Cambridge A Level Mathematics 9709 — 2015 Oct/Nov Paper 5 · Variant 3

1 A particle is projected with speed 25 m s−1 at an angle of 50Å above the horizontal. Calculate the time after projection when the particle has speed 18 m s−1 and is rising. [4]

Mark scheme: 1 Vv 2 = 18 2 – (25cos50)2 M1 Finds vertical comp of velocity Vv = 8.1095656.. A1 8.1095656... = 25sin50 – gt M1 v = u – gt vertically t = 1.1(0) s A1 4

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2 One end of a light inextensible string of length 0.5 m is attached to a fixed point A. A particle P of mass 0.2 kg is attached to the other end of the string. P moves with constant speed in a horizontal circle with centre O which is 0.4 m vertically below A. (i) Show that the tension in the string is 2.5 N. [2] (ii) Find the speed of P. [3]

Mark scheme: 2 (i) Tcosθ = 0.2 g M1 Weight = vertical comp of tension 4.0 T × = 2 5.0 T = 2.5 N AG A1 2 2.0 v2 (ii) 2.5sinθ = M1 Horiz comp of tension and r accn = v2/r 3.0 2.0 v 2 2.5 × = A1 5.0 3.0 v = 1.5 ms–1 A1 3

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3 A particle P is projected with speed V m s−1 at an angle of 1Å above the horizontal from a point O on horizontal ground. At the instant 4 s after projection the particle passes through the point A, where OA = 40 m and the line OA makes an angle of 30Å with the horizontal. Calculate V and 1. [5]

Mark scheme: 3 4Vcosθ = 40cos30 B1 42 g 4Vsinθ – = 40sin30 B1 2 V 2 = (10cos30)2 + 252 or M1 25 tanθ = 10 cos 30 V = 26.5 A1 θ = 70.9 A1 5

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4 0.5 m O P A particle P of mass 0.4 kg moves with constant speed in a horizontal circle on the smooth inner surface of a fixed hollow hemisphere with centre O and radius 0.5 m (see diagram). (i) Given that the speed of the particle is 4 m s−1 and its angular speed is 10 rad s−1, calculate the angle between OP and the vertical. [2] (ii) Given instead that the magnitude of the force exerted on P by the hemisphere is 6 N, calculate (a) the angle between OP and the vertical, [2] (b) the angular speed of P. [3]

Mark scheme: 4 (i) 4 = 10r B1 v = ωr, r ˂ 0.5 θ = 53.1° B1 2 From sinθ = 0.4/0.5 (ii) (a) 0.4g = 6cosθ M1 θ = 48.2° A1 2 (b) 6sinθ = 0.4ω2 × 0.5sinθ M1 Accn = ω2 × 0.5sinθ A1 Using cv(48.2), allow any acute θ ω = 5.48 rad s–1 A1 3

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5 A particle P of mass 0.5 kg is projected vertically upwards from a point on a horizontal surface. A resisting force of magnitude 0.02v2 N acts on P, where v m s−1 is the upward velocity of P when it is a height of x m above the surface. The initial speed of P is 8 m s−1. (i) Show that, while P is moving upwards, vdv = −10 −0.04v2. [2] dx (ii) Find the greatest height of P above the surface. [3] (iii) Find the speed of P immediately before it strikes the surface after descending. [4]

Mark scheme: 5 (i) dv 2 0.5v = –0.5g – 0.02v M1 dx dv 2 v = –10 – 0.04v AG A1 2 dx 0 v x 2 ∫0 (ii) ∫a − 10 − .0 04v dv = dx M1 Separates the variables and attempts to integrate 2 0 x = [− ln(10 + .0 04 v /) .0 08 ] 8 M1 Uses limits or finds constant x = 2.85 A1 3 dv 2 (iii) v = 10 – 0.04v B1 dx v v .2 85 2 dv = ∫0dx M1 Integrates new acceleration ∫0 10 − .0 04v ln [(10 − .004v 2 /) 10 ] / − ( .008) = .285 M1 Uses earlier answer as distance v = 7.14 ms–1 A1 4

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6 d m 0.2 m G h m An object is formed by joining a hemispherical shell of radius 0.2 m and a solid cone with base radius 0.2 m and height h m along their circumferences. The centre of mass, G, of the object is d m from the vertex of the cone on the axis of symmetry of the object. The object rests in equilibrium on a horizontal plane, with the curved surface of the cone in contact with the plane (see diagram). The object is on the point of toppling. 0.04 (i) Show that d = h + . [3] h (ii) It is given that the cone is uniform and of weight 4 N, and that the hemispherical shell is uniform and of weight W N. Given also that h = 0.8, find W. [6]

Mark scheme: 6 (i) dcosθ = h/cosθ M1 θ = semi-vertical angle h cos θ = (0.2 2 + h 2 ) h d = ( h 2 /( .004 + h 2 )) M1 .004 d = h + AG A1 3 h (ii) 0.6 × 4 + 0.9W = d(4 + W) M1 Table of moments idea A1 2.0 2 d = 0.8 + B1 0.85 8.0 2.4 + 0.9W = 0.85(4 + W) M1 0.05W = 1 A1 W = 20 A1 6 12 5e

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7 A particle P of mass M kg is attached to one end of a light elastic string of natural length 0.8 m and modulus of elasticity 12.5 N. The other end of the string is attached to a fixed point A. The particle is released from rest at A and falls vertically until it comes to instantaneous rest at the point B. The greatest speed of P during its descent is 4.4 m s−1 when the extension of the string is e m. (i) Show that e = 0.64M. [2] (ii) Find a second equation in e and M, and hence find M. [6] (iii) Calculate the distance AB. [3]

Mark scheme: 125.e 7 (i) Mg = M1 Uses T = λe/l 8.0 e = 0.64M AG A1 2 (ii) Mg(0.8 + e) = M1 PE/KE/EE conservation M × 44 2 12 5.e 2 + A1 2 ( 2 × 8.0) 10M(0.8 + 0.64M) = M1 12 5.( .064 M ) 2 2 9.68M + A1 8M+6.4M 2= 9.68M +3.2M 6.1 8+6.4M = 9.68 + 3.2M M1 Attempt to solve equation in M M = 0.525 A1 6 12 5.( d − 8.0) 2 (iii) 0.525gd = M1 PE/EE balance ( 2 × 8.0) 0.672d = d 2 – 1.6d + 0.64 M1 d = 1.94 A1 3

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