Cambridge A Level Mathematics 9709 — 2008 May/June Paper 4 · Variant 1

1 A particle slides down a smooth plane inclined at an angle of α◦to the horizontal. The particle passes through the point A with speed 1.5 m s−1, and 1.2 s later it passes through the point B with speed 4.5 m s−1. Find (i) the acceleration of the particle, [2] (ii) the value of α. [2]

Mark scheme: 1 (i) [4.5 = 1.5 + 1.2a] M1 For using v = u + at Acceleration is 2.5 ms-2 A1 [2] (ii) M1 For using (m)gsin α ° = (m)a α = 14.5 A1 [2]

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2 A block is being pulled along a horizontal floor by a rope inclined at 20◦to the horizontal. The tension in the rope is 851 N and the block moves at a constant speed of 2.5 m s−1. (i) Show that the work done on the block in 12 s is approximately 24 kJ. [3] (ii) Hence find the power being applied to the block, giving your answer to the nearest kW. [1]

Mark scheme: 2 (i) Distance is 2.5x12m or power = 851cos20° x 2.5 B1 [WD = 851x30cos20°] M1 For using WD = Tdcosα (or Pt) Work done is 24 kJ A1 [3] AG (ii) Power is 2 kW B1 [1]

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3 F N O ° 10 N 13 N Three horizontal forces of magnitudes F N, 13 N and 10 N act at a fixed point O and are in equilibrium. The directions of the forces are as shown in the diagram. Find, in either order, the value of θ and the value of F. [5]

Mark scheme: 3 M1 For resolving forces in i and j directions F 13 or sketching a triangle of forces (with 10, h 13 and F shown) [Fcosθ° = 10, Fsin θ° = 13; 10 ] [tanθ° = 13/10, 269 sinθ° = 13] M1 For an equation in θ only θ = 52.4 A1 [F2 = 102 + 132, Fcos52.4o = 10] M1 For an equation in F only F = 16.4 A1 [5] Alternative scheme for candidates who use scale drawing: M1 For scale drawing of correct triangle M1 For measuring θ and finding a value in the range [51, 54] θ = 52.4 A1 M1 For measuring F and finding a value in the range [15.5, 17.5] F = 16.4 A1 [5]

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4 O 2.4 m 50° A C B OABC is a vertical cross-section of a smooth surface. The straight part OA has length 2.4 m and makes an angle of 50◦with the horizontal. A and C are at the same horizontal level and B is the lowest point of the cross-section (see diagram). A particle P of mass 0.8 kg is released from rest at O and moves on the surface. P remains in contact with the surface until it leaves the surface at C. Find (i) the kinetic energy of P at A, [2] (ii) the speed of P at C. [2] The greatest speed of P is 8 m s−1. (iii) Find the depth of B below the horizontal through A and C. [3]

Mark scheme: 4 (i) [KE = Loss of PE = 0.8g(2.4sin50o), For using KE = PE loss = mgh or KE = ½ 0.8 x 2(gsin50o)2.4] M1 KE = ½ mv2 and v2 = 2as Kinetic energy at A is 14.7J A1 [2] (ii) [14.7 = ½ mv2] M1 For using KE at C = KE at A = ½ mv2 Speed at C is 6.06ms-1 A1ft [2] ft v = (2.5 KE)½ (iii) [½ m82 = mgH, ½ m82 – ½ m6.062 = mgh] M1 For using the principle of conservation of energy h = 3.2 – 2.4sin50o or 10h = ½ (82 – 6.062) A1ft ft 10h = ½ (82 – vC2) Depth is 1.36m A1 [3] SR in (iii) (max. mark 1/3) For depth = 1.36 from v2 = u2 + 2gs B1 GCE A/AS LEVEL – May/June 2008 9709 04

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5 B 3 m C A A block B of mass 0.6 kg and a particle A of mass 0.4 kg are attached to opposite ends of a light inextensible string. The block is held at rest on a rough horizontal table, and the coefficient of friction between the block and the table is 0.5. The string passes over a small smooth pulley C at the edge of the table and A hangs in equilibrium vertically below C. The part of the string between B and C is horizontal and the distance BC is 3 m (see diagram). B is released and the system starts to move. (i) Find the acceleration of B and the tension in the string. [6] (ii) Find the time taken for B to reach the pulley. [2]

Mark scheme: 5 (i) F = 0.5(0.6g) B1 M1 For applying Newton’s second law to A or to B 0.4g – T = 0.4a A1 Alternative to either of the above equations:- T – F = 0.6a A1 0.4g – F = (0.4 + 0.6)a B1 SR in lieu of the previous 3 marks (max. mark 1/3) 0.4g – T = 0.4ga and T – F = 0.6ga B1 M1 For substituting for F and solving for a or for T Acceleration is 1ms-2 and tension is 3.6N A1 [6] (ii) M1 For using s = (0) + ½ at2 Time taken is 2.45s A1ft [2] ft t = (6/a)½ 2

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6 A particle P of mass 0.6 kg is projected vertically upwards with speed 5.2 m s−1 from a point O which is 6.2 m above the ground. Air resistance acts on P so that its deceleration is 10.4 m s−2 when P is moving upwards, and its acceleration is 9.6 m s−2 when P is moving downwards. Find (i) the greatest height above the ground reached by P, [3] (ii) the speed with which P reaches the ground, [2] (iii) the total work done on P by the air resistance. [4]

Mark scheme: 6 (i) M1 For using 0 = u2 + 2as, or 0 = u + at and s = ut + ½ at2, or 0 = u + at and s = (u + 0)t/2 0 = 5.22 – 2x10.4s1 or s1 = 5.2x0.5 - ½ 10.4x0.52 or s1 = (5.2 + 0)x0.5/2 A1 Greatest height is 7.5m A1 [3] (ii) [v2 = 2x9.6x7.5, v = 9.6x1.25, For using v2 = 0 + 2as, or v = 2x7.5/1.25] M1 s = ½ at2 and v = at, or s = ½ at2 and 0 + v = 2s/t Speed is 12ms-1 A1 [2] (iii) PE loss = 0.6g x 6.2 (= 37.2) or Initial total energy = 0.6gx6.2 + ½ 0.6x5.22 (= 45.312) or Energy loss upward = ½ 0.6x5.22 – 0.6gx1.3 (= 0.312) B1 KE gain = ½ 0.6(122 – 5.22) (= 35.088) or Final total energy = ½ 0.6x122 (= 43.2) Energy loss downward = - ½ 0.6x122 + 0.6gx7.5 (=1.8) B1ft ft ans (ii) For using WD = PE loss from the start – KE gain from the start or WD = Initial total energy – final total energy [WD = 37.2 – 35.088 or 45.312 – 43.2 or M1 WD = energy loss upward + 0.312 + 1.8] energy loss downward Work done is 2.11(2) J A1 [4] Accept exact or 3sf GCE A/AS LEVEL – May/June 2008 9709 04 Alternatively [0.6g + Rup = 0.6x10.4 or 0.6g - Rdown = M1 For applying Newton’s second law to the 0.6x9.6] upward motion or to the downward motion, and attempting to find Rup or Rdown Rup = 0.24 or Rdown = 0.24 A1 May be implied by final answer. M1 For using WD(upward) = 1.3Rup or WD(downward) = ans(i)Rdown Work done is 2.11(2) J A1ft [4] ft ans (i)

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7 v (m s –1) t (s) O 10 30 80 An object P travels from A to B in a time of 80 s. The diagram shows the graph of v against t, where v m s−1 is the velocity of P at time t s after leaving A. The graph consists of straight line segments for the intervals 0 ≤t ≤10 and 30 ≤t ≤80, and a curved section whose equation is v = −0.01t2 + 0.5t −1 for 10 ≤t ≤30. Find (i) the maximum velocity of P, [4] (ii) the distance AB. [9]

Mark scheme: 7 (i) (dv/dt) = -0.02t + 0.5 or B1 v = -0.01[(t – T)2 – 100V] where T = 25 and V = 5.25 (or equivalent) M1 For solving dv/dt = 0 or for selecting t = T or vmax = V May be implied when vmax = V is selected and T is 25 in the ‘B1’ expression for v t = 25 A1 Maximum velocity is 5.25ms-1 A1 [4] (ii) M1 For integrating v(t) s2 = -0.01t3/3 + 0.5t2/2 - t A1 M1 For using limits 10 and 30 s2 = (-90 + 225 – 30) – (-10/3 + 25 – 10) (= 93.3m) A1 M1 For evaluating v(10) and v(30) v(10) = 3 and v(30) = 5 A1 M1 For evaluating s1 and s3 s1 = ½ 3x10 and s3 = ½ 5x50 A1ft ft incorrect values of v(10) and/or v(30) Distance is 233m A1ft [9] ft 140 + s2 (depends on the 1stM1) SR for candidates who treat the first line segment as part of the curve in part (ii) (max. mark 6/9) Integration M1 A1 as scheme s1 + s2 = 105 A1 v(30) = 5 B1 s3 = ½ 5x50 B1ft Distance is 230m A1ft (ft 125 + s1 + s2)

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