Mathematics 9709/42 February/March 2026: question paper and mark scheme

Mark scheme

1 [ R = 200 g ] F = 0.8 × 200 g [ = 1600] *B1 4000 *B1 For P = Fv . D= OE seen at any point in working. v 4000 DM1 Using Newton’s second law with a = 0 . = 0.8 × 200g Dependent on both previous B marks. v v = 2.5 A1 OE. 4

Mark scheme

2(a) [Velocity after 10 s = ] 0.5 × 10 [ = 5] B1 v 2 = their 52 + 2 × 1.5 × 13 M1 Use of suvat equation(s) to get an expression for v 2 or v , using their 5 ≠ 0 . v =8 A1 3 2(b)(i) 1 1 B1FT FT their value of v. Distance for first stage s2 = 0 + × × 102 [ = 25] 2 2 OR Distance for third stage s2 = 5 × their 8 [ = 40] [ 25 +13 + 40 + 28] = 106 m B1FT FT their value of v. 5v + 66 . 2 Trace ID: PC-5E761E6444 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. PapaCambridge Downloaded from PapaCambridge - https://papacambridge.com/ 2(b)(ii) their 8 − their 5 B1FT FT their value of v and their 5 ≠ 0 Time for second stage t = [ t = 2] Must get an expression for t in either 1.5 OR second or fourth stage of motion. 28 8 Time for 4th stage t = ×2 [ t = 7] May see deceleration of 7 in working for their 8 + 0 final phase. 106 53 B1 Allow AWRT 4.42. 10 + 5 + 2 + 7 = 12 m s –1 106 Allow . 24 4.416666…. 2 Trace ID: PC-5E761E6444 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. PapaCambridge Downloaded from PapaCambridge - https://papacambridge.com/

Mark scheme

3 Attempt to resolve in any direction to form an equation *M1 With correct number of terms. Allow sin/cos mix. Allow sign errors. 20sin θ − F − 2 F sin 30 = 0 [ 20sin θ − 2 F = 0] A1 OE. CWO. 20cosθ − 2 F cos30 = 0 20cos θ − 3F = 0 A1 OE. For attempt to find θ DM1 F + 2F sin 30 −1 2 F + 2F sin 30 θ = tan −1 = tan . −1 2 θ = tan −1 = tan 2F cos30 3 2F cos30 3 From equations with the correct number their F + 2 × ( their F ) sin 30 of relevant terms. OR θ = sin −1 20 2 × ( their F ) cos30 OR θ = cos−1 20 For attempt to find F DM1 From equations with the correct number 20sin ( their θ ) of relevant terms. F= 1 + 2cos30 20cos ( their θ ) OR F = 2cos30 202 OR F = (1 + 2sin 30 )2 + ( 2cos30 )2 7 A1 OE θ = 49.1066... AWRT 49.1. θ = 49.1 F = 20 = 7.56 F = 7.55928… AWRT 7.56. 7 Trace ID: PC-5E761E6444 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. PapaCambridge Downloaded from PapaCambridge - https://papacambridge.com/ 3 Alternative using triangle of forces For attempt at correct cosine rule (M1) Using any numerical angle other than 90°. 202 = F 2 + ( 2F ) − 2 × F × 2F cos120 (A1) 2 20 7 (A1) OE F = 7.55928… AWRT 7.56. F= or F = 7.56 7 For attempt at correct sine rule (M1) 20 F 2F (A1FT) FT their F only. Any two of the three = = sin120 sin (θ − 30 ) sin ( 90 − θ ) needed. θ = 49.1 (A1) θ = 49.1066... AWRT 49.1 6 Trace ID: PC-5E761E6444 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. PapaCambridge Downloaded from PapaCambridge - https://papacambridge.com/

Mark scheme

4 1 B1 For either correct. KE A = × 1600 × 102 [ = 80000] 1 2 Do not allow ×1600 × (12 − 10 ) . 2 1 2 KE B = ×1600 × 122 [ = 115200] NB Difference = 35200. 2 PEchange = ±1600g ( 250 − 200) [ = ±800000] B1 Allow 50 ±1600 g × 2000 × sin1.4 ( 32543738 ) . OR PEchange = ±1600 g × × 2000 [ = ±800000] 2000 WD against resistance = ±150 × 2000 [ = ±300000] B1 1 1 M1 Attempt at work energy equation with Work done by braking force = ×1600 ×102 − ×1600 ×122 + 1600 g ( 250 − 200 ) − 150 × 2000 correct number of relevant terms; 2 2 dimensionally correct. 1 1 50 Allow using 2 rather than 2000. OR Work done by braking force = ×1600 × 102 − × 1600 × 122 + 1600 g × × 2000 − 150 × 2000 Allow sign errors, but must be a 2 2 2000 difference in KE terms. Allow with BF × 2000 for work done by braking force. Allow 1600 g × 2000 × sin 1.4 ( 32543738) . 1 ×1600 × (12 − 10 ) . 2 M0 if using 2 Work done by braking force = 464 800 J A1 Allow 465 000 J. DO NOT ISW. Allow 463 000 from use of angle of 1.43 (463379.7522) and 464000 from use of angle of 1.432 (464496.4145). 5 Trace ID: PC-5E761E6444 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. PapaCambridge Downloaded from PapaCambridge - https://papacambridge.com/ 4 Special case for use of constant acceleration and N2L 11 M1 Must have a correct equation in a and a 122 = 102 + 2 × 2000a a = = 0.011 N2L equation with correct number of 1000 terms, but allow sign errors and sin/cos 50 AND 1600 g × − 150 − BF = 1600 × ( their a ) [ BF = 232.4] mix. Must be an equation in BF only. 2000 Allow using 2 rather than 2000 in finding a. Work done by braking force = [ 232.4 × 2000] 464 800 J A1 Allow 465 000 J. 2 Trace ID: PC-5E761E6444 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. PapaCambridge Downloaded from PapaCambridge - https://papacambridge.com/

Mark scheme

5(a) 3 B1 RP = 0.3g cos30 = 2 3 = 2.598076211 or RQ = 0.2 g cos30 = 3 = 1.732050808 5 or RP +Q = ( 0.2 + 0.3) g cos30 = 2 3 = 4.330127019 Attempt at Newton’s second law for either particle or the system M1 Must have correct number of terms. Allow sign errors. Allow sin/cos mix. Allow g missing. Allow with their F or just F . 12 − T − 0.2 g sin30 − 0.4 × 0.2 g cos30 = 0.2a A2 For any two correct equations. 2 A1 for any one equation. 12 − T − 1 − 5 3 = 0.2a Allow with their a . Allow with theirT . 3 Or T − 0.3g sin 30 − 0.4 × 0.3g cos30 = 0.3a T − 1.5 − 3 = 0.3a 5 Or 12 − ( 0.2 + 0.3) g sin 30 − 0.4 × ( 0.2 + 0.3) g cos30 = ( 0.2 + 0.3) a 12 − 2.5 − 3 = 0.5a Attempt to solve for a or T M1 Must have correct number of terms. Allow sign errors. Allow sin/cos mix. Allow g missing. a = 15.5 m s–2 or a = 19 − 2 3 AND T = 7.2 N A1 a = 15.5358983 . AWRT 15.5. Allow T = 7.19 from use of a = 15.5 in 0.3 kg N2L equation, CWO. Allow T = 7.21 from use of a = 15.5 in 0.2 kg N2L equation, CWO. 6 Trace ID: PC-5E761E6444 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. PapaCambridge Downloaded from PapaCambridge - https://papacambridge.com/ 5(b) −0.3g sin30 − 0.4 × 0.3g cos30 = 0.3a [a = −8.46410 or a = −5 − 2 3] *M1 For use of Newton’s second law. Must have correct number of terms. Allow sign errors. Allow sin/cos mix. Allow with m or mass omitted. M0 for use of 0.2 or 0.5. 0 = 2 + ( their − 8.46) t or 0 = −2 + (their −8.46 ) t DM1 For getting an equation in t only using their a ≠ 10 . Must lead to a positive value of t , so the sign of u and the sign of a must be different. Or other complete method. 10 − 4 3 A1 t = 0.23629 AWRT 0.236. t = 0.236 or t = 13 10 − 4 3 BEWARE s = = 0.236 . 13 ALTERNATIVE USING ENERGY 1 (*M1) Dimensionally correct work energy × 0.3 × 22 = 0.3g sin 30 × h + 0.4 × 0.3g cos30 × h equation with 3 relevant terms. Allow 2 sign errors, allow sin/cos mix. 10 − 4 3 → h = = 0.236 Allow with m or mass omitted. 13 M0 for use of 0.2 or 0.5. 1 (DM1) For getting an equation in t only using their 0.236 = ( 2 + 0) t their h . 2 Must lead to a positive value of t . Or other complete method. 10 − 4 3 (A1) t = 0.23629 AWRT 0.236. t = 0.236 or t = 13 10 − 4 3 BEWARE s = = 0.236 . 13 3 Trace ID: PC-5E761E6444 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. PapaCambridge Downloaded from PapaCambridge - https://papacambridge.com/

Mark scheme

6(a)(i) 4m × 6.5 [ +0] = 4m × 2 + 3mv M1 For attempt at use of conservation of momentum. 3 non-zero terms, allow sign errors. Do not allow with m missing. v=6 A1 Must see at least one correct intermediate step eg 3mv =, 3v = , but if v = then must be an expression on RHS. AG CWO. If using mg in momentum equation, then M1A0 max. 2 6(a)(ii) 1 1 1 M1 Attempt at KE equation. Correct number × 4m × 6.52 − × 4m × 22 − × 3m × 62 = 45 of terms. Allow ±45 but do not allow any 2 2 2 other sign errors. [84.5m − 8m − 54m = 45 22.5m = 45] m=2 A1 Do not allow m = −2 becoming m = 2 2 Trace ID: PC-5E761E6444 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. PapaCambridge Downloaded from PapaCambridge - https://papacambridge.com/ 6(b) Q hits wall after 1.5 s *B1 9 OE, e.g. . 6 Distance travelled by P up to time at which Q hits wall s [ = 2 × 1.5] = 3 m *B1FT FT their 1.5 s. After Q hits wall sP = ±2t AND sQ = ±0.5t *B1 Allow relative speed = ± ( 2 + 0.5 ) . 2t + 0.5t = 9 − their 3 [⟹ t = 2.4 s] DM1 For equating distances to get an equation that leads to a positive t , with 2t and 0.5t having the same sign if on the same side of the equation. Dependent on all previous B marks Distance from wall [ = 0.5 × 2.4] = 1.2 m A1 OE. Alternative method using equal times Q hits wall after 1.5 s (*B1) 9 OE, e.g. . 6 Distance travelled by P up to time at which Q hits wall s [ = 2 × 1.5] = 3 m (*B1FT) FT their 1.5 s. x 6− x (*B1) tP = ± AND tQ = ± 0.5 2 x 6− x (DM1) For equating times to get an equation that = 0.5 2 x 6− x leads to a positive x , with and 0.5 2 6− y y having the same sign if on opposite sides OR = of the equation. 0.5 2 Dependent on all previous B marks. Trace ID: PC-5E761E6444 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. PapaCambridge Downloaded from PapaCambridge - https://papacambridge.com/ 6(b) Distance from wall = 1.2 m (A1) OE. 5

Mark scheme

7(a) Attempt to differentiate M1 − 1 Of the form 1 − kt where k ≠ ±7 . 2 Use of v = at is M0. Expect 1 1 dv 1 −1 7 − =1− 7 × t 2 =1− t 2 . dt 2 2 At t = 4 acceleration = − 0.75 ms–2 A1 Do not ISW. OE. Must be negative. CWO. SC for no differentiation seen At t = 4 acceleration = − 0.75 ms–2 (B1) Do not ISW. OE. Must be negative. 2 Trace ID: PC-5E761E6444 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. PapaCambridge Downloaded from PapaCambridge - https://papacambridge.com/ 7(b) 1 B1 For setting given v expression equal to t − 7t 2 + 12 = 0 zero. May be implied, for example, by at least one correct value of t . t =9 B1 t = 16 B1 If other t values given on the range, then B1B1B0 max. 3 Trace ID: PC-5E761E6444 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. PapaCambridge Downloaded from PapaCambridge - https://papacambridge.com/ 7(c) Attempt to integrate given v expression once only *M1 The power of t must increase by 1 in at least 1 term, with a change of coefficient in the same term. Use of s = vt is M0 1 1+1 1 7 2 +1 1 14 3 A1 Allow un-simplified. ( s =) t − 1 t + 12t ( +c ) = t 2 − t 2 + 12t ( +c ) 1+1 +1 2 3 2 1 2 14 2 3 DM1 Using their t value(s) where both are Use limits 0, their 9 or their 9, their 16 or their 16, 25 correctly in their t − t + 12t 0 < t < 25 , using limits correctly in their 3 2 3 term s . Allow un-simplified. 1 2 14 2 1 3 1 DM1 For all three, using their t values where 0 to 9: × 9 − × 9 + 12 × 9 − 0 = 22 2 − 0 = 22 2 both are 0 < t < 25 , using limits correctly. 2 3 Allow un-simplified. 1 14 3 1 14 3 1 1 1 7 9 to 16: ± × 162 − × 16 2 + 12 × 16 − × 92 − × 9 2 + 12 × 9 = ± 21 − 22 = μ1 = ± 2 3 2 3 3 2 6 6 1 14 3 1 14 3 1 1 5 47 16 to 25: × 252 − × 25 2 + 12 × 25 − × 162 − × 16 2 + 12 × 16 = 29 − 21 = 7 = 2 3 2 3 6 3 6 6 1 A1 OE. AWRT 31.5. Total distance = 31 m 2 SC for no integration seen 7 47 (B1) OE. 21.5 or ± or 6 6 Trace ID: PC-5E761E6444 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. PapaCambridge Downloaded from PapaCambridge - https://papacambridge.com/ 7(c) 1 (B1) OE. AWRT 31.5. Total distance = 31 m 2 5 Trace ID: PC-5E761E6444 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. PapaCambridge Downloaded from PapaCambridge - https://papacambridge.com/