Cambridge A Level Mathematics 9709 — 2025 Oct/Nov Paper 4 · Variant 2

2 A railway locomotive of mass 240 000 kg is towing a coach of mass 36 000 kg down a hill inclined at an angle of sin -1 0.04 to the horizontal. The driving force produced by the locomotive is 450 000 N and there are resistances to motion of 120 000 N on the locomotive and 15 000 N on the coach. The coupling between the locomotive and the coach is light, rigid and parallel to the hill. Find the acceleration of the locomotive and the tension in the coupling. [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 Attempt at Newton’s second law at least once *M1 Correct number of terms, allow sign errors, allow sin/cos mix, allow g missing. 450000 + 240000 g  0.04 − 120000 − T = 240000a A2 A1 for any equation correct. A2 for any 2 equations correct.  450000 + 96000 − 120000 − T = 240000a    Allow sin2.3 or better for 0.04.  426000 − T = 240000a  A candidate may be using theira in the either of the T + 36000 g  0.04 − 15000 = 36000a first 2 equations if they have used the 3rd equation to  T + 14400 − 15000 = 36000a  find a .    T − 600 = 36000a  450000 + ( 240000 + 36000 ) g  0.04 − 120000 − 15000 = ( 240000 + 36000 ) a  450000 + 110400 − 120000 − 15000 = 276000a     425400 = 276000 a  Solving for a or T, from equation(s) with the correct number of relevant terms DM1 Allow g missing, terms should be components where necessary. Must get to a = or T = Acceleration = 1.54 m s-2 and Tension = 56100 N A1 709 Allow a = , a = 1.54130 , T = 56086.9 460 Allow with use of sin2.3. If using a = 1.54 , then T = 56400 from locomotive equation or T = 56040 from coach equation. If using a = 1.541 , then T = 56160 from locomotive equation or T = 56076 from coach equation. 5

More questions on Kinematics of motion in a straight line

4 A particle P of mass 0.1 kg is projected vertically upwards with speed 30 m s -1 from horizontal ground. At the same instant a particle Q of mass 0.4 kg is projected vertically upwards with speed 10 m s -1 from a height of 15 m above the ground. P and Q move in the same vertical line. (a) Find the height above the ground at which P and Q collide. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ When P and Q collide, they coalesce. (b) Find the speed of the combined particle at the instant that it reaches the ground. [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 4(a) 1 2 *M1 1 2 [Height gained by P ( Ps ) =]30t − gt For use of s = ut + at at least once with a =  g and 2 2 1 2 u = 30 or u = 10 . [Height lost by Q = ]10t − gt ( sQ ) Allow this M1 only if using a found value of t . 2 1 2 1 2 DM1 For use of s P = sQ  15 with Ps and s Q of the correct Meet when 30t − gt = 10t − gt + 15 2 2 form which would lead to a linear equation in t (so in 1 2 the expressions for Ps and s Q the signs of the gt 2 terms must be the same). 3 A1 OE. t = or 0.75 CWO. 4 315 A1 Allow 19.7 or better. Height = 19.6875 m or m CWO. 16 DO NOT ISW. 4 4(b)  3   3  *B1FT For either expression for the speed of P or Q before 30 − 10 −  =  2.5  v P =    g  g   = 22.5  vQ =    3 4 4     impact FT theirt = and/or FT their height. 4 2 315 OR  vP =   30 − 2 g  19.6875  = 22.5 = = 19.6875 . 16  19.6875 − 15 ) = 2.5  vQ =   10 2 − 2 g  ( The 2 M marks in 4(a) must have been awarded. 0.1 22.5 + 0.4  2.5 = ( 0.1 + 0.4 )  vPQ DM1 Use of conservation of momentum; correct number of non-zero terms, allow sign errors. 22.5 and 2.5 coming from a correct method. Do not allow with v P = 30 or with vQ =  10. If using mg rather than m , then do not allow subsequent A marks. v PQ = 6.5 A1 SOI, allow to 2sf. v 2 = their 6.5 2 + 2 g  their19.6875 DM1 Dependent on previous M1 and B1. Complete method to get an equation in speed or OR 2 (speed)2. 0 = their 6.5 + 2 ( − g ) s =s 2.1125 , May see t = 2.738061302 . height above ground = 2.1125 + 19.6875 = 21.8 hence v 2 = 0 2 + 2 g  21.8 Speed = 20.9 m s-1 or 2 109 m s-1 A1 20.88061303 Use of g in momentum equation can be awarded B1M1A0M1A0, 3 marks max. 5

More questions on Momentum

5 20 N 35° 5 kg 10° A block of mass 5 kg is being pulled straight down a line of greatest slope of a rough plane by a force of magnitude 20 N. The plane is inclined at an angle of 10° to the horizontal and the 20 N force acts at an angle of 35° above the line of greatest slope of the plane (see diagram). The coefficient of friction between the block and the plane is 0.4. The speed of the block when it passes a point O is 2 m s -1. Find the speed of the block when it has moved 3 m down the plane from O. [6] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................. ............................................................................................................................

Mark scheme: 5 For resolving in any direction to form an equation *M1 Correct number of relevant dimensionally correct terms; allow sin/cos mix; allow sign errors. R = 5 g cos10 − 20sin35  = 37.76885892  A1 OE. For reference F = 15.10754357 20cos35 + 5 g sin10 − F = 5a  25.06544977 − F = 5 a  A1 OE, allow with their possibly incorrect F . Allow F − 20cos35 − 5 g sin10 = 5a . For use of F = 0.4 R to get an equation in a only DM1 From both dimensionally correct equations with the correct number of relevant terms. For reference a = 1.99 1.99158124  . v 2 = 22 + 2  ( their 1.99 )  3 DM1 Dependent on both previous M marks. M0 if their calculated a is negative from using N2L down the plane. M0 if their calculated a is positive from using N2L up the plane. Their calculated 1.99 and 3 must be same sign. For use of v 2 = u 2 + 2 as or any complete method to get an equation in v or 2v , where v 2  0 . Speed = 3.99 m s−1 A1 3.993680939 5 Alternative for Q5 using energy For resolving perpendicular to the plane to form an equation for normal *M1 Correct number of relevant dimensionally correct reaction terms; allow sin/cos mix; allow sign errors. R = 5 g cos10 − 20sin35  = 37.76885892  A1 OE. 1 2 1 2 *M1 Attempt at Work-energy equation. Correct number of 20cos35 +3 5 g sin10 +3 5 2 − F =3 5 v relevant terms, dimensionally correct, allow sin/cos 2 2 2 mix, allow sign errors, allow their F .  49.1491+ 26.0472+ 10 − 3 F = 2.5v    A1 Allow their possibly incorrect F . Note: 60cos10 comes from 3  0.4 ( 5 g cos10 ) . For use of F = 0.4 R to get an equation in v only DM1 Dependent on both previous M marks. Attempt at Work-energy equation. Correct number of relevant terms, dimensionally correct. Speed = 3.99 m s−1 A1 3.993680939 6

More questions on Energy, work and power

6 A particle starts from rest at a point O. The acceleration of the particle at time t s after leaving O is a m s -2, where - 21 a = 2 ( t + 1) - 1 for t H 0 . Find the distance that the particle travels from O until the time at which its acceleration is zero. [8] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 6  − 1 1  B1 − 1 Acceleration = 0 when 2 or claim that 2 = 2 t + 1) t + 1) 2 −=1 0  ( t + 1)  2 (  t = 3 CWO. Those who use 2 (   1 1 − − 2 − 1 score 0 marks. 2 ( t + 1) 2 − 1 is ( 2t + 2 ) 1 *M1 1 − For attempt at integration of ( a = ) 2 ( t + 1) 2 − 1 v = at is M0. Of the form a ( t + 1) 2 − t  + c  , where a  2 . 2 −+1 1 1 A1 Allow un-simplified.  v =  ( t + 1) 2 − t  + c  = 4 ( t + 1) 2 − t  + c  1 − + 1 2  1  DM1 For attempt to find c using v = 0 at t = 0 . 2 −−t 4 v = 0 at t = 0  c = −4  v = 4 ( t + 1)   Must get a value for c . M0 for stating v = 0 at t = 0    c = 0 with no working. Attempt to integrate theirv, which has come from integration DM1 Increase power by 1 and change of coefficient on the same term from a 3 term expression in t to get an 3 2 + qt + rt , where expression of the form p ( t + 1) 2 p  0 , q  0 and r  0 . s = vt is M0. Dependent on previous 2 M marks. 4 1 +1 1 1+1 8 3 1 2 A1 Allow un-simplified.  s =  ( t + 1) 2 − t − 4t  + k  = ( t + 1) 2 − t − 4t  + k  Condone using +c again. 1 1 + 1 3 2 + 1 2  8 32 1 2   8  DM1 Use of limits 0 and their positive 3 correctly, their 3  4 − =   3 −4 3  must have come from considering given a = 0 . −  − 0 − 0    2  3   3  Dependent on all 3 previous M marks. 8 OR evaluate k from using s = 0 when t = 0 AND substitute t = their 3 If evaluating k , then k = − if correct. 3 6 A1 SC, If no integrating v seen then SCB1 for 2.17 so Distance = 13m or 2.17m 6 B1M1A1M1SCB1 5 marks max 8

More questions on Kinematics of motion in a straight line

7 30 N 20 N P P 30° 20° a° 45° 15 N S N 25 N Fig. 7.1 Fig. 7.2 Four coplanar forces of magnitudes 20 N, 30 N, 15 N and 25 N act at a point P in the directions shown in Fig. 7.1. The forces act in a vertical plane. The resultant of these forces has magnitude S N and acts at an angle a° below the horizontal as shown in Fig. 7.2. (a) Find the value of S and the value of a. [6] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ A small ring of mass 0.6 kg is threaded on a rough straight horizontal wire. The four forces shown in Fig. 7.1 act on the ring and are in the same vertical plane as the wire. The ring starts from rest and takes 3 s to travel a distance of 2 m along the wire. (b) Find the coefficient of friction between the ring and the wire. [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................

Mark scheme: 7(a) For attempt to resolve in any direction to form an expression or equation *M1 Correct number of terms; allow sign errors and sin/cos mix ONLY.  X = ( 30cos30 + 15cos45 − 20cos20 ) A1 OE. OR  S cos= ( 30cos30 + 15cos45 − 20cos20 ) 17.79351142 Y = ( 25 + 15sin 45 − 20sin 20 − 30sin30 ) A1 OE. OR  S sin= ( 25 + 15sin 45 − 20sin 20 − 30sin30 ) 13.76619885 2 2 DM1 For attempt to find force from equations with the S = ( their 17.7935) + ( their 13.76661) correct number of relevant terms (only sign errors and their 17.7935  their 13.76661 sin/cos mix allowed). OR S = OR S = cos ( their 37.7278 ) sin ( their 37.7278) −1  their 13.7661  DM1 Allow reciprocal in tan−1 only. = tan    their 17.7935  For attempt to find angle from equations with the correct number of relevant terms (only sign errors and −1  their 17.7935   −1  their 13.7661  OR = cos   OR = sin   sin/cos mix allowed).  their 22.4970    their 22.4970   [S =] 22.5 = 37.7 A1 S = 22.497050406 = 37.72782933 Condone = 37.8 . Allow S = −22.5 becoming 22.5 with no explanation. Allow = −37.7 becoming 37.7 with no explanation. A0 for = −37.7 ONLY, A0 for S = −22.5 ONLY 6 7(b) ( their 17.7935−) F = 0.6 a *M1 Using N2L with correct number of dimensionally correct terms, allow sign errors, allow sin/cos mix. OR 30cos30 + 15cos45 − 20cos20 − F = 0.6a their 17.7935 must have come from horizontal component from 7(a) with only sign errors and sin/cos mix allowed. May be using their F and/or theira which may be incorrect. OR ( their S ) cos ( their −) F = 0.6  a so in this case OR ( their S ) cos ( their −) F = 0.6 a 17.8025 − F = 0.6 a  dependent on all 3 M marks in part (a) and must be using cosine. 1 2  4  *M1 1 2 2 = 0 + a  3 a = Use of s = ut + at or other complete method to get   2  9  2 an equation in a , using s = 2, u = 0 and t = 3 . R = their (13.7661+) 0.6 g  = 19.7661 *B1FT their 13.7661 must have come from vertical component from 7(a) with only sign errors and sin/cos OR R = 25 + 15sin45 − 20sin20 − 30sin30 + 0.6 g mix allowed. OR R = ( their S ) sin ( their ) + 0.6 g so in this case OR R = ( their S ) sin ( their ) + 0.6 g  R = 13.759 + 0.6 g  dependent on all 3 M marks in 7(a) and must be using sine  4  DM1 Dependent on all previous marks. Use of F = R to ( their 17.7935 ) −  19.7661 = 0.6  their   9  get an equation in  only. = 0.887 A1 0.886711 Condone 0.886. Condone 0.888. 7(b) Alternative scheme for using energy 1 2 *M1 Work energy equation with correct number of  0.6  v = ( their 17.7935 ) −2 F  2 dimensionally correct terms, allow sign errors, allow 2 sin/cos mix. their 17.7935 must have come from horizontal 1 2 OR  0.6  v = ( 30cos30 + 15cos45 − 20cos20 )  2 − F  2 component from 7(a) with only sign errors and sin/cos 2 mix allowed. May be using their F and/or theirv which may be incorrect. 1 2  0.6  v = ( their S ) cos ( their )  2 − F  2 , so in 2 1 2 OR  0.6  v = ( their S ) cos ( their )  2 − F  2 this case dependent on all 3 M marks in 7(a) and must 2 be using cosine. 1  4  *M1 1 v = Use of s = ( u + v ) t or other complete method to get 2 = ( 0 + v )  3   2  3  2 an equation in v , using s = 2 , u = 0 and t = 3 . R = their (13.7661+) 0.6 g  = 19.7661 *B1FT their 13.7661 must have come from vertical component from 7(a) with only sign errors and sin/cos mix OR R = 25 + 15sin45 − 20sin20 − 30sin30 + 0.6 g allowed. Or R = ( their S ) sin ( their ) + 0.6 g so in this case OR R = ( their S ) sin ( their ) + 0.6 g  R = 13.759 + 0.6 g  dependent on all M marks in 7(a) and must be using sine. 2 DM1 Dependent on all previous marks. Use of F = R to 1  4   0.6  their 17.7935 ) −2 19.7661   2  their  = ( get an equation in  only. 2  3  = 0.887 A1 0.886711 Condone 0.886. Condone 0.888. 5

More questions on Forces and equilibrium