Cambridge A Level Mathematics 9709 — 2013 Oct/Nov Paper 5 · Variant 1

1 A particle P of mass 0.3 kg is attached to one end of a light inextensible string of length 0.6 m. The other end of the string is attached to a fixed point O of a smooth horizontal plane. P moves on the plane at constant speed 5 m s−1 in a circle with centre O. Calculate the tension in the string. [2]

Mark scheme: 1 T = 0.3 × 5 2 / 0.6 M1 Uses acc n = v 2 /r T = 12.5 N A1 [2] [2]

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2 B A C O 0.6 m A uniform frame consists of a semicircular arc ABC of radius 0.6 m together with its diameter AOC, where O is the centre of the semicircle (see diagram). (i) Calculate the distance of the centre of mass of the frame from O. [4] The frame is freely suspended at A and hangs in equilibrium. (ii) Calculate the angle between AC and the vertical. [2]

Mark scheme: 2 (i) OG(arc) = 0.6sin(π / 2)/(π / 2) B1 0.38197... (0.6π + 2 × 0.6)d M1 Moment equation = 2 × 0.6 × 0 + 0.6π × 0.382 A1 d = 0.233 m A1 [4] 0.2333.. (ii) tanθ = 0.233/0.6 M1 θ = 21.2 / 21.3 o or 0.371 radians A1ft [2] tan −(cv(i)/0.6)1 [6] 2 M1 N2L t diff t i

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3 A particle P of mass 0.8 kg moves along the x-axis on a horizontal surface. When the displacement of P from the origin O is x m the velocity of P is v m s−1 in the positive x-direction. Two horizontal forces act on P. One force has magnitude 4e−x N and acts in the positive x-direction. The other force has magnitude 2.4x2 N and acts in the negative x-direction. (i) Show that vdv = 5e−x −3x2. [2] dx (ii) The velocity of P as it passes through O is 6 m s−1. Find the velocity of P when x = 2. [5]

Mark scheme: 3 (i) 0.8vdv/dx = 4e − x – 2.4x 2 M1 N2L, terms different signs vdv/dx = 5e − x – 3x 2 AG A1 [2] − x 2 M1 Attempts integration (ii) ∫ vdv = ∫ (5e – 3x )dx v 2 / 2 = –5e − x – 3x 3 / 3 (+c) A1 Accept c omitted x = 0, v = 6, hence c = 23 B1 Or uses limits 0 and 2 v 2 / 2 = –5e − 2 – 3x2 3 / 3 + 23 M1 Puts x = 2 in v(x) expression v = 5.35 ms − 1 A1 [5] v = 5.352.. [7]

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4 A small ball B is projected from a point O with speed 14 m s−1 at an angle of 60Å above the horizontal. (i) Calculate the speed and direction of motion of B for the instant 1.8 s after projection. [5] The point O is 2 m above a horizontal plane. (ii) Calculate the time after projection when B reaches the plane. [3]

Mark scheme: 4 (i) V(vert) = 14sin60 – 1.8g B1 –5.8756.. V 2 = (–)5.8756 2 + (14cos60) 2 M1 V = 9.14 ms − 1 A1 9.1391.. tanθ = (–)5.8756/(14cos60) M1 θ = 40(.0) o below horizontal A1 [5] (ii) –2 = (14sin60)t – gt 2 /2 M1 –2 = ut – gt 2 /2 used vertically 5t 2 – 12.124t – 2 = 0 M1 Solves correct 3 term quadratic t = 2.58 s A1 [3] [8] GCE A LEVEL – October/November 2013 9709 51

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5 A 0.4 m P 0.3 m B A particle P of mass 0.2 kg is attached to a fixed point A by a light inextensible string of length 0.4 m. A second light inextensible string of length 0.3 m connects P to a fixed point B which is vertically below A. The particle P moves in a horizontal circle, which has its centre on the line AB, with the angle APB = 90Å (see diagram). (i) Given that the tensions in the two strings are equal, calculate the speed of P. [5] (ii) It is given instead that P moves with its least possible angular speed for motion in this circle. Find this angular speed. [3]

Mark scheme: 5 (i) M1 Resolves vertically, 3 forces Tx(4/5) – Tx(3/5) = 0.2g A1 T = 10 A1 Maybe implied Tx(4/5) + Tx(3/5) = 0.2v 2 / (0.4 × 3 / 5) M1 Resolves horizontally. N2L v = 4.1(0) ms − 1 A1 [5] (ii) Tx(4 / 5) = 0.2g B1 T = 2.5 Tx(3 / 5) = 0.2ω 2 x(0.4 × 3 / 5) M1 N2L horizontally, single force ω =5.59 rads − 1 A1 [3] [8]

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6 B 1.5 m 0.8 m A C E 30° D ABCD is the cross-section through the centre of mass of a uniform rectangular block of weight 260 N. The lengths AB and BC are 1.5 m and 0.8 m respectively. The block rests in equilibrium with the point D on a rough horizontal floor. Equilibrium is maintained by a light rope attached to the point A on the block and the point E on the floor. The points E, A and B lie in a straight line inclined at 30Å to the horizontal (see diagram). (i) By taking moments about D, show that the tension in the rope is 146 N, correct to 3 significant figures. [5] (ii) Given that the block is in limiting equilibrium, calculate the coefficient of friction between the block and the floor. [4] [Question 7 is printed on the next page.]

Mark scheme: 6 (i) 0.8T = 260 × (DG) × cosθ M1 Moments about D DG = 1.7/2, θ = (30+D) M1 Both needed Angle BDC = 28 o DA1 D = 28.072.. 0.8T = 260 × (1.7/2) × cos58.07 A1ft ftcv(DG≠0.8,1.5.1.7,θ ≠30,28) T = 146 N AG A1 [5] OR Moment of weight M1 =(260cos30) × 0.75 – (260sin30) × 0.4 DA1 Difference of moments of perp components (116.87…) M1 Moments about D 0.8T = 116.87.. A1 Needs no evaluation T = 146 N AG A1 (ii) F r = 146cos30 B1 126.52.. R = 260 + 146cos60 B1 333.04.. µ =(146cos30)/(260+146sin30) M1 Denominator not 260 µ = 0.38(0) A1 [4] [9] GCE A LEVEL – October/November 2013 9709 51

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7 A particle P of mass 0.4 kg is attached to one end of a light elastic string of natural length 0.8 m and modulus of elasticity 32 N. The other end of the string is attached to a fixed point O. The particle is released from rest at O. (i) Calculate the distance OP at the instant when P first comes to instantaneous rest. [4] A horizontal plane is fixed at a distance 1 m below O. The particle P is again released from rest at O. (ii) Calculate the speed of P immediately before it collides with the plane. [3] (iii) In the collision with the plane, P loses 96% of its kinetic energy. Calculate the distance OP at the instant when P first comes to instantaneous rest above the plane, given that this occurs when the string is slack. [3]

Mark scheme: 7 (i) M1 PE/EE balance 0.4gd = 32(d–0.8) 2 /(2 × 0.8) A1 20d 2 – 36d + 12.8 = 0 M1 Solves 3 term quadratic d = 1.31 m only A1 [4] Other value 0.4876.. OR 0.4g(0.8 + e) = 32e 2 /(2 × 0.8) M1 PE/EE balance 20e 2 –4e + 3.2 = 0 A1 e = 0.5(1) (also –3.12) M1 Solves 3 term quadratic d = 1.31 m only A1 M1 EE/KE/PE balance (ii) 0.4v 2 /2 A1 = 0.4g × 1 – 32(1–0.8) 2 /(2 × 0.8) v = 4 ms − 1 A1 [3] B1ft ftcv(v(ii) × (1–0.96) = 0.2v(ii) (iii) Rebound v = 0.8 0 = 0.4 × 0.8 2 / 2 + 32 × 0.2 2 / 1.6–0.4gh M1 EE/PE/KE balance, h = 0.232 OP (=1–h) = 0.768 m A1 [3] [10]

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