Cambridge A Level Mathematics 9709 — 2014 Oct/Nov Paper 4 · Variant 3

1 A car of mass 1400 kg moves on a horizontal straight road. The resistance to the car’s motion is constant and equal to 800 N and the power of the car’s engine is constant and equal to P W. At an instant when the car’s speed is 18 m s−1 its acceleration is 0.5 m s−2. (i) Find the value of P. [3] The car continues and passes through another point with speed 25 m s−1. (ii) Find the car’s acceleration at this point. [2]

Mark scheme: 1 (i) DF = P ÷ 18 B1 [P ÷ 18 – 800 = 1400 × 0.5] M1 For using DF – R = ma P = 27000 A1 3 (ii) [1080 – 800 = 1400a] M1 For using DF = P ÷ 25 and DF – R = ma Acceleration is 0.2 ms–2 A1 2

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2 Q P A B The tops of each of two smooth inclined planes A and B meet at a right angle. Plane A is inclined at angle to the horizontal and plane B is inclined at angle to the horizontal, where sin = 63 and 65 sin = 1665. A small smooth pulley is fixed at the top of the planes and a light inextensible string passes over the pulley. Two particles P and Q, each of mass 0.65 kg, are attached to the string, one at each end. Particle Q is held at rest at a point of the same line of greatest slope of the plane B as the pulley. Particle P rests freely below the pulley in contact with plane A (see diagram). Particle Q is released and the particles start to move with the string taut. Find the tension in the string. [5]

Mark scheme: 2 M1 For applying Newton’s 2nd law to P or to Q 0.65 × 10 × (63/65) – T = 0.65a or A1 T – 0.65 × 10 × (16/65) = 0.65a T – 0.65 × 10 × (16/65) = 0.65a or B1 0.65 × 10 × (63/65) – T = 0.65a or 0.65 × 10 × (63 – 16)/65 = 2 × 0.65a [T – 1.6 = 6.3 – T] or [T = 6.3 – 0.65 × (47/13)] or [T = 1.6 + 0.65 × (47/13)] M1 For eliminating a Tension is 3.95 N A1 5

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3 O W N 7 N 8 N Each of three light inextensible strings has a particle attached to one of its ends. The other ends of the strings are tied together at a point O. Two of the strings pass over fixed smooth pegs and the particles hang freely in equilibrium. The weights of the particles and the angles between the sloping parts of the strings and the vertical are as shown in the diagram. It is given that sin = 0.8 and cos = 0.6. (i) Show that W cos = 3.8 and find the value of W sin . [3] (ii) Hence find the values of W and . [3]

Mark scheme: 3 (i) [Wcosα + 7 × 0.6 = 8] M1 For resolving forces acting at O vertically Wcosα = 3.8 (cwo) A1 AG Wsinα = 5.6 B1 3 (ii) For using W2 = (Wsinα)2 + (Wcosα)2 or tanα = (Wsinα ÷ Wcosα) M1 W = 6.77 or α = 55.8 A1 α = 55.8 or W = 6.77 B1 3

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4 A particle P starts from rest and moves in a straight line for 18 seconds. For the first 8 seconds of the motion P has constant acceleration 0.25 m s−2. Subsequently P’s velocity, v m s−1 at time t seconds after the motion started, is given by v = −0.1t2 + 2.4t −k, where 8 ≤t ≤18 and k is a constant. (i) Find the value of v when t = 8 and hence find the value of k. [2] (ii) Find the maximum velocity of P. [2] (iii) Find the displacement of P from its initial position when t = 18. [3]

Mark scheme: 4 (i) v(8) = 0.25 × 8 = 2 B1 2 = –6.4 + 19.2 – k k = 10.8 B1 2 ft (12.8 – v) (ii) [dv/dt = –0.2t + 2.4 (= 0 when t = 12) M1 For finding t when vmax = –0.1 × 144 + 2.4 × 12 – 10.8] dv/dt = 0 and substituting into v(t) Maximum speed is 3.6 ms–1 A1 2 ft (14.4 – incorrect k) (iii) Displacement s1 = ½ 0.25 × 82 (= 8) B1 [Displacement M1 For using displacement 18 18 d t (− 1.0t 2 + 4.2t − 10 8. ) s2 = [–0.1t3/3 + 1.2t2 – 10.8t]8 s 2 = ∫8 (=26.7)] Displacement is 34.7 m A1 3

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5 A box of mass 8 kg is on a rough plane inclined at 5 to the horizontal. A force of magnitude P N acts on the box in a direction upwards and parallel to a line of greatest slope of the plane. When P = 7X the box moves up the line of greatest slope with acceleration 0.15 m s−2 and when P = 8X the box moves up the line of greatest slope with acceleration 1.15 m s−2. Find the value of X and the coefficient of friction between the box and the plane. [8]

Mark scheme: 5 [P – 8gsin5o – F = 8a] M1 For using Newton’s 2nd law (either case) 7X – 8gsin5o – F = 8 × 0.15 and 8X – 8gsin5o – F = 8 × 1.15 A1 X = 8 A1 M1 For obtaining a numerical expression for F F = 56 – 8gsin5o – 8 × 0.15 or F = 64 – 8gsin5o – 8 × 1.15 or F = 56 × 1.15 – 64 × 0.15 – 8gsin5o or F = 47.8(275...) A1 ft X either from error for one term in X/F equation or from error in solution of correct X/F equations R = 8gcos5o (= 79.695...) B1 F [µ = 47.8÷79.7] M1 For using µ = R Coefficient is 0.600 (accept 0.6) A1 8

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6 v m s−1 4 Particle P O t s 1.0 1.4 1.8 P Q Particle Q h m −4 Fig. 1 Fig. 2 Particles P and Q have a total mass of 1 kg. The particles are attached to opposite ends of a light inextensible string which passes over a smooth fixed pulley. P is held at rest and Q hangs freely, with both straight parts of the string vertical. Both particles are at a height of h m above the floor (see Fig. 1). P is released from rest and the particles start to move with the string taut. Fig. 2 shows the velocity-time graphs for P’s motion and for Q’s motion, where the positive direction for velocity is vertically upwards. Find (i) the magnitude of the acceleration with which the particles start to move and the mass of each of the particles, [5] (ii) the value of h, [1] (iii) the greatest height above the floor reached by particle P. [2] [Question 7 is printed on the next page.]

Mark scheme: 6 (i) M1 For using the gradient property for acceleration Acceleration is 4 ms–2 A1 M1 For applying Newton’s 2nd law to both particles or using the formula (M + m)a = (M – m)g and for using m + M = 1 For T – mg = 4m and (1 – m)g – T = 4(1 – m) or 4 = (1 – m – m)g A1 P has mass 0.3 kg and Q has mass 0.7 kg A1 5 (ii) For using the area property of the graph or h = ½ at2 to obtain h = 2 B1 1 (iii) Distance travelled upwards by B1 P = ½ 1.4 × 4 Height is 4.8 m B1 2

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7 35 N s−1 m 4 A m 12.5 O A small block of mass 3 kg is initially at rest at the bottom O of a rough plane inclined at an angle to the horizontal, where sin = 0.6 and cos = 0.8. A force of magnitude 35 N acts on the block at an angle above the plane, where sin = 0.28 and cos = 0.96. The block starts to move up a line of greatest slope of the plane and passes through a point A with speed 4 m s−1. The distance OA is 12.5 m (see diagram). (i) For the motion of the block from O to A, find the work done against the frictional force acting on the block. [4] (ii) Find the coefficient of friction between the block and the plane. [3] At the instant that the block passes through A the force of magnitude 35 N ceases to act. (iii) Find the distance the block travels up the plane after passing through A. [4]

Mark scheme: 7 (i) 42 = 02 + 2a × 12.5 a = 0.64 B1 [35 × 0.96 – 3g × 0.6 – F = 3 × 0.64] M1 For using Newton’s 2nd law to find F F = 13.68 A1 WD against F = 13.68 × 12.5 = 171 J B1 4 (ii) Rfrom O to A = 3g × 0.8 – 35 × 0.28 B1 [µ = 13.68 ÷ 14.2 (= 0.96338)] M1 For using µ = F ÷ R Coefficient is 0.963 (accept 0.96) A1 3 (iii) [–3g × 0.6 – 0.96338 × (3g × 0.8) = 3a] M1 For applying Newton’s 2nd law to the block to find a Acceleration is –13.7 ms–2 A1 [0 = 16 + 2(–13.7)s] M1 For using v2 = u2 + 2as to find s Distance travelled is 0.584 m A1 4 Alternative for part (i) (i) Gain in KE = ½ 3 × 42 ( = 24 J) B1 Gain in PE = 3g × 12.5 × 0.6 ( = 225 J) B1 [WD = 35 × 12.5 × 0.96 – ½ 3 × 42 – M1 For using WD against F 3g × 12.5 × 0.6] = WD by applied force – KE gain – PE gain WD against F is 171 J A1 4 Alternative for part (iii) WD against F = 0.96(338..) × 3g × 0.8s B1 M1 For using KE loss = PE gain + WD against friction ½ 3 × 42 = 3gs(0.6) + 0.96(338..) × 3g × 0.8s A1 Distance travelled is 0.584 m A1 4

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