Momentum — Paper 3 · A Level Mathematics - Further 9231

6 A particle P of mass m is moving with speed u on a fixed smooth horizontal surface. The particle strikes a fixed vertical barrier. At the instant of impact the direction of motion of P makes an angle a with the barrier. The coefficient of restitution between P and the barrier is e. As a result of the impact, the direction of motion of P is turned through 90°. 1 (a) Show that tan 2a = . [3] e … … … … … … … … … … … … … … … … … … … … … … … … The particle P loses two-thirds of its kinetic energy in the impact. (b) Find the value of a and the value of e. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 6(a) Let components of velocity (parallel to plane and perpendicular) after impact be (x, y) cos sin α α = = y v eu B1 sin cos α α = = x v u B1 Divide: 1 tan tan α α = e : tan2 α = 1. e B1 3 Question Answer Marks 6(b) 2 2 1 3 = v u B1 2 2 cos 1 sin 3 α α  =     u u M1 ( ) 2 tan 3 α = M1 60 α = ° A1 1 3 = e A1 Alternative method for 6(b) KE after impact = ( ) 2 2 1 2 + m x y ( ) ( ) ( ) 2 2 2 1 cos sin 2 α α = + m u e u M1 From (a) ( ) sin 1/ 1 α = + e and ( ) cos / 1 α = + e e B1 KE = 2 2 1 2 1 1   +   + +   e e mu e e = 2 1 2 mu e A1 This is equal to 2 1 1 3 2 × mu so 1 3 = e M1 tan 3, 60 α α = = ° A1 5

This question in 9231/31 May/June 2020

6 A particle P of mass m is moving with speed u on a fixed smooth horizontal surface. The particle strikes a fixed vertical barrier. At the instant of impact the direction of motion of P makes an angle a with the barrier. The coefficient of restitution between P and the barrier is e. As a result of the impact, the direction of motion of P is turned through 90°. 1 (a) Show that tan 2a = . [3] e … … … … … … … … … … … … … … … … … … … … … … … … The particle P loses two-thirds of its kinetic energy in the impact. (b) Find the value of a and the value of e. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 6(a) Let components of velocity (parallel to plane and perpendicular) after impact be (x, y) cos sin α α = = y v eu B1 sin cos α α = = x v u B1 Divide: 1 tan tan α α = e : tan2 α = 1.e B1 3 Question Answer Marks 6(b) 2 2 1 3 = v u B1 2 2 cos 1 sin 3 α α  =     u u M1 ( ) 2 tan 3 α = M1 60 α = ° A1 1 3 = e A1 Alternative method for 6(b) KE after impact = ( ) 2 2 1 2 + m x y ( ) ( ) ( ) 2 2 2 1 cos sin 2 α α = + m u e u M1 From (a) ( ) sin 1/ 1 α = + e and ( ) cos / 1 α = + e e B1 KE = 2 2 1 2 1 1   +   + +   e e mu e e = 2 1 2 mu e A1 This is equal to 2 1 1 3 2 × mu so 1 3 = e M1 tan 3, 60 α α = = ° A1 5

This question in 9231/32 May/June 2020

5 A B a° (90 – a)° u m m u Two uniform smooth spheres A and B of equal radii each have mass m. The two spheres are each moving with speed u on a horizontal surface when they collide. Immediately before the collision A’s direction of motion makes an angle of a° with the line of centres, and B’s direction of motion is perpendicular to that of A (see diagram). The coefficient of restitution between the spheres is e. Immediately after the collision, B moves in a direction at right angles to the line of centres. 1 + e (a) Show that tana = . [4] 1 - e … … … … … … … … … … … … … … … … … (b) Given that tan a = 2 , find the speed of A after the collision. [4] … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 5(a) Let w be speed of A along line of centres after collision cos sin α α ← = − + mw mu mu M1 0 ( cos sin α α − = + w e u u ) M1 Rearrange: ( ) ( ) sin cos α α − = + u eu u eu M1 1 tan 1 α + = − e e . AG A1 4 5(b) 1 tan 2 3 α =  = e B1 1 1 2 3 5 5 5   = + =     u w u M1 Speed = ( ) 2 2 sinα + w u M1 = 2 2 4 5 5 + = u u u A1 4

This question in 9231/33 May/June 2020

6 Two smooth spheres A and B have equal radii and masses m and 2m respectively. Sphere B is at rest on a smooth horizontal floor. Sphere A is moving on the floor with velocity u and collides directly with B. The coefficient of restitution between the spheres is e. (a) Find, in terms of u and e, the velocities of A and B after the collision. [3] … … … … … … … … … … Subsequently, B collides with a fixed vertical wall which makes an angle i with the direction of motion of B, where tan i = 34 . The coefficient of restitution between B and the wall is 2.3 Immediately after B collides with the wall, the kinetic energy of A is 325 of the kinetic energy of B. (b) Find the possible values of e. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 6(a) = + B1 Momentum equation (with m) v w eu − = B1 Restitution with consistent signs ( )1 3 u v e = + ( ) 1 2 3 u w e = − B1 Both correct. 3 Question Answer Marks Guidance 6(b) Perpendicular to plane: sin y ev θ = Parallel to plane: cos x v θ = B1 Both Speed of B = 2 2 x y + = 2 2 2 4 2 3 ( ) . 5 3 5 v     +           (= 2 5 v ) M1 Speed of B KE of B = ( ) 2 2 1 4 .2 . 1 2 5 9 u m e + M1 KE of B in terms of u . 1 2 and 2m needed KE of A = ( ) 2 2 1. . 1 2 2 9 u m e − So ( ) ( ) 2 2 2 2 1 5 1 4 . . 1 2 . .2 . 1 2 9 32 2 5 9 u u m e m e − = + M1 A1 Relate the two KEs ( ) ( ) 2 2 4 1 2 1 e e − = + or 2 15 18 3 0 e e − + = M1 Rearrange and simplify to quadratic ( ) 1 2 1 2 e e + = ± − 1, 1 5 e = A1 Both values 7

This question in 9231/31 Oct/Nov 2020

2 A B a 2m m u Two uniform smooth spheres A and B of equal radii have masses 2m and m respectively. Sphere B is at rest on a smooth horizontal surface. Sphere A is moving on the surface with speed u and collides with B. Immediately before the collision, the direction of motion of A makes an angle a with the line of centres of the spheres, where tan a = 43 (see diagram). The coefficient of restitution between the spheres is 1.3 Find the speed of A after the collision. [5] … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 2 Speeds v and w after collision 2 2 cos mv mw mu α + = M1 Momentum equation with m. Correct masses, allow sin instead of cos cos w v eu α −= M1 Restitution, with consistent signs ( ) 1 1 3 1 1 cos 2 . 2 3 3 5 3 3 v u e u u α   = − = − =     A1 Square of speed of A = ( ) 2 2 1 sin 3u u α  +     M1 Uses correct speed perpendicular to motion 2 2 1 4 3 5 u u     = +         Speed = 13 15u (= 0.867u) A1 5

This question in 9231/32 Oct/Nov 2020

6 Two smooth spheres A and B have equal radii and masses m and 2m respectively. Sphere B is at rest on a smooth horizontal floor. Sphere A is moving on the floor with velocity u and collides directly with B. The coefficient of restitution between the spheres is e. (a) Find, in terms of u and e, the velocities of A and B after the collision. [3] … … … … … … … … … … Subsequently, B collides with a fixed vertical wall which makes an angle i with the direction of motion of B, where tan i = 34 . The coefficient of restitution between B and the wall is 2.3 Immediately after B collides with the wall, the kinetic energy of A is 325 of the kinetic energy of B. (b) Find the possible values of e. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 6(a) = + B1 Momentum equation (with m) v w eu − = B1 Restitution with consistent signs ( )1 3 u v e = + ( ) 1 2 3 u w e = − B1 Both correct. 3 Question Answer Marks Guidance 6(b) Perpendicular to plane: sin y ev θ = Parallel to plane: cos x v θ = B1 Both Speed of B = 2 2 x y + = 2 2 2 4 2 3 ( ) . 5 3 5 v     +           (= 2 5 v ) M1 Speed of B KE of B = ( ) 2 2 1 4 .2 . 1 2 5 9 u m e + M1 KE of B in terms of u . 1 2 and 2m needed KE of A = ( ) 2 2 1. . 1 2 2 9 u m e − So ( ) ( ) 2 2 2 2 1 5 1 4 . . 1 2 . .2 . 1 2 9 32 2 5 9 u u m e m e − = + M1 A1 Relate the two KEs ( ) ( ) 2 2 4 1 2 1 e e − = + or 2 15 18 3 0 e e − + = M1 Rearrange and simplify to quadratic ( ) 1 2 1 2 e e + = ± − 1, 1 5 e = A1 Both values 7

This question in 9231/33 Oct/Nov 2020

6 A B a b m m u u Two uniform smooth spheres A and B of equal radii each have mass m. The two spheres are each moving with speed u on a horizontal surface when they collide. Immediately before the collision, A’s direction of motion makes an angle a with the line of centres, and B’s direction of motion makes an angle b with the line of centres (see diagram). The coefficient of restitution between the spheres is 13 and 2 cos b = cos a . (a) Show that the direction of motion of A after the collision is perpendicular to the line of centres. [4] … … … … … … … … … … … … … … … … … The total kinetic energy of the spheres after the collision is 34 mu 2 . (b) Find the value of a. [4] … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 6(a) Along line of centres, speeds 1 v and 2 v 1 2 cos cos α β + = − mv mv mu mu M1 Momentum (condone missing masses). ( ) 2 1 cos cos β α − = + v v eu M1 Restitution. Both correct, masses seen. A1 1 0 = v so A has no speed along line of centres: moves perpendicular to line of centres A1 AG. 4 6(b) 2 1 ( cos cos 2 α β = = v u u ) KE of B after collision is ( ) ( ) 2 2 2 1 sin 2 β + m v u KE of A after collision = ( ) 2 1 sin 2 α m u M1 Both components. Add both KEs and equate to 2 3 4 mu M1 Simplify to equation in sinα M1 1 sin 2 α = , 45 α = ° A1 4

This question in 9231/31 May/June 2021

6 A B a b m m u u Two uniform smooth spheres A and B of equal radii each have mass m. The two spheres are each moving with speed u on a horizontal surface when they collide. Immediately before the collision, A’s direction of motion makes an angle a with the line of centres, and B’s direction of motion makes an angle b with the line of centres (see diagram). The coefficient of restitution between the spheres is 13 and 2 cos b = cos a . (a) Show that the direction of motion of A after the collision is perpendicular to the line of centres. [4] … … … … … … … … … … … … … … … … … The total kinetic energy of the spheres after the collision is 34 mu 2 . (b) Find the value of a. [4] … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 6(a) Along line of centres, speeds 1 v and 2 v 1 2 cos cos α β + = − mv mv mu mu M1 Momentum (condone missing masses). ( ) 2 1 cos cos β α − = + v v eu M1 Restitution. Both correct, masses seen. A1 1 0 = v so A has no speed along line of centres: moves perpendicular to line of centres A1 AG. 4 6(b) 2 1 ( cos cos 2 α β = = v u u ) KE of B after collision is ( ) ( ) 2 2 2 1 sin 2 β + m v u KE of A after collision = ( ) 2 1 sin 2 α m u M1 Both components. Add both KEs and equate to 2 3 4 mu M1 Simplify to equation in sinα M1 1 sin 2 α = , 45 α = ° A1 4

This question in 9231/32 May/June 2021

6 A B i u m km Two uniform smooth spheres A and B of equal radii have masses m and km respectively. Sphere A is moving with speed u on a smooth horizontal surface when it collides with sphere B which is at rest. Immediately before the collision, A’s direction of motion makes an angle i with the line of centres (see diagram). The coefficient of restitution between the spheres is 1.3 4u cos i (a) Show that the speed of B after the collision is . [3] 3 ( 1 + k) … … … … … … … … … … … … … … … … … … … … 70% of the total kinetic energy of the spheres is lost as a result of the collision. (b) Given that tan i = 13 , find the value of k. [6] … … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 6(a) Let velocities of A and B along line of centres after collision be 1 v and 2 v . 1 2 cosθ + = mv kmv mu . 2 1 1 cos 3 θ − = v v u M1 Restitution, consistent signs, correct way up. Solve: ( ) 2 4 cos 3 1 θ = + u v k A1 AG shown convincingly. 3 6(b) ( ) ( ) 1 3 cos 3 1 θ − = + k u v k B1 Or equivalent, may be unsimplified. Use velocity of A with both components. B1 ( ) 2 2 1 sinθ + v u seen. ( ) ( ) 2 2 2 2 2 1 1 1 3 1 sin 2 2 10 2 kmv m v u mu θ + + = × M1 KE after = 30% KE before (all terms present). M0 if incorrect masses. Substitute from part (a) and for θ. M1 Eliminate trigonometric terms, must be KE equation, in terms of k only. ( ) ( ) 2 2 3 16 2 1 − + = + k k k , 2 6 7 0 − − = k k M1 Obtain simplified quadratic equation in k . 7 = k A1 6

This question in 9231/33 May/June 2021

7 A B c b a u P C The smooth vertical walls AB and CB are at right angles to each other. A particle P is moving with speed u on a smooth horizontal floor and strikes the wall CB at an angle a. It rebounds at an angle b to the wall CB. The particle then strikes the wall AB and rebounds at an angle c to that wall (see diagram). The coefficient of restitution between each wall and P is e. (a) Show that tan b = e tan a . [3] … … … … … … … … … (b) Express c in terms of a and explain what this result means about the final direction of motion of P. [4] … … … … … … … … … … … … … As a result of the two impacts the particle loses 89 of its initial kinetic energy. (c) Given that a + b = 90° , find the value of e and the value of tana. [4] … … … … … … … … … … … … … … … … … … … … Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown. … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 7(a) α β = M1 sin sin α β = eu v M1 Divide: tan tan β α = e A1 AG. Must see divide OE. 3 7(b) ( ) sin cos sin β γ α = = v w eu M1 ( ) cos sin cos β γ α = = ev w eu M1 Divide: tan 1/ tan γ α = : 90 γ α = ° − *A1 After second rebound, direction of motion is parallel to initial path. DB1 4 7(c) Final KE = ( ) ( ) ( ) 2 2 1 sin cos 2 α α + m eu eu 2 2 1 2   =     me u M1 Energy expression in terms of u. So 2 2 2 1 1 1 2 9 2 = × me u mu giving 1 3 = e A1 Part (a) gives ( ) tan 90 tan α α − = e M1 So tan 3 α = A1 4

This question in 9231/31 Oct/Nov 2021

5 A B u 60° m 32 m u Two uniform smooth spheres A and B of equal radii have masses m and 32 m respectively. The two spheres are each moving with speed u on a horizontal surface when they collide. Immediately before the collision A’s direction of motion is along the line of centres, and B’s direction of motion makes an angle of 60° with the line of centres (see diagram). The coefficient of restitution between the spheres is 2.3 (a) Find the angle through which the direction of motion of B is deflected by the collision. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … (b) Find the loss in the total kinetic energy of the system as a result of the collision. [3] … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 5(a) Let speeds of A and B along line of centres after collision be 1v and 2v 1 2 3 3 cos60 2 2 4   + = − ° + =     u mv mv mu mu M1 Momentum with masses correct. ( ) 2 1 2 cos60 ( 3 − = − − ° − = v v u u u ) M1 Restitution, with consistent signs on LHS. 1 2 1 1 2 2   = − =     v u v u A1 Perpendicular to line of centres, speed of B is 3 sin60 2 ° = u u B1 Direction of B is now 60° above line of centres. M1 Angle of deflection is 60°. A1 FT FT (120° – their direction of B angle) 6 5(b) KE before = 2 2 2 1 1 3 5 . 2 2 2 4 + = m mu u mu B1 KE after = 2 2 2 1 1 3 3 . 2 2 2 2 2 2           + +                   u m u u m 2 7 8 mu       = B1 FT FT only their speeds from (a) Loss in KE = 2 3 8 mu B1 3

This question in 9231/32 Oct/Nov 2021

7 A B c b a u P C The smooth vertical walls AB and CB are at right angles to each other. A particle P is moving with speed u on a smooth horizontal floor and strikes the wall CB at an angle a. It rebounds at an angle b to the wall CB. The particle then strikes the wall AB and rebounds at an angle c to that wall (see diagram). The coefficient of restitution between each wall and P is e. (a) Show that tan b = e tan a . [3] … … … … … … … … … (b) Express c in terms of a and explain what this result means about the final direction of motion of P. [4] … … … … … … … … … … … … … As a result of the two impacts the particle loses 89 of its initial kinetic energy. (c) Given that a + b = 90° , find the value of e and the value of tana. [4] … … … … … … … … … … … … … … … … … … … … Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown. … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 7(a) α β = M1 sin sin α β = eu v M1 Divide: tan tan β α = e A1 AG. Must see divide OE. 3 7(b) ( ) sin cos sin β γ α = = v w eu M1 ( ) cos sin cos β γ α = = ev w eu M1 Divide: tan 1/ tan γ α = : 90 γ α = ° − *A1 After second rebound, direction of motion is parallel to initial path. DB1 4 7(c) Final KE = ( ) ( ) ( ) 2 2 1 sin cos 2 α α + m eu eu 2 2 1 2   =     me u M1 Energy expression in terms of u. So 2 2 2 1 1 1 2 9 2 = × me u mu giving 1 3 = e A1 Part (a) gives ( ) tan 90 tan α α − = e M1 So tan 3 α = A1 4

This question in 9231/33 Oct/Nov 2021

6 Two uniform smooth spheres A and B of equal radii have masses m and km respectively. The two spheres are on a horizontal surface. Sphere A is travelling with speed u towards sphere B which is at rest. The spheres collide. Immediately before the collision, the direction of motion of A makes an angle a with the line of centres. The coefficient of restitution between the spheres is 1.2 3u cos a (a) Show that the speed of B after the collision is and find also an expression for the speed of 2 ( 1 + k) A along the line of centres after the collision, in terms of k, u and a. [4] … … … … … … … … … … … … … … … … … … … … … … … … After the collision, the kinetic energy of A is equal to the kinetic energy of B. (b) Given that tan a = 23 , find the possible values of k. [5] … … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 6(a) Let v and w be speeds after collision: cos mv kmw mu    M1 Momentum along line of centres. 1 cos 2 w v u    M1 NEL consistent signs. Add to give   3 cos 2 1 u k   A1 AG Convincing working. Substitute back or re-solve:     2 cos 2 1 k u v k     A1 Accept without modulus sign. 4 Question Answer Marks Guidance 6(b)     2 2 2 cos ( sin ) 2 1 k u u k             B1 For speed of A (SOI). Equal KE after collision:       2 2 2 2 cos 1 3 cos 1 ( sin ) 2 2 1 2 2 1 k u u km m u k k                                         2 2 2 2 2 9 cos 4 1 sin 2 cos k k k            M1 Equate KEs. Use 2 tan : 3      2 2 16 1 2 9 4 4 81 k k k k k       M1 2 25 85 52 0 k k    leading to    5 4 5 13 0 k k    M1 Obtain quadratic and attempt to solve. 4 13 or 5 5 k  A1 5

This question in 9231/31 May/June 2022

6 Two uniform smooth spheres A and B of equal radii have masses m and km respectively. The two spheres are on a horizontal surface. Sphere A is travelling with speed u towards sphere B which is at rest. The spheres collide. Immediately before the collision, the direction of motion of A makes an angle a with the line of centres. The coefficient of restitution between the spheres is 1.2 3u cos a (a) Show that the speed of B after the collision is and find also an expression for the speed of 2 ( 1 + k) A along the line of centres after the collision, in terms of k, u and a. [4] … … … … … … … … … … … … … … … … … … … … … … … … After the collision, the kinetic energy of A is equal to the kinetic energy of B. (b) Given that tan a = 23 , find the possible values of k. [5] … … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 6(a) Let v and w be speeds after collision: cos mv kmw mu    M1 Momentum along line of centres. 1 cos 2 w v u    M1 NEL consistent signs. Add to give   3 cos 2 1 u k   A1 AG Convincing working. Substitute back or re-solve:     2 cos 2 1 k u v k     A1 Accept without modulus sign. 4 Question Answer Marks Guidance 6(b)     2 2 2 cos ( sin ) 2 1 k u u k             B1 For speed of A (SOI). Equal KE after collision:       2 2 2 2 cos 1 3 cos 1 ( sin ) 2 2 1 2 2 1 k u u km m u k k                                         2 2 2 2 2 9 cos 4 1 sin 2 cos k k k            M1 Equate KEs. Use 2 tan : 3      2 2 16 1 2 9 4 4 81 k k k k k       M1 2 25 85 52 0 k k    leading to    5 4 5 13 0 k k    M1 Obtain quadratic and attempt to solve. 4 13 or 5 5 k  A1 5

This question in 9231/32 May/June 2022

6 C b u i 60° A B AB and BC are two fixed smooth vertical barriers on a smooth horizontal surface, with angle ABC = 60° . A particle of mass m is moving with speed u on the surface. The particle strikes AB at an angle i with AB. It then strikes BC and rebounds at an angle b with BC (see diagram). The coefficient of restitution between the particle and each barrier is e and tan i = 2 . The kinetic energy of the particle after the first collision is 40% of its kinetic energy before the first collision. (a) Find the value of e. [4] … … … … … … … … … … … … … … … … (b) Find the size of angle b. [4] … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 6(a) Energy loss: 2 2 2 2 1 2 1 2 , 2 5 2 5 mv mu v u    B1 Energy loss. cos cos v u    sin sin v eu    B1 Both. Combine to form equation in e only 2 2 1 4 5 5 5 e    M1     2 2 2 cos sin v u eu     1 2 e  A1 4 6(b) tan tan e    , so tan 1, 45      B1 For 2nd collision   cos cos 180 60 w v        sin sin 180 60 w ev      M1 Both. May be implied by the A1.   tan tan 120 e their     M1 Divide to find . 61.8   A1 4

This question in 9231/33 May/June 2022

6 A B u a 5 8 u m km Two uniform smooth spheres A and B of equal radii have masses m and km respectively. The two spheres are moving on a horizontal surface with speeds u and 58 u respectively. Immediately before the spheres collide, A is travelling along the line of centres, and B’s direction of motion makes an angle a with the line of centres (see diagram). The coefficient of restitution between the spheres is 23 and tan a = 34 . After the collision, the direction of motion of B is perpendicular to the line of centres. (a) Find the value of k. [4] … … … … … … … … … … … … … … … … … … … … … (b) Find the loss in the total kinetic energy as a result of the collision. [4] … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 6(a) Let speed of A after collision be → v A and speed of B M1 perpendicular to line of centres be v 5 Along line of centres: mu − km u cos= mv A Momentum. 8  5  M1 NEL NEL: 0 − v A = e  u cos+ u   8  5 2  5  M1 Solve. So u − ku cos= −  u cos+ u  8 3  8  Substitute for cos, to give k = 4 A1 4 6(b) 5 3 B1 vB = u sin = Velocity perpendicular to line of centres 8 8u v A = −u B1 FT 2 M1 NOTE: KE before and after for A is unchanged. 1 2 1  5  1 2 25 2 41 2 KE before = mu + km u = mu + mu = mu   2 2  8  2 32 32 Both. 1 2 1 2 1 2 9 2 25 2 KE after = mv A + kmvB = mu + 2m u = mu 2 2 2 64 32 2  41 25  1 2 A1 Loss = mu  −  = mu  32 32  2 4

This question in 9231/31 Oct/Nov 2022

7 u 2u A B α β 1 m m 2 Two uniform smooth spheres A and B of equal radii have masses m and 12 m respectively. The two spheres are moving on a horizontal surface when they collide. Immediately before the collision, sphere A is travelling with speed u and its direction of motion makes an angle a with the line of centres. Sphere B is travelling with speed 2u and its direction of motion makes an angle b with the line of a + b = 90° . centres (see diagram). The coefficient of restitution between the spheres is 58 and (a) Find the component of the velocity of B parallel to the line of centres after the collision, giving your answer in terms of u and a. [4] … … … … … … … … … … … … … … … … … … The direction of motion of B after the collision is parallel to the direction of motion of A before the collision. (b) Find the value of tana. [5] … … … … … … … … … … … … … … … … … … … … … … … … … … Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 7(a) Let v , w be speeds of A and B along line of centres after collision M1 1 1 mv + mw = mu cos− m.2u cos  Momentum: masses correct, opposite signs on RHS. 2 2 w −=v e(2u cos + u cos) M1 NEL: LHS signs must be consistent with momentum equation, same sign for both terms on RHS. + = 90, so cos = sin M1 Solve to find an expression of the correct form. Use this fact and solve to find w 2  1 13  A1 w = u  sin+ cos 3  4 8  4 7(b) Perpendicular to line of centres, speed of B is B1 2u sin = 2u cos After, velocity of B makes angle  with line of centres, so B1 2u cos tan= w sin 2u cos M1* Obtain homogeneous equation in cos and sin or an equation in tan = giving cos 2  1 13  u  sin+ cos 3  4 8  2 1 2 13 DM1 Obtain quadratic and solve to find values of tan 3 ( cos) = ( sin) + sincos 4 8 2 ( tan) 2 + 13tan− 24 = 0 , ( 2tan − 3 )( tan + 8 ) = 0 3 A1 tan= 2 5

This question in 9231/32 Oct/Nov 2022

6 A B u a 5 8 u m km Two uniform smooth spheres A and B of equal radii have masses m and km respectively. The two spheres are moving on a horizontal surface with speeds u and 58 u respectively. Immediately before the spheres collide, A is travelling along the line of centres, and B’s direction of motion makes an angle a with the line of centres (see diagram). The coefficient of restitution between the spheres is 23 and tan a = 34 . After the collision, the direction of motion of B is perpendicular to the line of centres. (a) Find the value of k. [4] … … … … … … … … … … … … … … … … … … … … … (b) Find the loss in the total kinetic energy as a result of the collision. [4] … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 6(a) Let speed of A after collision be → v A and speed of B M1 perpendicular to line of centres be v 5 Along line of centres: mu − km u cos= mv A Momentum. 8  5  M1 NEL NEL: 0 − v A = e  u cos+ u   8  5 2  5  M1 Solve. So u − ku cos= −  u cos+ u  8 3  8  Substitute for cos, to give k = 4 A1 4 6(b) 5 3 B1 vB = u sin = Velocity perpendicular to line of centres 8 8u v A = −u B1 FT 2 M1 NOTE: KE before and after for A is unchanged. 1 2 1  5  1 2 25 2 41 2 KE before = mu + km u = mu + mu = mu   2 2  8  2 32 32 Both. 1 2 1 2 1 2 9 2 25 2 KE after = mv A + kmvB = mu + 2m u = mu 2 2 2 64 32 2  41 25  1 2 A1 Loss = mu  −  = mu  32 32  2 4

This question in 9231/33 Oct/Nov 2022

2 P u a i A particle P of mass m is moving with speed u on a fixed smooth horizontal surface. It collides at an angle a with a fixed smooth vertical barrier. After the collision, P moves at an angle i with the barrier, where tan i = 12 (see diagram). The coefficient of restitution between P and the barrier is e. The particle P loses 20% of its kinetic energy as a result of the collision. Find the value of e. [5] … … … … … … … … … … … … … … …

5 marks

Mark scheme: 2 Parallel to wall cos cos v u Perpendicular to wall sin sin v eu    M1 Both Dividing, 1 2tan e   A1 AEF KE reduced by 20%, so   1 4 1 2 2 2 2 2 2 5 2 cos sin mu e mu      M1 Dimensionally correct equation in u or v , but not both. Must have either  or , but not both. Must see 4 5 on the correct side of the equation. Eliminate e: 4 cos 5  A1 2 3 e  A1 Question Answer Marks Guidance 2 Alternative method for question 2 Parallel to wall cos cos v u    Perpendicular to wall sin sin v eu    M1 Both     5 2 5 5 2 5 sin , cos sin , cos 5 5 5 5 v v u u e               A1   2 2 2 2 2 2 2 2 4 5 5 v v u u cos u sin e           A1 AEF, e.g. 2 2 1 4 5 v e        2 2 2 2 1 4 1 2 1 4 2 5 2 5 5 v mv mu m e                 M1 Dimensionally correct equation in v . Must have either  or , but not both. Must see 4 5 or 2 5 on the correct side of the equation. 2 3 e  A1 5

This question in 9231/31 May/June 2023

2 P u a i A particle P of mass m is moving with speed u on a fixed smooth horizontal surface. It collides at an angle a with a fixed smooth vertical barrier. After the collision, P moves at an angle i with the barrier, where tan i = 12 (see diagram). The coefficient of restitution between P and the barrier is e. The particle P loses 20% of its kinetic energy as a result of the collision. Find the value of e. [5] … … … … … … … … … … … … … … …

5 marks

Mark scheme: 2 Parallel to wall cos cos v u Perpendicular to wall sin sin v eu    M1 Both Dividing, 1 2tan e   A1 AEF KE reduced by 20%, so   1 4 1 2 2 2 2 2 2 5 2 cos sin mu e mu      M1 Dimensionally correct equation in u or v , but not both. Must have either  or , but not both. Must see 4 5 on the correct side of the equation. Eliminate e: 4 cos 5  A1 2 3 e  A1 Question Answer Marks Guidance 2 Alternative method for question 2 Parallel to wall cos cos v u    Perpendicular to wall sin sin v eu    M1 Both     5 2 5 5 2 5 sin , cos sin , cos 5 5 5 5 v v u u e               A1   2 2 2 2 2 2 2 2 4 5 5 v v u u cos u sin e           A1 AEF, e.g. 2 2 1 4 5 v e        2 2 2 2 1 4 1 2 1 4 2 5 2 5 5 v mv mu m e                 M1 Dimensionally correct equation in v . Must have either  or , but not both. Must see 4 5 or 2 5 on the correct side of the equation. 2 3 e  A1 5

This question in 9231/32 May/June 2023

4 2u u A B 30° m m Two identical smooth uniform spheres A and B each have mass m. The two spheres are moving on a smooth horizontal surface when they collide with speeds u and 2u respectively. Immediately before the collision, A’s direction of motion makes an angle of 30° with the line of centres, and B’s direction of motion is perpendicular to the line of centres (see diagram). After the collision, A and B are moving in the same direction. The coefficient of restitution between the spheres is e. (a) Find the value of e. [5] … … … … … … … … … … … … … … … … … … (b) Find the loss in the total kinetic energy of the spheres as a result of the collision. [3] … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 4(a) Let speeds of A and B along line of centres after collision be VA and VB cos30    A B V V u (1) M1 Allow sign errors, allow missing m. cos30     A B V V eu (2) M1 Signs on LHS must be consistent with (1). Speeds perpendicular to line of centres after collision are sin30 u and 2u Moving in same direction, so sin30 2   A B V V u u (3) B1 SOI 4  B A V V Use 4  B A V V in (1): 5 cos30   A V u From (2): 3 cos30   A V eu then Combine to find equation in e only. M1 A complete method to find equation in e only 3 5 e  A1 Question Answer Marks Guidance 4(a) Alternative method for question 4(a) Let speeds of A and B along line of centres after collision be VA and VB cos30    A B V V u (1) M1 Allow sign errors, allow missing m. cos30     A B V V eu (2) M1 Signs on LHS must be consistent with (1). Speeds perpendicular to line of centres after collision are sin30 u and 2u Moving in same direction, so sin30 2   A B V V u u (3) B1 SOI 4  B A V V Solve (1) and (2):   1 1 cos30 2    A V u e ,   1 1 cos30 2    B V u e Substitute in (3) to find equation in e only . M1 Note: 4 3, 3 10 10   A B u u V V 3 5 e  A1 5 4(b) KE after =   2 2 2 2 1 1 ( 2 2 2 2                  A B u m V m u V ) B1 Correct expression for KE for one of the spheres, after collision, with both components. KE for A after = 2 7 50 mu or KE for B after = 2 56 25 mu or KE loss for A = 2 9 25 mu or KE gain for B = 2 6 25 mu B1 Implied by total KE after = 2 119 50 mu . Total loss in KE = 2 3 25 mu B1 Term   2 1 2 2 m u may be omitted from KE of B before and after. 3

This question in 9231/33 May/June 2023

1 u u A B 60° i m 2m Two uniform smooth spheres A and B of equal radii have masses m and 2m respectively. The two spheres are moving with equal speeds u on a smooth horizontal surface when they collide. Immediately before the collision, A’s direction of motion makes an angle of 60° with the line of centres, and B’s direction of motion makes an angle i with the line of centres (see diagram). The coefficient of restitution between the spheres is e. After the collision, the component of the velocity of A along the line of centres is v and B moves perpendicular to the line of centres. Sphere A now has twice as much kinetic energy as sphere B. i - 1) . [1] (a) Show that v = 12 u ( 4 cos … … … … … … … (b) Find the value of cosi. [4] … … … … … … … … … … … … … … … … … … … (c) Find the value of e. [2] … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: Question Answer Marks Guidance 1(a) PCLM: mv = −mu cos60+ 2mu cos B1 First line must be seen. 1 1 v = − u + 2u cos= v = u ( 4cos− 1) AG 2 2 1 1(b) 1 2 2 B1 KE of A = m v + ( u sin60 ) ( ) 2 KE of A = 2  KE of B, so M1 1 2 2 1 2 m v + ( u sin60 ) = 2   2m ( u sin) ( ) 2 2 2 M1 Use result of (a) and rearrange.  1  3 2 2 2 u = 4u ( sin) Obtain 3-term quadratic.  u ( 4cos− 1)  +  2  4 8cos 2 − 2cos− 3 = 0 3 1 3 A1 cos= , − but angle is acute, so cos= 4 2 4 4 1(c) NEL: v = e ( u cos60 + u cos) M1  1 3  A1 v = u , u = e u + u    2 4  4 e = 5 2

This question in 9231/31 Oct/Nov 2023

4 u B A i d m m d d Two smooth vertical walls meet at right angles. The smooth sphere A, with mass m, is at rest on a smooth horizontal surface and is at a distance d from each wall. An identical smooth sphere B is moving on the horizontal surface with speed u at an angle i with the line of centres when the spheres collide (see diagram). After the collision, the spheres take the same time to reach a wall. The coefficient of restitution between the spheres is 1.2 (a) Find the value of tani. [4] … … … … … … … … … … … … … … … … … (b) Find the percentage loss in the total kinetic energy of the spheres as a result of this collision. [3] … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 4(a) After collision, A has velocity vA towards wall on right and B has component B1 of velocity towards lower wall of u sin . Same distance and time, so vA = u sin . Along line of centres: M1 Both, consistent signs, must be cos. PCLM: mv A + mvB = mu cos NEL: v A − v B = eu cos 2sin = (1 + e ) cos M1 Eliminating vA and vB to find an equation in . Condone common factor of u. 3 1 Note: v A = u cos and vB = u cos. 4 4 tan= 3 A1 4 4 4(b) 1 2 2 2 B1 Both components of velocity of B needed. Final KE = m ( v A + ( u sin) + vB ) 2 1 2 1 2 2 2 1 2  9 9 1  M1 vA, vB, substituted, ft their final KE with both = mu Loss = mu − m  1 − − −  components of velocity of B included. ( v A + ( u sin) + vB ) 2 2 2  25 25 25  3 2 A1 Loss = mu 25 Percentage loss = 24% 3

This question in 9231/32 Oct/Nov 2023

1 u u A B 60° i m 2m Two uniform smooth spheres A and B of equal radii have masses m and 2m respectively. The two spheres are moving with equal speeds u on a smooth horizontal surface when they collide. Immediately before the collision, A’s direction of motion makes an angle of 60° with the line of centres, and B’s direction of motion makes an angle i with the line of centres (see diagram). The coefficient of restitution between the spheres is e. After the collision, the component of the velocity of A along the line of centres is v and B moves perpendicular to the line of centres. Sphere A now has twice as much kinetic energy as sphere B. i - 1) . [1] (a) Show that v = 12 u ( 4 cos … … … … … … … (b) Find the value of cosi. [4] … … … … … … … … … … … … … … … … … … … (c) Find the value of e. [2] … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: Question Answer Marks Guidance 1(a) PCLM: mv = −mu cos60+ 2mu cos B1 First line must be seen. 1 1 v = − u + 2u cos= v = u ( 4cos− 1) AG 2 2 1 1(b) 1 2 2 B1 KE of A = m v + ( u sin60 ) ( ) 2 KE of A = 2  KE of B, so M1 1 2 2 1 2 m v + ( u sin60 ) = 2   2m ( u sin) ( ) 2 2 2 M1 Use result of (a) and rearrange.  1  3 2 2 2 u = 4u ( sin) Obtain 3-term quadratic.  u ( 4cos− 1)  +  2  4 8cos 2 − 2cos− 3 = 0 3 1 3 A1 cos= , − but angle is acute, so cos= 4 2 4 4 1(c) NEL: v = e ( u cos60 + u cos) M1  1 3  A1 v = u , u = e u + u    2 4  4 e = 5 2

This question in 9231/33 Oct/Nov 2023

1 u A B i m 5m Two smooth uniform spheres A and B of equal radii have masses m and 5m respectively. Sphere A is moving on a smooth horizontal surface with speed u when it collides with sphere B which is at rest on the surface. Immediately before the collision, A’s direction of motion makes an angle of i with the line of centres. After the collision, the kinetic energies of A and B are equal. The coefficient of restitution between the spheres is 1.2 Find the value of tani. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 1 Along line of centres, PCLM: 5 B A mv mv mu NEL: 1 2 cos B A v v u    M1 Signs consistent with PCLM equation. cos , cos 4 4     B A u u v v A1 Perpendicular to line of centres: speed of A is sin u B1   2 2 2 1 1 cos sin 5 cos 2 4 2 4                           u u m u m M1 Equate final kinetic energies, 3 terms, correct masses.   2 4 5 cos   , 2 1 2 5 cos , tan     A1 6

This question in 9231/31 May/June 2024

1 u A B i m 5m Two smooth uniform spheres A and B of equal radii have masses m and 5m respectively. Sphere A is moving on a smooth horizontal surface with speed u when it collides with sphere B which is at rest on the surface. Immediately before the collision, A’s direction of motion makes an angle of i with the line of centres. After the collision, the kinetic energies of A and B are equal. The coefficient of restitution between the spheres is 1.2 Find the value of tani. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 1 Along line of centres, PCLM: 5 B A mv mv mu NEL: 1 2 cos B A v v u    M1 Signs consistent with PCLM equation. cos , cos 4 4     B A u u v v A1 Perpendicular to line of centres: speed of A is sin u B1   2 2 2 1 1 cos sin 5 cos 2 4 2 4                           u u m u m M1 Equate final kinetic energies, 3 terms, correct masses.   2 4 5 cos   , 2 1 2 5 cos , tan     A1 6

This question in 9231/32 May/June 2024

3 3u 2u A B θ m m The diagram shows two identical smooth uniform spheres A and B of equal radii and each of mass m. The two spheres are moving on a smooth horizontal surface when they collide with speeds 2u and 3u respectively. Immediately before the collision, A’s direction of motion makes an angle i with the line of centres and B’s direction of motion is perpendicular to that of A. After the collision, B moves perpendicular to the line of centres. The coefficient of restitution between the spheres is 1. 3 (a) Find the value of tani. [3] … … … … … … … … … … … … … … … … … … … (b) Find the total loss of kinetic energy as a result of the collision. [2] … … … … … … … … … … … … … … … (c) Find, in degrees, the angle through which the direction of motion of A is deflected as a result of the collision. [2] … … … … … … … … … … …

7 marks

Mark scheme: 3(a) PCLM along line of centres: −mv = m2u cos− m3u sin B1 Must include m, must have minus sign on RHS, accept positive or negative v. If velocity of B after collision is included, it must be equated to zero before this mark is awarded. NEL: v = eu ( 3sin+ 2cos) M1 Must have plus sign on RHS, accept positive or negative v (sign of v does not need to be consistent with PCLM equation) If velocity of B after collision is included, it must be equated to zero before this mark is awarded. 4 A1 Correct work only, except possibly missing m. Eliminate v: 6sin= 8cos, tan= 3 3 3(b) Only change in KE is along line of centres M1 1 2 2 1 2 Loss = m ( 2u cos) + ( 3u sin) − mv ( ) 2 2 1 2  36 144  12 6  2  72 2 A1 (Note that tan= 23 leads to final answer 1336 mu 2 ) mu  + −  −   = mu 2  25 25  5 5   25   2 Alternative method for question 3(b) Alternative method, using total KE M1 1 2 1 2 Loss in KE = [ m ( 2u ) + m ( 3u ) ] 2 2 Or equivalent, with all necessary terms present  1 2 1 2 1 2  −  mv + m ( 2u sin) + m ( 3u cos)   2 2 2  13 2 181 2 72 2 A1 mu − mu = mu 2 50 25 2 3(c) 6 8 M1 [Components of velocity of A after collision are  u  u so] angle between 5 5 line of centres and A’s direction is . Angle of deflection = 180 – 2tan-1 4/3 = 73.7 A1FT FT their answer to part (a) 2

This question in 9231/32 Oct/Nov 2024

6 A B u θ θ 2u m m Two identical uniform smooth spheres A and B, each with mass m, are moving on a horizontal surface with speeds 2u and u respectively when they collide. Immediately before the collision, the spheres are moving parallel to each other in opposite directions such that their directions of motion each make an angle i with the line of centres (see diagram). As a result of the collision, B moves in a direction which is perpendicular to its initial direction of motion. The coefficient of restitution between the spheres is e. (a) Find an expression for tani in terms of e. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … As a result of the collision, A moves in a direction which is perpendicular to the line of centres. (b) Find the value of i. [2] … … … … … … … … … … … … …

8 marks

Mark scheme: 6(a) Let v and w denote the horizontal components of the velocities for A and B M1 Allow sign errors, m must be present. respectively after the collision. 2mu cos− mu cos= mw + mv w − v = e (3u cos) M1 Restitution equation: directions must be consistent with those used in momentum equation (signs must be consistent). 1 w = 12 ( 3eu cos + u cos)  = 2 (3e + 1)u cos A1 Combining to find equation with w, e, u, . SOI 1 2 ( 3e + 1) u cos M1 OE, using perpendicular motion of B. = tan u sin tan 2= 12 (3e + 1) M1 Obtaining an expression for tan in terms of e . 1 A1 CWO, must reject negative term if seen. tan= 2 (3e + 1) 6 6(b) w = 3eu cos= u cos M1 Using v = 0 to find a value for e. e = 13 tan= 1 , = 45 A1 Alternative method for question 6(b) w M1 v = 0, w = u cos and tan u sin= u cos= u sintan, tan 2 = 1, = 45 A1 2

This question in 9231/33 May/June 2025

7 A particle P is projected from a point O with speed U at an angle 45° above the horizontal and moves freely under gravity. (a) State the vertical and horizontal components of velocity at time t. [1] … … … … … … … At time T, particle P is moving at an angle of 60° below the horizontal. U (b) Show that T = ` 2 + 6j. [3] 2g … … … … … … … … … … … … … … … … … At time T, the particle strikes a smooth horizontal plane at a point which is a horizontal distance D from O and a vertical distance H below O. (c) Find the ratio H : D. [4] … … … … … … … … … … … After striking the horizontal plane, P rebounds with speed w. The coefficient of restitution between P and the plane is 2. 3 (d) Find w in terms of U. [3] … … … … … … … … … … … … Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 7(a) yv = U sin45 − gT B1 vx = U cos45 1 7(b) v y B1 SOI = − tan60 v x  U  M1 Substitute values for trigonometric ratios to obtain − gT   a linear equation in U and T only. 2  = − 3  U − 2 gT = − 3 U  U   2  U 2U U A1 AG, shown convincingly. T = 1 + 3 = 1 + 3 = 2 + 6 ( ) ( ) ( ) 2 g 2 g 2 g 3 7(c) U B1 D in terms of U and T . D =  T 2 U 1 2 B1 H in terms of U and T . H =  T − 2 gT With or without modulus sign. 2 U 1 2 M1 OE 1 1  T − 2 gT 2 + 6 Obtain a numerical value, may be un-simplified. 2 U ( 2 2 − 12 ( ) ) 1 − 3 2 = = Division or ratios. 1 U  T 2 2 2 With or without modulus sign. 2 H : D = 1 :1 + 3 A1 OE (for example H : D = 3 − 1:2 ), CWO 7(c) Alternative method for question 7(c) U  U  B1 D in terms of U and g . 2 + 6 D =  ( )  2  2 g  U  U  1  U  2  U 2  B1 H in terms of U and g . 2 + 6 2 + 6 = −  With or without modulus sign. H =  ( )  − 2 g  ( )   2  2 g   2 g   2 g  2 M1 OE U  U  1  U  2 + 6 2 + 6 Obtain a numerical value, may be un-simplified.  ( )  − 2 g  ( )  2  2 g   2 g  1 − 3 Division or ratios. = U  U  2 With or without modulus sign. 2 + 6  ( )  2  2 g  H : D = 1 :1 + 3 A1 OE (for example H : D = 3 − 1:2 ), CWO Alternative method for question 7(c) U U B1 U 2 D = 2 + 6 1 + 3 ( ) D = ( ) 2 2 g 2 g 2 2 B1  6   2    U  =  U  − 2 gH      2   2  U 2 M1 (  ) H = 2 g H : D = 1 :1 + 3 A1 OE (for example H : D = 3 − 1:2 ), CWO 4 7(d) 2 3  2  B1 OE, accept un-simplified. wy =  U  =  U  Allow negative sign. 3 2  3  2 M1 2 2  U  2 2 1 2 w = w y +    = 3 U + 2 U   2  2 1 7 A1 CWO w = U 3 + 2 = U 6 3

This question in 9231/33 May/June 2025

6 u A B i 5m 4m Two uniform smooth spheres A and B of equal radii have masses 5m and 4m respectively. Sphere A is moving with speed u on a horizontal surface when it collides with sphere B which is at rest. Immediately before the collision, A’s direction of motion makes an angle i with the line of centres (see diagram). The coefficient of restitution between the spheres is e. (a) Show that the speed of B after the collision is 5 u ( 1 + e)cos i . [3] 9 … … … … … … … … … … … … … … … … … … … … After the collision the kinetic energy of A is equal to the kinetic energy of B. (b) Given that tan i = 2 , find the value of e. [6] 3 … … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 6(a) Speeds for A and B along line of centres after the collision are M1 Momentum equation: m must be present. 1v and 2v , respectively. Allow minus sign between terms on LHS. 5mv1 + 4mv2 = 5mu cos v2 − v1 = eu cos M1 Restitution equation: directions must be consistent with those used in momentum equation (signs must be consistent).  A1 AG, shown convincingly. v2 = 95 u (1 + e ) cos 3  B1 May be found in part 6(a) but credit it in part 6(b).6(b) v1 = 19 u ( 5 − 4e ) cos Does not need to be simplified. Speed of V perpendicular to line of centres = u sin. *B1 Seen anywhere. 1 2 2 1 2 M1 Equate KEs after collision. = 2 (4 m) v2 1 2 (5m) ( v1 + ( u sin) ) Allow 2 m to be missing on both sides but must have the 5 and 4. KE of A must have two terms. Allow sin/cos mix. 5 1 2 2 4 5 2 DM1 Substitute expressions for speeds AND numerical values for = 2 ( 9 u (1 + e ) cos) ( 9 u ( 5 − 4e ) cos) + ( u sin ) ) 2 ( trigonometric functions. Accept un-simplified.  2 2  2 This mark is dependent on the second B1.       3 2 3       5 1 100   + u  5 u (1 + e )     2 9 u ( 5 − 4e ) = 24         OE may be seen. 5 u 2 ( 5 − 4e ) 2 + 20u 2 = 9 u 2 (1 + e ) 2     9  13 13 13            9    4e2 + 80e − 41 = 0 DM1 AEF, obtain 3-term quadratic equation in e. This mark is dependent on the second B1.

This question in 9231/34 May/June 2025

1 Two uniform smooth spheres A and B of equal radii have masses 4m and m respectively. Sphere B is at rest on a smooth horizontal surface. Sphere A is moving on the surface with speed u and collides directly with sphere B. After the collision, the momentum of A is three times the momentum of B. Find the value of the coefficient of restitution e. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: Question Answer Marks Guidance 1 Let v A and v B be the speeds of A and B respectively after the collision. M1 Conservation of momentum, masses correct. 4 mv A + mv B = 4 mu −v A + vB = eu M1 NEL, signs must be consistent with momentum equation. v A = 34 u , v B = u A1 Use momentum condition after the collision to obtain expressions for vA and vB in terms of u. e = 14 A1 Alternative method for question 1 Let v A and v B be the speeds of A and B respectively after the collision. M1 Conservation of momentum, masses correct. 4 mv A + mv B = 4 mu −v A + vB = eu M1 NEL, signs must be consistent with momentum equation. 4 u (1 + e ) A1 Combine to obtain expressions for v A and v B in terms of u v A = 15 u ( 4 − e ) , v B = 5 and e. 1 5 u ( 4 − e ) 1 A1 Use momentum condition after the collision to solve for e. = 3, e = 4 4 5 u (1 + e ) 4

This question in 9231/31 Oct/Nov 2025

6 17ag P 60° a O A particle P of mass m is attached to one end of a light inextensible string of length a. The other end of the string is attached to a fixed point O. Initially P is held with the string taut and making an angle of 60° with the upward vertical through O. The particle P is projected perpendicular to the string in a downwards direction with speed 17ag . It then starts to move along a circular path in a vertical plane with centre O (see diagram). At the lowest point of its path, vertically below O, the particle P collides with a stationary particle Q. (a) Find, in terms of a and g, an expression for the speed of P immediately before the collision with Q. [2] … … … … … … … … … … … … … … … … … As a result of the collision, P rebounds and moves back along a circular path with centre O. The string becomes slack when P reaches the point on the circle vertically above O. (b) Find, in terms of a and g, an expression for the speed of P immediately after the collision with Q. [3] … … … … … … … … … … … The mass of particle Q is km and the collision between P and Q is perfectly elastic. (c) Find the value of k. [3] … … … … … … … … … … … …

8 marks

Mark scheme: 6(a) Let v be the speed of P before the collision. M1 Dimensionally correct energy equation with a GPE term and Energy conservation before the collision: two KE terms. 1 2 m (17ag ) + 2 3 mag = 12 mv 2 A1 v = 20ag 2 6(b) Let wP be speed of P after the collision and V be its speed at the top. B1 Dimensionally correct equation. 2 T = 0 must be seen or implied. mV N2L: = ( T ) + mg V = ag    a 1 2 1 2 M1 Dimensionally correct energy equation with a GPE term and Energy equation: 2 mwP = 2 mV + 2mga two KE terms. A1 wP = 5ag 3 6(c) Let wQ the speed of Q after the collision. B1 Momentum conserved (must see masses). mv = kmwQ − mwP wP + wQ = ev  = v  B1 NEL with consistent signs. OR OR Use conservation of kinetic energy. 1 2 mv 2 = 12 mwP2 + 12 kmwQ2 k = 3 B1 Must have e = 1 seen or implied. 3

This question in 9231/31 Oct/Nov 2025

5 Two uniform smooth spheres, A and B, of equal radii are on a horizontal surface. They have masses of m and km respectively. Sphere A is moving with speed u at an angle a with the line of centres when it collides with sphere B which is stationary. Immediately after the collision, sphere A moves with speed v at an angle 2a with the line of centres (see diagram). A v B 2α α u m km It is given that tan a = 3 . 4 (a) Find v in terms of u. [2] … … … … … … (b) Find the coefficient of restitution between the spheres in terms of k. [4] … … … … … … … … … … … … … … … … … … … … … … … … … (c) Find the range of possible values of k. [3] … … … … … … … … … … … … … …

9 marks

Mark scheme: 5(a) u sin= v sin ( 2) B1 May be implied by correct answer.   5 3 u = 2425 v  v = 85 u B1 2 5(b) mu cos= mv cos ( 2) + kmw B1 PCLM equation, must include masses.   54 u = 257 v + kw, kw = 85 u  B1 NEL equation. eu cos= w − v cos ( 2)   54 eu = w − 257 v  5 4 5 7 M1 Combine and substitute for v and w to obtain = e +  equation in e and k only. 8k 5 8 25 25 7 A1 25 −k7 e = − AEF, for example . 32 k 32 32 k Not dependent on first B mark. 4 5(c) 25 7 M1 Form inequality using their expression for e. 0 ≤ − ≤ 1 Allow equality or strict inequality. 32k 32 7 k ≤ 25 ≤ 39k M1 Rearrange at least one side to the form ak ≤ b or ak ≥ b . Allow equality or strict inequality. 39 25 ≤ k ≤ 257 A1 CAO 3

This question in 9231/32 Oct/Nov 2025

1 Two uniform smooth spheres A and B of equal radii have masses 4m and m respectively. Sphere B is at rest on a smooth horizontal surface. Sphere A is moving on the surface with speed u and collides directly with sphere B. After the collision, the momentum of A is three times the momentum of B. Find the value of the coefficient of restitution e. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: Question Answer Marks Guidance 1 Let v A and v B be the speeds of A and B respectively after the collision. M1 Conservation of momentum, masses correct. 4 mv A + mv B = 4 mu −v A + vB = eu M1 NEL, signs must be consistent with momentum equation. v A = 34 u , v B = u A1 Use momentum condition after the collision to obtain expressions for vA and vB in terms of u. e = 14 A1 Alternative method for question 1 Let v A and v B be the speeds of A and B respectively after the collision. M1 Conservation of momentum, masses correct. 4 mv A + mv B = 4 mu −v A + vB = eu M1 NEL, signs must be consistent with momentum equation. 4 u (1 + e ) A1 Combine to obtain expressions for v A and v B in terms of u v A = 15 u ( 4 − e ) , v B = 5 and e. 1 5 u ( 4 − e ) 1 A1 Use momentum condition after the collision to solve for e. = 3, e = 4 4 5 u (1 + e ) 4

This question in 9231/33 Oct/Nov 2025

6 17ag P 60° a O A particle P of mass m is attached to one end of a light inextensible string of length a. The other end of the string is attached to a fixed point O. Initially P is held with the string taut and making an angle of 60° with the upward vertical through O. The particle P is projected perpendicular to the string in a downwards direction with speed 17ag . It then starts to move along a circular path in a vertical plane with centre O (see diagram). At the lowest point of its path, vertically below O, the particle P collides with a stationary particle Q. (a) Find, in terms of a and g, an expression for the speed of P immediately before the collision with Q. [2] … … … … … … … … … … … … … … … … … As a result of the collision, P rebounds and moves back along a circular path with centre O. The string becomes slack when P reaches the point on the circle vertically above O. (b) Find, in terms of a and g, an expression for the speed of P immediately after the collision with Q. [3] … … … … … … … … … … … The mass of particle Q is km and the collision between P and Q is perfectly elastic. (c) Find the value of k. [3] … … … … … … … … … … … …

8 marks

Mark scheme: 6(a) Let v be the speed of P before the collision. M1 Dimensionally correct energy equation with a GPE term and Energy conservation before the collision: two KE terms. 1 2 m (17ag ) + 2 3 mag = 12 mv 2 A1 v = 20ag 2 6(b) Let wP be speed of P after the collision and V be its speed at the top. B1 Dimensionally correct equation. 2 T = 0 must be seen or implied. mV N2L: = ( T ) + mg V = ag    a 1 2 1 2 M1 Dimensionally correct energy equation with a GPE term and Energy equation: 2 mwP = 2 mV + 2mga two KE terms. A1 wP = 5ag 3 6(c) Let wQ the speed of Q after the collision. B1 Momentum conserved (must see masses). mv = kmwQ − mwP wP + wQ = ev  = v  B1 NEL with consistent signs. OR OR Use conservation of kinetic energy. 1 2 mv 2 = 12 mwP2 + 12 kmwQ2 k = 3 B1 Must have e = 1 seen or implied. 3

This question in 9231/33 Oct/Nov 2025