Cambridge A Level Mathematics 9709 — 2017 May/June Paper 5 · Variant 1

1 A particle is projected with speed 20 m s−1 at an angle of 60Å above the horizontal. Calculate the time after projection when the particle is descending at an angle of 40Å below the horizontal. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 tan40 = v / 20cos60 M1 v = 10tan40 ( = 8.3909...) A1 –10tan40 = 20sin60 – gt M1 Uses v = u + at vertically t = 1.27 s A1 Total: 4 7 = 0.35λ / 0.25 M1 Uses T = λx / L

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2 A 7 N B P 0.6 m Fig. 1 One end of a light inextensible string is attached to a fixed point A. The other end of the string is attached to a particle P of mass m kg which hangs vertically below A. The particle is also attached to one end of a light elastic string of natural length 0.25 m. The other end of this string is attached to a point B which is 0.6 m from P and on the same horizontal level as P. Equilibrium is maintained by a horizontal force of magnitude 7 N applied to P (see Fig. 1). (i) Calculate the modulus of elasticity of the elastic string. 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A P 0.3 m 30Å B Fig. 2 P is released from rest by removing the 7 N force. In its subsequent motion P first comes to instantaneous rest at a point where BP = 0.3 m and the elastic string makes an angle of 30Å with the horizontal (see Fig. 2). (ii) Find the value of m. 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Mark scheme: 2(i) λ = 5 A1 Total: 2 2(ii) EE = 2 0.35 × 5 / (2 × 0.25) or 2 0.05 × 5 / (2 × 0.05) B1 Uses EE = λ 2x / 2L PE = mg × 0.3sin30 B1 mg × 0.3sin30 = 2 0.35 × 5 / (2 × 0.25) − 2 0.05 × 5 / (2 × 0.25) M1 Sets up a 3 term energy equation involving EE, KE and PE m = 0.8 A1 Total: 4

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3 0.56 m O 0.28 m An object is made from a uniform solid hemisphere of radius 0.56 m and centre O by removing a hemisphere of radius 0.28 m and centre O. The diagram shows a cross-section through O of the object. (i) Calculate the distance of the centre of mass of the object from O. [4] [The volume of a hemisphere is 230r3.] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The object has weight 24 N. A uniform hemisphere H of radius 0.28 m is placed in the hollow part of the object to create a non-uniform hemisphere with centre O. The centre of mass of the non-uniform hemisphere is 0.15 m from O. (ii) Calculate the weight of H. 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Mark scheme: 3(i) CofM of hemisphere = 3 8 × 0.56 or 3 8 × 0.28 [ 2 3π × 3 0.56 – 2 3π × 3 0.28 ]X = 2 3 π × 3 0.56 × 3 8 × 0.56 – 2 3 π × 3 0.28 × 3 8 × 0.28 M1A1 Take moments about O X = 0.225 m A1 Total: 4 3(ii) 24 × 0.225 + W(3 × 0.28 / 8) = (24 + W) × 0.15 M1A1 Attempts to take moments about O W = weight of uniform hemi-sphere W = 40 N A1 Total: 3 B1

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4 A particle is projected from a point O on horizontal ground. The initial components of the velocity of the particle are 10 m s−1 horizontally and 15 m s−1 vertically. At time t s after projection, the horizontal and vertically upwards displacements of the particle 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 the particle. 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The horizontal ground is at the top of a vertical cliff. The point O is at a distance d m from the edge of the cliff. The particle is projected towards the edge of the cliffand does not strike the ground before it passes over the edge of the cliff. (ii) Show that d is less than 30. 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(iii) Find the value of x when the particle is 14 m below the level of O. 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Mark scheme: 4(i) x = 10t or y = 2 gt / 2 y = 15x / 10 – g(x / 10 2) / 2 M1A1 Attempts to eliminate t y = 1.5x – 0.05 2x A1 Total: 4 Question Answer Marks Guidance 4(ii) 0 = 1.5x – 0.05 2x M1 Substitute y = 0 into the trajectory equation x = 30 A1 Total: 2 4(iii) –14 = 1.5x – 2 0.05x M1 Sets up a quadratic equation and attempts to solve it x = 37.5 A1 Total: 2 OG = 2 × 0.7sin(π / 2) / (3π / 2) (= 0.297)

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5 B P m 0.9 m 0.7 30Å A A uniform semicircular lamina of radius 0.7 m and weight 14 N has diameter AB. The lamina is in a vertical plane with A freely pivoted at a fixed point. The straight edge AB rests against a small smooth peg P above the level of A. The angle between AB and the horizontal is 30Å and AP = 0.9 m (see diagram). (i) Show that the magnitude of the force exerted by the peg on the lamina is 7.12 N, correct to 3 significant figures. 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(ii) Find the angle with the horizontal of the force exerted by the pivot on the lamina at A. 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Mark scheme: 5(i) B1 0.9R = 14(0.7cos30 – 0.297sin30) M1A1 Attempts to take moments about A R = 7.12 N A1 Total: 4 5(ii) H = 7.12sin30 and V = 14 − Rcos30 M1 Resolves horizontally and vertically tanθ = (14 – 7.12cos30) / (7.12sin30) M1 Uses tanθ = V / H, where θ is the required angle θ = 65.6 A1 Total: 3

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6 A particle P of mass 0.15 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 A. The particle P moves in a horizontal circle which has its centre vertically below A, with the string inclined at 1Å to the vertical and AP = 0.5 m. (i) Find the angular speed of P and the value of 1. 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(ii) Calculate the difference between the elastic potential energy stored in the string and the kinetic energy of P. 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Mark scheme: 6(i) B1 Uses T = λx / L 3sinθ = 0.15 2 ω (0.5sinθ) M1 Uses Newton's Second Law horizontally ω = 6.32 rad 1 s− A1 Tcosθ = 0.15g (cosθ = 0.5) M1 Resolves vertically θ = 60 A1 Total: 5 6(ii) v = 6.32 × 0.5sin60 B1 FT Uses v = rω and r = 0.5sin60 KE = 0.15(6.32 × 0.5sin60 2) / 2 (=0.5625J) B1 Difference = 0.5625 – 12 × 0. 21 / (2 × 0.4) M1 Uses EE = λ 2x / (2L) Difference = 0.4125 J A1 Total: 4 B1 Uses F = µ R

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7 A particle P of mass 0.5 kg is at rest at a point O on a rough horizontal surface. At time t = 0, where t is in seconds, a horizontal force acting in a fixed direction is applied to P. At time t s the magnitude of the force is 0.6t2 N and the velocity of P away from O is v m s−1. It is given that P remains at rest at O until t = 0.5. (i) Calculate the coefficient of friction between P and the surface, and show that dv = 1.2t2 −0.3 for t > 0.5. [3] dt ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Express v in terms of t for t > 0.5. 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(iii) Find the displacement of P from O when t = 1.2. 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Mark scheme: 7(i) µ = 0.6 × 0. 2 5 / (0.5 g) ( = 0.03) 0.5dv / dt = 0.6 2t – 0.03 × 0.5g M1 Uses Newton's Second Law horizontally dv / dt = 1.2 2t – 0.3 A1 Total: 3 Question Answer Marks Guidance 7(ii) dv ∫ = (∫1.2 2t – 0.3) dt v = 0.4 3t – 0.3t ( + c) M1 Separates the variables and attempts to integrate t = 0.5, v = 0 hence c = 0.1 M1 Attempts to find c v = 0.4 3t – 0.3t + 0.1 A1 Total: 3 7(iii) dx ∫ = (∫0.4 3t – 0.3t + 0.1) dt x = 0.1 4t – 0.15 2t + 0.1t ( + c) M1 Attempts to integrate t = 0.5, x = 0 hence c = –0.01875 M1 Finds c or substitutes the limits x(1.2) = 0.0926(1) A1 Total: 3

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