Cambridge A Level Mathematics 9709 — 2012 Oct/Nov Paper 4 · Variant 1

1 An object is released from rest at a height of 125 m above horizontal ground and falls freely under gravity, hitting a moving target P. The target P is moving on the ground in a straight line, with constant acceleration 0.8 m s−2. At the instant the object is released P passes through a point O with speed 5 m s−1. Find the distance from O to the point where P is hit by the object. [4]

Mark scheme: 1 [125 = ½ 10t2 M1 For using h = ½ gt2 t = 5 s A1 [s = 5 × 5 ½ 0.8 × 52] M1 For using s = ut + ½ at2 Distance is 35 m A1 4 d

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2 A 0.3 kg B 0.2 kg Particles A and B, of masses 0.3 kg and 0.2 kg respectively, are attached to the ends of a light inextensible string. A is held at rest on a rough horizontal table with the string passing over a small smooth pulley at the edge of the table. B hangs vertically below the pulley (see diagram). The system is released and B starts to move downwards with acceleration 1.6 m s−2. Find (i) the tension in the string after the system is released, [2] (ii) the frictional force acting on A. [3]

Mark scheme: 2 (i) [0.2g – T = 0.2 × 1.6] M1 For applying Newton’s 2nd law to B Tension is 1.68 N A1 2 (ii) M1 For applying Newton’s 2nd law to A T – F = 0.3 × 1.6 A1 Frictional force is 1.2 N A1ft 3 ft T – 0.48

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3 0.6 N a P a A particle P of mass 0.5 kg rests on a rough plane inclined at angle α to the horizontal, where sin α = 0.28. A force of magnitude 0.6 N, acting upwards on P at angle α from a line of greatest slope of the plane, is just sufficient to prevent P sliding down the plane (see diagram). Find (i) the normal component of the contact force on P, [2] (ii) the frictional component of the contact force on P, [3] (iii) the coefficient of friction between P and the plane. [2]

Mark scheme: 3 (i) [R + 0.6sinα = 0.5g cosα ] M1 For resolving forces perpendicular to the plane Normal component is 4.63(2) N A1 2 (ii) M1 For resolving forces parallel to a line A1 of greatest slope F + 0.6cosα = 0.5g sinα Frictional component is 0.824 N A1 3 (iii) M1 For using µ = F/R Coefficient is 0.178 A1 ft 2

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4 R N 12 N 8 N q° 10° 25° 2 N Three coplanar forces of magnitudes 8 N, 12 N and 2 N act at a point. The resultant of the forces has magnitude R N. The directions of the three forces and the resultant are shown in the diagram. Find R and θ. [7]

Mark scheme: 4 M1 For resolving forces in the ‘x’ and ‘y’ directions X = 12cos25o – 8cos10o (= 2.9972....) A1 Y = 12sin25o + 8sin10o – 2 (= 4.4606....) A1 M1 For using R2 = X2 + Y2 R = 5.37 A1 M1 For using tanθ = X/Y θ = 33.9 A1 7 2 2

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5 Particle P travels along a straight line from A to B with constant acceleration 0.05 m s−2. Its speed at A is 2 m s−1 and its speed at B is 5 m s−1. (i) Find the time taken for P to travel from A to B, and find also the distance AB. [3] Particle Q also travels along the same straight line from A to B, starting from rest at A. At time t s after leaving A, the speed of Q is kt3 m s−1, where k is a constant. Q takes the same time to travel from A to B as P does. (ii) Find the value of k and find Q’s speed at B. [5]

Mark scheme: 5 (i) [5 = 2 + 0.05t or 25 = 4 + 2 × 0.05(AB)] M1 For using v = u + at or v2 = u2 + 2as Time taken is 60 s (or Distance is 210 m) A1 Distance is 210 m (or Time taken is 60 s) B1 3 (ii) s = kt4/4 (+C) B1 C = 0 (may be implied by its absence) B1 [210 = k × 604/4] M1 For using s = 210 when t = 60 k = 7/108000 or 0.0000648 A1 Speed of Q at B is 14 ms-1 B1ft 5 ft k × 603 GCE AS/A LEVEL – October/November 2012 9709 41 2 2

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6 C B 8 m s–1 3.0 m 2.7 m P A D The diagram shows the vertical cross-section ABCD of a surface. BC is a circular arc, and AB and CD are tangents to BC at B and C respectively. A and D are at the same horizontal level, and B and C are at heights 2.7 m and 3.0 m respectively above the level of A and D. A particle P of mass 0.2 kg is given a velocity of 8 m s−1 at A, in the direction of AB (see diagram). The parts of the surface containing AB and BC are smooth. (i) Find the decrease in the speed of P as P moves along the surface from B to C. [4] The part of the surface containing CD exerts a constant frictional force on P, as it moves from C to D, and P comes to rest as it reaches D. (ii) Find the speed of P when it is at the mid-point of CD. [5]

Mark scheme: 2 2 6 (i) ½ mvB = ½ mvA – mg × 2.7 M1 For using the principle of 2 2 and ½ mvc = ½ mvA – mg × 3 A1 conservation of energy from A to B or from A to C 2 2 [vB = 82 – 20 × 2.7, vC = 82 – 20 × 3] M1 For substituting for vA to find vB – vC ½ Loss of speed = 10 – 2 = 1.16 ms–1 A1 4 2 (ii) Work done = ½ 0.2 × 2 + 0.2 × g × 3 M1 For using: (= 6.4) A1 WD against friction (C to D) = KE at C + loss of PE (C to D) M1 For using WD against friction (M to D) = KE at M + loss of PE (M to D) 2 ½ (0.4 + 6) = ½ 0.2vM + 0.2g × 1.5 A1 Speed at midpoint is 1.41 ms–1 A1 5

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7 A car of mass 1200 kg moves in a straight line along horizontal ground. The resistance to motion of the car is constant and has magnitude 960 N. The car’s engine works at a rate of 17 280 W. (i) Calculate the acceleration of the car at an instant when its speed is 12 m s−1. [3] The car passes through the points A and B. While the car is moving between A and B it has constant speed V m s−1. (ii) Show that V = 18. [2] At the instant that the car reaches B the engine is switched off and subsequently provides no energy. The car continues along the straight line until it comes to rest at the point C. The time taken for the car to travel from A to C is 52.5 s. (iii) Find the distance AC. [5]

Mark scheme: 7 (i) DF = 17280/12 (= 1440 N) B1 [DF – R = ma 1440 – 960 = 1200a] M1 For using Newton’s 2nd law Acceleration is 0.4 ms–2 A1 3 (ii) [17280/V – 960 = 0] M1 For using P/v – R = 0 V = 18 A1 2 AG (iii) For BC, –960 = 1200a (a = –0.8) B1 M1 For using 0 = 18 +at and 0 = 182 + 2as for BC tBC = (0 – 18)/(–0.8) and sBC = (0 – 182)/(–1.6) (= 22.5 s and 202.5 m) A1 Distance AB = 18(52.5 – 22.5) B1 Distance is AC is 742.5 m A1 5 Accept 742 or 743

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