Cambridge A Level Mathematics 9709 — 2015 Oct/Nov Paper 4 · Variant 2

5 2.5 m P 0.7 m Q A smooth inclined plane of length 2.5 m is fixed with one end on the horizontal floor and the other end at a height of 0.7 m above the floor. Particles P and Q, of masses 0.5 kg and 0.1 kg respectively, are attached to the ends of a light inextensible string which passes over a small smooth pulley fixed at the top of the plane. Particle Q is held at rest on the floor vertically below the pulley. The string is taut and P is at rest on the plane (see diagram). Q is released and starts to move vertically upwards towards the pulley and P moves down the plane. (i) Find the tension in the string and the magnitude of the acceleration of the particles before Q reaches the pulley. [5] At the instant just before Q reaches the pulley the string breaks; P continues to move down the plane and reaches the floor with a speed of 2 m s−1. (ii) Find the length of the string. [3] [Questions 6 and 7 are printed on the next page.]

Mark scheme: 5 (i) For applying Newton 2nd law to P or to Q or for applying N2 to the system M1 7 Any two correct 0.5g × – T = 0.5a Allow sin 16.3 for 7 / 25 25 T – 0.1g = 0.1a 1.4 – 1 = 0.6a A1 For eliminating T and obtaining 2 a = ms–2 B1 3 M1 For substituting for a to find T Tension is 1.07 N A1 5 Allow T = 16 / 15 N (ii)  2  For using v2 = u2 + 2as to find the speed of [v2 = 2 ×   × 0.7] the particles immediately before the string  3  breaks M1 2 For applying v2 = u2 + 2as for the motion of [22 = 2 × × 0.7 + 2 × 0.28g × s] P when the string is slack and s is the 3 distance travelled by P after the break until it reaches the floor M1 Length of string = 2.5 – s = 1.95 m A1 3 Allow length = 41 / 21 m

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6 0.195 N 1 1 0.195 N Fig. 1 Fig. 2 A small ring of mass 0.024 kg is threaded on a fixed rough horizontal rod. A light inextensible string is attached to the ring and the string is pulled with a force of magnitude 0.195 N at an angle of 1 with the horizontal, where sin 1 = 13.5 When the angle 1 is below the horizontal (see Fig. 1) the ring is in limiting equilibrium. (i) Find the coefficient of friction between the ring and the rod. [6] When the angle 1 is above the horizontal (see Fig. 2) the ring moves. (ii) Find the acceleration of the ring. [4]

Mark scheme: 6 (i) [0.195 cos θ = F] M1 For resolving forces horizontally 12 F = 0.195cos 22.6 = 0.195 × 13 9 A1 = 0.18 = 50 [R = 0.24 + 0.195 sin θ] M1 For resolving forces vertically R = 0.24 + 0.195sin 22.6 = 5 0.24 + 0.195 × = 0.315 13 63 = A1 200 M1 For using µ = F / R Coefficient µ = 4 / 7 or 0.571 A1 6 (ii) R = 0.24 – 0.195sin 22.6 5 = 0.24 – 0.195 × 13 33 = 0.165 = B1 200 For using Newton’s second law for motion M1 along the rod 12  4  0.195 × –   × 0.165 13  7  = 0.024a A1 Acceleration is 3.57 ms–2 A1 4 Allow acceleration = 25 / 7

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7 A car of mass 1600 kg moves with constant power 14 kW as it travels along a straight horizontal road. The car takes 25 s to travel between two points A and B on the road. (i) Find the work done by the car’s engine while the car travels from A to B. [2] The resistance to the car’s motion is constant and equal to 235 N. The car has accelerations at A and B of 0.5 m s−2 and 0.25 m s−2 respectively. Find (ii) the gain in kinetic energy by the car in moving from A to B, [5] (iii) the distance AB. [3]

Mark scheme: 7 (i) [WD = 14000 × 25] M1 For using P = WD÷∆t Work done is 350 kJ or 350 000 J A1 2 (ii) For using DF = P / v and Newton’s 2nd law to find the speed of the car at A or at B M1 14000 / vA – 235 = 1600 × 0.5 → vA = 2800 / 207 vA = 13.53 ms–1 A1 14000 / vB – 235 = 1600 × 0.25 → vB = 2800 / 127 vB = 22.05 ms–1 A1 [KE gain = For using KE gain 2 1 1 2 1600(22.052 – 13.532)] M1 = 2 m(vB – vA2) KE gain = 242.5 kJ or 242 500 J A1 5 (iii) For using WD by DF M1 = KE gain + resistance × AB 350 000 = 242 500 + 235 × AB A1 Distance AB is 457 m A1 3

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