T into N2L and obtain linear A = B +kmg, 5x −3x =4k 1 2 4a 8a A B expression in x and x in terms of a and k AND use A B 15a+2ak 25a−2ak x A = 4 OR x B = 4 the expression and x A +x B =10a to obtain an expression for x or x only AND substitute their x 15a+2ak 57a−2ak A B A Either BP=18a−x =18a− = into BP=18a−x or their x into BP=8a+x . A 4 4 A B B 25a−2ak 57a−2ak OR BP=8a+x =8a+ = A1 AG, shown convincingly. A 4 4 5 6(b) 6mg(18a−8a)2 5mg(14a−4a)2 B1 At least one correct EPE term. +18kmga= +8kmga 28a 24a M1 Dimensionally correct energy equation with two EPE terms and at least one GPE term (no KE terms). A1 Fully correct energy equation. k = 5 A1 2 4 © Cambridge University Press & Assessment 2025 Page 17 of 18

6(a) EITHER: (B1 Correct expression for the EPE. 15x2 3x2   22.5 15x2 M1 Conservation of energy with exactly 3 terms (GPE 1.5g(2.5+x)= + 11.5v2 and KE terms must be correct). Dimensionally 22.5 2 correct – allow sign errors and slips in the EPE term (but must be of the form © Cambridge University Press & Assessment 2026 Page 18 of 22 k 1 x 2 ). 37.5+15x=3x2 +0.75v2 A1) AG – at least one line of working from application of conservation of energy to given answer. Any error v2 =50+20x−4x2 =−4(x2 −5x−12.5) seen is A0. OR: (B1 15 dv 1.5g− x=1.5v 2.5 dx  dv  v =10−4x    dx  12 v 2 = 1 0 x − 2 x 2 + c x = 0 , v 2 = 2  1 0  2 .5 = 5 0  c = 2 5 M1 Integrate correctly and explicitly use the correct initial condition to find c. As a minimum, must see x = 0 , v 2 = 5 0  c = 2 5 (OE, for example, may find that the constant for their expression is 50). 1v2 =10x−2x2+25v2 =−4(x2−5x−12.5) A1) AG 2 Any error seen is A0. Available marks 3 Trace ID: PC-76076FFBDF Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/3a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/31 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 6(b) 6.83 only B1 AWRT 6.83 (6.8301270…) or © Cambridge University Press & Assessment 2026 Page 19 of 22 52 ( 1 + 3 ) . If −1.830127... seen it must be explicitly rejected. If working shown, then it must be correct. 1 6(c) [Maximum occurs when] x = 2 .5 B1 The correct answer for v can imply this mark. Must be clear that 2.5 is the extension at maximum speed.  v 2 = 7 5   v = 8 .6 6 m s–1 B1 AWRT 8.66 (8.660254…) or exact, for example, 5 3 . 2 Trace ID: PC-76076FFBDF Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/3a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/31 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

6(a) EITHER: (B1 Correct expression for the EPE. 15x2 3x2   22.5 © Cambridge University Press & Assessment 2026 Page 18 of 22 1 .5 g ( 2 .5 + x ) = 1 2 5  x 2 2 .5 + 12  1 .5  v 2 M1 Conservation of energy with exactly 3 terms (GPE and KE terms must be correct). Dimensionally correct – allow sign errors and slips in the EPE term (but must be of the form k 1 x 2 ). 3 7 .5 + 1 5 x = 3 x 2 + 0 .7 5 v 2  v 2 = 5 0 + 2 0 x − 4 x 2 = − 4 ( x 2 − 5 x − 1 2 .5 ) A1) AG – at least one line of working from application of conservation of energy to given answer. Any error seen is A0. OR: 1 .5 g − 1 2 5 .5 x = 1 .5 v d d v x   v d d v x = 1 0 − 4 x  (B1 12 v 2 = 1 0 x − 2 x 2 + c x = 0 , v 2 = 2  1 0  2 .5 = 5 0  c = 2 5 M1 Integrate correctly and explicitly use the correct initial condition to find c. As a minimum, must see x = 0 , v 2 = 5 0  c = 2 5 (OE, for example, may find that the constant for their expression is 50). 12 v 2 = 1 0 x − 2 x 2 + 2 5  v 2 = − 4 ( x 2 − 5 x − 1 2 .5 ) A1) AG Any error seen is A0. Available marks 3 Trace ID: PC-852B15CDF8 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/3a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/32 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 6(b) 6.83 only B1 AWRT 6.83 (6.8301270…) or © Cambridge University Press & Assessment 2026 Page 19 of 22 52 ( 1 + 3 ) . If − 1 .8 3 0 1 2 7 ... seen it must be explicitly rejected. If working shown, then it must be correct. 1 6(c) [Maximum occurs when] x = 2 .5 B1 The correct answer for v can imply this mark. Must be clear that 2.5 is the extension at maximum speed. v2 =75v=8.66 m s–1 B1 AWRT 8.66 (8.660254…) or exact, for example,   5 3 . 2 Trace ID: PC-852B15CDF8 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/3a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/32 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

6(a) 1.6mgx B1 Correct use of Hooke’s law. T = 10a © Cambridge University Press & Assessment 2026 Page 17 of 20 T = 0 .2 m g  45  = 0 .1 6 m g  M1 Use of N2L, must be considering same mass on both sides. Allow sin/cos mix. Allow sign error. Must replace s i n  with a numerical value. 0 .1 6 m g = 1 .6 1 m 0 g a x A1 AG, set up correct equation and find length. x = a OP =10a−a=9a. 3 6(b) 12  0 .3 m v 2 = 1 0 a  45  0 .3 m g OR v 2 = 2  1 0 a  g  45 *M1 Energy equation before collision. Same mass on both sides (allow m instead of 0.3m). Correct number of terms, dimensionally correct. Substitute value for s i n  (may be done after m cancelled). Allow sin/cos mix. OR Use kinematics to find an equation containing v, a and g, including substituting value for s i n . Allow sin/cos mix. v = 4 a g A1 Award A1 for v2 =16ag. 0.3m'4 ag'=(0.3m+0.2m)w DM1 Dimensionally correct momentum equation, masses must be present, with their v substituted. w=2.4 ag A1 OE, for example, 1 0 .2 .5 a g . SC B1 for correct answer if m is missing throughout. 4 Trace ID: PC-202C151F9E Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/3a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/33 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 6(c) x defined as the distance from O (2 GPE terms): © Cambridge University Press & Assessment 2026 Page 18 of 20 1 2  0 .5 m  " 2 .4 2 a g " + 0 .5 m g  9 a  4 5 + 1 .6 m 2 0 g a a 2 = 0 .5 + m 1 g .6  m x g  ( 1 2 0 4 5 0 a a − x ) 2 OR x defined as the final extension (2 GPE terms): 1 2  0 .5 m  " 2 .4 2 a g " + 0 .5 m g  ( x − a )  4 5 + 1 .6 m 2 0 g a a 2 = 1 .6 2 m 0 g a x 2 OR x defined as the change in extension (one GPE term): 1 2  0 .5 m  " 2 .4 2 a g " + 0 .5 m g x  4 5 + 1 .6 m 2 0 g a a 2 = 1 .6 m g 2 ( 0 x a + a ) 2 B1 At least one correct EPE. M1 Dimensionally correct energy equation with 2 EPE terms, one KE term and one/two GPE term(s) depending on how x is defined. Allow sign errors. Allow sin/cos mix. Values for trigonometric functions must be substituted. Allow slips in the EPE terms but must be of the form k 1 x 2 . Do not allow sign errors in the EPE terms. Mass of particle needs to be present. Simplified equations for reference: 3 2 6 5 m a g + 2 5 m g ( 9 a − x ) + 2 m 2 g 5 a a 2 = 2 m g ( 1 2 0 5 a a − x ) 2 OR 3 2 6 5 m a g + 2 5 m g ( x − a ) + 2 m 2 g 5 a a 2 = 2 m 2 g 5 x a 2 OR 3 2 6 5 m a g + 2 5 m g x + 2 m 2 g 5 a a 2 = 2 m g ( 2 x 5 + a a ) 2 A1FT All terms correct. FT their w 2 .  x =  3 a A1 If seen, must reject alternative solutions outside [0, 10a]. 4 Trace ID: PC-202C151F9E Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/3a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/33 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

7(a) © Cambridge University Press & Assessment 2026 Page 22 of 25 R = m g  11 23  = 1 2 0 1 3 m  B1 Correct expression for the normal contact force. Allow 23 or better (22.61986…) for angle. Allow R = m g c o s ( s i n − 1 ( 51 3 ) ) . Angle must be substituted. T 11 23 m g 51 3 m g m 54 g  −  − =  M1 Use of N2L parallel to the slope. Must contain the correct number of dimensionally correct relevant terms. Allow sin/cos mix, allow sign errors. Must include a component of the weight not mass. Allow with T or their possibly incorrect T (but all other values must be substituted). Normal contact force must be either 11 23 m g or 51 3 m g (OE, using an angle of 23 or better). 6mg3a B1 Correct unsimplified expression for T using T = = 9mg 8a  4  Hooke’s law. 9mg−12mg− 5 mg=m5g A1 Allow AWRT 0.667 (but not 0.666).   4 13 13 4 = 2 3 4 Trace ID: PC-051B44EFC7 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/3a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/34 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 7(b) EITHER: METHOD 1 (SEE SC ON PAGE 25 FOR THOSE WHO ASSUME THAT P COMES TO REST AFTER THE STRING IS SLACK) © Cambridge University Press & Assessment 2026 Page 23 of 25 1 2  6 m 8 a g  ( 3 a ) 2  = 2 78 m g a  (*B1 1 6mg B0 for  3a2 unless recovered. 2 8a 1 2  6 m g 8 ( 3 a a ) 2 = 1 2  6 m 8 a g  ( 3 a − d ) 2 + 1 5 3  m g d + 2 3  1 1 2 3 m g d  or 1 6mg(3a)2 1 6mg 5 212   =  e2 + mg(3a−e)+  mg(3a−e) 2 8a 2 8a 13 313  or 1 2  6 m g 8 ( 3 a a ) 2 = 1 2  6 m g ( x 8 − a 8 a ) 2 + 1 5 3 m g (1 1 a − x ) + 2 3  1 1 2 3 m g (1 1 a − x )  *M1 Work-energy equation using their  with the correct number of dimensionally correct relevant terms. Allow sign errors. Allow sin/cos mix on the GPE and work done by friction. Final EPE must be dimensionally correct, and its extension X must be consistent with their GPE and work done by friction terms, but allow other slips in their EPE term. If their X in EPE implies that X  0 , then M0. A1FT Fully correct equation. FT their value of . 3 d 8 a 2 − 5 4 d = 0  d = 1 0 3 a [  3 a s o P m o v i n g w h e n t h e s t r i n g i s s la c k ] A1 Correctly obtaining or (3e+a)(e−3a)=0e=−1a 3 d = 1 0 3 a or e = − 13 a or x = 2 33 a from ( 3 x − 2 3 a ) ( x − 1 1 a ) = 0 1 2  6 m 8 a g  ( 3 a ) 2 = m g d  1 5 3 + 2 3   1 1 2 3 m g d  1 6mg or  (3a)2 =mgd or 2 8a 1 2  6 m 8 a g  ( 3 a ) 2 = m g ( 3 a − e )  1 5 3 + 2 3   1 1 2 3 m g ( 3 a − e )  DM1 OE, work-energy equation with correct number of dimensionally correct relevant terms. Allow sign errors. Allow sin/cos mix on the GPE and OE work done by friction (but these two terms must contain a component of weight). Must be using correct EPE at this stage.  d = 2 78 a  O C = 1 1 a − 2 78 a =  6 18 a A1) WWW 7.625a (condone 7.63a). For reference, e=−3aOC =11a−3a−3a= 61a. 8 8 8 Trace ID: PC-051B44EFC7 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/3a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/34 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 7(b) OR: (*B1 Correct unsimplified expression for the initial METHOD 2 (Showing KE or velocity is > 0 when the string goes slack) EPE. 1 6mg  (3a)2 = 27mga 2 8a  8  1 6mg 5 2 12  2  8a (3a)2 =KE slack +3amg 13 + 3  13 mg3a  or © Cambridge University Press & Assessment 2026 Page 24 of 25 1 2  6 m 8 a g  ( 3 a ) 2 = K E sla c k + 3 m g a *M1 Work-energy equation using their  with the correct number of dimensionally correct relevant terms. Allow sign errors. Allow sin/cos mix on the GPE and work done by friction (but these two terms must contain a component of weight and the correct distance of 3a). EPE must be of the form k 1 ( 3 a ) 2 where k 1 is dimensionally correct, but allow other slips. May see K E sla c k = 12 m v 2 . A1FT Fully correct equation. FT their value of .  K E = 38 m g a or v 2 = 34 g a [  0 s o P i s m o v i n g w h e n t h e s t r i n g i s s la c k ] . A1 WWW, correct expression for the KE when the string goes slack OE. No comment required that this is positive (OE). 1 6mg 5 2 12   (3a)2 =mgd +  mgd 2 8a 13 3 13  or 1 2  6 m 8 a g  ( 3 a ) 2 = m g d DM1 Work-energy equation with correct number of dimensionally correct relevant terms – allow sign errors. Allow sin/cos mix on the GPE and work done by friction (but these two terms must contain a component of weight). Must be using correct EPE at this stage.  d = 2 78 a  O C = 1 1 a − 2 78 a =  6 18 a A1) WWW 7.625a (condone 7.63a). Trace ID: PC-051B44EFC7 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/3a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/34 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 7(b) OR: SC FOR THOSE WHO ASSUME THAT P COMES TO REST AFTER THE STRING IS SLACK © Cambridge University Press & Assessment 2026 Page 25 of 25 1 2  6 m 8 a g  ( 3 a ) 2  = 2 78 m g a  (*B1 1 6mg B0 for  3a2 unless recovered. 2 8a 1 2  6 m 8 a g  ( 3 a ) 2 = m g d  1 5 3 + 2 3   1 1 2 3 m g d  DM1 OE, work-energy equation with correct number of dimensionally correct relevant terms. Allow sign errors. Allow sin/cos mix on the GPE and or work done by friction (but these two terms must 1 6mg  (3a)2 =mgd contain a component of weight). Must be using 2 8a the correct expression for EPE and their  from 7(a).  d = 2 78 a  O C = 1 1 a − 2 78 a =  6 18 a A1 7.625a (condone 7.63a) Maximum of three marks for the SC. The maximum number of marks available to candidates who assume that P comes to rest after the string is slack with no further work is 3 marks as above. However, full marks available for a candidate who explicitly shows why P is still in motion when the string goes slack. An example of such an approach is as follows. [Sum of work done by friction to travel 3a up the plane and the GPE is given by] M1 For sum of the two terms with sin/cos mix only mg3a 5 +2 (12mg3a ) =3mga and their . 13 3 13 A1FT FT their . [As] 2 78 m g a  3 m g a , [P is still in motion.] A1) Correct comparison with the EPE. Available marks 6 Trace ID: PC-051B44EFC7 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/3a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/34 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/