Cambridge A Level Mathematics 9709 — 2018 Oct/Nov Paper 4 · Variant 3

1 A B 70Å P N R 45Å A small smooth ring R of mass 0.2 kg is threaded onto a light inextensible string ARB. The two ends of the string are attached to points A and B on a sloping roof inclined at 45Å to the horizontal. A horizontal force of magnitude P N, acting in the plane ARB, is applied to the ring. The section BR of the string is perpendicular to the roof and the section AR of the string is inclined at 70Å to the horizontal (see diagram). The system is in equilibrium. Find the tension in the string and the value of P. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 M1 T = 1.21 N (1.21447…) A1 [P + Tcos70 = Tcos45] M1 Resolving horizontally P = 0.443 (0.443389…) A1 4

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2 50 N 20Å A block is pushed along a horizontal floor by a force of magnitude 50 N which acts at an angle of 20Å to the horizontal (see diagram). The coefficient of friction between the block and the floor is 0.3. Given that the speed of the block is constant, find the mass of the block. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 B1 [F = 0.3(mg + 50sin20)] M1 Use of F = µR M1 Resolving horizontally 50cos20 – 0.3(mg + 50sin20) = 0 A1ft ft R (R containing term in m) m = 14.0 kg (13.9514…) A1 5

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3 A particle of mass 1.2 kg moves in a straight line AB. It is projected with speed 7.5 m s−1 from A towards B and experiences a resistance force. The work done against this resistance force in moving from A to B is 25 J. (i) Given that AB is horizontal, find the speed of the particle at B. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) It is given instead that AB is inclined at 30Å below the horizontal and that the speed of the particle at B is 9 m s−1. The work done against the resistance force remains the same. Find the distance AB. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(i) M1 v = 3.82 m s–1 (3.81881…) A1 [2] 3(ii) 1.2gdsin30 B1 Correct expression for PE [½ × 1.2 × 7.52 – 25 + 1.2gdsin30 = ½ × 1.2 × 92] M1 For 4 term work / energy equation d = 6.64 m (6.64166…) A1 3

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4 A runner sets offfrom a point P at time t = 0, where t is in seconds. The runner starts from rest and accelerates at 1.2 m s−2 for 5 s. For the next 12 s the runner moves at constant speed before decelerating uniformly over a period of 3 s, coming to rest at Q. A cyclist sets offfrom P at time t = 10 and accelerates uniformly for 10 s, before immediately decelerating uniformly to rest at Q at time t = 30. (i) Sketch the velocity-time graph for the runner and show that the distance PQ is 96 m. [4] v (m s−1) t (s) 0 10 20 30 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the magnitude of the acceleration of the cyclist. 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Mark scheme: 4(i) B1 v = 6 m s–1, t = 5 s and t = 17 s B1 Correct trapezium with key values [½ × 6 × (12 + 20)] or [½ × 5 × 6 + 12 × 6 + ½ × 3 × 6] M1 Use of trapezium area or use of suvat formulae Total distance = 96 m A1 AG 4 Question Answer Marks Guidance 4(ii) [½ × 20 × v = 96] M1 Uses area of triangle = 96 or uses s = ut + ½ at2 to form equation in a v = 9.6 m s–1 or 48 = ½ a (10)2 A1 Acceleration = 9.6 / 10 = 0.96 m s–2 A1 3

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5 P Q 0.3 kg 0.5 kg h m Two particles P and Q, of masses 0.3 kg and 0.5 kg respectively, are attached to the ends of a light inextensible string. The string passes over a fixed smooth pulley with the particles hanging freely below it. Q is held at rest with the string taut at a height of h m above a horizontal floor (see diagram). Q is now released and both particles start to move. The pulley is sufficiently high so that P does not reach it at any stage. The time taken for Q to reach the floor is 0.6 s. (i) Find the acceleration of Q before it reaches the floor and hence find the value of h. 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Q remains at rest when it reaches the floor, and P continues to move upwards. (ii) Find the velocity of P at the instant when Q reaches the floor and the total time taken from the instant at which Q is released until the string becomes taut again. 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Mark scheme: 5(i) M1 use of a = (mQ – mP)g / (mP + mQ) T – 0.3g = 0.3a and 0.5g – T = 0.5a or a = (0.5g – 0.3g) / (0.5 + 0.3) A1 [0.5g – 0.3g = 0.8a] M1 Solve for a a = 2.5 A1 [h = 0 + ½ × 2.5 × 0.62] M1 For use of s = ut + ½at2 h = 0.45 A1 6 Question Answer Marks Guidance 5(ii) Velocity of P when Q reaches floor = 0 + 0.6 × 2.5 = 1.5 m s–1 B1ft ft a from (i) × 0.6 [0 = 1.5 – gt → t = …] (t = 0.15) M1 Use of suvat to find time to highest point Total time = 2 × 0.15 + 0.6 = 0.9 s A1 3

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6 A van of mass 3200 kg travels along a horizontal road. The power of the van’s engine is constant and equal to 36 kW, and there is a constant resistance to motion acting on the van. (i) When the speed of the van is 20 m s−1, its acceleration is 0.2 m s−2. Find the resistance force. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ When the van is travelling at 30 m s−1, it begins to ascend a hill inclined at 1.5Å to the horizontal. The power is increased and the resistance force is still equal to the value found in part (i). (ii) Find the power required to maintain this speed of 30 m s−1. 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(iii) The engine is now stopped, with the van still travelling at 30 m s−1, and the van decelerates to rest. Find the distance the van moves up the hill from the point at which the engine is stopped until it comes to rest. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(i) B1 [36000 / 20 – R = 3200 × 0.2] M1 Use of Newton’s Second Law R = 1160 N A1 [3] 6(ii) Driving force F = 3200gsin1.5 + 1160 M1 Resolving along plane [Power = (3200gsin1.5 + 1160) × 30] M1 Use of P = Fv Power = 59900 W (59929.87…) A1 3 Question Answer Marks Guidance 6(iii) [– (3200gsin1.5 + 1160) = 3200a] M1 Use of Newton’s Second Law (a = –0.62426…) A1 [02 = 302 + 2as] M1 Use of v2 = u2 + 2as to find s Distance s = 721 m (720.84…) A1 4 OR: 6(iii) [3200gsin1.5s] or [½ × 3200 × 900] M1 For PE gain or KE loss 3200gsin1.5s and ½ × 3200 × 900 A1 For PE gain and KE loss [½ × 3200 × 900 = 1160s + 3200gsin1.5s] M1 For work / energy equation Distance s = 721 m (720.84…) A1 4

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7 A particle moves in a straight line. The particle is initially at rest at a point O on the line. At time t s after leaving O, the acceleration a m s−2 of the particle is given by a = 25 −t2 for 0 ≤t ≤9. (i) Find the maximum velocity of the particle in this time period. 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(ii) Find the total distance travelled until the maximum velocity is reached. 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The acceleration of the particle for t > 9 is given by a = −3t−12. (iii) Find the velocity of the particle when t = 25. 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Mark scheme: 7(i) B1 [v = 25t – ⅓t3] M1 Use of integration [Max speed = 25 × 5 – ⅓× 53] M1 Substitution for t Max speed = 83⅓ m s–1 A1 4 Question Answer Marks Guidance 7(ii) [s = 12½t2 – 1/12t4] M1 Use of integration Distance = 260 m (260.4166…) A1 2 7(iii) At t = 9, v = 25 × 9 – ⅓ × 93 = –18 B1ft ft v from (i) [ s = 25 1 2 9 3 −   −     ∫ t dt = 1 2 6   −     t ] M1 Use of integration [Change in velocity from t = 9 to t = 25 = 1 2 6   −     t = –6 × 5 + 6×3 = –12] M1 Substituting limits Velocity at t = 25 is –18 – 12 = –30 m s–1 A1 4 OR: 7(iii) At t = 9, v = 25 × 9 – 1/3 × 93 = –18 B1ft ft v from (i) [ s = ∫ –3t –½ dt = –6t ½ (+ C)] M1 Use of integration [t = 9, v = –18 → C = 0, t = 25, v = –6 × 25 ½ ] M1 Finds C and substitutes t = 25 Velocity at t = 25 is –30 m s–1 A1 4

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