Cambridge A Level Mathematics 9709 — 2019 May/June Paper 5 · Variant 2

1 A small ball is projected from a point O on horizontal ground at an angle of 30Å above the horizontal. At time t s after projection the vertically upwards displacement of the ball from O is 14t −kt2 m, where k is a constant. (i) State the value of k. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Show that the initial speed of the ball is 28 m s−1. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) Find the horizontal displacement of the ball from O when t = 3. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 1(i) 5 2 = = g k B1 Use the trajectory equation from the formula sheet 1 1(ii) Vsin30 = 14 M1 Use the trajectory equation from the formula sheet V = 28 ms–1 AG A1 2 1(iii) x = 28cos30 × 3 M1 Use horizontal motion. Allow their V for M1 x = 72.7 m A1 2

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2 B C D 0.3 m F 0.7 m E A G A uniform lamina ABCEFG is formed from a square ABDG by removing a smaller square CDFE from one corner. AB = 0.7 m and DF = 0.3 m (see diagram). Find the distance of the centre of mass of the lamina from A. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 2 2 0.35 0.35 + and Smaller square: Area = 0.32, CoM = ( ) 2 2 0.15 0.15 + 0.09, 0.21213... from D or E ( ) ( ) 0.49 0.09 0.09 0.98 0.045 0.49 0.495 − + − = × AX M1A1 Attempt to take moments about A AX = 0.431 m A1 4 Alternative method for question 2 (0.49 × 0.35) = (0.09 × 0.55) + 0.4X → X = 0.305 M1 Take moments about AG or AB X = Y = 0.305 B1 Question Answer Marks Guidance 2 ( ) 2 2 0.305 0.305 = + AX M1 Use Pythagoras’s theorem AX = 0.431 A1 4

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3 A particle P of mass 0.4 kg is attached to a fixed point A by a light inextensible string of length 0.5 m. The point A is 0.3 m above a smooth horizontal surface. The particle P moves in a horizontal circle on the surface with constant angular speed 5 rad s−1. (i) Calculate the tension in the string. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the magnitude of the force exerted by the surface on P. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(i) B1 Use Pythagoras’s theorem 2 cos 0.4 5 0.4 θ = × × T M1 Use Newton’s Second Law 0.4 4, 5 0.5 × = = T T N A1 3 3(ii) 0.4 sinθ = − R g T M1 Resolve vertically. Allow for their T for M1 R = 1N A1 2

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4 A particle P of mass 0.5 kg is attached to one end of a light elastic string of natural length 0.8 m and modulus of elasticity 16 N. The other end of the string is attached to a fixed point O. The particle P is released from rest at the point 0.8 m vertically below O. When the extension of the string is x m, the downwards velocity of P is v m s−1 and a force of magnitude 25x2 N opposes the motion of P. (i) Show that, when P is moving downwards, vdv = 10 −40x −50x2. [2] dx ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) For the instant when P has its greatest downwards speed, find the kinetic energy of P and the elastic potential energy stored in the string. 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Mark scheme: 4(i) 2 d 16 0.5 0.5 25 d 0.8 v x v g x x = − − M1 Use Newton’s Second Law vertically 2 d 10 40 50 d v v x x x = − − AG A1 2 Question Answer Marks Guidance 4(ii) ( ) 2 d 10 40 50 d = − − ∫ ∫ v v x x x M1 Attempt to integrate 2 3 2 50 10 20 ( ) 2 3 v x x x c = − − + A1 0 = 10 – 40x – 50x2 M1 Put the acceleration equal to zero x = 0.2 (Ignore x = –1 if seen) A1 2 0.5 8 0.533J 2 15 = = v B1 Use 2 2 = mv KE ( ) 2 0.2 16 0.4J 2 0.8 × = × B1 Use ( ) 2 2 λ = x EE l 6

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5 A light elastic string has natural length a m and modulus of elasticity , N. When the length of the string is 1.6 m the tension is 4 N. When the length of the string is 2 m the tension is 6 N. (i) Find the values of a and ,. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ One end of the string is attached to a fixed point O on a smooth horizontal surface. The other end of the string is attached to a particle P of mass 0.2 kg. The particle P moves with constant speed on the surface in a circle with centre O and radius 1.9 m. (ii) Find the speed of P. 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Mark scheme: 5(i) 1.6 4 λ − = a a B1 Use λ   =     x T l twice ( ) 2 6 a a λ − = B1 ( ) ( ) 2 1.5 1.6 − = − a a M1 Attempt to solve the simultaneous equations Question Answer Marks Guidance 5(i) 0.4 = 0.5(a), a = 0.8 A1 λ = 4 A1 5 5(ii) ( ) 1.1 4 5.5 0.8 = × = T B1 FT Use λ = x T L , ft candidates λ and a 2 0.2 5.5 1.9 = v M1 Use Newton’s Second Law horizontally v = 7.23 ms–1 A1 3

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6 A particle is projected with speed 15 m s−1 at an angle of 1Å above the horizontal. At the instant 4 s after projection the speed of the particle is 30 m s−1. (i) Find 1. 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(ii) Show that at the instant 4 s after projection the particle is 33.75 m below the level of the point of projection and find the direction of motion at this instant. 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Mark scheme: 6(i) B1 Use horizontal and vertical motion (15cosθ)2 + (15sinθ – 4g)2 = 302 M1 Use Pythagoras’s theorem [225 – 1200sinθ + 1600 = 900] M1 Attempt to solve for θ θ = 50.4° A1 4 Question Answer Marks Guidance 6(i) Alternative Method ( ) ( ) 2 4 15sin 4 2 g h θ = × − B1 ( ) ( ) 2 2 15 30 2 2 = + m m mgh M1 Allow h not replaced M1 Attempt to eliminate h and attempt to solve for θ θ = 50.4° A1 4 6(ii) 2 1 15sin50.4 4 4 2 s g = × − × × M1 Use vertical motion. Allow their θ for first M1 s = 33.75 m AG A1 15cos50.4 cos 30 α = M1 Use trigonometry of a right angled triangle α = 71.4° below the horizontal A1 4 If g = 9.8 or 9.81 used then M1A0M1A0

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7 r m A B C Fig. 1 Fig. 1 shows an object made from a uniform wire of length 0.8 m. The object consists of a straight part AB, and a semicircular part BC such that A, B and C lie in the same straight line. The radius of the semicircle is r m and the centre of mass of the object is 0.1 m from line ABC. (i) Show that r = 0.2. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ A B C 7 N Fig. 2 The object is freely suspended at A and a horizontal force of magnitude 7 N is applied to the object at C so that the object is in equilibrium with ABC vertical (see Fig. 2). (ii) Calculate the weight of the object. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ [Question 7(iii) is printed on the next page.] The 7 N force is removed and the object hangs in equilibrium with ABC at an angle of 1Å with the vertical. (iii) Find 1. 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Mark scheme: 7(i) 2 π = r X 2 0.8 0.1 π π × = × r r M1 Take moments about ABC r = 0.2 A1 3 7(ii) AC = 0.8 + 2 × 0.2 – 0.2π (= 0.57168…) B1 0.1W = 7AC M1 AC must be a numerical value. Take moments about A W = 40(0.) N A1 3 7(iii) (0.8 – 0.2π + 0.2) [= 0.37168…] B1 ( ) ( ) ( ) ( ) 0.8 0.2 0.8 0.8 0.2 0.2 0.8 0.2 0.2 2 π π π π − = − × + × − + Y M1A1 Y = 0.310(338) A1 0.1 tan 0.310338 θ = M1 θ = 17.9 A1 Allow 17.8

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