Cambridge A Level Mathematics 9709 — 2016 May/June Paper 5 · Variant 3

1 A small ball is projected with speed 16 m s−1 at an angle of 45Å above the horizontal from a point on horizontal ground. Calculate the period of time, before the ball lands, for which the speed of the ball is less than 12 m s−1. [4]

Mark scheme: Part Qu Answer Marks Notes Marks 1 v 2 = 12 2 – (16cos45 2) M1 v = 4 A1 –4 = 4 – gt M1 t = 0.8 s A1 4

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2 C 15 N 0.8 m B 70Å m 0.8 A A uniform wire has the shape of a semicircular arc, with diameter AB of length 0.8 m. The wire is attached to a vertical wall by a smooth hinge at A. The wire is held in equilibrium with AB inclined at 70Å to the upward vertical by a light string attached to B. The other end of the string is attached to the point C on the wall 0.8 m vertically above A. The tension in the string is 15 N (see diagram). (i) Show that the horizontal distance of the centre of mass of the wire from the wall is 0.463 m, correct to 3 significant figures. [3] (ii) Calculate the weight of the wire. [2]

Mark scheme: 2 (i) OG = 0.4sin(π/2)/(π/2) B1 = 0.25464... d = OG cos70 + 0.4sin70 M1 d = 0.463 AG A1 3 (ii) 0.463W = 15 × 0.8cos35 M1 W = 21.2 N A1 2

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3 A particle P of mass 0.4 kg is released from rest at a point O on a smooth plane inclined at 30Å to the horizontal. When the displacement of P from O is x m down the plane, the velocity of P is v m s−1. A force of magnitude 0.8e−x N acts on P up the plane along the line of greatest slope through O. (i) Show that vdv = 5 −2e−x. [2] dx (ii) Find v when x = 0.6. [4]

Mark scheme: 3 (i) 0.4vdv/dx = 0.4 g sin30 – 0.8 e− x M1 vdv/dx = 5 – 2 e− x AG A1 2 (ii) ∫ vdv = ∫ (5 − 2 e− x ) dx M1 Separates the variables and attempts to integrate v 2 /2 = 5x + 2 e− x ( + c ) A1 M1 Uses limits or finds c (c = –2) v = 2.05 A1 4

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4 4.4 m 0.4 m 20Å A uniform solid cone has base radius 0.4 m and height 4.4 m. A uniform solid cylinder has radius 0.4 m and weight equal to the weight of the cone. An object is formed by attaching the cylinder to the cone so that the base of the cone and a circular face of the cylinder are in contact and their circumferences coincide. The object rests in equilibrium with its circular base on a plane inclined at an angle of 20Å to the horizontal (see diagram). (i) Calculate the least possible value of the coefficient of friction between the plane and the object. [2] (ii) Calculate the greatest possible height of the cylinder. [4]

Mark scheme: 4 (i) µ = Wsin20/(Wcos20) M1 µ = tan20 µ = 0.364 A1 2 (ii) Wx/2 + W(x+4.4/4) = 2WOG M1 Attempts to take moments A1 OG = distance to C from M OG = 0.4tan70 ( = 0.4/tan20) B1 x = 0.732 A1 4

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5 A particle is projected at an angle of 1Å below the horizontal from a point at the top of a vertical cliff 26 m high. The particle strikes horizontal ground at a distance 8 m from the foot of the cliff2 s after the instant of projection. Find (i) the speed of projection of the particle and the value of 1, [6] (ii) the direction of motion of the particle immediately before it strikes the ground. [3] [Questions 6 and 7 are printed on the next page.]

Mark scheme: 5 (i) vcosθ = 8/2 B1 –26 = –2vsinθ – g 2 2 /2 M1 Accept with sign errors vsinθ = 3 A1 v 2 = (+/–3 2) + 4 2 or tanθ = 3/4 M1 v = 5 m s−1 A1

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6 A C 0.4 m 0.3 m B O 0.3 m A light inextensible string passes through a small smooth bead B of mass 0.4 kg. One end of the string is attached to a fixed point A 0.4 m above a fixed point O on a smooth horizontal surface. The other end of the string is attached to a fixed point C which is vertically below A and 0.3 m above the surface. The bead moves with constant speed on the surface in a circle with centre O and radius 0.3 m (see diagram). (i) Given that the tension in the string is 2 N, calculate (a) the angular speed of the bead, [3] (b) the magnitude of the contact force exerted on the bead by the surface. [2] (ii) Given instead that the bead is about to lose contact with the surface, calculate the speed of the bead. [4]

Mark scheme: 6 (i) (a) 2cos45 + 2 × 3/5 = 0.4 ω 2 × 0.3 M1 Uses N2L with 2 components of T and A1 accn = 0.3 ω 2 ω = 4.67 rad s−1 A1 3 (i) (b) R + 2sin45 + 2 × 4/5 = 0.4 g M1 R = 0.986 N A1 2 (ii) Tsin45 + T(4/5) = 0.4 g M1 T = 2.65 A1 2.654 Tcos45 + T(3/5) = 0.4 v 2 /0.3 M1 v = 1.61 m s−1 A1 4

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7 A particle P is attached to one end of a light elastic string of natural length 1.2 m and modulus of elasticity 12 N. The other end of the string is attached to a fixed point O on a smooth plane inclined at an angle of 30Å to the horizontal. P rests in equilibrium on the plane, 1.6 m from O. (i) Calculate the mass of P. [2] A particle Q, with mass equal to the mass of P, is projected up the plane along a line of greatest slope. When Q strikes P the two particles coalesce. The combined particle remains attached to the string and moves up the plane, coming to instantaneous rest after moving 0.2 m. (ii) Show that the initial kinetic energy of the combined particle is 1 J. [4] The combined particle subsequently moves down the plane. (iii) Calculate the greatest speed of the combined particle in the subsequent motion. [5]

Mark scheme: 7 (i) 12(1.6–1.2)/1.2 = mgsin30 M1 Uses T = λext/l m = 0.8 kg A1 2 (ii) PE change = 1.6 B1 2 × ans(i) B1 Both EE terms correct IKE + 12 × 0.4 2 /2.4 = M1 KE/PE/EE balance 2 Both EE terms correct 1.6 × 0.2gsin30 + 12 × 0.2 /2.4 IKE = 1 J AG A1 4 (iii) 12e/1.2 = 1.6 g sin30 M1 λe × t/l = new weight component e = 0.8 A1 May be stated without explanation 1.6 v 2 /2 + 12 × 0.8 2 /2.4 = M1 Must use new equilibrium position 2 A1 1.6g × 0.6sin30 + 12 × 0.2 /2.4 v = 1.5 m s−1 A1 5

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