Cambridge A Level Mathematics 9709 — 2018 Oct/Nov Paper 5 · Variant 1

1 A small ball B is projected with speed 30 m s−1 at an angle of 60° to the horizontal from a point on horizontal ground. Find the time after projection when the speed of B is 25 m s−1 for the second time. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 M1 v = vertical velocity at the required point v = (±) 20 A1 –20 = 30sin60 – gt M1 Use v = u + at vertically t = 4.6(0) s A1 4

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2 m 0.3 0.2 m B A A uniform object is made by attaching the base of a solid hemisphere to the base of a solid cone so that the object has an axis of symmetry. The base of the cone has radius 0.3 m, and the hemisphere has radius 0.2 m. The object is placed on a horizontal plane with a point A on the curved surface of the hemisphere and a point B on the circumference of the cone in contact with the plane (see diagram). (i) Given that the object is on the point of toppling about B, find the distance of the centre of mass of the object from the base of the cone. 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(ii) Given instead that the object is on the point of toppling about A, calculate the height of the cone. [3] [The volume of a cone is 30r2h.1 The volume of a hemisphere is 30r3.]2 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(i) cosθ = 0.2/0.3 B1 Axis makes an angle θ with the horizontal tanθ = x/0.3 M1 x = 0.335(41..) A1 3 2(ii) M1 Attempt to take moments about A (π0.32h/3)×(h/4) = (2π0.23/3)(3×0.2/8) A1 h = 0.231 A1 3

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3 A particle P of mass 0.4 kg is attached to a fixed point O by a light elastic string of natural length 0.5 m and modulus of elasticity 20 N. The particle P is released from rest at O. (i) Find the greatest speed of P in the subsequent motion. 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(ii) Find the distance below O of the point at which P comes to instantaneous rest. 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Mark scheme: 3(i) 20e/0.5 = 0.4g M1 Use T = λx/L e = 0.1 A1 0.4v2/2 = 0.4g(0.5 + 0.1) – 20×0.12/(2×0.5) M1 Attempt to set up a 3 term energy equation v = 11 = 3.32 A1 4 3(ii) 0.4g(5 + x) = 20x2/(2×0.5) M1 Attempt to set up a 2 term energy equation [0 = 20x2 – 4x – 2] [ x = 0.432] M1 Attempt to solve a 3 term quadratic equation Distance below O = (0.5 + 0.432) = 0.932 m A1 3

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4 x m v m s−1 P O A 1.6 m A particle P of mass 0.5 kg is projected along a smooth horizontal surface towards a fixed point A. Initially P is at a point O on the surface, and after projection, P has a displacement from O of x m and velocity v m s−1. The particle P is connected to A by a light elastic string of natural length 0.8 m and modulus of elasticity 16 N. The distance OA is 1.6 m (see diagram). The motion of P is resisted by a force of magnitude 24x2 N. (i) Show that vdv = 32 −40x −48x2 while P is in motion and the string is stretched. 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The maximum value of v is 4.5. (ii) Find the initial value of v. 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Mark scheme: 4(i) T = 16(1.6 – 0.8 – x)/0.8 ( = 16 – 20x) B1 Use T = λx/L 0.5vdv/dx = 16(1.6 – 0.8 – x)/0.8 – 48x2 M1 Use Newton's Second Law horizontally vdv/dx = 32 – 40x – 48x2 AG A1 3 Question Answer Marks Guidance 4(ii) 48x2 + 40x – 32 = 0 M1 Put acceleration = 0 for maximum velocity x = 0.5 A1 vdv ∫ = ∫(32 – 40x – 48x2)dx (v2/2 = 32x – 40x2/2 – 48x3/3 + c) M1 Attempt to integrate the equation from part (i) 4.52/2 = 32×0.5 – 20 ×0.52 – 16×0.53 + c, c = 1.125 M1 Substitute x = 0.5 , v = 4.5 to find c v = 1.5 A1 Use x = 0 5

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5 A particle P of mass 0.1 kg is attached to one end of a light inextensible string of length 0.5 m. The other end of the string is attached to a fixed point A. The particle P moves in a circle which has its centre O on a smooth horizontal surface 0.3 m below A. The tension in the string has magnitude T N and the magnitude of the force exerted on P by the surface is R N. (i) Given that the speed of P is 1.5 m s−1, calculate T and R. 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(ii) Given instead that T = R, calculate the angular speed of P. 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Mark scheme: 5(i) 0.1×1.52/0.4 = Tcosθ M1 Note r = 0.4, cosθ = 0.8, sinθ = 0.6 Use Newton's Second Law horizontally T = 0.703 A1 R = 0.1g – Tsinθ M1 Resolve vertically for P R = 0.578 A1 4 Question Answer Marks Guidance 5(ii) T + Tsinθ = 0.1g M1 Resolve vertically for P T = 0.625 A1 0.1ω 2×0.4 = 0.625cosθ M1 Use Newton's Second Law horizontally ω = 3.54 rad s–1 A1 4

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6 E 0.3 m A F 0.3 m 0.6 m D G B C 0.7 m Fig. 1 Fig. 1 shows the cross-section ABCDE through the centre of mass G of a uniform prism. The cross- section consists of a rectangle ABCF from which a triangle DEF has been removed; AB = 0.6 m, BC = 0.7 m and DF = EF = 0.3 m. (i) Show that the distance of G from BC is 0.276 m, and find the distance of G from AB. 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B A G E 2 N C D Fig. 2 The prism is placed with CD on a rough horizontal surface. A force of magnitude 2 N acting in the plane of the cross-section is applied to the prism. The line of action of the force passes through G and is perpendicular to DE (see Fig. 2). The prism is on the point of toppling about the edge through D. (ii) Calculate the weight of the prism. 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Mark scheme: 6(i) 0.375y = 0.42×0.6/2 –0.045(0.6 – 0.3/3) M1 Take moments about BC y = 0.276 m AG A1 0.375x = 0.42×0.7/2 – 0.045(0.7 – 0.3/3) M1 Take moments about AB x = 0.32 m A1 5 6(ii) M1 Attempt to take moments about D 2cos45× (0.7 – 0.32) = 2cos45× (0.3 – 0.276) + W(0.3 – 0.276) A1 W = 21(.0) N A1 3

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7 24 m s−1 15Å 45Å O A small object is projected with speed 24 m s−1 from a point O at the foot of a plane inclined at 45° to the horizontal. The angle of projection of the object is 15° above a line of greatest slope of the plane (see diagram). At time t s after projection, the horizontal and vertically upwards displacements of the object from O are x m and y m respectively. (i) Express x and y in terms of t, and hence find the value of t for the instant when the object strikes the plane. 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(ii) Express the vertical height of the object above the plane in terms of t and hence find the greatest vertical height of the object above the plane. 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Mark scheme: 7(i) x = (24cos60)t B1 Use horizontal motion y = (24sin60)t – gt 2/2 B1 Use vertical motion (24cos60)t = (24sin60)t –gt 2/2 M1 Recognise that x = y t = 1.76 A1 4 7(ii) h = (24sin60)t – gt 2/2 – (24cos60)t B1 M1 Attempt to differentiate dh/dt = 24(sin60 – cos60) – gt A1 24(sin60 – cos60) – gt = 0, t = 0.878(46..) M1 Equate dh/dt = 0 to find t h = 3.86 m A1 5

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