Momentum

6(a) Along line of centres, speeds v and v M1 Momentum (condone missing masses). 1 2 mv +mv = mucosα−mucosβ 1 2 ( ) v −v =eu cosβ+ cosα M1 Restitution. 2 1 Both correct, masses seen. A1 v = 0 so A has no speed along line of centres: moves A1 AG. 1 perpendicular to line of centres 4 6(b) 1 M1 Both components. (v = u cosα=ucosβ ) 2 2 ( ) KE of B after collision is 1 m v 2 +( usinβ)2 2 2 KE of A after collision = 1 m ( usinα)2 2 3 M1 Add both KEs and equate to mu2 4 Simplify to equation in sinα M1 sinα= 1 , α=45° A1 2 4 © UCLES 2021 Page 11 of 13

6(a) Along line of centres, speeds v and v M1 Momentum (condone missing masses). 1 2 mv +mv = mucosα−mucosβ 1 2 ( ) v −v =eu cosβ+ cosα M1 Restitution. 2 1 Both correct, masses seen. A1 v = 0 so A has no speed along line of centres: moves A1 AG. 1 perpendicular to line of centres 4 6(b) 1 M1 Both components. (v = u cosα=ucosβ ) 2 2 ( ) KE of B after collision is 1 m v 2 +( usinβ)2 2 2 KE of A after collision = 1 m ( usinα)2 2 3 M1 Add both KEs and equate to mu2 4 Simplify to equation in sinα M1 sinα= 1 , α=45° A1 2 4 © UCLES 2021 Page 11 of 13

6(a) Let velocities of A and B along line of centres after collision be v M1 Momentum, must include m, allow cos/sin mix. 1 and v . 2 mv + kmv = mucosθ . 1 2 1 M1 Restitution, consistent signs, correct way up. v −v = ucosθ 2 1 3 4ucosθ A1 AG shown convincingly. Solve: v 2 = 3 ( 1+k ) 3 6(b) ( 3−k ) ucosθ B1 Or equivalent, may be unsimplified. v = 1 3 ( 1+k ) Use velocity of A with both components. B1 v2+( usinθ)2 seen. 1 1 kmv 2 + 1 m ( v2 +( usinθ)2 ) = 3 × 1 mu2 M1 KE after = 30% KE before (all terms present). 2 2 2 1 10 2 M0 if incorrect masses. Substitute from part (a) and for θ . M1 Eliminate trigonometric terms, must be KE equation, in terms of k only. ( 3−k )2 +16k=2 ( 1+k )2 , k2 −6k −7=0 M1 Obtain simplified quadratic equation in k . k =7 A1 6 © UCLES 2021 Page 13 of 16

7(a) ucosα=vcosβ M1 eusinα=vsinβ M1 Divide: tanβ=etanα A1 AG. Must see divide OE. 3 7(b) vsinβ=wcosγ (=eusinα) M1 evcosβ=wsinγ (=eucosα) M1 Divide: tanγ=1/tanα: γ=90°−α *A1 After second rebound, direction of motion is parallel to initial path. DB1 4 7(c) Final KE = 1 m ( ( eusinα)2 +( eucosα)2 )  = 1 me2u2  M1 Energy expression in terms of u.   2  2  1 1 1 1 A1 So me2u2 = × mu2 giving e= 2 9 2 3 Part (a) gives tan ( 90−α)=etanα M1 So tanα= 3 A1 4 © UCLES 2021 Page 12 of 12

5(a) Let speeds of A and B along line of centres after collision be v and M1 Momentum with masses correct. 1 v 2 3 3  u mv + mv =− mucos60°+mu = 1 2 2 2   4   v −v =− 2(−ucos60°−u ) (=u) M1 Restitution, with consistent signs on LHS. 2 1 3  1  1 A1   v 1 =− 2 u  v 2 = 2 u 3 B1 Perpendicular to line of centres, speed of B is usin60°= u 2 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) 1 1 3m 5 B1 KE before = mu2 + . u2 = mu2 2 2 2 4 1 u 2 1 3m  u 2  3u 2  7  B1 FT FT only their speeds from (a) KE after = m  + .    +       = mu2  2 2 2 2 2  2    8    3 B1 Loss in KE = mu2 8 3 © UCLES 2021 Page 10 of 12

7(a) ucosα=vcosβ M1 eusinα=vsinβ M1 Divide: tanβ=etanα A1 AG. Must see divide OE. 3 7(b) vsinβ=wcosγ (=eusinα) M1 evcosβ=wsinγ (=eucosα) M1 Divide: tanγ=1/tanα: γ=90°−α *A1 After second rebound, direction of motion is parallel to initial path. DB1 4 7(c) Final KE = 1 m ( ( eusinα)2 +( eucosα)2 )  = 1 me2u2  M1 Energy expression in terms of u.   2  2  1 1 1 1 A1 So me2u2 = × mu2 giving e= 2 9 2 3 Part (a) gives tan ( 90−α)=etanα M1 So tanα= 3 A1 4 © UCLES 2021 Page 12 of 12

6(a) Let v and w be speeds after collision: M1 Momentum along line of centres. mvkmwmucos 1 M1 NEL consistent signs. wv ucos 2 3ucos A1 AG Add to give 21k Convincing working. Substitute back or re-solve: A1 Accept without modulus sign. 2kucos v 21k 4 © UCLES 2022 Page 11 of 14 6(b) 2kucos 2 B1 For speed of A (SOI). (usin)2    21k    Equal KE after collision: M1 Equate KEs. 1 3ucos 2 1  2kucos 2 km    m(usin)2    2  21k 2   21k         9kcos2 41k2sin2 2k2cos2   2 M1 Use tan : 3 16  12kk2 9  44kk2 81k 25k2 85k520 leading to 5k45k130 M1 Obtain quadratic and attempt to solve. 4 13 A1 k  or 5 5 5 © UCLES 2022 Page 12 of 14

6(a) Let v and w be speeds after collision: M1 Momentum along line of centres. mvkmwmucos 1 M1 NEL consistent signs. wv ucos 2 3ucos A1 AG Add to give 21k Convincing working. Substitute back or re-solve: A1 Accept without modulus sign. 2kucos v 21k 4 © UCLES 2022 Page 11 of 14 6(b) 2kucos 2 B1 For speed of A (SOI). (usin)2    21k    Equal KE after collision: M1 Equate KEs. 1 3ucos 2 1  2kucos 2 km    m(usin)2    2  21k 2   21k         9kcos2 41k2sin2 2k2cos2   2 M1 Use tan : 3 16  12kk2 9  44kk2 81k 25k2 85k520 leading to 5k45k130 M1 Obtain quadratic and attempt to solve. 4 13 A1 k  or 5 5 5 © UCLES 2022 Page 12 of 14

6(a) Let v be speed of rebound from 1st collision: B1 Energy loss. 1 2 1 2 Energy loss: mv2   mu2, v2  u2 2 5 2 5 vcosucos B1 Both. vsineusin Combine to form equation in e only M1 v2 ucos 2 eusin 2 2 1 4  e2 5 5 5 1 A1 e 2 4 6(b) tanetan, so tan1,  45 B1 For 2nd collision M1 Both. wcosvcos18060 May be implied by the A1. wsinevsin18060 tanetan120their  M1 Divide to find . 61.8 A1 4 © UCLES 2022 Page 11 of 13

6(a) Let speed of A after collision be →v and speed of B M1 A perpendicular to line of centres be v 5 Along line of centres: mu−km ucos=mv Momentum. 8 A 5  M1 NEL NEL: 0−v A =e 8 u cos+u  5 25  M1 Solve. So u− kucos=−  u cos+u 8 38  Substitute for cos, to give k=4 A1 4 6(b) 5 3 B1 v = usin = u B 8 8 Velocity perpendicular to line of centres v =−u B1 FT A 2 M1 NOTE: KE before and after for A is unchanged. 1 1 5  1 25 41 KE before = mu2 + km  u  = mu2 + mu2 = mu2 2 2 8  2 32 32 Both. 1 1 1 9 25 KE after = mv 2 + kmv 2 = mu2 +2m u2 = mu2 2 A 2 B 2 64 32 41 25 1 A1 Loss = mu2  − = mu2 32 32 2 4 © UCLES 2022 Page 10 of 12

7(a) Let v, w be speeds of A and B along line of centres after collision M1 1 1 mv+ mw=mucos− m.2ucos Momentum: masses correct, opposite signs on RHS. 2 2 w−v=e(2ucos+ucos) 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 © UCLES 2022 Page 14 of 15 7(b) Perpendicular to line of centres, speed of B is B1 2usin=2ucos After, velocity of B makes angle  with line of centres, so B1 2ucos tan= w sin 2ucos M1* Obtain homogeneous equation in cos and sin or an equation in tan = giving cos 2 1 13  u sin+ cos 3 4 8  1 13 DM1 Obtain quadratic and solve to find values of tan 3(cos)2 = (sin)2 + sincos 4 8 2(tan)2 +13tan−24=0, (2tan−3)(tan+8)=0 3 A1 tan= 2 5 © UCLES 2022 Page 15 of 15

6(a) Let speed of A after collision be →v and speed of B M1 A perpendicular to line of centres be v 5 Along line of centres: mu−km ucos=mv Momentum. 8 A 5  M1 NEL NEL: 0−v A =e 8 u cos+u  5 25  M1 Solve. So u− kucos=−  u cos+u 8 38  Substitute for cos, to give k=4 A1 4 6(b) 5 3 B1 v = usin = u B 8 8 Velocity perpendicular to line of centres v =−u B1 FT A 2 M1 NOTE: KE before and after for A is unchanged. 1 1 5  1 25 41 KE before = mu2 + km  u  = mu2 + mu2 = mu2 2 2 8  2 32 32 Both. 1 1 1 9 25 KE after = mv 2 + kmv 2 = mu2 +2m u2 = mu2 2 A 2 B 2 64 32 41 25 1 A1 Loss = mu2  − = mu2 32 32 2 4 © UCLES 2022 Page 10 of 12

2 Parallel to wall vcosucos M1 Both Perpendicular to wall vsineusin 1 A1 AEF Dividing, e 2tan KE reduced by 20%, so 1mu2 cos2e2sin2   41mu2 M1 Dimensionally correct equation in u or v, but not 2 5 2 both. Must have either  or , but not both. Must see 4 on the correct side of the equation. 5 4 A1 Eliminate e: cos 5 2 A1 e 3 © UCLES 2023 Page 7 of 15 2 Alternative method for question 2 Parallel to wall vcosucos M1 Both Perpendicular to wall vsineusin  5 2 5 5v 2 5v A1 sin , cos  usin , ucos 5 5 5e 5   4v2 v2 A1 v2 1  u2   u2cos2u2sin2   5  5e2 AEF, e.g. 5   4 e2   1 4 1 2 v2  1  M1 Dimensionally correct equation in v. 2 mv2   5  2 mu2   5 m 5   4 e2   Must have either  or , but not both. 4 2 Must see or on the correct side of the 5 5 equation. 2 A1 e 3 5 © UCLES 2023 Page 8 of 15

2 Parallel to wall vcosucos M1 Both Perpendicular to wall vsineusin 1 A1 AEF Dividing, e 2tan KE reduced by 20%, so 1mu2 cos2e2sin2   41mu2 M1 Dimensionally correct equation in u or v, but not 2 5 2 both. Must have either  or , but not both. Must see 4 on the correct side of the equation. 5 4 A1 Eliminate e: cos 5 2 A1 e 3 © UCLES 2023 Page 7 of 15 2 Alternative method for question 2 Parallel to wall vcosucos M1 Both Perpendicular to wall vsineusin  5 2 5 5v 2 5v A1 sin , cos  usin , ucos 5 5 5e 5   4v2 v2 A1 v2 1  u2   u2cos2u2sin2   5  5e2 AEF, e.g. 5   4 e2   1 4 1 2 v2  1  M1 Dimensionally correct equation in v. 2 mv2   5  2 mu2   5 m 5   4 e2   Must have either  or , but not both. 4 2 Must see or on the correct side of the 5 5 equation. 2 A1 e 3 5 © UCLES 2023 Page 8 of 15

4(a) Let speeds of A and B along line of centres after collision be V and V M1 Allow sign errors, allow missing m. A B V V ucos30 (1) A B V V eucos30 (2) M1 Signs on LHS must be consistent with (1). A B Speeds perpendicular to line of centres after collision are usin30 and 2u B1 SOI V V V 4V Moving in same direction, so A  B (3) B A usin30 2u Use V 4V in (1): 5V ucos30 M1 A complete method to find equation in e only B A A From (2): 3V eucos30 then A Combine to find equation in e only. e 3 A1 5 © UCLES 2023 Page 8 of 14 4(a) Alternative method for question 4(a) Let speeds of A and B along line of centres after collision be V and V M1 A B V V ucos30 (1) Allow sign errors, allow missing m. A B V V eucos30 (2) M1 Signs on LHS must be consistent with (1). A B Speeds perpendicular to line of centres after collision are usin30 and 2u B1 SOI V V V 4V Moving in same direction, so A  B (3) B A usin30 2u 1 1 M1 u 4u Solve (1) and (2): V  u1ecos30, V  u1ecos30 Note: V  3, V  3 A 2 B 2 A 10 B 10 Substitute in (3) to find equation in e only . e 3 A1 5 5 4(b) KE after = 1 2 m    V A 2     u 2    2    1 2 m (2u2 V B 2) B1 C af o te rr r e c c o t l e li x s p io re n s , s w io i n th f b o o r t K h E c o f m or p o o n n e e n o t f s . t he spheres,   7 56 B1 119 KE for A after = mu2 or KE for B after = mu2 Implied by total KE after = mu2. 50 25 50 9 6 or KE loss for A = mu2 or KE gain for B = mu2 25 25 3 B1 1 Total loss in KE = mu2 Term m2u2 may be omitted from KE of B 25 2 before and after. 3 © UCLES 2023 Page 9 of 14

1(a) PCLM: mv=−mucos60+2mucos B1 First line must be seen. 1 1 v=− u+2ucos=v= u(4cos−1) AG 2 2 1 1(b) KE of A = 1 m ( v2 +(usin60)2) B1 2 KE of A = 2  KE of B, so M1 1 m ( v2 +(usin60)2) =2 1 2m(usin)2 2 2 2 M1 Use result of (a) and rearrange.   1 u(4cos−1)   + 3 u2 =4u2(sin)2 Obtain 3-term quadratic. 2  4 8cos2−2cos−3=0 3 1 3 A1 cos= , − but angle is acute, so cos= 4 2 4 4 1(c) NEL: v=e(ucos60+ucos) M1 1 3  A1 v=u, u=e  u+ u  2 4  4 e= 5 2 © UCLES 2023 Page 6 of 13

4(a) After collision, A has velocity v towards wall on right and B has component B1 A of velocity towards lower wall of u sin . Same distance and time, so v = u sin . A Along line of centres: M1 Both, consistent signs, must be cos. PCLM: mv +mv =mucos A B NEL: v −v =eucos A B 2sin=(1+e)cos M1 Eliminating v and v to find an equation in . A B Condone common factor of u. 3 1 Note: v = ucos and v = ucos. A 4 B 4 tan= 3 A1 4 4 © UCLES 2023 Page 8 of 15 4(b) Final KE = 1 m ( v2 +(usin)2 +v2 ) B1 Both components of velocity of B needed. 2 A B Loss = 1 2 mu2 − 1 2 m ( v A 2 +(usin)2 +v B 2 ) = 1 2 mu2    1− 2 9 5 − 2 9 5 − 2 1 5    M1 v co A , m v p B , o  n e s n u t b s s o ti f t u v t e e l d o , c f i t t y t h o e f i r B f i i n n a c l l u K d E ed w . ith both 3 A1 Loss = mu2 25 Percentage loss = 24% 3 © UCLES 2023 Page 9 of 15

1(a) PCLM: mv=−mucos60+2mucos B1 First line must be seen. 1 1 v=− u+2ucos=v= u(4cos−1) AG 2 2 1 1(b) KE of A = 1 m ( v2 +(usin60)2) B1 2 KE of A = 2  KE of B, so M1 1 m ( v2 +(usin60)2) =2 1 2m(usin)2 2 2 2 M1 Use result of (a) and rearrange.   1 u(4cos−1)   + 3 u2 =4u2(sin)2 Obtain 3-term quadratic. 2  4 8cos2−2cos−3=0 3 1 3 A1 cos= , − but angle is acute, so cos= 4 2 4 4 1(c) NEL: v=e(ucos60+ucos) M1 1 3  A1 v=u, u=e  u+ u  2 4  4 e= 5 2 © UCLES 2023 Page 6 of 13

1 Along line of centres, PCLM: 5mv mv mucos M1 Must include correct masses. B A NEL: v B v A  1 2 ucos M1 Signs consistent with PCLM equation. u u A1 v  cos, v  cos B 4 A 4 Perpendicular to line of centres: speed of A is usin B1 1 m     u cos  2 usin2    1 5m  u cos  2 M1 E m q a u ss a e te s . f inal kinetic energies, 3 terms, correct 2   4   2 4    cos2  4 , cos 2 , tan 1 A1 5 5 2 6 © Cambridge University Press & Assessment 2024 Page 6 of 13

1 Along line of centres, PCLM: 5mv mv mucos M1 Must include correct masses. B A NEL: v B v A  1 2 ucos M1 Signs consistent with PCLM equation. u u A1 v  cos, v  cos B 4 A 4 Perpendicular to line of centres: speed of A is usin B1 1 m     u cos  2 usin2    1 5m  u cos  2 M1 E m q a u ss a e te s . f inal kinetic energies, 3 terms, correct 2   4   2 4    cos2  4 , cos 2 , tan 1 A1 5 5 2 6 © Cambridge University Press & Assessment 2024 Page 6 of 13

1 1 1 1 M1 Masses must be included. Allow sign errors. Along line of centres, PCLM: 2m ucosmu 2m ucosm u 2 2 4 5 A1 cos 8 1 1 1  M1 Allow sign errors, e must be on correct side. NEL: ucos ue  ucosu  2 4 2  3 A1 AEF e 7 4 © Cambridge University Press & Assessment 2024 Page 6 of 15

7(a) When P strikes plane, velocity is →ucos, M1 3 u 5 3 3 Before impact: parallel to inclined plane ucos, perpendicular to plane usin 5 5 3 3 A1 After impact: components ucos (parallel) and eusin (perpendicular) 5 5 3 3 M1 Since velocity is vertical after impact, tan= ucos/ eusin 5 5 tan=1/etan, e tan2 =1 A1 AG 4 © Cambridge University Press & Assessment 2024 Page 11 of 12 7(b) (usin)2 8u2 M1A1 Note: alternative methods. Greatest height of P before impact: H = = 2g 25g 3 M1 After impact, vertical speed of P is u (cos)2 +e2(sin)2 5 3 M1 Use V2 =U2 +2as to greatest height, equal to H 16 9 u2 ( (cos)2 +e2(sin)2) =2g 3 H 25 16 1 e 1 M1 Use part (a): tan= , cos= , sin= e 1+e 1+e Substitute to find e 1 A1 3e2 +2e−1=0, e= 3 6 © Cambridge University Press & Assessment 2024 Page 12 of 12

3(a) PCLM along line of centres: −mv=m2ucos−m3usin 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 © Cambridge University Press & Assessment 2024 Page 9 of 19 3(b) Only change in KE is along line of centres M1 Loss = 1 m ( (2ucos)2 +(3usin)2) − 1 mv2 2 2 1 36 144 12 6 2 72 A1 (Note that tan= 2 leads to final answer 36mu2) mu2 + − −  = mu2 3 13 2 25 25  5 5  25   2 Alternative method for question 3(b) Alternative method, using total KE M1 1 1 Loss in KE = [ m(2u)2 + m(3u)2 ] 2 2 Or equivalent, with all necessary terms present − 1 mv2 + 1 m(2usin)2 + 1 m(3ucos)2   2 2 2  13 181 72 A1 mu2 − mu2 = mu2 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 © Cambridge University Press & Assessment 2024 Page 10 of 19

7(a) When P strikes plane, velocity is →ucos, M1 3 u 5 3 3 Before impact: parallel to inclined plane ucos, perpendicular to plane usin 5 5 3 3 A1 After impact: components ucos (parallel) and eusin (perpendicular) 5 5 3 3 M1 Since velocity is vertical after impact, tan= ucos/ eusin 5 5 tan=1/etan, e tan2 =1 A1 AG 4 © Cambridge University Press & Assessment 2024 Page 11 of 12 7(b) (usin)2 8u2 M1A1 Note: alternative methods. Greatest height of P before impact: H = = 2g 25g 3 M1 After impact, vertical speed of P is u (cos)2 +e2(sin)2 5 3 M1 Use V2 =U2 +2as to greatest height, equal to H 16 9 u2 ( (cos)2 +e2(sin)2) =2g 3 H 25 16 1 e 1 M1 Use part (a): tan= , cos= , sin= e 1+e 1+e Substitute to find e 1 A1 3e2 +2e−1=0, e= 3 6 © Cambridge University Press & Assessment 2024 Page 12 of 12

6(a) 912mv2 = 1mv2 3v =v  B1 KE equation linking v and v . 2 2 A 2 B A B A B Needs to be KE equation with 1 and m. 2 2mu=2mv +mv B1 Conservation of momentum. A B Must see m present. v = 2u v =6u B1 AG, shown convincingly.  A 5  B 5 If a quadratic equation is formed, must show method for solving and obtain a single answer. 3 6(b) Let w denote the speed after the second collision. B1 OE, for example 5cos=2cos. Perpendicular to wall: v cos=wcos B Parallel to wall: e v sin=wsin B1 OE, for example 2sin= 5sin. B e2v2sin2=w2sin2 576sin2=144sin2 *M1 Obtain an equation involving only one of , . B 625 125  Eliminate  or  by squaring both sides and v2cos2=w2cos2 36cos2=144cos2 B 25 125  adding. 576sin2+36 ( 1−sin2 ) =144 DM1 Use sin2+cos2=1 to reduce to equation 625 25 125 with just sine or cosine of 1 variable. sin2= 5 sin=1 5 A1 Award for any of the following:  9 3 cos= 2 or cos= 1 5or sin= 2. 3 3 3 sin(+)=1 A1 Exact answer. 6 © Cambridge University Press & Assessment 2025 Page 14 of 17

6(a) 912mv2 = 1mv2 3v =v  B1 KE equation linking v and v . 2 2 A 2 B A B A B Needs to be KE equation with 1 and m. 2 2mu=2mv +mv B1 Conservation of momentum. A B Must see m present. v = 2u v =6u B1 AG, shown convincingly.  A 5  B 5 If a quadratic equation is formed, must show method for solving and obtain a single answer. 3 6(b) Let w denote the speed after the second collision. B1 OE, for example 5cos=2cos. Perpendicular to wall: v cos=wcos B Parallel to wall: e v sin=wsin B1 OE, for example 2sin= 5sin. B e2v2sin2=w2sin2 576sin2=144sin2 *M1 Obtain an equation involving only one of , . B 625 125  Eliminate  or  by squaring both sides and v2cos2=w2cos2 36cos2=144cos2 B 25 125  adding. 576sin2+36 ( 1−sin2 ) =144 DM1 Use sin2+cos2=1 to reduce to equation 625 25 125 with just sine or cosine of 1 variable. sin2= 5 sin=1 5 A1 Award for any of the following:  9 3 cos= 2 or cos= 1 5or sin= 2. 3 3 3 sin(+)=1 A1 Exact answer. 6 © Cambridge University Press & Assessment 2025 Page 14 of 17

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. 2mucos−mucos=mw+mv w−v=e(3ucos) M1 Restitution equation: directions must be consistent with those used in momentum equation (signs must be consistent). w=1(3eucos+ucos) =1(3e+1)ucos A1 Combining to find equation with w, e, u, . SOI   2 2 1(3e+1)ucos M1 OE, using perpendicular motion of B. 2 =tan usin tan2= 1(3e+1) M1 Obtaining an expression for tan in terms of e. 2 A1 CWO, must reject negative term if seen. tan= 1(3e+1) 2 6 6(b) w=3eucos=ucos M1 Using v = 0 to find a value for e. e= 1 3 tan=1,  =45 A1 Alternative method for question 6(b) w M1 v=0, w=ucos and =tan usin ucos=usintan, tan2=1,  =45 A1 2 © Cambridge University Press & Assessment 2025 Page 14 of 17

6(a) Speeds for A and B along line of centres after the collision are M1 Momentum equation: m must be present. v and v , respectively. Allow minus sign between terms on LHS. 1 2 5mv +4mv =5mucos 1 2 v −v =eucos M1 Restitution equation: directions must be consistent with those used 2 1 in momentum equation (signs must be consistent). v = 5u(1+e)cos A1 AG, shown convincingly. 2 9 3 6(b) v =1u(5−4e)cos B1 May be found in part 6(a) but credit it in part 6(b). 1 9 Does not need to be simplified. Speed of V perpendicular to line of centres = usin. *B1 Seen anywhere. 1(5m) ( v2 +(usin)2) = 1(4m)v 2 M1 Equate KEs after collision. 2 1 2 2 Allow 1m to be missing on both sides but must have the 5 and 4. 2 KE of A must have two terms. Allow sin/cos mix. 5 ( (1u(5−4e)cos )2 +(usin)2 ) = 4(5u(1+e)cos )2 DM1 Substitute expressions for speeds AND numerical values for 2 9 2 9 trigonometric functions. Accept un-simplified.  2 2 2 This mark is dependent on the second B1.   3    2     3  5  1u(5−4e)   +u    = 4  5u(1+e)   5u2(5−4e)2 +20u2 =100u2(1+e)2 OE may be seen. 2  9  13   13   2  9  13 9 9    4e2+80e−41=0 DM1 AEF, obtain 3-term quadratic equation in e. This mark is dependent on the second B1. © Cambridge University Press & Assessment 2025 Page 15 of 17

1 Let v and v be the speeds of A and B respectively after the collision. M1 Conservation of momentum, masses correct. A B 4mv +mv =4mu A B −v +v =eu M1 NEL, signs must be consistent with momentum equation. A B v = 3u, v =u A1 Use momentum condition after the collision to obtain A 4 B expressions for v and v in terms of u. A B e= 1 A1 4 Alternative method for question 1 Let v and v be the speeds of A and B respectively after the collision. M1 Conservation of momentum, masses correct. A B 4mv +mv =4mu A B −v +v =eu M1 NEL, signs must be consistent with momentum equation. A B v = 1u(4−e), v = 4u(1+e) A1 Combine to obtain expressions for v and v in terms of u A 5 B 5 A B and e. 1u(4−e) A1 Use momentum condition after the collision to solve for e. 5 =3, e= 1 4u(1+e) 4 5 4 © Cambridge University Press & Assessment 2025 Page 9 of 17

5(a) usin=vsin(2) B1 May be implied by correct answer. 3u= 24v   5 25 v= 5u B1 8 2 5(b) mucos=mvcos(2)+kmw B1 PCLM equation, must include masses. 4u= 7 v+kw, kw=5u   5 25 8 eucos=w−vcos(2) B1 NEL equation. 4eu=w− 7 v   5 25 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 25−7k e= − AEF, for example . 32k 32 32k 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 7k≤25≤39k M1 Rearrange at least one side to the form ak≤b or ak≥b. Allow equality or strict inequality. 25≤k≤25 A1 CAO 39 7 3 © Cambridge University Press & Assessment 2025 Page 13 of 17

1 Let v and v be the speeds of A and B respectively after the collision. M1 Conservation of momentum, masses correct. A B 4mv +mv =4mu A B −v +v =eu M1 NEL, signs must be consistent with momentum equation. A B v = 3u, v =u A1 Use momentum condition after the collision to obtain A 4 B expressions for v and v in terms of u. A B e= 1 A1 4 Alternative method for question 1 Let v and v be the speeds of A and B respectively after the collision. M1 Conservation of momentum, masses correct. A B 4mv +mv =4mu A B −v +v =eu M1 NEL, signs must be consistent with momentum equation. A B v = 1u(4−e), v = 4u(1+e) A1 Combine to obtain expressions for v and v in terms of u A 5 B 5 A B and e. 1u(4−e) A1 Use momentum condition after the collision to solve for e. 5 =3, e= 1 4u(1+e) 4 5 4 © Cambridge University Press & Assessment 2025 Page 9 of 17

7 Perpendicular to the wall: eusin=vsin B1 PP =uT B1 1 2 PP =vkT 2 3 d d M1 sin= sin= PP PP 1 2 2 3 d d M1 Eliminate PP and PP . sin= sin= 1 2 2 3 uT vkT eud vd M1 = uT vkT 1 A1 e= k 6 © Cambridge University Press & Assessment 2025 Page 18 of 18

4(a) © Cambridge University Press & Assessment 2026 Page 14 of 22 v 1 = 11 23 e u and v 2 = 51 3 u or v s i n 11 23 e u  = and v c o s 51 3 u  = B1 v 1 is the velocity component of P perpendicular to the barrier after impact and v 2 is velocity component of P parallel to the barrier after impact. May see LHS of v 1 equation as s i n  v or c o s  v and similar for v , but must have opposite trigonometric ratios. 2 Allow vsin=eusin(67) or better and c o s c o s ( 6 7 )  v = u  or better. 12 m v 2 = 41 3  12 m u 2   v 2 = 41 3 u 2  *B1 Correct equation relating kinetic energies. This mark can be implied by seeing v2 = 4 u2. 13 ( 11 23 e u ) 2 + ( 51 3 u ) 2 = 41 3 u 2 DM1 Correct use of 2 s i n 2 2 c o s 2 2   v + v = v to form an equation in e (and possibly u) only. Allow minor slips, for example, e rather than e 2 but other terms must have been squared. One component must be either 12eu or 5 eu and the 13 13 other must be either 51 3 u or 11 23 u (or using sin(67), cos(67), or better). Allow consistent sin/cos mix only. B1 B1 DM1 can be implied by ( 11 23 e ) 2 + ( 51 3 ) 2 = 41 3 . e=1 3 ONLY A1 AWRT 0.433 or better (0.4330127…). 4 4 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/ 4(b) tan= 1 312 tan= 3 3=1.03923... *B1FT FT their e provided 4 5  5  © Cambridge University Press & Assessment 2026 Page 15 of 22 0  e  1 to find an expression for tan or . If correct, may see an angle of 46.102… (which may be seen on a diagram). Or for c o s 52 6 1 3  = or s i n 32 6 3 9  = . tan(60−)=0.247... DB1 Follow through their t a n  or  (the angle between P and the first barrier after the first collision). If exact and correct, may see t a n ( 6 0 ) 17 3  − = . OR for finding the angle between the second barrier and P before the second collision – if correct, this angle is 13.897… May be seen on a diagram. s i n s i n ( 6 0 )   w = e v − c o s c o s ( 6 0 )   w = v − M1 Allow, for example, w for 1 s i n  w and w 2 for c o s  w – both equations must be correct. Their value for  does not need to be substituted for this mark. Allow e or their value for e. Can be implied by tan=etan(60−). tan=etan(60−)= 1 31 3= 3  =6.12 A1 AWRT 6.12 (6.1155035…) or 6.11 WWW.  7 4 28  4 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/

4(a) © Cambridge University Press & Assessment 2026 Page 14 of 22 v 1 = 11 23 e u and v 2 = 51 3 u or v s i n 11 23 e u  = and v c o s 51 3 u  = B1 v 1 is the velocity component of P perpendicular to the barrier after impact and v is velocity component 2 of P parallel to the barrier after impact. May see LHS of v 1 equation as s i n  v or c o s  v and similar for v 2 , but must have opposite trigonometric ratios. Allow s i n s i n ( 6 7 )  v = e u  or better and c o s c o s ( 6 7 )  v = u  or better. 12 m v 2 = 41 3  12 m u 2   v 2 = 41 3 u 2  *B1 Correct equation relating kinetic energies. This mark can be implied by seeing v 2 = 41 3 u 2 . ( 11 23 e u ) 2 + ( 51 3 u ) 2 = 41 3 u 2 DM1 Correct use of 2 s i n 2 2 c o s 2 2   v + v = v to form an equation in e (and possibly u) only. Allow minor slips, for example, e rather than e 2 but other terms must have been squared. One component must be either 11 23 e u or 51 3 e u and the other must be either 51 3 u or 11 23 u (or using sin(67), cos(67), or better). Allow consistent sin/cos mix only. B1 B1 DM1 can be implied by (12e)2 +(5)2 = 4. 13 13 13 e=1 3 ONLY A1 AWRT 0.433 or better (0.4330127…). 4 4 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/ 4(b) © Cambridge University Press & Assessment 2026 Page 15 of 22 t a n 14 3 1 25 t a n 35 3 1 .0 3 9 2 3 ...   =    = =  *B1FT FT their e provided 0  e  1 to find an expression for tan or . If correct, may see an angle of 46.102… (which may be seen on a diagram). Or for c o s 52 6 1 3  = or s i n 32 6 3 9  = . t a n ( 6 0 ) 0 .2 4 7 ...  − = DB1 Follow through their t a n  or  (the angle between P and the first barrier after the first collision). If exact and correct, may see tan(60−)=1 3. 7 OR for finding the angle between the second barrier and P before the second collision – if correct, this angle is 13.897… May be seen on a diagram. s i n s i n ( 6 0 )   w = e v − c o s c o s ( 6 0 )   w = v − M1 Allow, for example, w 1 for s i n  w and w 2 for c o s  w – both equations must be correct. Their value for  does not need to be substituted for this mark. Allow e or their value for e. Can be implied by t a n t a n ( 6 0 )   = e − . t a n e t a n ( 6 0 ) 17 3 14 3 32 8 6 .1 2     = − =  =   =  A1 AWRT 6.12 (6.1155035…) or 6.11 WWW. 4 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/

5(a) © Cambridge University Press & Assessment 2026 Page 15 of 20 35 u A = 45 v A *B1 Velocity perpendicular to line of centres. Substitute values of s i n  and sin. OE, for example, 3 u A = 4 v A . 1mu 2 −1m(3u )2  DB1 Calculate the percentage loss in KE of A. 2 A 2 4 A 100=43.75(%) OE, for example,  1mu 2   2 A  1 74 5 . Condone 43.8. 2 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/ 5(b) © Cambridge University Press & Assessment 2026 Page 16 of 20 72 5 u B = 35 v B  7 u B = 1 5 v B  B1 Velocity perpendicular to line of centres. Values for trigonometric functions must be substituted. mu cos−mu cos=−mv cos+mv cos M1 Dimensionally correct momentum equation (m required). A B A B Allow sin/cos mix only. leading to Values for trigonometric functions must be substituted. m (4u ) −m (24u ) =−mv 3+mv 4. 5 A 25 B A 5 B 5 Allow 37 or better (36.8698…) for , . Allow 16 or better (16.2602…) for . Allow 53 or better (53.131…) for . OE, for example, m (4u ) −m (24u ) =−m (3u ) 3+m (7 u ) 4 . 5 A 25 B 4 A 5 15 B 5 May be implied by 1 5 u A = 1 6 u B , 7 v A = 1 2 v B . ( c o s c o s ) c o s c o s     e u A + u B = v A + v B M1 Use Newton’s experimental law to form an equation. Allow sin/cos mix only. leading to Values for trigonometric functions must be substituted. e (4u +24u ) =3v +4v 5 A 25 B 5 A 5 B Allow 37 or better (36.8698…) for , . Allow 16 or better (16.2602…) for . Allow 53 or better (53.131…) for . Signs for approach velocities must be consistent with those for velocities before the collision in momentum equation. OE, for example, e (4u +24u ) = 9 u +28u . 5 A 25 B 20 A 75 B e= 8 =0.471 A1 Substitute expressions for 17 u A , v A , u B , v B and simplify. e = − 8 8 2 6 3  = − 0 .3 3 5  from correct working B1M1M1A0. 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/

4(a) Case 1: © Cambridge University Press & Assessment 2026 Page 13 of 25 m u = k m v A − m v B 0 .2 u = v A + v B Case 2: m u = k m v A + m v B 0 .2 u = v A − v B M1 PCLM: must include m (but not g). NEL: must be 0.2 multiplied by u. RHS of one equation must be the sum of v A and v B and the RHS of the other equation must be a difference of v and v . For example, allow for A B case 2: 0 .2 u = v B − v A for M1. Case 1: v B =  0 .2 k k + − 1 1  u or v B =  k − 6 5  v A or 6v =(k−5)v B A Case 2: v B =  1 − k 0 + .2 1 k  u or v B =  5 − 6 k  v A or 6 v B = ( 5 − k ) v A A1 Any equivalent (possibly unsimplified) correct expression for v B in terms of u or a correct equation in v and A v B (allow with m present but not u). For reference, v A = 5 ( k 6 + 1 ) u . Case 1: 0 .2 k − 1  0  k  5 Case 2: 1 − 0 .2 k  0  k  5 A1 AG WWW, set correct expression for v B  0 (case 1) or v B  0 (case 2) followed by k > 5. Accept (with equivalents for case 2), for example,  0 .2 k k + − 1 1  u  0  k  5 or 0 .2 k k + − 1 1  0  k  5 , OE with v B in terms of v A , for example, ( v B = )  k − 6 5  v A  0  k  5 . Award the final two A1 marks for v A = 1 1 .2 + u k  1 1 .2 + u k − 0 .2 u  0  1 + k  6 so k  5 with the first A1 mark for ( v B = ) 1 1 .2 + u k − 0 .2 u . A0 for any errors seen. 3 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/ 4(b) © Cambridge University Press & Assessment 2026 Page 14 of 25 w B = 0 .8 v B  = 0 .8  0 .2 k k + − 1 1 u  B1 Multiply their expression for v B (in any form) by  0.8 or for quote w B =  0 .8 v B provided it is clear. d 2d d + = v w v B B A    0 .2 k d k + − 1 1  u +  0 .8   2 d 0 .2 k k + − 1 1  u  =  1 k d .2 + 1  u  M1 d 2d d For correctly stating or applying + = v w v B B A with their expressions. Must be an equation in terms of speeds and distances only (not time). For reference, if correct, then may see  0 .2 k 1 k + − 1 1  +  0 .8  2 0 .2 k k + − 1 1  = 1 k 1 .2 + 1 OE or 0 .2 1 k − 1 + 0 .8  ( 2 0 .2 k − 1 ) = 5 6 . Candidates may derive the result that v B = 3 .5 v A (possibly leading to k 6 − 5 = 2 7 ) as part of their working. k = 2 6 A1 WWW 3 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/