Cambridge A Level Mathematics 9709 — 2018 May/June Paper 5 · Variant 2

1 A B 12 m O 20 m A small ball B is projected from a point O on horizontal ground towards a point A 12 m above the ground. 0.9 s after projection B has travelled a horizontal distance of 20 m and is vertically below A (see diagram). (i) Find the angle and the speed of projection of B. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Calculate the distance AB when B is vertically below A. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 1(i) θ ( = 30.96) = 31(.0)° A1 Vcos30.96 = 20 0.9 M1 Use horizontal motion. Allow their θ for the M mark. V = 25.9 m 1 −s A1 Total: 4 1(ii) H = 25.9sin31 × 0.9 – g × 2 0.9 2 ( = 7.948) M1 Use s = ut + 1 2 a 2t vertically. H is the height above the ground. Allow their V and θ for the M mark. AB ( = 12 – 7.95) = 4.05 m A1 Allow AB = 4.06 Total: 2 EPE = 24( ) B1 Correct EPE term. Note x = OP

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2 One end of a light elastic string is attached to a fixed point O. The other end of the string is attached to a particle P of mass 0.4 kg. The string has natural length 0.6 m and modulus of elasticity 24 N. The particle is released from rest at O. Find the two possible values of the distance OP for which the particle has speed 1.5 m s−1. [6] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 2 0.6 − x /(2 × 0.6) 0.4 × 2 1.5 /2 = 0.4gx – 24( ) 2 0.6 − x /(2 × 0.6) [20 2 x – 28x + 7.65 = 0 or equivalent] M1 Attempt to find a 3 term energy equation M1 Attempt to solve the 3 term quadratic equation OP = 1.0279 m, 0.372 m (reject) A1 Correct answer chosen 0.4 × 2 1.5 /2 = 0.4gx M1 Note the particle is moving upwards and the string is slack OP = 0.1125 m A1 Total: 6 Question Answer Marks Guidance 2 Alternative method EPE = 24 2 x /(2 × 0.6) B1 x is the extension 0.4 × 2 1.5 /2 = 0.4g(x + 0.6) – 24 2 x /(2 × 0.6) [20 2 x – 4x – 1.95 = 0 or equivalent ] M1 Attempt to find a 3 term energy equation M1 Attempt to solve the 3 term quadratic equation [ x = 0.42787, – 0.22787 .reject] OP = 0.6 + 0.42787 = 1.0279 A1 0.4 × 2 1.5 /2 = 0.4g( x + 0.6) [x = – 0.4875] M1 Note the particle is moving upwards and the string is slack OP = 0.6 – 0.4875 = 0.1125 A1 Total: 6 d = xsinθ/2 – acosθ or equivalent

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3 A 1 a B x C ABC is an object made from a uniform wire consisting of two straight portions AB and BC, in which AB = a, BC = x and angle ABC = 90Å. When the object is freely suspended from A and in equilibrium, the angle between AB and the horizontal is 1 (see diagram). (i) Show that x2 tan 1 −2ax −a2 = 0. 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(ii) Given that tan 1 = 1.25, calculate the length of the wire in terms of a. 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Mark scheme: 3(i) B1 Note d is the distance of the C of M of BC from the vertical through A a(acosθ)/2 = x(xsinθ/2 – acosθ) M1 Take moments about A 2 x tanθ – 2ax – 2 a = 0 AG A1 Total: 3 3(ii) 1.25 2 x – 2ax – 2 a = 0 [x = 2a and x = – 2a/5] M1 Attempts to solve the equation Length ( = 2a + a ) = 3a A1 Total: 2

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4 A particle P is projected from a point O on horizontal ground with initial speed 20 m s−1 and angle of projection 30°. At the instant t s after projection, the horizontal and vertically upwards displacements of P from O are x m and y m respectively. (i) Express x and y in terms of t and hence find the equation of the trajectory of P. 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P is at the same height above the ground at two points which are a horizontal distance apart of 15 m. (ii) Calculate this height. 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Mark scheme: 4(i) x = (20cos30)t or 10 3 t y = (20sin30)t – 1 2 g 2t or 10t – 5 2t B1 Use vertical motion y = (20sin30)[x/(20cos30)] – 5[x/(20cos30) 2] M1 Attempt to eliminate t y = x/ 3 – 2 x /60 or 0.577x – 0.0167 2 x A1 Total: 4 4(ii) x/ 3 – 2 x /60 = (x+15)/ 3 – ( ) 2 15 + x /60 M1 Simplifies to 0 = 15/ 3 – (30x+225)/60 x = 9.821 A1 y = 4.06(25) m A1 Total: 3 Alternative method 0.577x – 0.0167 2 x = 0.577(x+15) – 0.0167( ) 2 15 + x M1 x = 9.775 A1 y = 4.044 A1 Total: 3

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5 0.4 m 0.6 m 0.5 m 0.8 m x m A uniform object is made by joining a solid cone of height 0.8 m and base radius 0.6 m and a cylinder. The cylinder has length 0.4 m and radius 0.5 m. The cylinder has a cylindrical hole of length 0.4 m and radius x m drilled through it along the axis of symmetry. A plane face of the cylinder is attached to the base of the cone so that the object has an axis of symmetry perpendicular to its base and passing through the vertex of the cone. The object is placed with points on the base of the cone and the base of the cylinder in contact with a horizontal surface (see diagram). The object is on the point of toppling. (i) Show that the centre of mass of the object is 0.15 m from the base of the cone. 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(ii) Find x. [4] [The volume of a cone is 30r2h.]1 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(i) tanθ = x /0.6 M1 x = 0.15 m AG A1 Total: 3 5(ii) (π 2 0.6 × 0.8/3) × (0.8/4) – [π( 2 0.5 – 2 x ) × 0.4] × (0.4/2) = [π 2 0.6 × 0.8/3 + π( 2 0.5 – 2 x ) × 0.4] x M1 A1 Attempts to take moments about the base of the cone using their x Note x =0.15 Correct equation for the A mark. M1 Attempts to solve the equation x = 0.464 A1 Note 2 x = 0.216 Total: 4

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6 A 0.6 m 0.3 m P 0.6 m B A particle P of mass 0.2 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 A. The particle P is also attached to one end of a second light inextensible string of length 0.6 m, the other end of which is attached to a fixed point B vertically below A. The particle moves in a horizontal circle of radius 0.3 m, which has its centre at the mid-point of AB, with both strings straight (see diagram). (i) Calculate the least possible angular speed of P. 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The string AP will break if its tension exceeds 8 N. The string BP will break if its tension exceeds 5 N. (ii) Find the greatest possible speed of P. 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Mark scheme: 6(i) cosθ = 0.5 and sinθ = 3 /2 B1 θ is the angle that AP makes with the horizontal. Note tanθ = 3 Tsinθ = 0.2 g M1 Resolve vertically for P. Note tension in BP is zero Tcosθ = 0.2 2 ω × 0.3 M1 Use Newton's Second Law horizontally ω = 4.39 rad 1 −s A1 Total: 4 Question Answer Marks Guidance 6(ii) A T sinθ = 0.2 g + B T sinθ M1 Resolve vertically for P A T sinθ = 0.2 g + 5sinθ M1 Use B T = 5 A T = 7.309 A1 5cosθ + 7.309cosθ = 0.2 2 v /0.3 M1 Use Newton's Second Law horizontally v = 3.04 m 1 −s A1 Total: 5

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7 A particle P of mass 0.2 kg is released from rest at a point O above horizontal ground. At time t s after its release the velocity of P is v m s−1 downwards. A vertically downwards force of magnitude 0.6t N acts on P. A vertically upwards force of magnitude ke−t N, where k is a constant, also acts on P. dv (i) Show that = 10 −5ke−t + 3t. 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(ii) Find the greatest value of k for which P does not initially move upwards. 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(iii) Given that k = 1, and that P strikes the ground when t = 2, find the height of O above the ground. 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Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown. ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................ ........................................................................................................................................................................

Mark scheme: 7(i) 0.2dv/dt = 0.2 g + 0.6t – k −t e dv/dt = 10 + 3t – 5 −t ke AG A1 Total: 2 7(ii) dv/dt = 10 – 5k 0e = 0 M1 Recognise that dv/dt = 0 when t = 0 M1 Attempts to solve the equation 7(ii) k = 2 A1 Total: 3 Question Answer Marks Guidance 7(iii) ∫dv = (∫10 + 3t – 5k −t e )dt M1 Attempts to integrate the equation from part i with k not replaced [v = 10t + 3 2t /2 + 5 −t e + c, v = 0, t = 0 so c = – 5] v = 10t + 3 2t /2 + 5 −t e – 5 A1 ∫dx = (∫10t + 3 2t /2 + 5 −t e – 5)dt x = 5 2t + 3t /2 – 5 −t e – 5t + c M1 Attempts to integrate again. Allow their k or just k not replaced x = 0, t = 0, so c = 5 and substitutes t = 2 x = 5 × 2 2 + 3 2 /2 – 5 2 − e – 5 × 2 + 5 M1 Height = 18.3 m A1 Total: 5

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