Cambridge A Level Mathematics 9709 — 2011 May/June Paper 4 · Variant 1

1 A car of mass 700 kg is travelling along a straight horizontal road. The resistance to motion is constant and equal to 600 N. (i) Find the driving force of the car’s engine at an instant when the acceleration is 2 m s−2. [2] (ii) Given that the car’s speed at this instant is 15 m s−1, find the rate at which the car’s engine is working. [2]

Mark scheme: 1 (i) [DF – 600 = 700 × 2] M1 For using Newton’s second law (3 terms needed) Driving force is 2000 N A1 [2] (ii) [P = 2000 × 15] M1 For using P = Fv Rate of working is 30000 W (or 30 kW) A1ft [2]

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2 A load of mass 1250 kg is raised by a crane from rest on horizontal ground, to rest at a height of 1.54 m above the ground. The work done against the resistance to motion is 5750 J. (i) Find the work done by the crane. [3] (ii) Assuming the power output of the crane is constant and equal to 1.25 kW, find the time taken to raise the load. [2]

Mark scheme: 2 (i) Gain in PE = 1250g × 1.54 ( = 19250 J) B1 [WD = 1250g × 1.54 + 5750] M1 For using WD by crane = Gain in PE + WD against resistance Work done is 25000 J (or 25 kJ) A1 [3] (ii) [1250 = 25000 / T] M1 for using P = ∆(WD) / ∆t Time is 20 s A1ft [2] ft Ans(i) ÷ 1250

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3 A q q 15.5 N R B A small smooth ring R of weight 8.5 N is threaded on a light inextensible string. The ends of the string are attached to fixed points A and B, with A vertically above B. A horizontal force of magnitude 15.5 N acts on R so that the ring is in equilibrium with angle ARB = 90◦. The part AR of the string makes an angle θ with the horizontal and the part BR makes an angle θ with the vertical (see diagram). The tension in the string is T N. Show that T sin θ = 12 and T cos θ = 3.5 and hence find θ. [6]

Mark scheme: 3 M1 For resolving forces horizontally or vertically (3 terms needed) Tcosθ + Tsinθ = 15.5 A1 AEF –Tcosθ + Tsinθ = 8.5 A1 AEF DM1 For solving for Tsinθ and Tcosθ Tsinθ = 12 and Tcosθ = 3.5 A1 AG θ = 73.7o (or 1.29c) B1 [6]

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4 A block of mass 11 kg is at rest on a rough plane inclined at 30◦to the horizontal. A force acts on the block in a direction up the plane parallel to a line of greatest slope. When the magnitude of the force is 2X N the block is on the point of sliding down the plane, and when the magnitude of the force is 9X N the block is on the point of sliding up the plane. Find (i) the value of X, [3] (ii) the coefficient of friction between the block and the plane. [4]

Mark scheme: 4 (i) M1 For resolving forces parallel to the plane (either case) – 3 terms needed 2X + F = 11gsin30o and A1 9X – F = 11gsin30o X = 10 A1 [3] (ii) F = 35 B1 May be implied. R = 11gcos30o B1 DM1 For using µ = F/R Coefficient is 0.367 A1ft [4]

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5 A train starts from rest at a station A and travels in a straight line to station B, where it comes to rest. The train moves with constant acceleration 0.025 m s−2 for the first 600 s, with constant speed for the next 2600 s, and finally with constant deceleration 0.0375 m s−2. (i) Find the total time taken for the train to travel from A to B. [4] (ii) Sketch the velocity-time graph for the journey and find the distance AB. [3] (iii) The speed of the train t seconds after leaving A is 7.5 m s−1. State the possible values of t. [1]

Mark scheme: 5 (i) v(600) = 0.025 × 600 B1 M1 For using 0 = v(600 + 2600) – 0.0375t3 and v(600 + 2600) = v(600) 0 = 15 – 0.0375t3 A1 Total time is 3600 s A1 [4] (ii) For correct graph M1 Shape only [d = ½ (2600 + 3600) × 15 or d = ½ 0.025 × 6002 + 2600 × 15 + A1ft For method of finding distance ½ 0.0375 × 4002] Distance is 46500 A1ft [3] (iii) Values of t are 300 and 3400 B1 [1] GCE AS/A LEVEL – May/June 2011 9709 41 ∫dt

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6 A particle travels in a straight line from a point P to a point Q. Its velocity t seconds after leaving P The distance PQ is 64 m. is v m s−1, where v = 4t −116t3. (i) Find the time taken for the particle to travel from P to Q. [5] (ii) Find the set of values of t for which the acceleration of the particle is positive. [4]

Mark scheme: 6 (i) M1 For using s = ∫vdt s = 2t2 – t4/64 (+ C) A1 [t4 – 128t2 + 642 = 0] M1 For attempting to solve s(t) = 64 (t2 – 64)2 = 0 A1 Time taken is 8 s A1 [5] (ii) M1 For using a = dv/dt a = 4 – 3t2/16 A1 8 8 a is positive for 0 < t < or B2 [4] SR: Allow B1 for t < 3 3 0 < t < 4.62 8 SR: B1 for 0 ≤ t ≤ or 4.62 3

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7 Loads A and B, of masses 1.2 kg and 2.0 kg respectively, are attached to the ends of a light inextensible string which passes over a fixed smooth pulley. A is held at rest and B hangs freely, with both straight parts of the string vertical. A is released and starts to move upwards. It does not reach the pulley in the subsequent motion. (i) Find the acceleration of A and the tension in the string. [4] (ii) Find, for the first 1.5 metres of A’s motion, (a) A’s gain in potential energy, (b) the work done on A by the tension in the string, (c) A’s gain in kinetic energy. [3] B hits the floor 1.6 seconds after A is released. B comes to rest without rebounding and the string becomes slack. (iii) Find the time from the instant the string becomes slack until it becomes taut again. [4]

Mark scheme: 7 (i) M1 For applying Newton’s second law to A or to B T – 12 = 1.2a and 20 –T = 2a A1 Accept (2 – 1.2)g = (2.0 + 1.2)a as an alternative for one of these equations Acceleration is 2.5 ms-2 B1 Tension is 15 N A1 [4] (ii) (a) PE gain = 12 × 1.5 = 18 J B1 (b) WD on A = 15 × 1.5 = 22.5J B1 (c) Gain in KE = ans(b) – ans(a) = 4.5 J B1ft [3] alt: KE = ½ 1.2(2 × 2.5 × 1.5) = 4.5J (iii) v = 1.6 × 2.5 B1ft M1 For using v = u – gt t = 0.4 s A1 May be implied Total time taken is 0.8 s A1 [4]

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