Cambridge A Level Mathematics 9709 — 2012 May/June Paper 5 · Variant 3

1 A particle P is projected with speed 25 m s−1 at an angle of 30◦above the horizontal from a point O on horizontal ground. Calculate the distance OP at the instant 2 s after projection. [4]

Mark scheme: 1 OX = (25cos30) × 2 B1 43.3 OY = (25sin30) × 2 – g × 22/2 B1 5 OP2 = 43.32 + 52 M1 OP = 43.6 m A1 [4] [4]

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2 C F N 0.7 m O 2 rad B A The diagram shows a uniform object ABC of weight 3 N in the form of an arc of a circle with centre O and radius 0.7 m. The angle AOC is 2 radians. The object rests in equilibrium with A on a horizontal surface and C vertically above A. Equilibrium is maintained by a horizontal force of magnitude F N applied at C in the plane of the object. Calculate F. [4]

Mark scheme: 2 OG = (0.7sin1)/1 B1 0.589 M1 Moments about A. Accept uncancelled form +/–3 × (0.589 – 0.7cos1) = A1 candidate’s value of 0.589 F × (0.7sin1) × 2 F = 0.537 N A1 [4] [4]

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3 A particle P of mass 0.2 kg is projected horizontally from a fixed point O, and moves in a straight line on a smooth horizontal surface. A force of magnitude 0.4x N acts on P in the direction PO, where x m is the displacement of P from O. (i) Given that P comes to instantaneous rest when x = 2.5, find the initial kinetic energy of P. [4] (ii) Find the value of x on the first occasion when the speed of P is 2 m s−1. [2]

Mark scheme: 3 (i) 0.2vdv/dx = –0.4x M1 Newton’s Second Law, – sign essential v2/2 = –2x2/2 (+ c) A1 Accept uncancelled form 0 = –2 × 2.52/2 + c →c = 6.25 M1 KE = 0.2 × 6.25 = 1.25 J A1 [4] v = 3.54 ms–1 (ii) 22/2 = –2x2/2 + 6.25 M1 v = 2 in accurate integral attempt at limits or finding arbitrary constant e.g. in (i) x = 2.06 A1 [2] [6]

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4 S 0.4 m A small sphere S of mass m kg is moving inside a fixed smooth hollow cylinder whose axis is vertical. S moves with constant speed in a horizontal circle of radius 0.4 m and is in contact with both the plane base and the curved surface of the cylinder (see diagram). (i) Given that the horizontal and vertical forces exerted on S by the cylinder have equal magnitudes, calculate the speed of S. [3] S is now attached to the centre of the base of the cylinder by a horizontal light elastic string of natural length 0.25 m and modulus of elasticity 13 N. The sphere S is set in motion and moves in a horizontal circle with constant angular speed ω rad s−1 and is in contact with both the plane base and the curved surface of the cylinder. (ii) It is given that the magnitudes of the horizontal and vertical forces exerted on S by the cylinder are equal if ω = 8. Calculate m. [3] (iii) For the value of m found in part (ii), find the least possible value of ω for the motion. [2]

Mark scheme: 4 (i) Vertical force = 10m B1 May be implied 10m = mv2/0.4 M1 Newton’s Second Law radially v = 2 ms–1 A1 [3] (ii) T = 13 × (0.4 – 0.25)/0.25 B1 T = 7.8 N m × 82 × 0.4 = 7.8 + 10m M1 Newton’s Second Law radially, 2 horizontal forces m = 0.5 A1 [3] m(25.6 – 10) = 7.8 (iii) 7.8 = m × ω2 × 0.4 M1 Newton’s Second Law radially, no horizontal reaction ω = 6.24 A1 [2] ( 39 ) [8] GCE AS/A LEVEL – May/June 2012 9709 53

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5 A light elastic string has natural length 3 m and modulus of elasticity 45 N. A particle P of mass 0.6 kg is attached to the mid-point of the string. The ends of the string are attached to fixed points A and B which lie on a line of greatest slope of a smooth plane inclined at 30◦to the horizontal. The distance AB is 4 m, and A is higher than B. (i) Calculate the distance AP when P rests on the slope in equilibrium. [3] P is released from rest at the point between A and B where AP = 2.5 m. (ii) Find the maximum speed of P. [4] (iii) Show that P is at rest when AP = 1.6 m. [2]

Mark scheme: 5 (i) M1 Uses T = 45ext/1.5 45e/1.5 = 45(1 – e)/1.5 ± 0.6gsin30 A1 Note either portion may be e AP (= 0.55 + 1.5) = 2.05 m A1 [3] (ii) M1 KE/EE/PE energy conservation 45 × 12/(2 × 1.5) = A1 3 correct EE terms 45 × 0.552/(2 × 1.5) + 45 × 0.452/(2 × 1.5) + 0.6g × 0.45sin30 + 0.6v2/2 A1 Correct equation v = 4.5 ms–1 A1 [4] (iii) M1 EE/PE conservation 45 × 12/(2 × 1.5) = 45(1.6 – 1.5)2/(2 × 1.5) + 45(4 – 1.6 – 1.5)2/(2 × 1.5) + 0.6 × 10(2.5 – 1.6)sin30 A1 [2] Total energy = 15 [9]

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6 E D A h m a m B 0.5 m C A uniform lamina ABCDE consists of a rectangle BCDE and an isosceles triangle ABE joined along their common edge BE. For the triangle, AB = AE, BE = a m and the perpendicular height is h m. For the rectangle, BC = DE = 0.5 m and CD = BE = a m (see diagram). (i) Show that the distance in metres of the centre of mass of the lamina from BE towards CD is 3 −4h2 [4] 12 + 12h. The lamina is freely suspended at E and hangs in equilibrium. (ii) Given that DE is horizontal, calculate h. [2] (iii) Given instead that h = 0.5 and AE is horizontal, calculate a. [3]

Mark scheme: 6 (i) M1 Table of moments idea (ah/2+0.5a)x A1 Correct sum of parts = (ah/2)(–h/3)+(0.5a)(0.5/2) 0.5a(1 + h)x = 0.5a(0.25 – h2/3) M1 Must include cancelling of a x = (3 – 4h2)/(12 + 12h) AG A1 [4] (ii) 3 – 4h2 = 0 M1 Uses x = 0 h = 0.866 A1 [2] (iii) tanθ = x/(a/2) = (a/2)/h M1 Correct trigonometry (2/18)/(a/2) = (a/2)/0.5 DM1 Ratios accurately substituted a = 0.471 A1 [3] [9] GCE AS/A LEVEL – May/June 2012 9709 53 1

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7 The equation of the trajectory of a projectile is y = 0.6x −0.017x2, referred to horizontal and vertically upward axes through the point of projection. (i) Find the angle of projection of the projectile, and show that the initial speed is 20 m s−1. [3] (ii) Find the speed and direction of motion of the projectile when it is at a height of 5.2 m above the level of the point of projection for the second time. [7]

Mark scheme: 7 (i) θ = 31(.0)° B1 θ = tan–10.6 0.017 = 10/[2(vcos31)]2 M1 v = 20 AG A1 [3] (ii) v2 = M1 Accept v2 = 202 – 2g × 5.2 (20cos31)2 + [(20sin31)2 – 2g × 5.2] v = 17.2 ms–1 A1 0.017x2 – 0.6x + 5.2 = 0 M1 Solves 3 term quadratic equation x = 20 A1 Ignore smaller root if shown dy/dx = 0.6 – 0.017(2x) M1 tan α = 0.6 – 0.017(2 × 20) A1 α = 4.6° below the horizontal A1 [7] 4.57° OR 7 (i) θ = 31(.0)° B1 θ = tan–10.6 0.017 = 10/[2(vcos31)]2 M1 v = 20 AG A1 [3] (ii) 5.2 = 20sin31 –10t2/2 M1 Sets up and solves a 3 term quadratic equation t = 1.17 (t = 1.166..) A1 Ignore smaller root if shown v vert = (–)1.37(2) A1 From v = 20sin31 – 10t v2 = 17.1(5)2 + 1.37(2)2 M1 Or uses method in (ii) above 17.1(5) is horizontal velocity component v = 17.2 ms–1 A1 tan α = 1.37(2)/17.1(5) M1 α = 4.6° below the horizontal A1 [7] 4.57° [10]

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