Cambridge A Level Mathematics 9709 — 2023 Oct/Nov Paper 4 · Variant 3

1 A particle is projected vertically upwards from horizontal ground with a speed of ums−1. The particle has height sm above the ground at times 3 seconds and 4 seconds after projection. Find the value of u and the value of s. [3] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 Use of suvat to find expressions for s M1 𝑠= 3𝑢−5 × 9 A1 𝑠= 4𝑢−5 × 16 𝑢= 35, 𝑠= 60 A1 Alternative method for Question 1: Use of suvat to find expressions for u at max height M1 0 = 𝑢−10 × 3.5 A1 𝑢= 35, 𝑠= 60 A1 3

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2 1.2 kg 0.004 kg A machine for driving a nail into a block of wood causes a hammerhead to drop vertically onto the top of a nail. The mass of the hammerhead is 1.2kg and the mass of the nail is 0.004kg (see diagram). The hammerhead hits the nail with speed vms−1 and remains in contact with the nail after the impact. The combined hammerhead and nail move immediately after the impact with speed 40ms−1. (a) Calculate v, giving your answer as an exact fraction. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) The nail is driven 4cm into the wood. Find the constant force resisting the motion. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(a) Attempt at conservation of momentum M1 [1.2𝑣= (1.2 + 0.004) × 40] 602 A1 oe 𝑣= 15 2 2(b) 2 2 M1 Use of a ‘suvat’ method to get an equation in a. Allow 0 = ( 40 ) + 2  0.04  a  a = −20000 sign errors. Allow 20000 . 0 + 40 or 0.04 = t gets t = 0.002, so 0 = 40 + 0.002 a  a = −20000  Do not allow 4 in place of 0.04. 2 602 Allow use of 40.1 or for velocity in place of 40. 15 Attempt to use Newton’s Second Law vertically. M1 Must have the correct number of relevant terms. Allow −  R + ( 1.2 + 0.004 ) g = (1.2 + 0.004 ) a  sign errors, but terms including masses must be effectively added. Do not allow any mass other than (1.2  − R + 12.04 = 1.204 a  + 0.004). 602301 A1 WWW. R = 24 100 N [24 092.04 = ] 25 Note: use of wrong sign for g leads to answers 24 067.96 which gets max M1M1A0. Note: Missing weight term gets 24 080 which gets Max M1M0A0. 3 2(b) Alternative method for Question 2(b) using energy [Change in PE =] 1.204 g  0.04  = 0.4816  B1 602 Allow use of 40.1 or for velocity in place of 40. 15 1 2 or [change in KE =]  1.204  ( 40 )  = 963.2  B0 for kinetic energy, if extra kinetic energy terms 2 present. 1 2 M1 Attempt at work energy equation. Must have correct 1.204 g  0.04 +  1.204  ( 40 ) = 0.04 R 2 number of relevant terms. dimensionally correct; allow sign errors. Do not allow 4 in place of 0.04. 602 Allow use of 40.1 or for velocity in place of 40. 15 602301 A1 WWW R = 24 100 N [24 092.04 = ] 25 Note: use of wrong sign for g leads to answers 24 067.96 which gets max B1M1A0. Note: Missing potential energy term gets 24 080, which gets maximum of B1M0A0.

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3 A block of mass 8kg slides down a rough plane inclined at 30Å to the horizontal, starting from rest. The coefficient of friction between the block and the plane is -. The block accelerates uniformly down the plane at 2.4ms−2. (a) Draw a diagram showing the forces acting on the block. [1] (b) Find the value of -. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Find the speed of the block after it has moved 3m down the plane. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(a) Correct force diagram with 3 forces in the correct directions. B1 No labels required on the 3 forces and ignore wrong labels. Arrows needed. Allow either or both components of weight if fully labelled. Allow sin/cos mix. If forces are not connected to the block, then the line of action of each force must go through the block. 1 3(b) R = 8 g cos30  = 40 3 = 69.282  B1 Resolving perpendicular to the plane.   Resolving parallel to the plane and attempt to apply Newton’s second law. M1* 3 terms. Allow sign errors, sin/cos mix. Allow g  8 g sin30 − F = 8  2.4  F = 20.8  missing, otherwise dimensionally correct. Use of F = R to get an equation in  only. DM1 Allow g missing in either or both of F and R. Allow sign errors, consistent sin/cos mix. 8 g sin30 − 8 gcos30 = 8  2.4 40 − 40 3= 19.2  R must be a single component of a force.   Allow the 3 masses to be cancelled.  20.8 20.8  A1 13 3 104 3 = 0.3  0   May first see or  Allow exact value or oe.  40 3 69.282   75 600 4 3(c) B1 3.79473… (3.8 without a more accurate value seen gets 2 6 10 [ v = 2  2.4  3  greatest speed =] 3.79 ms–1 = B0 and should be annotated SF). 5 1

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4 A car has mass 1600kg. (a) The car is moving along a straight horizontal road at a constant speed of 24ms−1 and is subject to a constant resistance of magnitude 480N. Find, in kW, the rate at which the engine of the car is working. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The car now moves down a hill inclined at an angle of 1 to the horizontal, where sin 1 = 0.09. The engine of the car is working at a constant rate of 12kW. The speed of the car is 24ms−1 at the top of the hill. Ten seconds later the car has travelled 280m down the hill and has speed 32ms−1. (b) Given that the resistance is not constant, use an energy method to find the total work done against the resistance during the ten seconds. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(a) P M1 P P = 480  24 or, e.g. − 480 = 0 For − F = 0 or P = Fv oe. 24 v P = 11.52 [kW] A1 Allow 11.5 M1A0 for 11 520 or 11 500. 2 4(b) 1 2 B1 For either correct.  = 460800   KE before =  1600  24  2 1 2 Do not allow  1600  ( 32 − 24 ) . 1 2 2  = 819200   KE after =  1600  32  2  PE loss  = 1600 g  280  0.09  = 1600 g  25.2 = 403200  B1 Allow 1600 g  280  sin5.16 or 1600 g  280  sin5.2 but not simply 1600 g  280  sin (unless implied by correct final answer). Total WD = 12000  10  = 120000  B1 WD oe, e.g. 12 000 = . 10 4(b) Work done against resistance = or 280F = or WD = or W = oe M1 Attempt at work energy equation with 5 relevant terms (4 relevant terms plus work done against resistance); 1 2 1 2 12000  10 + 1600 g  280  0.09 −  1600  32 +  1200  24 dimensionally correct. Allow sign errors. 2 2 M0 for use of constant acceleration.  = 120000 + 403200 − 819200 + 460800  1 2 Do not allow  1600  ( 32 − 24 ) . 2 WD = 164 800 [J] A1 Or 164.8 kJ CAO but condone 165 kJ or 165 000 [J] Not from use of constant acceleration or Newton’s second law. ISW attempt to find force after correct WD found. 5

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5 A 1Å B P N 80 N A light string AB is fixed at A and has a particle of weight 80N attached at B. A horizontal force of magnitude PN is applied at B such that the string makes an angle 1Å to the vertical (see diagram). (a) It is given that P = 32 and the system is in equilibrium. Find the tension in the string and the value of 1. 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(b) It is given instead that the tension in the string is 120N and that the particle attached at B still has weight 80N. Find the value of P and the value of 1. 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Mark scheme: 5(a) Attempt to resolve in one direction and form equation. M1 Must have correct number of relevant terms (forces must have components as required). Allow sin/cos mix. Allow sign errors. If only one equation shown and it involves 32, it must be 32, not P. T sin= 32 and T cos= 80 A1 For both horizontal and vertical, or both parallel and or perpendicular. 0 = 80sin− 32cos and T = 80cos+ 32sin Attempt to solve for T or  M1 2 2 Must get to T or ; e.g. T = 32 + 80 or −1 32  −1 80   = tan   . Condone, e.g.  = tan   .  80   32  Must come from equations with correct number of relevant terms. T = 86.2 [N 86.1626….] or 16 29 or 7424 and = 21.8 [21.801…] A1 For both. 4 5(a) Alternative method using triangle of forces T 2 = 802 + 322  −2 80  32cos90 or T sin= 32 or T cos= 80 M1 For any of the five; allow sign errors. 80tan= 32 or T = 80cos+ 32sin oe A1 For any two equations. Attempt to solve for T or  M1 2 2 Must get to T or ; e.g. T = 32 + 80 or −1 32   = tan   .  80  T = 86.2 [N 86.1626….] or 16 29 or 7424 and = 21.8 [21.801…] A1 Alternative Triangle of forces method using sine rule T 32 80 M1 For any two. = = sin90 sin () sin ( 90 − ) A1 For all three. Attempt to solve for T or  M1 −1 32  e.g.  = tan   .  80  T = 86.2 [N 86.1626….] or 16 29 or 7424 and = 21.8 [21.801…] A1 For both. 5(b) Attempt to resolve in one direction and form equation M1 Must have correct number of relevant terms (forces must have components as required). Allow sin/cos mix. Allow sign errors. Must use 120, not T. 120sin= P and 120cos= 80 A1 For both horizontal and vertical, or both parallel and or 0 = 80sin− P cos and 120 = 80cos+ P sin perpendicular. Attempt to solve for P or  M1 2 2 Must get to P or ; e.g. P = 120 − 80 or −1 80   = cos   .  120  Must come from equations with correct number of relevant terms. A1 For both; allow P = 89.5 (from 120sin48.2 ). P = 89.4 89.4427 or 40 5 or 8000 = 48.2  48.1896 4 Alternative method using triangle of forces 1202 = P2 + 802  −2 80  P cos90 or 120sin= P or 120cos= 80 M1 For any of the five; allow sign errors. or 80tan= P or 120 = 80cos+ P sin oe A1 For any two equations. Attempt to solve for P or  M1 2 2 Must get to P or ; e.g. P = 120 − 80 or −1 80   = cos   , oe.  120  P = 89.4 89.4427  or 40 5 or 8000 = 48.2  48.1896  A1 For both; allow P = 89.5 (from 120sin48.2) . 5(b) Alternative Triangle of forces method using sine rule 120 P 80 M1 For any two. = = sin90 sin () sin ( 90 − ) A1 For all three. Attempt to solve for P or  M1 Must get to P or ; −1 80  −1 80  e.g. = 90 − sin   or  = cos   .  120   120  A1 For both; allow P = 89.5 (from 120sin48.2) . P = 89.4 89.4427 or 40 5 or 8000 = 48.2  48.1896

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6 A particle moves in a straight line. At time t s, the acceleration, ams−2, of the particle is given by a = 36 −6t. The velocity of the particle is 27ms−1 when t = 2. (a) Find the values of t when the particle is at instantaneous rest. 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(b) Find the total distance the particle travels during the first 12 seconds. 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Mark scheme: 6(a) Attempt to integrate a M1* The power of t must increase by 1 with a change of coefficient in the 2t term. Do not penalise missing c. Use of v = at scores M0. 2 6t 2 A1 Condone an integral sign in front of correct answer.  v =  36t − 3t  + c  or  v =  36t −  + c  2 0 = 36t − 3t 2 − 33 DM1 Use t = 2 and v = 27 to find c . 2 Must get to c = and set 3 term quadratic equal to zero.  27 = 36 −2 3 2 + c  c = −33   Solve 0 = 36t − 3t 2 − 33 to get t = 1 and t = 11 A1 Allow t = 1 or t = 11 ; t = 1 , t = 11 oe. 4 6(b) 2 M1* The power of t must increase by 1 with a change of Attempt to integrate an expression of the form at + bt  +c  with non-zero a coefficient in the same term. Use of s = vt scores M0. and b. 36t 2 3t 3 2 3 If correct  s =  − − 33t  + c ' or  s =18t − t − 33t  +c ' 2 3 Attempt to evaluate their 18t 2 − t 3 − 33t  for t = 0 to t = 1 or DM1 Attempt using their limits (at least one strictly between 0   and 12) correctly. t = 1 to t = 11 or t = 11 to t = 12 0 to 1: –16 – 0 = –16 1 to 11: 484 – (–16) = 500 11 to 12: 468 – 484 = –16 For all three DM1 Allow 11 to 12 implied by symmetry instead of found separately. Distance = 16 + 500 + 16  = 532 m A1 4

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7 B A 3.3 kg 2.4 kg 1 m 1Å Particles A and B, of masses 2.4kg and 3.3kg respectively, are connected by a light inextensible string that passes over a smooth pulley which is fixed to the top of a rough plane. The plane makes an angle of 1Å with horizontal ground. Particle A is on the plane and the section of the string between A and the pulley is parallel to a line of greatest slope of the plane. Particle B hangs vertically below the pulley and is 1m above the ground (see diagram). The coefficient of friction between the plane and A is -. (a) It is given that 1 = 30 and the system is in equilibrium with A on the point of moving directly up the plane. Show that - = 1.01 correct to 3 significant figures. 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(b) It is given instead that 1 = 20 and - = 1.01. The system is released from rest with the string taut. Find the total distance travelled by A before coming to instantaneous rest. You may assume that A does not reach the pulley and that B remains at rest after it hits the ground. 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Mark scheme: 7(a) Resolving for both particles or for the system to form equation(s) M1* Must have correct number of terms. Allow sign errors. Allow sin/cos mix. Allow g missing. M0 if acceleration included unless subsequently equated to zero. Masses must be appropriate for their equation(s). Forces must have components (or not) as required. Either T − F − 2.4 g sin30 = 0 AND 3.3 g − T = 0 A1 Both correct or system correct. May get F = 21. Can be with a wrong non-zero F. Or 3.3 g − F − 2.4 g sin30 = 0 R = 2.4 g cos30  = 12 3 = 20.7846 B1   Use of F = R to get an equation in only DM1 Must be from F dimensionally correct and single term R which is equal to a component the 2.4 kg weight. Allow  3.3 g − 2.4 g cos30 − 2.4 g sin30 = 0  consistent sin/cos mix but must be different components of weight. F and R must be numerical expressions. = 1.01 [sight of 1.01036… or 1.0104] A1 AG perhaps from one of 3.3 g − 24sin30 33 − 12 21 7 3 21 = = = = = 2.4 g cos30 12 3 12 3 12 20.7846 21 = 20.8 Do not allow unless evidence of 30 substituted for . E.g.: sight of 1.01036… or 1.0104. 5 7(b) Using Newton’s second law for both particles or the system M1* Must have correct number of terms. Allow sign errors. Allow sin/cos mix. Allow g missing. Masses must be appropriate for their equation(s). Forces must have components (or not) as required. Either 3.3 g − T = 3.3a and T − F − 2.4 g sin20 = 2.4a A1 Both correct or system equation correct. T − 22.778 − 8.208 = 2.4 a  or T − 30.986 = 2.4 a  Can be with a wrong non-zero F. or 3.3 g − F − 2.4 g sin 20 = ( 2.4 + 3.3 ) a  2.013367 = 5.7a  F = 1.01  2.4 g cos20  = 22.778  B1 For correct expression for F. Attempt to solve for a a = 0.353 [0.353222…] DM1 Using their F Must get to ‘a =’. If sin/cos mix must be consistent. v 2 = 2  0.353  1 [= 0.706444…] or  v = 0.841 A1FT FT their value of a  g to get an expression for v 2 or 1 2 v. Or 1 = 0 +  0.353t =t 2.3795  v = 0.353  2.38 2 Can be implied by awrt 0.84 for v or awrt 0.71 for v 2 . This mark does not depend on previous A or B mark, but both Ms must have been awarded. Using Newton’s second law on A after B reaches the ground M1* Must have correct number of terms. Allow sign errors. − F − 2.4 g sin20 = 2.4a Allow sin/cos mix. Allow g missing.  a = −12.911  −1.01  2.4 g cos20 − 2.4 g sin 20 = 2.4 a    −22.78814 − 8.20848 = 2.4a  Use of suvat to find s DM1 Using their a  g .  0 = their 0.8412 + 2  their − 12.911 s =s 0.027358. Must get to ‘s =’. May find and use t = 0.0651.   7(b) Total distance = 1.03 m A1 8 Alternative method using energy for first 5 marks 1 2 2 B1 = 2.85v   KE gained =   ( 2.4 + 3.3 ) v   2  PE lost =  3.3 g −1 2.4 g  1sin 20  = 24.791 = B1 Allow omission of 1 in either or both terms. [Friction =] 1.01  2.4 g cos20  = 22.778 B1 For correct expression for F. 1 2 M1 For attempt at energy equation. Allow sign errors, allow  ( 2.4 + 3.3 ) v = 3.3 g −1 2.4 g  1sin20 − 1.01  2.4 g cos20  1 2 sin/cos mix but must have sin/cos where needed. Correct 2 number of terms, dimensionally correct. Or 2.85v = 24.791− 22.778 Allow omission of 1 in any or all the three relevant terms. Must have cos 20 and sin 20. To get a correct expression for v 2 A1 Can be implied by awrt 0.84 for v or awrt 0.71 for v 2 if  v 2 = 0.706444 or v = 0.841 expression not seen.   7(b) Alternative method using energy for final 3 marks 1 2 M1 Using their v 2 . KE =  2.4  0.841 2 1 2 M1 For attempt at 3 term energy equation and solved to get 1.01  2.4 g cos20 +s 2.4 g sin20 =s  2.4  0.841 2 to ‘s =’. Allow sign errors, allow consistent sin/cos mix   s = 0.027358.. but must have sin/cos where needed. Correct number of terms, dimensionally correct. Total distance = 1.03 m A1

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