Cambridge A Level Mathematics 9709 — 2013 Oct/Nov Paper 5 · Variant 3

1 A particle P of mass 0.1 kg is attached to one end of a light elastic string of natural length 0.4 m and modulus of elasticity 12 N. The other end of the string is attached to a fixed point O on a smooth horizontal surface. P moves on the surface in a horizontal circle with centre O and radius 0.6 m. Calculate the speed of P. [3]

Mark scheme: 1 T = 12 × (0.6–0.4)/0.4 M1 Uses T = λ x /L, (=6) 3 6 = 0.1v 2 /0.6 M1 N2L 1 force and RA − 1 A1 [3] v = 6 ms

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2 A particle P of mass 0.5 kg is released from rest at a point O and falls vertically. When P has downward displacement x m from O, the velocity of P is v m s−1. A resisting force of magnitude 0.015x2 N acts on P. (i) Show that vdv = 10 −0.03x2. [2] dx (ii) Find the value of x when the velocity of P is greatest. [1] (iii) Calculate the greatest value of v. [4]

Mark scheme: 2 (i) 0.5vdv/dx = 0.5g – 0.015x 2 M1 N2L 2 forces vdv/dx = 10 – 0.03x 2 AG A1 [2] (ii) 18.3 m B1 [1] (10 / .003) = 18.257.. M1 Attempts to integrate (iii) ∫vdv = ∫(10–0.03x 2 )dx v 2 /2 = 10x – 0.03x 3 /3 (+c) A1 Accept omission of c v 2 /2 = 10× 18.3 – 0.03×18.3 3 /3 M1 Uses ans (ii) in formula for v 2 v = 15.6 ms − 1 A1 [4] 7

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3 5 rad s–1 0.4 m P A particle P of mass 0.5 kg moves in a horizontal circle on the smooth inner surface of a hollow cone which is fixed with its axis vertical and its vertex downwards. P moves with angular speed 5 rad s−1 in a circle of radius 0.4 m (see diagram). Show that the semi-vertical angle of the cone is 45 and calculate the magnitude of the force exerted on P by the surface of the cone. [6]

Mark scheme: 3 Rcosθ = 0.5 g (=5) B1 Resolving vertically Rsinθ = 0.5 × 5 2 × 0.4 (= 5) M1 Use of N2L horizontally with acc n = w 2 r tanθ =(0.5 × 5 2 × 0.4)/(0.5 g) M1 Eliminating R θ = 45 o AG A1 R = 0.5 g/cos45 M1 R 2 = (0.5 × 5 2 × 0.4) 2 + (0.5 g) 2 R = 7.07 N A1 [6] 7.071.. 6

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4 A particle P of mass 0.2 kg is projected horizontally with velocity 0.9 m s−1 from a point O on a rough horizontal surface. P moves in a straight line, and at time t s after projection the velocity of P is v m s−1. A force of magnitude 0.024t N acts on P in the direction OP. The coefficient of friction between P and the surface is 0.3. (i) Express the acceleration of P in terms of t, and hence show that, before P comes to rest, v = 0.06 t2 −50t + 15 . (ii) Find the value of t when P comes to rest. [2] (iii) Find the value of t when P subsequently begins to move again. [2]

Mark scheme: 4 (i) 0.2a = 0.024 t – 0.2 g × 0.3 M1 Uses N2L a = 0.12t – 3 A1 ∫dv M1 Integrates and finds c = ∫(0.12t – 3)dt, v = 0.12t 2 /2 – 3t + c, t = 0, v = 0.9 hence c = 0.9 v = 0.06(t 2 – 50t + 15) AG A1 [4] (ii) t 2 – 50t + 15 = 0 M1 Solves 3 term quadratic t = 0.302 A1 [2] Smaller +ve root only GCE A LEVEL – October/November 2013 9709 53 (iii) 0.024t = 0.3 × 0.2 g M1 Equates tractive force and friction force t = 25 A1 [2] 8

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5 The top of a vertical cliff is 20 m above sea level. A particle P is projected with speed 15 m s−1 at an angle of 30 above the horizontal from a point O at the top of the cliff. Calculate (i) the speed and direction of motion of P when it strikes the water, [4] (ii) the distance OP at the instant P strikes the water. [4]

Mark scheme: 5 (i) vv2 = (15sin30) 2 + 2 g × 20 M1 v v = 21.3600.. V 2 = (15cos30) 2 + (15sin30) 2 + M1 V orθ from components, V by energy 2 g × 20 V = 25 ms − 1 A1 θ (= tan −21.36/(15cos30)1 = 58.7 o A1 [4] 58.69.. (ii) –20 = (15sin30)t – gt 2 /2 M1 M1 maybe gained in (i) 2t 2 – 3t – 8 = 0⇒t(= 2.866..) = 2.89 A1 A1 maybe gained in (i) OR t = (15sin30)/10 + (25sin58.7)/10 M1 Separating rise and fall times t = 2.89 A1 OP 2 =20 2 + (15cos30 × 2.886) 2 M1 42.491.. OP = 42.5 m A1 [4] 8

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6 One end of a light elastic string of natural length 0.8 m and modulus of elasticity 50 N is attached to a fixed point O. A particle P of mass 0.4 kg is attached to the other end of the string. P is projected downwards with speed 1.5 m s−1 from a point 0.82 m vertically below O. (i) Find the greatest speed of P. [5] (ii) Show that P cannot reach O. [3]

Mark scheme: 6 (i) 0.4 g = 50e/0.8 M1 Uses T = λ × /L (e = 0.064) Moves down = 0.044 m A1 (0.8 + 0.064 – 0.82) 0.4 × 1.5 2 /2 + 0.4 g × 0.044 + M1 Sets up 2EE/2KE/PE equation 50(0.82–0.8) 2 /(2 × 0.8) 2 2 A1 =0.4v /2 + 50 × 0.064 /(2 × 0.8) v = 1.6(0) ms − 1 A1 [5] (ii) PE gain to reach O = 0.4 g × 0.82 B1 From initial position, (3.28J) KE + EE = 0.4 × 1.5 2 /2 M1 At initial position, (0.4625J) 2 A1 +50(0.82 – 0.8) /(2 × 0.8) Shows by evaluation that insufficient [3] 8 energy GCE A LEVEL – October/November 2013 9709 53

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7 0.4 m 0.5 m 0.4 m A uniform solid is made from a cylinder and a cone, both with radius 0.5 m and height 0.4 m. The circular base of the cone is attached to a circular face of the cylinder, with their circumferences coinciding. The solid rests in equilibrium with the circular face of the solid on a rough horizontal surface (see diagram). (i) Show that the centre of mass of the solid is 0.275 m above the surface. [3] The weight of the solid is 60 N. A horizontal force of increasing magnitude P N is applied to the vertex of the cone which causes the solid eventually to topple without sliding. (ii) Calculate the value of P for which the solid is on the point of toppling. [2] (iii) Find the least possible value for the coefficient of friction between the solid and the surface. [1] The force of magnitude P N is removed, and the solid is held with the curved surface of the cylinder in contact with the horizontal surface. The horizontal surface is then tilted so that it makes an angle of 30 with the horizontal. The solid is released, with its axis of symmetry parallel to a line of greatest slope and the conical portion pointing down the slope. (iv) Show that the solid does not slide, but does topple. [4]

Mark scheme: 7 (i) M1 Uses table of moments idea π × 0.5 2 × 0.4 × 0.2+π × 0.5 2 × A1 0.4 × 0.5/3 = (π × 0.5 2 × 0.4+π × 0.5 2 × 0.4/3)0G AG d = 0.275 m A1 [3] (ii) (0.4 + 0.4)F = 0.5 × 60 M1 Takes moments F = 37.5 A1 [2] (iii) µ (= 37.5/60) = 0.625 B1 [1] cv(F)/60 ft (iv) F/R=(60sin30)/(60cos30) (= 0.577..) M1 Or quotes tan30 p 0.625 0.577 p 0.625 (or µ ), no sliding AG A1 tanθ = (0.4 – 0.275)/0.5 M1 Or 0.5tan30 = 0.288.. θ = 14 o AG A1 [4] 0.4 – 0.29 p 0.275, topples 10

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