Cambridge A Level Mathematics 9709 — 2025 May/June Paper 4 · Variant 5

1 A box of mass 25 kg is pulled 12 m up a rough plane inclined at an angle of 8° to the horizontal. The box moves up a line of greatest slope against a frictional force of 50 N. The force pulling the box is parallel to the line of greatest slope. The box starts from rest and has speed 2.4 m s–1 at the end of the 12 m. Find the work done by the pulling force. [4] ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ...........................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 PE = 25 g  12sin8  = 3000sin8 = 417.519  B1 If not seen separately allow B1 for  ( 25 g sin8  50 )  12 . 1 2 B1 If the candidate mistakenly thinks that the distance is KE =   25  2.4  =  72  12m vertically upwards then allow only this mark. 2 1 2 M1 Correct number of relevant terms with component WD = 25 g  12sin8 +  25  2.4 + 50  12 included in PE term, dimensionally correct. Could 2 have 12F in place of WD . Allow sign errors; allow sin/cos mix. WD = 1090 J Allow to subsequently find F (ISW) but A0 if state ‘work done A1 1089.519303 Allow 1089 − 1090 J. = 90.793’ OE. Alternative method using Newton’s second law 2.4 2 = 0 2 + 2 a  12  a = 0.24  *M1 Or other suvat method using u = 0, v = 2.4 and s = 12 . F − 50 − 25 g sin8 = 25  their a *DM1 For use of Newton’s second law with their value for a. W Correct number of relevant terms with weight May have instead of F. component included. Allow sign errors; allow sin/cos 12 mix. May if correct see F − 50 − 34.793= 6 or F = 90.793 Work done = their F  12 DM1 W OE e.g. F = Dep on both M marks. 12 Work done = 1090 J A1 Allow 1089 − 1090 J. 4

More questions on Energy, work and power

2 1.5 kg 0.005 kg A machine for driving a nail into a block of wood causes a hammerhead to drop vertically onto the top of the nail. The mass of the hammerhead is 1.5 kg and the mass of the nail is 0.005 kg (see diagram). The hammerhead hits the nail with speed 32 m s–1 and remains in contact with the nail after the impact. (a) Calculate the speed with which the combined hammerhead and nail move immediately after the impact. Give your answer correct to 3 decimal places. [2] ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... There is a constant force resisting the motion of magnitude 25 000 N. (b) Calculate the distance the nail is driven into the wood. [3] ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...................................................................................................................................................

Mark scheme: 2(a) 1.5  32 = (1.5 + 0.005 ) v M1 Conservation of momentum. Must have three non-zero terms. Allow sign errors. Must have correct masses with relevant velocities. Note: M1A0 if g included with the masses speed = 31.894 m s-1 Allow ‘31.894 = 31.9’ A1 ISW. Must be given to 3dp as specified in question. 2 2(b) (1.5 + 0.005 ) g − 25000 = (1.5 + 0.005 ) a  → a =−16601.29568  *M1 Use of N2L with correct number of relevant terms; allow sign errors, but masses must be added, not subtracted. Mass must be (1.5 + 0.005 ) . 2 DM1 Use of constant acceleration to get an equation in s 0 = ( their 31.894 ) + 2  ( their −16601.29568 ) s using their negative a and their speed (allow rounded to 3sf or better). Must use suvat correctly. s = 0.0306 m A1 0.030636983… Note: answer 0.0306184 from omitting weight terms in N2L giving a = 16611.29.. . This gets M0M0A0. Use of speed = 31.9 gives s = 0.030649 which gets full credit if correctly obtained. Alternative method using energy PE ‒ work done against friction =  ( (1.5 + 0.005 ) g − 25000 ) s  =24984.95s  B1 ( (1.5 + 0.005 ) g − 25000 ) s = − 1  (1.5 + 0.005 )  their 31.894 2 M1 Allowbe correct.sign Correcterrors includingnumber ofin termsPE butandall dimensionallymasses must 2 correct. Using their speed (allow rounded to 3sf or better). s = 0.0306 m A1 0.030636983… 3

More questions on Energy, work and power

3 A train travels 4.8 km between two stations, A and B. The train starts from rest at A and accelerates at a constant rate until it reaches a speed of 30 m s–1. It then travels at this constant speed for T seconds, before decelerating at a constant rate, coming to rest at B. The total time for the journey is 180 s. Find the value of T and hence find the distance moved by the train while travelling at the constant speed of 30 m s–1. [5] ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ...........................................................................................................................................................................

Mark scheme: 3 s1 = 0.5  30  t1 s3 = 0.5  30  t3 s2 = 30  T *M1 Use of suvat for displacement s= for at least two of the 900 900 three or for s1+ 3 = 0.5  30  (180 − T ) . Allow s1 = s3 = s2 = 30  T 2 a1 2 a3 Must not use the same variable for all three. DM1 Adding displacements in terms of times and equating 4800 = 15t1 + 30t 2 + 15t3 = 15 ( 180 − T ) + 30T  to 4800. Allow if use 4.8 rather than 4800. 4800 = 15 (180 − T ) + 30T oe A1 For correct equation in any time variable. T = 140 A1 Condone the assumption that magnitude of acceleration and deceleration are equal (which should come to 1.5) and allow full credit if everything else correct. Allow X or any other variable for time Distance 4200m B1FT FT their T 30  theirT Alternative method for question 3 Use of a trapezium (may be indicated by diagram) M1 If no correct subsequent working, then allow M1 for Or 0.5  30  (180 + T ) trapezium drawn with height 30, base 180 and top of length T (do not need to indicate that area = 4800). 0.5  30  (180 + T ) = 4800 B2 OE. T = 140 A1 Allow X or any other variable for time. Distance 4200m B1FT FT their T 30  theirT . 5

More questions on Kinematics of motion in a straight line

4 51 N a O x a 17 N 34 N Coplanar forces of magnitudes 17 N, 51 N and 34 N act at a point O in the directions shown in the diagram, 15 where tan a = . 8 (a) Find the magnitude and direction of the resultant of the three forces. 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The force of magnitude 51 N is replaced by a force of magnitude P N acting in the same direction. The resultant of the three forces now acts in the positive x-direction. (b) Find the value of P. 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Mark scheme: 4(a) Resolving either direction. Each force must have a component *M1 With correct number of relevant terms. Allow consistent sin/cos mix but must have more than simply 15 8 sinand / or cos. Allow sign errors. Do not allow Note: sin = , cos= 17 17 ‘forces to the left = forces to the right’ e.g. 8 8 15 May see X =  ( 51cos+ 34cos− 17sin) 51  + 34  = 17  unless subsequently 17 17 17 and Y =  ( 51sin− 34sin− 17cos) but no marks until values of sin ‘corrected’. and cos used.  −115  Allow sin  tan  etc. for M1A1A1 if correct.  8   8 8 15  A1 Allow use of 62° or 28° or better [leading to  X = R cos =    51  + 34  − 17   =  ( 24 + 16 − 15 ) = 25  ±24.8949… or better].  17 17 17  Note: consistent sin/cos mix leads to 67 which gets M1A0A0.  15 15 8  A1 Allow use of 62° or 28° or better [leading to Y = R sin =    51  − 34  − 17   =  ( 45 − 30 − 8 ) = 7  ±7.0290… or better].  17 17 17  Note: consistent sin/cos mix leads to −7 which gets M1A0A0. 2 2 DM1 Attempt to solve for R Attempt to find R. R = 25 + 7 From equations with the correct number of relevant terms. All forces resolved as appropriate but allow 25 7 Or R = or consistent sin/cos muddle. (cos their ) (cos their ) Allow use of their  provided correctly derived from equations with the correct number of relevant terms. 7 25 Or R = or (sin their ) (sin their ) 4(a) − 1 7 − 1 25 DM1 Attempt to solve for .  = tan or  = tan 25 7 From equations with the correct number of relevant terms. All forces resolved as appropriate but allow −1 25 −1 7 consistent sin/cos muddle. Allow use of their R Or  = cos or cos or provided correctly derived from equations with the their R their R correct number of relevant terms. −1 7 −1 25 Or  = sin or sin their R their R R = 26 .0 N or 674 A1 Both correct Allow resultant sketched with angle indicated and = 15.6 abovepositive x − axis arrowed. If not specified in working, please check Allow 15.6right of positive x − axis , anticlockwise 15.6 from positive x original diagram to see if direction specified there axis ‘74.4° clockwise from the y-axis’ ‘74.4° to right of the y-axis’ instead. Allow any explanation of direction that could Allow 15.6North of East or ‘15.6above the east’. Bearing of [0]74.4°, be drawn uniquely. N74.4°E Do not allow ‘East north 15.6°’ Alternative method resolving in the directions of the 34N and 17N forces Resolving either direction. M1 With correct number of relevant terms. Allow consistent sin/cos mix but must have more than simply sinand / or cos. Allow sign errors. Do not allow ‘forces to the left = forces to the right’ unless subsequently correct.  (17 − 51cos33.855=) 25.3529  A1  ( 34 − 51cos56.144=) 5.5882 A1 2 2 M1 Attempt to solve for R. R = 25.3529 + 5.5882  −1 5.5882  −1 8 M1 OE.  = tan + tan Attempt to solve for . −25.3529  15 4(a) R = 26 .0 N = 15.6  abovepositive x − axis or otherwise as above A1 6 4(b) 15 15 8 15 15 8 15 M1 With correct number of relevant terms. Allow sign P − 34  − 17  = 0 or P = 34  + 17  or P = 38 or errors. Do not allow sin/cos mix. Must be an equation, 17 17 17 17 17 17 17 NOT just an expression. P − 34 8 = 17 15 May see P sin− 34sin− 17cos= 0 but no marks until values of sin and cos used P = 43.1 A1 646 Allow . 15 2

More questions on Forces and equilibrium

5 A car of mass 1500 kg is travelling along a straight horizontal road. (a) It is given that there is a constant resistance to motion. The engine of the car is working at 24 kW while the car is travelling at a constant speed of 32 m s–1. The power is now increased to 28 kW. Find the acceleration of the car at the instant it is travelling at a speed of 36 m s–1. 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(b) It is given instead that the resistance to the motion of the car is ( 340 + 4v)N when the speed of the car is v m s–1. When the engine is working at 20 kW, the car is travelling at constant speed. Find this constant speed. 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Mark scheme: 5(a) 24000 B1 OE e.g. 24000 = D  32 R = D =  = 750  32 Attempt at N2L M1 Use of N2L with 3 relevant terms; allow sign errors. 28000 Allow R or their R. − R = 1500  a 36 28000 24000 A1 − = 1500  a or 777.777− 750 = 1500  a 36 32 −2 1 − 2 A1 a = 0.0185ms or ms 54 4 5(b) 20000 *M1 Use of power = F  v OE, 20000 = ( 340 + 4v )  v or = 340 + 4v v with F given in question but allow miscopy of F. Allow if candidate uses 20 instead of 20000. Attempt to solve a three-term quadratic to obtain at least one value of v . DM1 Correct quadratic is 4v 2 + 340v − 20000 = 0 . Must be at least one correct real value for their 3 Three-term quadratic. If no solving seen, must have three-terms and one correct value for their three-term quadratic to be awarded this mark. Allow if candidate uses 20 instead of 20000. Speed = 40 m s–1 (only) A1 No solving needed for A1 if no wrong working. 3

More questions on Kinematics of motion in a straight line

6 A particle P moves in a straight line starting from a point O. At time t s after leaving O, the velocity, v m s–1, of P is given by v = ( 15 - 2 t) 2 . (a) Find the values of t when the velocity of P is 100 m s–1. 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(b) Show that there is a particular value of t for which the velocity and acceleration of the particle are both zero. 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(c) Find the displacement of P from O at the time that the velocity and acceleration of the particle are both zero. 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Mark scheme: 6(a) 5 B1 t = 2.5 or 2 25 B1 t = 12.5 or 2 2 6(b)  a = (2 15 − 2t )( −2 ) = 8t − 60  *M1 For attempt at differentiation. Must have expression of the form pt + q with p  0, p  4, q  0 . p and q need v not be simplified. Use of a = scores M0. t 2 DM1 No need for working to solve either equation. 8t − 60 = 0 AND (15 − 2t ) = 0 AND attempt to solve at least one of these Could have one of the two equations with their 7.5 equations. Allow 225 − 60t + 4t 2 = 0 Allow their a = 0. substituted rather than t. t = 7.5 (twice) A1 AG CWO Any error seen A0. Allow substitution of 7.5 for t instead into second equation after first one solved. 2 i.e. (15 −2 7.5 ) = 0 OR 8  7.5 − 60 = 0 OE. Candidates who solve a = v would only get the first M1 unless they show that 7.5 gives zero for either acceleration or velocity. 3 6(c) For attempt at integration *M1 The power of t must increase by 1 with a change of  4t − 60t + 225 dt  s = ( 2 coefficient in the same term. Use of s = vt scores M0. ) 3 E.g. s = a (15 − 2t ) for a  1or a  0 4 3 2 A1 Correct integral, allow un-simplified.  s =  t − 30t + 225t  + c  c = 0  If constant wrong can still get A1. 3 1 3 1 or  s = (15 − 2t )   + c  c = 562.5  3 −2 7.5 DM1 Correct use of limits (their t from part (b) and zero)  4 3 2  4 3 2 s = t − 30t + 225t =  7.5 − 30  7.5 + 225  7.5  −0  and their integral (but allow one mistake in    3  0 3 calculation).  1 3 1  7.5 1 3 1 Could instead find c = 562.5 then substitute their t = or s = 0 − (15 −2 0 )   ( 15 − 2t )   = 7.5.  3 −2  0 3 −2 If they go on to add another displacement area to correctly obtained value allow M1A0. 1125 A1 Allow DM1A1 without working if correct answer after s = 562.5 or or 563 finding correct integral. 2 SCB1 for correct answer with no attempt at integration seen. If a candidate includes + c in their final answer they do not get the final A1 mark. 4

More questions on Kinematics of motion in a straight line

7 B A m kg 6.5 kg a Two particles A and B of masses 6.5 kg and m kg respectively are connected by a light inextensible string that passes over a smooth pulley. The pulley is fixed at the top of a rough slope which is at an angle of 5 a to the horizontal ground, where tan a = . A is on the rough slope and B hangs below the pulley (see 12 diagram). The coefficient of friction between the slope and A is 0.4. (a) Given that the system is in equilibrium, find the set of possible values of m. 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(b) It is given instead that m = 12 and the particles are released from rest with the string taut. Use an energy method to find the speed of the particles when each particle has moved 0.6 m. You may assume that this occurs before A reaches the pulley or B reaches the ground. 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Mark scheme: 7(a) T = mg or T − mg = 0 B1 12 B1 Allow cos23 or better but must have more than simply R = 6.5 g   = 60  cos. Must be identified as reaction here or when 13 friction term is formed. 5 *M1 Resolving along the plane, either case to form an T − 6.5 g  + F = 0 or T − 25 + F = 0 equation. Must be component of weight and g must be 13 present. Allow any non-zero single term F. Allow any Can instead use system equation mg = 25 − F non-zero single term T in terms of mg . 5 T − 6.5 g  − F = 0 or T − 25 − F = 0 If either of the system equations seen, then first B1 can 13 be implied (even if friction or component of 6.5g Can instead use system equation mg = 25 + F missing). Allow sin/cos mix (need not be consistent) but must have more than simply sinand / or cos with = 23 or better. A1 For both correct. Must have more than simply sinand / or cos but = 23 or better. Use F = 0.4 R  = 24  to get an equation in m only DM1 Where R is a component of weight. m = 0.1 or m = 4.9 A1 0.1 m 4.9 or m  :0.1 m 4.9 or  0.1,4.9  A1 7 7(b)  5  B1 Can be implied from equation with appropriate signs. PEchange =   12 g  0.6 − 6.5 g  0.6  =  ( 72 − 15 ) = 57  5  13  Must have either or sin23 not just sin. 13 1 2 B1 Can be implied from equation with appropriate signs KEchange =   (12 + 6.5 ) v 2 12 B1 12 WD friction = 0.4  65   0.6  = 24  0.6 = 14.4  Must have either or cos23 not just cos 13 13 5 12 2 M1 Attempt at work energy equation with correct number 12 g  0.6 − 6.5 g  0.6  − 0.4  65   0.6 = 0.5  (12 + 6.5 ) v of relevant terms; dimensionally correct. PE term must 13 13 2 consist of two parts, with components as required. 72 − 15 − 14.4 = 0.5  (12 + 6.5 ) v Allow sign errors. Do NOT allow sin/cos mix. 71  0.6 = 0.5  (12 + 6.5 ) v 2 42.6 = 0.5  (12 + 6.5 ) v 2 42.6 852 A1 2.1460… v = = 2.15 Allow v = 9.25 185 Alternative method finding tension then using energy . Use of Newton’s second law for A AND B *M1 Must have correct number of relevant terms. Allow For B: 12 g − T = 12 a sign errors. Do not allow sin/cos mix. Forces must have components (or not) as required. 5 12 For A: T − 6.5 g  − 0.4  6.5 g  = 6.5a 5 13 13 Must have either or sin23 not just sin and 13 likewise with cos. Must not use their value of T from part (a). 2736 A1 T = = 73.945 37 7(b) 2736 DB1 Energy from tension =  0.6 37 1 2 2736 DM1 For either For B:  12v = 12 g  0.6 −  0.6 2 37 1 2 5 12 2736 For A:  6.5v + 6.5 g   0.6 + 0.4  6.5 g   0.6 = 0.6  2 13 13 37 42.6 852 A1 If no marks scored allow SCB1 for work done v = = 2.15 Allow v = 12 9.25 185 = 0.4  65   0.6  = 24  0.6 = 14.4  . 13 Alternative method using energy but treating the particles separately 1 2 *B1 Must not use their value of T from part(a). For B:  12v = (12 g − T )  0.6 2 Attempt to find an energy equation for particle A *M1 Must have correct number of terms. Allow sign errors. Do NOT allow sin/cos mix. 5 Must have either or sin23 not just sin and 13 likewise with cos. Must not use their value of T from part (a). 1 2  5 12  A1 5  6.5v =  T − 6.5 g  − 0.4  6.5 g    0.6 Must have either or sin23 not just sin and 2  13 13  13 likewise with cos. Must not use their value of T from part (a). For attempt to solve to get to ‘v =’ or ' v 2 = ' (by eliminating T) DM1 7(b) 42.6 852 A1 If no solving seen allow M1A1 for correct answer or v = = 2.15 Allow v = M1 if correct for their simultaneous equations. 9.25 185 If no marks scored allow SCB1 for work done 12 = 0.4  65   0.6  = 24  0.6 = 14.4  . 13 Alternative method finding acceleration then forces then using energy. Use of Newton’s second law for system or for A AND B *M1 Must have correct number of relevant terms Allow 5 12 sign errors. Do not allow sin/cos mix. For system: 12 g − 6.5 g  − 0.4  6.5 g  = (12 + 6.5 ) a Forces must have components (or not) as required. 13 13 120 − 25 − 24 = 18.5a  Must have either 5 or sin23 not just sin and 13 For B: 12 g − T = 12 a 120 − T = 12 a  likewise with cos. 5 12 Must not use their value of T from part (a). For A: T − 6.5 g  − 0.4  6.5 g  = 6.5a T − 25 − 24 = 6.5a  13 13 142 A1 a = = 3.8378 37 142 1704 142 923 DB1 FB = 12  = or FA = 6.5  = AND any one of 37 37 37 37  EN A = 923  0.6 or  EN B = 1704  0.6 or  EN Total = 1704 + 923   0.6 37 37  37 37  1 2  923  1 2  1704  DM1  6.5v =    0.6 or  12v =    0.6 2  37  2  37  1 2  1704 923  or  (12 + 6.5 ) v =  +   0.6 2  37 37  7(b) 42.6 852 A1 If no marks scored allow SCB1 for work done v = = 2.15 Allow v = 12 9.25 185 = 0.4  65   0.6  = 24  0.6 = 14.4  13 5

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