Cambridge A Level Mathematics 9709 — 2013 Oct/Nov Paper 4 · Variant 1

1 X a P F N A particle P of mass 0.3 kg is attached to one end of a light inextensible string. The other end of the string is attached to a fixed point X. A horizontal force of magnitude F N is applied to the particle, which is in equilibrium when the string is at an angle ! to the vertical, where tan ! = 8 (see diagram). 15 Find the tension in the string and the value of F. [4]

Mark scheme: 1 [Tcosα = mg] M1 For resolving forces vertically Tension is 3.4 N A1 [F = Tsinα] M1 For resolving forces horizontally F = 1.6 A1 4

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2 30 N a B x b 40 N A block B lies on a rough horizontal plane. Horizontal forces of magnitudes 30 N and 40 N, making angles of ! and " respectively with the x-direction, act on B as shown in the diagram, and B is moving in the x-direction with constant speed. It is given that cos ! = 0.6 and cos " = 0.8. (i) Find the total work done by the forces shown in the diagram when B has moved a distance of 20 m. [2] (ii) Given that the coefficient of friction between the block and the plane is 58, find the weight of the block. [3]

Mark scheme: 2 (i) [WD = 30 × 20 × 0.6 + 40 × 20 × 0.8] M1 For using WD = Fdcosθ Work done is 1000 J A1 2 (ii) For applying F = µW and Newton’s 2nd law M1 with a = 0 30 × 0.6 + 40 × 0.8 – 0.625W = 0 A1 Weight is 80 N A1 3 d

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3 A cyclist exerts a constant driving force of magnitude F N while moving up a straight hill inclined at an angle ! to the horizontal, where sin ! = 325.36 A constant resistance to motion of 32 N acts on the cyclist. The total weight of the cyclist and his bicycle is 780 N. The cyclist’s acceleration is −0.2 m s−2. (i) Find the value of F. [4] The cyclist’s speed is 7 m s−1 at the bottom of the hill. (ii) Find how far up the hill the cyclist travels before coming to rest. [2]

Mark scheme: 3 (i) For applying Newton’s 2nd law to the M1 bicycle/cyclist F – 780 × (36÷325) – 32 A2 (A2 for all correct, A1 for one error, A0 for = 78 × (–0.2) more than one error) F = 103 (102.8 exact) A1 4 (ii) [0 = 72 + 2(–0.2)s] M1 For using 0 = u2 + 2as Distance is 122.5 m (accept 122 or 123) A1 2 d

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4 Particles P and Q are moving in a straight line on a rough horizontal plane. The frictional forces are the only horizontal forces acting on the particles. (i) Find the deceleration of each of the particles given that the coefficient of friction between P and the plane is 0.2, and between Q and the plane is 0.25. [2] At a certain instant, P passes through the point A and Q passes through the point B. The distance AB is 5 m. The velocities of P and Q at A and B are 8 m s−1 and 3 m s−1, respectively, both in the direction AB. (ii) Find the speeds of P and Q immediately before they collide. [5]

Mark scheme: 4 (i) For using Newton’s 2nd law, [– µmg = ma] M1 F = µR and R = mg Decelerations of P and Q are 2 ms–2 and 2.5 ms–2. A1 2 (ii) For using s = ut + ½ at2 M1 and sP = sQ + 5 8t – t2 = 3t – 1.25t2 +5 A1 t = √120 – 10 (=0.95445…) A1 For using v = u + at for both P and Q M1 Speed of P = 6.09 ms–1, speed of Q = 0.614 ms–1 A1 5 GCE AS/A LEVEL – October/November 2013 9709 41

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5 A lorry of mass 15 000 kg climbs from the bottom to the top of a straight hill, of length 1440 m, at a constant speed of 15 m s−1. The top of the hill is 16 m above the level of the bottom of the hill. The resistance to motion is constant and equal to 1800 N. (i) Find the work done by the driving force. [4] On reaching the top of the hill the lorry continues on a straight horizontal road and passes through a point P with speed 24 m s−1. The resistance to motion is constant and is now equal to 1600 N. The work done by the lorry’s engine from the top of the hill to the point P is 5030 kJ. (ii) Find the distance from the top of the hill to the point P. [3]

Mark scheme: 5 (i) Gain in PE =15000g × 16 B1 WD against resistance = 1800 × 1440 B1 For using:– WD by driving force = Gain in PE M1 + WD against resistance Work done is 4.99x106 J A1 4 (ii) For using :– WD by engine = Increase in KE + WD against resistance M1 5030 000 = ½ 15 000(242 – 152) + 1600d A1 Distance is 1500 m A1 3 d

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6 A B 0.52 m Particles A and B, of masses 0.3 kg and 0.7 kg respectively, are attached to the ends of a light inextensible string. The string passes over a fixed smooth pulley. A is held at rest and B hangs freely, with both straight parts of the string vertical and both particles at a height of 0.52 m above the floor (see diagram). A is released and both particles start to move. (i) Find the tension in the string. [4] When both particles are moving with speed 1.6 m s−1 the string breaks. (ii) Find the time taken, from the instant that the string breaks, for A to reach the floor. [5] [Question 7 is printed on the next page.]

Mark scheme: 6 (i) For applying Newton’s 2nd law to A or to B M1 T – 0.3g = 0.3a or 0.7g – T = 0.7a A1 0.7g – T = 0.7a or T – 0.3g = 0.3a or 0.7g – 0.3g = (0.7 + 0.3)a B1 Tension is 4.2 N A1 4 (ii) a = 4 B1 May be scored in (i) staut = 1.62/(2 × 4) (= 0.32) B1 [(0.52 + 0.32) = –1.6t + 5t2] M1 For using s = ut + ½ gt2 For solving the resultant quadratic [(t – 0.6)(5t + 1.4) = 0] M1 equation. Time taken is 0.6 s A1 5 Alternative Marking Scheme for the last three marks 02 = 1.62 – 2gsup, For using kinematic formulae to find tup tup = 2sup/(1.6 + 0) (= 0.16) M1 2 0.52 + staut + sup = 0 + ½ gtdown For using kinematic formulae to find tdown (tdown = 0.44) M1 Time taken = tup + tdown = 0.6 s B1 GCE AS/A LEVEL – October/November 2013 9709 41

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7 A particle P starts from rest at a point O and moves in a straight line. P has acceleration 0.6t m s−2 at time t seconds after leaving O, until t = 10. (i) Find the velocity and displacement from O of P when t = 10. [5] After t = 10, P has acceleration −0.4t m s−2 until it comes to rest at a point A. (ii) Find the distance OA. [7]

Mark scheme: 7 (i) For integrating 0.6t and using v(0) = 0 (may be implied by absence of constant of M1 integration) v(t) = 0.3t2 A1 For integrating v(t) and using s(0) = 0 (may be implied by absence of constant of integration) M1 s(t) = 0.1t3 A1 Velocity is 30 ms–1 and displacement is 100 m A1 5 (ii) For integrating –0.4t and using v(10) = 30 M1 v(t ) = –0.2t2 + 50 A1 At A, –0.2t2 + 50 = 0 t = √250 B1 For integrating v(t) and using s(10) = 100 M1 s(t) = –t3/15 + 50t – 1000/3 A1 M1 For finding s(√250) Distance OA is 194 m A1 7

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