Cambridge A Level Mathematics 9709 — 2025 May/June Paper 4 · Variant 2

1 A crate is being pushed in a straight line along a horizontal surface by a force of magnitude 25 N inclined at 20° above the horizontal. The crate moves a distance of 12 m in 8 seconds with constant speed. (a) Find the constant speed of the crate. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the work done by the 25 N force. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) Find the power at which the 25 N force is working. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1(a) 1.5 m s–1 B1 3 12 OE e.g. , . 2 8 1 1(b) Work done = 25cos20  12 M1 For 25cos20  12 or 25sin 20  12 or 25cos70  12 or 25sin70  12 . 282 J A1 281.9077… 2 1(c) 35.2 W B1FT FT their (b) divided by 8 or FT 25cos20their(a) . 282 Allow 35.3 (from ). 8 Note: 25  1.5 is B0. Do not accept 35.2 kW. 1

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2 Two particles P and Q, of masses 0.2 kg and 0.1 kg respectively, are free to move in a straight line on a smooth horizontal plane. P is projected towards Q with speed 5 ms -1. At the same instant, Q is projected away from P with speed 2 ms -1. When P collides with Q, the particles coalesce. Find the kinetic energy lost during the collision. [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 0.2 +5 0.1=2 ( 0.2 + 0.1) v *M1 Attempt at conservation of momentum; correct number of terms but allow sign errors – use of mg scores M1A0. v = 4 A1  1 2 1 2 1 2  DM1 Allow sign errors only; correct number of terms; Loss in KE =   0.2  5 +  0.1 2 −  ( 0.2 + 0.1)  ( their 4 )   dimensionally correct.  2 2 2    =  ( 2.5 + 0.2 − 2.4 )  [KE lost =] 0.3 J A1 Allow −0.3 . Use of mg in momentum scores max M1A0M1A0. 4

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3 2.6 m 35 N 2.4 m P A particle P of mass m kg is attached to one end of a light inextensible string of length 2.6 m. The other end of the string is attached to a fixed point on a horizontal ceiling, and the string is taut. The particle is held in equilibrium by a force of magnitude 35 N, acting in a vertical plane which is perpendicular to the ceiling and contains the string. The force acts in a direction perpendicular to the string (see diagram). The tension in the string is T N and the vertical distance of P from the ceiling is 2.4 m. Find, in either order, the value of m and the value of T. [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 3 12 5 cos or sin (where  is the angle 13 13 between the string and vertical). Candidates may use the complementary angle to  . Resolving in any direction to form an equation M1 Resolving either vertically/horizontally or parallel/perpendicular to the string. Correct number of terms, allow sign errors, allow sin/cos mix; forces that need resolving should be resolved. M0 for using an angle other than 23 (or better) or 67 (or better). 12 5 A1 Allow T cos+ 35sin= mg with = 23 or T  + 35  = mg 13 13 better (22.61986495) substituted OR T = mg cos with = 23 or better  0.9230 T + 35  0.3846 = mg  (22.61986495) substituted.   Allow with their T or m if already found.  0.9230 T + 13.461= mg  12 OR T = mg  T = mg  0.9230 13 5 12 A1 Allow T sin= 35cos with = 23 or better T  = 35  (22.61986495) substituted 13 13 OR mg sin= 35 with = 23 or better T  0.3846 = 35  0.9230   (22.61986495) substituted.    T  0.3846 = 32.3076   5 OR mg  = 35  mg  0.3846 = 35 13 3 T = 84 and m = 9.1 A1 Allow T = 84.1 and m = 9.11 . Allow AWRT 84.0 for T and 9.10 for m . Special Case for use of Lami mg 35 T M1 Attempt at one pair but allow with 2 ‘correct’ = = angles but with the wrong force. sin90 sin (180 − 22.6 ) sin ( 90 + 22.6 ) A1 For one correct pair with = 23 or better (22.61986495). A1 For all three correct with = 23 or better (22.61986495). T = 84 and m = 9.1 A1 Allow T = 84.1 and m = 9.11 . Allow AWRT 84.0 for T and 9.10 for m . 4

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4 A car is travelling along a straight horizontal road. The car passes through a point A, on the road travelling at a speed of 15 ms -1 , and then accelerates uniformly at 0.4 ms -2 for 30 seconds. The car then moves at constant speed for 3T seconds, where T 1 30 . The car then decelerates uniformly at 0.2 ms -2 and after a further T seconds passes through a point B on the road. (a) On the given axes, sketch a velocity-time graph for the motion of the car between points A and B. [2] v (m s–1) t (s) O The distance from A to B is 2750 m. (b) Find the value of T. 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The car continues its journey from B, decelerating uniformly at 0.5 ms -2 until it comes to rest at a point C on the road. (c) Find the total distance from A to C. 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Mark scheme: 4(a) Correct three-line segments for v-t graph B1 First line segment with positive gradient starting on the positive vertical axis, second line segment horizontal (parallel to t -axis) and third line segment with negative gradient, stopping before the horizontal axis. If the third line continues and touches the t -axis, there must be some indication that speed of the car at B is not 0. Values/expressions labelled on axes correctly B1 15 correct on vertical axis; 30, 30 + 3T and 30 + 4T correctly labelled on horizontal axis. 2 4(b) B1 Speed after 30 seconds = 15 + 0.4  30  = 27  Speed at B is their 27 + ( −0.2 ) T *M1 Use of v u at using their speed after 30 seconds for u, t = T and a 0.2 . Attempt at distance from A to B and equate to 2750 *M1 Using the combined area below the three line segments; using their 27. 1 1 A1 Correct (un-simplified) equation for T. (15 + 27 )  30 + 27  3T + ( 27 − 0.2T ) + 27  T = 2750 2 2 0.1T 2 − 108T + 2120 = 0  T = DM1 Re-arranging and attempting to solve their three- term quadratic equation in T. If method seen, must be using correct formula OR if factorising, two terms must be correct for their three-term quadratic when expanding brackets. If no method seen, must have at least one correct value for their three-term quadratic for this mark. T = 20 only A1 If T = 1060 also stated, then must be rejected. 6 4(c) *M1 Use of v = u + at using their speed after 30 Speed at B is ( their 27 ) + ( −0.2 )  ( their 20 ) = 23 seconds from (b), their T from (b) and a = 0.2 . Use of v 2 = u 2 + 2as with v = 0 and a = −0.5 Distance from B to C is s where 0 2 = ( their 23 ) 2 + 2 −( 0.5 )  s and attempt to DM1 and their speed at B for u (if correct s = 529). solve for s Total distance is [529 + 2750 =] 3279 m A1 Condone 3280 m. 3

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5 B 4 kg 5 kg C A 3 kg 30° One end of a light inextensible string is attached to a particle A of mass 3 kg. The other end of the string is attached to a particle B of mass 4 kg. Particle A is in contact with a rough plane inclined at 30° to the horizontal, and particle B is in contact with a smooth horizontal plane. A second light inextensible string is attached to B. The other end of this second string is attached to a particle C of mass 5 kg which hangs vertically. Both strings are taut and pass over small smooth pulleys that are fixed at the ends of the horizontal plane. The part of the string from A to the pulley is parallel to a line of greatest slope of the inclined plane, and A, B and C are in the same vertical plane (see diagram). The system is released from rest. In the subsequent motion, C moves vertically downwards with acceleration 2 ms -2 , and neither A nor B reach a pulley. (a) Find the tensions in each of the strings. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the coefficient of friction between A and the inclined plane. 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When the system has been in motion for 1.5 s, the string attached to A breaks. (c) Find the total distance that A travels up the plane from the instant that the system is released from rest to the instant that A comes to instantaneous rest. 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Mark scheme: 5(a) 5 g − TBC = 5  2 M1 Attempt at N2L for C – correct number of terms but allow sign errors (but must be using correct mass). TBC − TAB = 4  2 M1 Attempt at N2L for B – correct number of terms but allow sign errors (but must be using correct mass). Allow with their TBC . TBC = 40 N and TAB = 32 N A1 Both correct. 3 5(b) TAB − F − 3 g sin30 = 3  2 *M1 Attempt at N2L on A – correct number of terms but allow sign errors; allow sin/cos mix (but must be using correct mass). For reference: F = 11. R = 3g cos30 B1 Correct expression for normal contact force at A. their 32 −  3g cos30 − 3g sin30 = 3  2 and attempt to solve for  DM1 Use of F = R (where R is a component of weight) and their TAB to obtain an equation in  only and solve for . = 0.423 A1 11 3 . 45 4 5(c) 1 2 B1 Distance travelled by A in first 1.5 seconds is 2 1.5  = 2.25 2 When string breaks A is moving at a speed of 3 (m s–1) B1  26  *M1 Attempt at N2L for A – correct number of terms − F − 3 g sin30 = 3a a = −   but allow sign errors, and cos/sin mix.  3  2  27  DM1 Attempt at finding the distance travelled by A up s = 0 = 3 + 2  ( their a )  s   the plane after the string breaks using  52  2 2 v = u + 2as (or other complete method) with v = 0, u = 3 and their negative acceleration.  27  A1 36 Total distance travelled by A up the plane is 2.25 + = 2.77 m , 2.769230789 .    52  13 5

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7 A particle P of mass 3 kg is projected with a speed of 8 ms -1 up a line of greatest slope of a rough plane inclined at 30° to the horizontal. P is projected from a point A on the plane and comes to instantaneous rest at a point B on the plane. P then slides back down the plane. The coefficient of friction between P and the plane is 1 3 . 12 Using an energy method throughout, find the speed of P at the instant it returns to A. [6] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 7 Apply the work-energy principle for the motion from A to B to form an equation in *M1 Correct number of relevant terms, allow sign one variable only errors and sin/cos mix. Dimensionally correct, terms that need a component should have a component. A1 A1 for correct LHS. 1 2 3 3 8 − 3 gd sin30 =  3 g cos30  d (where d is the distance AB) 2 12 A1 A1 for correct RHS.  96 − 15d = 3.75d  For either d = 5.12 m or work done against friction is 19.2 J A1 Either value stated or clearly implied by later working. If using N2L then M0, but SCB1 only (see below). 1 2 1 2 2 DM1 Either consider the total work done against friction 1.5v = 96 − 38.4  3 v = 3 8 −2 ( their 19.2 )   from A to B and B to A or, consider motion from 2 2 B to A. In both cases must have the correct number of terms but allow sign errors and cos/sin mix. OR Dimensionally correct, terms that need a 1 2 2 component should have a component. 1.5v + 19.2 = 76.8  3 v + ( their 19.2 ) = 3 g  ( their 5.12 )  sin30   2 Dependent on previous M1 or previous SCB1. OR 1 2 3 3 v + ( their 5.12 )   3 g cos30 = 3 g  ( their 5.12 )  sin30 2 12 1.5v 2 + 19.2 = 76.8    7 v = 6.20 m s–1 ONLY A1 8 15 , 6.196773354  must be positive. 5 If using N2L then M0, but SCB1 only (see below). Allow 6.2 from CWO. A 3sf answer of 6.19 becoming 6.2 is A0. This mark is dependent on all previous 5 marks awarded or on the SCB1 and the previous M1. Special case for use of N2L for motion up the plane  1  *B1 3a =  3 g  3  cos30 + 3 g sin30 → a = 6.25    12  and then 02 = 82 + 2 ( −6.25 ) s →=s 5.12 Special case for use of N2L for motion down the plane 1 DB1 3a = 3 g sin30 − 3 g  3  cos30 → a = 3.75 12 and then v 2 = 20 + 2  3.75  5.12 →=v 6.20 6

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