Motion of a projectile

7(a) At greatest height 0=100sinθ− gt M1 t = 8 A1 Therefore times at height H are t=3 (and t =13) B1 1 M1 Substitute into H =100sinθt− gt2 2 H =195 A1 Alternative method to question 7(a) 1 M1 ↑H =100sinθt− gt2 2 And H =100sinθ( t+10 )− 1 g ( t+10 )2 A1 2 Subtract: 1000sinθ= 1 g ( 20t+100 ) M1 2 t=3 B1 H =195 A1 © UCLES 2021 Page 12 of 13 7(a) Alternative method to question 7(a) 1 B1 ↑H =100sinθt− gt2 2 M1 A1 (100sinθ)2−2gH Difference between roots = 1 g 2 Equate to 10 and rearrange to find H M1 H =195 A1 5 7(b) Time to required point = 15 s B1 ( ) ↑v=100sinθ−10×15 =−70 B1 Both components. →v=100cosθ=60 Magnitude = 92.2 B1 Angle below horizontal = tan -1 (70/60) = 49.4° B1 4 © UCLES 2021 Page 13 of 13

7(a) At greatest height 0=100sinθ− gt M1 t = 8 A1 Therefore times at height H are t=3 (and t =13) B1 1 M1 Substitute into H =100sinθt− gt2 2 H =195 A1 Alternative method to question 7(a) 1 M1 ↑H =100sinθt− gt2 2 And H =100sinθ( t+10 )− 1 g ( t+10 )2 A1 2 Subtract: 1000sinθ= 1 g ( 20t+100 ) M1 2 t=3 B1 H =195 A1 © UCLES 2021 Page 12 of 13 7(a) Alternative method to question 7(a) 1 B1 ↑H =100sinθt− gt2 2 M1 A1 (100sinθ)2−2gH Difference between roots = 1 g 2 Equate to 10 and rearrange to find H M1 H =195 A1 5 7(b) Time to required point = 15 s B1 ( ) ↑v=100sinθ−10×15 =−70 B1 Both components. →v=100cosθ=60 Magnitude = 92.2 B1 Angle below horizontal = tan -1 (70/60) = 49.4° B1 4 © UCLES 2021 Page 13 of 13

7(a) R2 M1 y = 0 in trajectory equation: Rtanθ−g =0 2u2 ( cosθ)2 ( R=)2u2sinθcosθ only A1 A u n 2 y si e n q 2 u θ ivale 2 n u t s 2 i t n a g n l θ e term expression, for example: g , , at least one intermediate line of g g sec2θ working, not just quoting a result. SC B1 using SUVAT. 2 7(b) u2sinθcosθ M1 Or use SUVAT. x=their and substitute in trajectory equation. g u2 ( sinθ)2 A1 Single term. H = 2g 2 7(c) 4H B1 AG Use R= and simplify: tanθ= 3, θ = 60° 3 1 © UCLES 2021 Page 14 of 16 7(d) dy gx M1 Differentiate with respect to x. =tanθ− dx u2 ( cosθ)2 x M1 dy tanθ− =±1 used Use =±1 as limiting case. 4 ( cosθ)2 dx x = 3 +1, x = 3 −1 A1 3 −1< x < 3 +1 A1 Strict inequality, exact values. Alternative method for question 7(d) 1 dy M1 Differentiate with respect to x. y= 3x− x2, = 3−x 2 dx dy M1 dy =±1 used Use =±1 as limiting case. dx dx x = 3 +1 , x = 3 −1 A1 3 −1< x < 3 +1 A1 Strict inequality, exact values. © UCLES 2021 Page 15 of 16 7(d) Alternative method for question 7(d) v =±v M1 Used, both cases considered. When moving at 45° to horizontal, x y v = 40 cosθ , v = 40sinθ−10t M1 x y 1 ( ) 1 ( ) t= 30− 10 , t= 30+ 10 10 10 x = 3 +1 , x = 3 −1 A1 3 −1< x < 3 +1 A1 Strict inequality, exact values. 4 © UCLES 2021 Page 16 of 16

5 At A: ↑usinθ−8g →ucosθ M1 Both. usinθ−8g A1 tanα= ucosθ At B: ↑usinθ−32g →ucosθ M1 Both. usinθ−32g A1 tanβ= ucosθ usinθ−8g usinθ−32g B1 Perpendicular directions, so tanα×tanβ=−1. × =−1 ucosθ ucosθ u2 −320u+25600=0 M1 Simplify to a quadratic in u. u=160 A1 7 © UCLES 2021 Page 10 of 12

1(a) Velocity: →ucosα B1 ↑usinα−gT B1 Allow 10 for g. Must be T. 2 1(b) ucosα sinα M1 FT Allow missing minus sign on RHS for M1. =− oe usinα−gT cosα FT from (a). u A1 T = gsinα 2 1(c) u B1 AG sinα<1 giving T > g 1 © UCLES 2021 Page 6 of 12

5 At A: ↑usinθ−8g →ucosθ M1 Both. usinθ−8g A1 tanα= ucosθ At B: ↑usinθ−32g →ucosθ M1 Both. usinθ−32g A1 tanβ= ucosθ usinθ−8g usinθ−32g B1 Perpendicular directions, so tanα×tanβ=−1. × =−1 ucosθ ucosθ u2 −320u+25600=0 M1 Simplify to a quadratic in u. u=160 A1 7 © UCLES 2021 Page 10 of 12

7(a) For Q: xucosT B1 35 B1 For P: x cosT 1 2 35 M1 Equate and attempt to rearrange. Collision, so cosT 1ucosT 2 35 3 2 A1 AG  T 1u T Shown convincingly. 2 5 5 4uT 21 5T 1 4 7(b) Vertical motion to collision: M1 A1 M1 for both expressions, one correct. 1 For Q: yusinT  gT2 2 35 1 For P: y sinT 1 g(T 1)2 2 2 1 1 35 4 1 M1 Equate and attempt to solve Equate: u T  gT2   T 1 g(T 1)2 5 2 2 5 2 14T 1 1 g  T2 2T 1T2  21 T 1 2 4 16T 3621T 21, 1 55T A1 T 3 4 © UCLES 2022 Page 13 of 14 7(c) x42 B1 y 24 M1 y24 (or 24 m below O) A1 Correct sign or in words. 3 © UCLES 2022 Page 14 of 14

7(a) For Q: xucosT B1 35 B1 For P: x cosT 1 2 35 M1 Equate and attempt to rearrange. Collision, so cosT 1ucosT 2 35 3 2 A1 AG  T 1u T Shown convincingly. 2 5 5 4uT 21 5T 1 4 7(b) Vertical motion to collision: M1 A1 M1 for both expressions, one correct. 1 For Q: yusinT  gT2 2 35 1 For P: y sinT 1 g(T 1)2 2 2 1 1 35 4 1 M1 Equate and attempt to solve Equate: u T  gT2   T 1 g(T 1)2 5 2 2 5 2 14T 1 1 g  T2 2T 1T2  21 T 1 2 4 16T 3621T 21, 1 55T A1 T 3 4 © UCLES 2022 Page 13 of 14 7(c) x42 B1 y 24 M1 y24 (or 24 m below O) A1 Correct sign or in words. 3 © UCLES 2022 Page 14 of 14

3(a) Components of velocity : 25cos 25sin2g B1 M1 A1 Expression for speed or square of speed. Speed = 25cos2 25sin2g2 25cos2 25sin2g2 152 M1 Attempt to solve and find value for sin 625100gsin4g2 225 800 4 A1 sin  1000 5 5 3(b) 225sin B1 Time of flight   = 4 (s)  g  225sin M1 Any equivalent method. Range = 25cos g Range = 60 (m) A1 CWO Alternative method for question 3(b) 4 1 B1 Equation of trajectory.. y x x2 3 45 Substitute y0 and solve M1 60 (m) A1 3 © UCLES 2022 Page 8 of 13

7(a) 1 M1 0=Vsin75t− gt2 2 2V A1 AG t = sin75 g 2 7(b) Vertical component of velocity is unchanged. B1 1 © UCLES 2022 Page 11 of 12 7(c)  15  B1 Horizontally to wall, →15=Vcos75t;  t =   Vcos75 eVcos75 B1 Speed after rebound. 15 M1 Time back to O T = eVcos75  3  t = T  5  Vertically for whole flight: M1 2V t+T = sin75 g 15 15 2V A1 + = sin75 Vcos75 eVcos75 g V2cos75sin75=20g M1 Multiply by 2: V2sin150=40g, V 2 =80g A1 ( ) V =4 5g =8.94 g 7 © UCLES 2022 Page 12 of 12

5(a) → x=ucos t B1 Result quoted from MF19 scores 0/4. 1 B1  y=usin t− gt2 2 2 M1 x 1  x  Eliminate t: y=usin  − g   cos 2 ucos  gx2 A1 Must be an intermediate line of working. y=xtan− (1+tan2) AG 2u2 4 5(b) 4 302  4 2 M1 Substituting values correctly. 20=30 −10 1+   3 2u2  3    u2 =625, u=25 A1 Substitute back into trajectory equation, M1 Obtain a 3-term quadratic. g302 36 20=30tan − sec2 =30tan − (1+tan2 ) 2.252 5 18tan2−75tan+68=0 4 M1 One solution is , (3tan−4)(6tan−17)=0 3 17 A1 Giving tan= 6 5 © UCLES 2022 Page 11 of 15

7(a) 1 M1 0=Vsin75t− gt2 2 2V A1 AG t = sin75 g 2 7(b) Vertical component of velocity is unchanged. B1 1 © UCLES 2022 Page 11 of 12 7(c)  15  B1 Horizontally to wall, →15=Vcos75t;  t =   Vcos75 eVcos75 B1 Speed after rebound. 15 M1 Time back to O T = eVcos75  3  t = T  5  Vertically for whole flight: M1 2V t+T = sin75 g 15 15 2V A1 + = sin75 Vcos75 eVcos75 g V2cos75sin75=20g M1 Multiply by 2: V2sin150=40g, V 2 =80g A1 ( ) V =4 5g =8.94 g 7 © UCLES 2022 Page 12 of 12

7(a) 800sin2 B1 H 80sin2 or g 40sin B1 T 4sin or g 2 7(b) 1 1 M1 A1 No extra terms. Between t T and t 3  H  103T2 4 2 1 M1 Substitute their expressions for H and T from Use results from part (a) 80sin2534sin2 part (a) and obtain a quadratic equation in sin 4 with no more than three terms. 4sin28sin30 1 A1 Single answer. NFWW. sin , 30 2 Alternative method for question 7 part (b) 3 1 M1 A1 120 sin45 H  y3403 sin 1032 4 2 3 M1 Substitute their expressions for H and T from Use results from (a): 80sin2120 sin45 part (a) and obtain a quadratic equation in sin 4 with no more than three terms. 4sin28sin30 1 A1 Single answer. NFWW. sin , 30 2 4 © UCLES 2023 Page 14 of 15 7(c) When t 3 speeds 40cos and 40sin103 B1 Square and add to find square of speed: v2   20 3 2 102 M1 Must be numerical. v2 1300, v10 13 [= 36.1] A1 3 © UCLES 2023 Page 15 of 15

7(a) 800sin2 B1 H 80sin2 or g 40sin B1 T 4sin or g 2 7(b) 1 1 M1 A1 No extra terms. Between t T and t 3  H  103T2 4 2 1 M1 Substitute their expressions for H and T from Use results from part (a) 80sin2534sin2 part (a) and obtain a quadratic equation in sin 4 with no more than three terms. 4sin28sin30 1 A1 Single answer. NFWW. sin , 30 2 Alternative method for question 7 part (b) 3 1 M1 A1 120 sin45 H  y3403 sin 1032 4 2 3 M1 Substitute their expressions for H and T from Use results from (a): 80sin2120 sin45 part (a) and obtain a quadratic equation in sin 4 with no more than three terms. 4sin28sin30 1 A1 Single answer. NFWW. sin , 30 2 4 © UCLES 2023 Page 14 of 15 7(c) When t 3 speeds 40cos and 40sin103 B1 Square and add to find square of speed: v2   20 3 2 102 M1 Must be numerical. v2 1300, v10 13 [= 36.1] A1 3 © UCLES 2023 Page 15 of 15

7(a) For aircraft, d 5T B1 1 B1 To point of collision. For ball, 4usin T  10T2 2 For ball, ucos T d 85T 8 B1 Eliminate u: *M1 u4 1 5  45T2 55T 8 4 T  10T2, u and u 5 2 4T 3T Dependent on LHS of second B1 being 4, 3  45T2 45T 8 expression involving only T 3T2 4T 40 DM1 Dependent on previous M1. Obtain and solve 3-term quadratic. T 2 A1 Single correct answer. 6 2 Note d 8 used leads to T  B1B1B0M1M1A0 3 © UCLES 2023 Page 13 of 14 7(b) usin10T M1 OE tan1 Accept ‘tan = ……’ ucos 8 A1 OE tan1 9 Direction is 41.6° below the horizontal A1 CAO Note: d 8 used leads to 20.9 above the horizontal. 3 © UCLES 2023 Page 14 of 14

5(a) gx2 M1 Use equation of trajectory: y=xtan− sec2: 2u2 3 1 (3a)2  1 a=3a −g 1+   8 3 2u2  9 5 5ga A1 At least one step of working. a= , u2 =8ga 8 u2 AG 2 © UCLES 2023 Page 10 of 13 5(b) 2usin B1 For P, time of flight T = g For P, range = Tucos B1 1 1 B1 For Q, time of flight = T, so range = TVcos 2 2 Equate: Vcos=2ucos (1) 1 M1 gT2 1 2 4Vsin For Q vertically: 0=Vsin T − , so T = 2 4 g Equate with result for P: A1 4Vsin 2usin 1 = so Vsin= usin (2) g g 2 From (1) and (2): V2 =u2   4(cos)2 + 1 (sin)2  M1  4   9 1 1  145 A1 V2 =u2  4 +   = u2  10 4 10 40 V = 29ag 7 © UCLES 2023 Page 11 of 13

6(a) → x=ucos t 1 B1 Both correct.  y=usin t− gt2 2 2 M1 Eliminate. x 1  x  Eliminate t: y=usin − g  ucos 2 ucosα gx2 A1 y=xtan− sec2 AG 2u2 3 © UCLES 2023 Page 11 of 15 6(b) 4 g322 16 B1 h=32tan− sec2 [= 32tan− sec2] 5 2402 5 g322 8 B1 h=32tan− sec2 [= 32tan− sec2] ( )2 5 2 40 2 8 5 16 M1 Equate expressions for h and obtain a 3-term quadratic in 32tan− sec2= (32tan− sec2) tan. 5 4 5 32t− 8( 1+t2) =40t−4 ( 1+t2) 5 3t2 −10t+3=0 1 A1 Both correct t=3, 3 h=80 M1 For using their value of t to work out one value of h. 80 h= A1 Both correct 9 6 © UCLES 2023 Page 12 of 15

5(a) gx2 M1 Use equation of trajectory: y=xtan− sec2: 2u2 3 1 (3a)2  1 a=3a −g 1+   8 3 2u2  9 5 5ga A1 At least one step of working. a= , u2 =8ga 8 u2 AG 2 © UCLES 2023 Page 10 of 13 5(b) 2usin B1 For P, time of flight T = g For P, range = Tucos B1 1 1 B1 For Q, time of flight = T, so range = TVcos 2 2 Equate: Vcos=2ucos (1) 1 M1 gT2 1 2 4Vsin For Q vertically: 0=Vsin T − , so T = 2 4 g Equate with result for P: A1 4Vsin 2usin 1 = so Vsin= usin (2) g g 2 From (1) and (2): V2 =u2   4(cos)2 + 1 (sin)2  M1  4   9 1 1  145 A1 V2 =u2  4 +   = u2  10 4 10 40 V = 29ag 7 © UCLES 2023 Page 11 of 13

3 usin605g, and ucos60 , or usin6015g , and ucos60 B1 usin605g M1* Accept equivalent for t15. If  is direction of velocity at t 5, tan ucos60 usin605g usin6015g M1dep Multiply two expressions involving relevant For perpendicular directions,   1 velocities and equate to – 1. ucos60 ucos60 Simplify: 3u2 75g2 10 3ug1u2 0, u2 100 3u75000 M1 Simplify to quadratic in u (may see g ). 4 4 u5 3g A1 OE. Accept 50 3 or 86.6. 5 © Cambridge University Press & Assessment 2024 Page 8 of 13

7(a) Energy: 1mu2  1mv2 mga1cos *M1 m must be present, dimensionally correct, no 2 2 missing terms. Allow sin instead of cos. Allow sign errors. mv2 B1 No reaction when P loses contact. N2L: mgcos a Eliminate v2 DM1 u2 2ag A1 AG cos 3ag 4 7(b) Vertical component of velocity of P when it leaves the sphere: vsin *B1 Must not come from u.  55ag      216   V2 vsin2 2ga1cos DM1 Use of 'v2 u2 2as. Allow sin for cos. Allow sign errors. 847ag A1 AEF V  216 3 © Cambridge University Press & Assessment 2024 Page 12 of 13 7(c) 1 847ag 55ag  M1 t       g 216 216   1  a a A1 847  55 1.48 6 6g g 2 © Cambridge University Press & Assessment 2024 Page 13 of 13

3 usin605g, and ucos60 , or usin6015g , and ucos60 B1 usin605g M1* Accept equivalent for t15. If  is direction of velocity at t 5, tan ucos60 usin605g usin6015g M1dep Multiply two expressions involving relevant For perpendicular directions,   1 velocities and equate to – 1. ucos60 ucos60 Simplify: 3u2 75g2 10 3ug1u2 0, u2 100 3u75000 M1 Simplify to quadratic in u (may see g ). 4 4 u5 3g A1 OE. Accept 50 3 or 86.6. 5 © Cambridge University Press & Assessment 2024 Page 8 of 13

7(a) Energy: 1mu2  1mv2 mga1cos *M1 m must be present, dimensionally correct, no 2 2 missing terms. Allow sin instead of cos. Allow sign errors. mv2 B1 No reaction when P loses contact. N2L: mgcos a Eliminate v2 DM1 u2 2ag A1 AG cos 3ag 4 7(b) Vertical component of velocity of P when it leaves the sphere: vsin *B1 Must not come from u.  55ag      216   V2 vsin2 2ga1cos DM1 Use of 'v2 u2 2as. Allow sin for cos. Allow sign errors. 847ag A1 AEF V  216 3 © Cambridge University Press & Assessment 2024 Page 12 of 13 7(c) 1 847ag 55ag  M1 t       g 216 216   1  a a A1 847  55 1.48 6 6g g 2 © Cambridge University Press & Assessment 2024 Page 13 of 13

6(a) Components of velocity are ucos, usingt B1 2 M1 Square and add components of velocity and ucos2 usingt2     3 4 u    equate to   3 u   2 . 4  7 A1 AG u2 100usin25000 At least one correct line of working seen. 16 3 © Cambridge University Press & Assessment 2024 Page 11 of 15 6(b) Let  be angle of direction of motion with horizontal at t5, then B1 Either way up. usin5g tan ucos  usin5g M1 Must be – 1 not +1. tan tan1, so tan   1 FT their expression for tan.  ucos  u50sin A1 Use in result from part (a) to form equation in u or sin M1 4 A1 Both. u2 1600, u40 and sin 5 Alternative method for question 6(b) 3 M1 ucos usin 4 4 4 A1 tan or sin 3 5 3 4 M1 A1 Allow sign error.  ucos u50 4 5 4 A1 Both seen. u40 and sin 5 © Cambridge University Press & Assessment 2024 Page 12 of 15 6(b) Alternative method for question 6(b) 3 M1 ucos usin 4 4 4 A1 tan or sin 3 5 Use in result from part (a) to form equation in u M1  1000 A1 u40  and   7  4 A1 Both seen. u40 only and sin 5 5 © Cambridge University Press & Assessment 2024 Page 13 of 15

1  1  M1 For one equation with one error. Use equation of trajectory with point (56, H) or 84, H  2  A1 Both correct. 5g 1 5g H =112− 562 or H =168− 842 2u2 2 2u2 Eliminate to find u or H M1 u=35 A1 H = 48 A1 5 Question Answer Marks Guidance

5(a) Use correct equation of trajectory: 4 a=3a 1 − g (3a)2   1+ 1  M1 No (implied) sight of trajectory equation M0. 5 3 2u2  9 4 5ga2 5ga 1 A1 At least one step of intermediate working must be a=a− , = , u2 =25ga seen. 5 u2 u2 5 AG 2 © Cambridge University Press & Assessment 2024 Page 12 of 19 5(b) For P, time of flight T and range R B1 For Q, time of flight ½ T and range R  2usin  10a Time of flight for P or Q. T = =    g  g 2 3 B1 Range for P. [From motion of P, R= 25ag =] 1 5a g 10 1 2R M1 Obtain an expression for vcos. May involve u For Q: →R=vcos T, vcos= and . 2 T 2 M1 Obtain an expression for vsin. May involve u 1 1 1  1  0=vsin T − g  T  , vsin= gT and . 2 2 2  4 2 2 M1 2R 1   5  Square and add: v2 =   +  gT   =90ag+ ag   T  4   8  725 A1 1 v2 = ag gT 8 tan= 4 = 1 2R 12 T © Cambridge University Press & Assessment 2024 Page 13 of 19 5(b) Alternative method for question 5(b) For P, time of flight T and range R M1 For Q, time of flight ½ T and range R Horizontal motion for P and Q T R=ucosT and R=(vcos) 2 Both. Vertical motion for P and Q M1 gT gT 1 usin= and vsin= Both, may come from using s=ut+ at2. 2 4 2 Equate two expressions for R: A1 6 vcos= u vcos=2ucos 10 Equate two expressions for vertical motion: A1 1 1 vsin= usin vsin= u 2 2 10  1   29  M1 Square and add: v2 =u2 4cos2+ sin2  = u2   4   8  29 725 A1 25ag = ag 8 8 6 © Cambridge University Press & Assessment 2024 Page 14 of 19

1  1  M1 For one equation with one error. Use equation of trajectory with point (56, H) or 84, H  2  A1 Both correct. 5g 1 5g H =112− 562 or H =168− 842 2u2 2 2u2 Eliminate to find u or H M1 u=35 A1 H = 48 A1 5 Question Answer Marks Guidance

2(a) 10=35t− g 352( 1+t2) B1 Use trajectory equation to set up quadratic in 2242 t=tan using (35,10). 6125t2−20160t+11885=0 M1 Form three-term quadratic in t and equate to 0. OE, for example 1225t2−4032t+2377=0. tan=0.769leading to =37.6. A1 If B0M0 earned, award B1 for correct answer. tan=2.52 leading to =68.4. A1 If B0M0 earned, award B1 for correct answer. 4 2(b) g M1 Complete method to find the range, x, using the Use trajectory equation with y = 0: 0=xtan− x2(1+tan2) 2242 larger of their two values found in part 2(a). 39.5 A1 Accept 39.4≤x≤39.5. Alternative method for question 2(b) Time of flight, T =4.8sin [= 4.4629…] M1 Complete method to find the range, x, using the x=24Tcos larger of their two values found in part 2(a). 39.5 A1 Accept 39.4≤x≤39.5. Alternative method for question 2(b) u2sin2 M1 Complete method to find the range, x, using the Use range formula: x= larger of their two values found in part 2(a). g 39.5 A1 Accept 39.4≤x≤39.5. 2 © Cambridge University Press & Assessment 2025 Page 10 of 17

2(a) 10=35t− g 352( 1+t2) B1 Use trajectory equation to set up quadratic in 2242 t=tan using (35,10). 6125t2−20160t+11885=0 M1 Form three-term quadratic in t and equate to 0. OE, for example 1225t2−4032t+2377=0. tan=0.769leading to =37.6. A1 If B0M0 earned, award B1 for correct answer. tan=2.52 leading to =68.4. A1 If B0M0 earned, award B1 for correct answer. 4 2(b) g M1 Complete method to find the range, x, using the Use trajectory equation with y = 0: 0=xtan− x2(1+tan2) 2242 larger of their two values found in part 2(a). 39.5 A1 Accept 39.4≤x≤39.5. Alternative method for question 2(b) Time of flight, T =4.8sin [= 4.4629…] M1 Complete method to find the range, x, using the x=24Tcos larger of their two values found in part 2(a). 39.5 A1 Accept 39.4≤x≤39.5. Alternative method for question 2(b) u2sin2 M1 Complete method to find the range, x, using the Use range formula: x= larger of their two values found in part 2(a). g 39.5 A1 Accept 39.4≤x≤39.5. 2 © Cambridge University Press & Assessment 2025 Page 10 of 17

7(a) v =Usin45−gT B1 y v =Ucos45 x 1 7(b) v B1 SOI y =−tan60 v x  U  M1 Substitute values for trigonometric ratios to obtain −gT   a linear equation in U and T only. 2  =− 3 U − 2gT =− 3U U    2  U ( ) 2U( ) U ( ) A1 AG, shown convincingly. T = 1+ 3 = 1+ 3 = 2+ 6 2g 2g 2g 3 7(c) U B1 D in terms of U and T . D= T 2 U B1 H in terms of U and T . H = T −1gT2 2 2 With or without modulus sign. U M1 OE 2 T −1 2 gT2 1 2 2−1 2 ( 1 2 U ( 2+ 6 )) 1− 3 Obtain a numerical value, may be un-simplified. = = Division or ratios. U 1 2 2 T 2 With or without modulus sign. 2 H:D=1 :1+ 3 A1 OE (for exampleH:D= 3−1:2), CWO © Cambridge University Press & Assessment 2025 Page 15 of 17 7(c) Alternative method for question 7(c) U  U ( ) B1 D in terms of U and g . D=  2+ 6  22g  U  U ( )  U ( ) 2  U2  B1 H in terms of U and g . H =  2+ 6 −1g 2+ 6  = −  With or without modulus sign. 22g  2 2g   2g  2 M1 OE U  U ( )  U ( )  2+ 6 −1g 2+ 6  Obtain a numerical value, may be un-simplified. 22g  2 2g  1− 3 Division or ratios. = U  U ( ) 2 With or without modulus sign.  2+ 6  22g  H:D=1 :1+ 3 A1 OE (for exampleH:D= 3−1:2), CWO Alternative method for question 7(c) D= U U ( 2+ 6 ) B1 D= U2 ( 1+ 3 ) 2 2g 2g 2 2 B1  6   2   U = U −2gH     2 2     U2 M1 ()H = 2g H:D=1 :1+ 3 A1 OE (for exampleH:D= 3−1:2), CWO 4 © Cambridge University Press & Assessment 2025 Page 16 of 17 7(d) 2 3  2  B1 OE, accept un-simplified. w = U = U Allow negative sign. y 3 2  3  2 M1  U  w2 =w2 + = 2U2 + 1U2 y   2    3 2  A1 CWO w=U 2+1 =U 7 3 2 6 3 © Cambridge University Press & Assessment 2025 Page 17 of 17

3(a)  x  B1 Explicitly stated. (→) x=(ucos) t or t =    ucos (  ) y=(usin) t−1gt2 B1 Explicitly stated. 2 2 M1 Eliminate t.  x   x  y=usin  −1g   Must see this substitution. ucos 2 ucos gx2 A1 AG, shown convincingly. y=xtan− sec2 Dependent on all 3 previous marks being awarded. 2u2 4 3(b) 8=64tan−128 ( 1+tan2 ) M1 May be un-simplified, must be an equation in a single 5 trigonometric function. 16tan2−40tan+21=0 M1 Obtain 3-term quadratic. AEF, for example 128tan2−320tan+168=0. tan= 3, 7 A1 Both answers required. 4 4 3 © Cambridge University Press & Assessment 2025 Page 11 of 17

7(a) dy 2gx M1 Valid attempt at differentiating. =tan− dx 2u2cos2 dy B1 =−1 dx −1= 4− 2 x M1 Form equation with –1 and attempt to find a value for x . 3 45 A A x =105 A1 525 A 2 52.5 or . g y = 35 A1 175 dy A 4 8.75 or . SC B1 for 35 OE from setting =+1. 2g 4 dx Alternative method for question 7(a) Horizontally: v =25cos=15 M1 Both equations, allow sign error. H Vertically: v =25sin−gt=20−gt V v B1 At A: V =−tan45=−1 v H 20−gt M1 With RHS = –1, find value for t and x or y =−1, t= 7 A A 15 2 x =105 A1 525 A 2 52.5 or . g y = 35 A1 175 dy A 4 8.75 or . SC B1 for 35 OE from setting =+1. 2g 4 dx 5 © Cambridge University Press & Assessment 2025 Page 16 of 17 7(b) Let v be the velocity before the collision at A. B1 u =ucos=15=v x x v =−v =−15 B1 y x B1 FT Speed: v2 +v2 =15 2 (ms–1) x y 3 7(c) Let w be the velocity after the collision. M1 Correct application of NEL. w=ev=5 2 3 A1 gx2 M1 Substitute values into trajectory equation and solve to find x. −35 =xtan45− , 9x2 −x−35 =0. OR 4 2 ( 5 2 )2 (cos45)2 5 4 Complete method to find x that firstly determines the time of 3 the flight after the particle strikes the barrier. OR 35=−5 21 2t+1gt2, t=3 4 3 2 2 2 x=(wcos45)t =53   3 2 x= 5 (metres) A1 2 4 © Cambridge University Press & Assessment 2025 Page 17 of 17

4(a) Horizontal velocities are the same: B1 B0 M0 M0 A0 for using U =U . q U cos=Ucos q d =(Usin)T −1gT2 + ( U sin ) T +1gT2 M1 Apply vertical equation of motion on both 2 q 2 particles and combine. Ucos  M1 Substitute U and simplify. d =(Usin)T + sinT q  cos  d =UT(sin+costan) A1 AG, shown convincingly. 4 4(b) Usin=gT B1 Correct equation for T. U2sin M1 Substitute their expression for T in given d = (sin+costan) result from part 4(a) OR complete method for g finding d in terms of U and g only. May include unevaluated trigonometric functions. Note: d =2UT and U =2gT . U2 A1 d = g 3 © Cambridge University Press & Assessment 2025 Page 12 of 17

7(a) dy 2gx M1 Valid attempt at differentiating. =tan− dx 2u2cos2 dy B1 =−1 dx −1= 4− 2 x M1 Form equation with –1 and attempt to find a value for x . 3 45 A A x =105 A1 525 A 2 52.5 or . g y = 35 A1 175 dy A 4 8.75 or . SC B1 for 35 OE from setting =+1. 2g 4 dx Alternative method for question 7(a) Horizontally: v =25cos=15 M1 Both equations, allow sign error. H Vertically: v =25sin−gt=20−gt V v B1 At A: V =−tan45=−1 v H 20−gt M1 With RHS = –1, find value for t and x or y =−1, t= 7 A A 15 2 x =105 A1 525 A 2 52.5 or . g y = 35 A1 175 dy A 4 8.75 or . SC B1 for 35 OE from setting =+1. 2g 4 dx 5 © Cambridge University Press & Assessment 2025 Page 16 of 17 7(b) Let v be the velocity before the collision at A. B1 u =ucos=15=v x x v =−v =−15 B1 y x B1 FT Speed: v2 +v2 =15 2 (ms–1) x y 3 7(c) Let w be the velocity after the collision. M1 Correct application of NEL. w=ev=5 2 3 A1 gx2 M1 Substitute values into trajectory equation and solve to find x. −35 =xtan45− , 9x2 −x−35 =0. OR 4 2 ( 5 2 )2 (cos45)2 5 4 Complete method to find x that firstly determines the time of 3 the flight after the particle strikes the barrier. OR 35=−5 21 2t+1gt2, t=3 4 3 2 2 2 x=(wcos45)t =53   3 2 x= 5 (metres) A1 2 4 © Cambridge University Press & Assessment 2025 Page 17 of 17

2(a) gx2 B1 Allow equation derived from first principles. Use trajectory equation from MF19: y=xtan− . 2u2cos2 Substitute tan=2, cos2=1 , g=10 and simplify. 5 10x2 25x2 AG, shown convincingly. y=2x− =2x− 2u21 u2 5 1 2(b) 2582 M1 Substitute correctly and solve equation from part 2(a). 12=28− u2 u2 =400 u=20 A1 25D2 M1 Solve the equation from part 2(a) with y=0. 0=2D− leading to D=0, D=32. 400 25D2 M1 Solve the equation from part 2(a) with y=7. 7=2D− leading to D=4, D=28. 400 28≤D≤32 only. A1 Accept strict inequalities. 5 © Cambridge University Press & Assessment 2025 Page 10 of 18

7 EITHER: s =10+53 [13] and hor 5 © Cambridge University Press & Assessment 2026 Page 20 of 22  s v e r =  45  5   4  (*B1 For both unsimplified distances – may see these two values on a diagram. 4 u t s i n 12 g t 2  = − and 1 3 c o s  = u t DM1 Applying s = u t + 12 a t 2 both horizontally and vertically. Allow sin/cos mix only. No sign errors, must be using a=−g. Dependent on previous B1 mark. ucos 4 usin−gt 3 B1 Use perpendicular condition to form a correct = or =− equation in u, t and gt−usin 3 ucos 4 −3ucos=4usin−4gt .  − 3  1 3 t  = 4  4 + 12t g t 2  − 4 g t M1 Eliminate two of 2gt2 =55   , ,  u t to obtain an equation in one variable only – dependent on all previous marks. Can be implied by t = 12 1 1 = 1 .6 5 8 3 1 ... u c o s 21 61 1 1 , u s i n 72 12 1 1 5 3 .8     = =   =  A1 AWRT 53.8 (53.78116…) or 0.939 (0.938658…). u = 1 3 .3 A1) AWRT 13.3 (13.26735…) or 7 74 44 5 . 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/ 7 OR: Alternative Method 1 s =10+53 [13] and hor 5 © Cambridge University Press & Assessment 2026 Page 21 of 22  s v e r =  45  5   4  (*B1 For both unsimplified distances – may see these two values on a diagram. 4=13tan− g132 ( 1+tan2 ) DB1 Correct trajectory equation in terms of u and  in any form. 2u2 tan− g13( 1+tan2 ) =−3 u2 4 4 2 t a n 4 1 3 (1 t a n 2 ) 3 2     u − g   + = − u  or 4 ucos = or 3 u2sin2−2g4 3 4 s i n c o s 1 c 3 o s    − = u u − u g M1 Differentiate general trajectory equation, set equal to − 34 and use their horizontal distance travelled to form an equation in u and  only. Allow a single slip in the differentiation of the trajectory equation. OR Use perpendicular condition to form a correct equation in u and . 1 6 9 2 6 ( 1 t a n t a n 2 8 ) 4 1 4 3 t a ( 1 n t a n 3 2 )     g + − = g  + + M1 Attempt to combine their two equations to arrive at an equation in either u or  only – dependent on all previous marks. May see, if correct, 1 6 9 1 6 t a n 2 1 3 t a n 42 95 76 [ 0 ]   − − = . t a n 75 12   =   5 3 .8  =  A1 AWRT 53.8 (53.78116…) or 0.939 (0.938658…). u = 1 3 .3 A1) AWRT 13.3 (13.26735…) or 7745 . 44 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/ 7 OR: Alternative Method 2 s =10+53 [13] and hor 5 © Cambridge University Press & Assessment 2026 Page 22 of 22  s v e r =  45  5   4  (B1 For both unsimplified distances – may see these two values on a diagram. 2 5ucos 3    =u2sin2−2104  4 5 91 6 u 2 c o s 2 u 2 s i n 2 8 0     = −  1 3 c o s 5 c o 4 s 3 5 0 s 1 i n 0    = u   u  + u  1 3 2 c 4 o 0 s ( 3 c o s 4 s i n )      = u +  M1 M1 for at least one equation in terms of u and only. Allow sin/cos mix on the  terms only (no other errors allowed). A1 For two different correct equations in u and only. Might see 35 54 u c o s u s i n 1 0 u 1 c 3 o s    −  = −  . 8 0 c 1 o 3 s 3 c o s 4 0 4 s i n 1 9 6 c o s 2 s i n 2      − =  +   −  M1 Attempt to combine their two equations to arrive at an equation in either u or  only – dependent on all previous marks. t a n 75 12   =   5 3 .8  =  A1 AWRT 53.8 (53.78116…) or 0.939 (0.938658…). u = 1 3 .3 A1) AWRT 13.3 (13.26735…) or 7745 . 44 Available marks 6 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/

7 EITHER: © Cambridge University Press & Assessment 2026 Page 20 of 22  s h o r =  1 0 + 5  35 [  1 3 ] and  s v e r =  45  5   4  (*B1 For both unsimplified distances – may see these two values on a diagram. 4 u t s i n 12 g t 2  = − and 1 3 c o s  = u t DM1 Applying s = u t + 12 a t 2 both horizontally and vertically. Allow sin/cos mix only. No sign errors, must be using a=−g. Dependent on previous B1 mark. ucos 4 = or gt−usin 3 s i n c o s 3 4   u u − g t = −  3 c o s 4 s i n 4     − u = u − g t B1 Use perpendicular condition to form a correct equation in u, t and .  − 3  1 3 t  = 4  4 + 12t g t 2  − 4 g t M1 Eliminate two of u,,t to obtain an equation in one 2gt2 =55   variable only – dependent on all previous marks. Can be implied by t=1 11=1.65831... 2 u c o s 21 61 1 1 , u s i n 72 12 1 1 5 3 .8     = =   =  A1 AWRT 53.8 (53.78116…) or 0.939 (0.938658…). u=13.3 A1) AWRT 13.3 (13.26735…) or 7 74 44 5 . 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/ 7 OR: Alternative Method 1 © Cambridge University Press & Assessment 2026 Page 21 of 22  s h o r =  1 0 + 5  35 [  1 3 ] and  s v e r =  45  5   4  (*B1 For both unsimplified distances – may see these two values on a diagram. 4 1 3 t a n 2 1 3 2 2 ( 1 t a n 2 )   = − g  u + DB1 Correct trajectory equation in terms of u and  in any form. t a n g u 1 2 3 ( 1 t a n 2 ) 34   −  + = − 4 2 t a n 4 1 3 (1 t a n 2 ) 3 2     u − g   + = − u  or 4 3 2 s i n c 2 o s 2 4   = u u − g  or 3 4 s i n c o s 1 c 3 o s    − = u u − u g M1 Differentiate general trajectory equation, set equal to − 34 and use their horizontal distance travelled to form an equation in u and  only. Allow a single slip in the differentiation of the trajectory equation. OR Use perpendicular condition to form a correct equation in u and . 1 6 9 2 6 ( 1 t a n t a n 2 8 ) 4 1 4 3 t a ( 1 n t a n 3 2 )     g + − = g  + + M1 Attempt to combine their two equations to arrive at an equation in either u or  only – dependent on all previous marks. May see, if correct, 1 6 9 1 6 t a n 2 1 3 t a n 42 95 76 [ 0 ]   − − = . tan= 71   52 5 3 .8  =  A1 AWRT 53.8 (53.78116…) or 0.939 (0.938658…). u = 1 3 .3 A1) AWRT 13.3 (13.26735…) or 7 74 44 5 . 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/ 7 OR: Alternative Method 2 © Cambridge University Press & Assessment 2026 Page 22 of 22  s h o r =  1 0 + 5  35 [  1 3 ] and  s v e r =  45  5   4  (B1 For both unsimplified distances – may see these two values on a diagram. 2 5ucos 3  =u2sin2−2104    4 5 91 6 u 2 c o s 2 u 2 s i n 2 8 0     = −  1 3 c o s 5 c o 4 s 3 5 0 s 1 i n 0    = u   u  + u  M1 M1 for at least one equation in terms of u and  only. Allow sin/cos mix on the  terms only (no other errors allowed).  u2cos  13= (3cos+4sin)  40 A1 For two different correct equations in u and only.   13 Might see −35ucos=usin−10 . 5 4 ucos 8 0 c 1 o 3 s 3 c o s 4 0 4 s i n 1 9 6 c o s 2 s i n 2      − =  +   −  M1 Attempt to combine their two equations to arrive at an equation in either u or  only – dependent on all previous marks. t a n 75 12   =   =53.8 A1 AWRT 53.8 (53.78116…) or 0.939 (0.938658…). u = 1 3 .3 A1) AWRT 13.3 (13.26735…) or 7 74 44 5 . Available marks 6 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/

7(a) © Cambridge University Press & Assessment 2026 Page 19 of 20 s i n  V y = u P − g t , and so 2 5 s i n 1 0  V y =  − t . At impact, V y = 0 , hence, 2 5  45 − 1 0 t = 0 , t = 2 B1 AG, minimum working . 2 5  45 − 1 0 t = 0 OR 2 5  g 45 = 2 1 0 0 . B0 if 20−10t=0 alone with no previous working shown. B0 if not using g = 10. 1 7(b) h=202−1g22 *M1 Use 2 s = u t + 12 a t 2 to find h. Substitute t = 2, u = 2 5  45 or 2 5  35 and a =  g . May see h = 2 5  45  2  12 g  2 2 or h = 2 5  35  2  12 g  2 2 or 0 = 2 0 2 − 2 g h 20 or h= 2. 2 Must be clear that h = 20 is the maximum height. 2 5 t Q − 12 g  t Q 2 = '2 0 ' DM1 Use s = u t + 12 a t 2 to find the time Q takes to get to h. Substitute u = 2 5 and a =  g . Allow their h. This line fully correct implies M1M1. t =1, T =2−t =1 A1 Exclude Q Q tQ = 4 . Must distinguish between T and tQ (for example, t = 1, T = 1 earns A1). 3 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(c) © Cambridge University Press & Assessment 2026 Page 20 of 20 c o s c o s  1 5    v = u P = leading to s i n c 1 o 5 s s i n    v =  *M1 Use the fact that horizontal component of the velocity is unchanged, meaning vcos=15, to find the vertical component of the velocity of P after the collision, V y . Implied by 1 5 t a n  V y = . 0 ( s i n ) 2 2   = v − g  y m a x  0 c 1 o 5 s s i n 2 2   =   − g  y m a x DM1 Use v 2 = u 2 + 2 a s . Substitute their V y and a =  g with vertical component of velocity at maximum height set to 0 to find h. y m a x 4 54 t a n 2  = A1 Greatest height above ground 2 0 4 54 t a n 2  = + A1 OE, for example, 2 0 1 1 .2 5 t a n 2  + . 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(a) © Cambridge University Press & Assessment 2026 Page 18 of 25 d 12 ( g s i n ) T 2  =  or 1mv2 =mgdsin 2 v 2 2 g d s i n   = *B1 Throughout this part allow t instead of T for full marks. First line of working must be a direct substitution into s=ut+1at2 (although u = 0 2 and v=(gsin)T 2gdsin=(gsin)T can be implied). Therefore, if the first line of working is 2 d g s i n T 2  =  then award B0. If working backwards from AG this is B0. Energy approach – must get to a correct equation in T for B1. 2d DB1 AG, at least one line of working from 2d =gsinT2 T = gsin or 2gdsin 2gdsin 2d T = = = gsin g2sin2 gsin d 12 ( g s i n ) T 2  =  to given answer OE. Allow d = 12 a T 2 with one line of working leading to T = 2 d a and then given answer provided a g s i n  = is explicitly stated. B1 B0 for a g s i n  = , therefore d = 12 a T 2 T 2 d a g 2 s d i n   = = . Energy: T 2 g g d s i s n i n g 2 s d i n    = = is B0. 2 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/ 6(b) EITHER: © Cambridge University Press & Assessment 2026 Page 19 of 25 d c o s u T c o s  d u T    =  = (*B1 Throughout this part allow t instead of T for 2d full marks. or dcos=u cos Stating d=uT implies this mark. gsin h u T s i n ( 2 ) 12 g T 2   − = − − or h u g 2 s d i n s i n ( 2 ) 1 2 g g 2 s d i n 2     − =  − −   *M1 Applying s = u t + 12 a t 2 (or other complete method) to find the vertical distance travelled by P. Allow s i n  for s i n ( 2 )   − . Allow y for the vertical height but not  d . Allow sign errors only. d s i n u g 2 s d i n s i n 1 2 g g 2 s d i n 2     − =  −   or u T s i n u T s i n 1 2 g T 2   − = − or d s i n d s i n 1 2 g T 2   − = − A1 OE, obtaining a second correct equation in two of d, u, and T. Can be implied by T 4 u s i g n  = or T 2 4 d s i g n  = . 4 d s i g n g 2 s d i n 2 4 s i n s i 2 n      =     = or s i n s i n s i 1 n    − = − DM1 Eliminate all given unknowns to obtain an equation in  only. =45 A1) OE, for example, 14 π or AWRT 0.785 (0.7853981…) radians. 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/ 6(b) OR: Alternative approach 1 – considering the motion of P parallel to the slope © Cambridge University Press & Assessment 2026 Page 20 of 25 d ( u c o s 2 ) T 12 ( g s i n ) T 2   = + (*M1 Throughout this part allow t instead of T for full marks. Applying s = u t + 12 a t 2 parallel to the slope. Allow sign errors only. Must be d on the LHS. A1 Fully correct equation. d ( u c o s 2 ) g 2 s d i n 1 2 ( g s i n ) g 2 s d i n 2     =  +     A1 Eliminate one of T, d or u to obtain a correct equation in the other two variables, for example, T2gsin 1 =(ucos2)T + (gsin)T2. 2 2  u T c o s 2 0   =  c o s 2 0  = DM1 Eliminate all given unknowns to obtain an equation in  only. 4 5  =  A1) OE, for example, 14 π or AWRT 0.785 (0.7853981…) radians. 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/ 6(b) OR: Alternative approach 2 – equation of the trajectory © Cambridge University Press & Assessment 2026 Page 21 of 25 d c o s u T c o s  d u T    =  = or d c o s u g 2 s d i n c o s    = (*B1 Throughout this part allow t instead of T for full marks. Stating d=uT implies this mark. Or for another correct equation in at least two of d, u and T. d s i n d c o s t a n g 2 ( u d 2 c c o o s s 2 ) 2      − =  − *M1 Applying the equation of the trajectory. Allow sign errors only. LHS must be d s i n   . g d 4 u 2 s i n    =  A1 Fully correct unsimplified equation. 4 u 2 s i n 2 2 u 2   =   2 s i n 2 1  = DM1 Eliminate all given unknowns to obtain an equation in  only. 4 5  =  A1) OE, for example, 14 π or AWRT 0.785 (0.7853981…) radians. Available marks 5 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/