Cambridge A Level Mathematics 9709 — 2013 May/June Paper 5 · Variant 1

1 A small ball is projected with speed 20 m s−1 at an angle of 45Å above the horizontal from a point O on horizontal ground. At time t s after projection, the horizontal and vertically upwards displacements of the ball from O are x m and y m respectively. (i) Express x and y in terms of t. [2] (ii) Show that the equation of the trajectory of the ball is y = x −140x2. [2] (iii) State the distance from O of the point at which the ball first strikes the ground. [1]

Mark scheme: 1 (i) x = (20cos45)t B1 Or sin45, 1/ 2 , 0.707 y = (20sin45)t – gt 2 /2 B1 [2] Or cos45, 1/ 2 , 0.707. (ii) y = (20sin45)(x/(20cos45) M1 Substitutes t = x/(20cos45) at least –g[x/(20cos45)] 2 /2 once y = x – x 2 /40 AG A1 [2] Only from g = 10 (iii) x = 40 m B1 [1] [5]

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2 A particle P of mass 0.4 kg is attached to one end of a light elastic string of natural length 1.2 m and modulus of elasticity 19.2 N. The other end of the string is attached to a fixed point A. The particle P is released from rest at the point 2.7 m vertically above A. Calculate (i) the initial acceleration of P, [3] (ii) the speed of P when it reaches A. [4]

Mark scheme: 2 (i) T = 19.2 ×(2.7 – 1.2)/1.2 B1 T = 24 N 0.4a = 0.4g + T M1 Newton’s Second Law with 3 terms a = 70 ms − 2 A1 [3] (ii) 19.2(2.7 – 1.2) 2 /(2 × 1.2) B1 Initial EE = 18 M1 For a 3 term energy equation 0.4v 2 /2 = 0.4g × 2.7 A1 + 19.2 × (2.7 – 1.2) 2 /(2 × 1.2 ) v = 12 ms − 1 A1 [4] [7]

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3 C 0.2 m B 4 N 0.3 m A A uniform object ABC is formed from two rods AB and BC joined rigidly at right angles at B. The rod AB has length 0.3 m and the rod BC has length 0.2 m. The object rests with the end A on a rough horizontal surface and the rod AB vertical. The object is held in equilibrium by a horizontal force of magnitude 4 N applied at B and acting in the direction CB (see diagram). (i) Find the distance of the centre of mass of the object from AB. [3] (ii) Calculate the weight of the object. [2] (iii) Find the least possible value of the coefficient of friction between the surface and the object. [2]

Mark scheme: 3 (i) M1 Table of values or a moment equation 0.2 × 0.1 + 0.3 × 0 = d(0.2+0.3) A1 Accept no mention of 0.3 × 0 d = 0.04 m A1 [3] (ii) 4 × 0.3 = 0.04W M1 Moments about A W = 30 N A1ft [2] ft 1.2/cv(d(i)) (iii) µ = 4/30 M1 4/cv(W(ii)) µ = 0.133 A1 [2] Accept 2/15 [7]

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5 A particle P is projected with speed 50 m s−1 at an angle of 40Å above the horizontal from a point O. For the instant 2.5 s after projection, calculate (i) the speed of P, [3] (ii) the angle between OP and the horizontal. [4]

Mark scheme: 5 (i) v y = 50sin40 – 2.5g B1 Vertical component speed (=7.139…) v 2 =(50sin40 – 2.5g) 2 + (50cos40) 2 M1 Uses Pythagoras with correct horizontal component v = 39(.0) ms − 1 A1 [3] (ii) x = 50cos40 × 2.5 B1 Horizontal displacement at 2.5s (=95.75..) y = 50sin40 × 2.5 – 2.5 2 g/2 B1 (=49.09..) tanθ = 49.09/95.75 M1 Appropriate ratio to find angle θ = 27.1 A1 [4] [7]

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6 A 60° 0.2 m P One end of a light inextensible string of length 0.2 m is attached to a fixed point A which is above a smooth horizontal table. A particle P of mass 0.3 kg is attached to the other end of the string. P moves on the table in a horizontal circle, with the string taut and making an angle of 60Å with the downward vertical (see diagram). (i) Calculate the tension in the string if the speed of P is 1.2 m s−1. [3] (ii) For the motion as described, show that the angular speed of P cannot exceed 10 rad s−1, and hence find the greatest possible value for the kinetic energy of P. [6]

Mark scheme: 6 (i) Radial acc n = 1.2 2 /(0.2cos30) B1 Radial acc n = 8.31..ms − 2 Tcos30 = 0.3 × 1.2 2 /(0.2cos30) M1 Component of tension = m × radial acc n T = 2.88 N A1 [3] (ii) (a) Tcos60 = 0.3g M1 Uses T max in limiting case when R = 0 T = 6 A1 May be implied 6cos30 = 0.3ω 2 (0.2cos30) M1 Component of max T = m × maximum radial acc n ω = 10 AG A1 [4] From g = 10 only OR Tcos30 = 0.3 × 10 2 (0.2cos30) M1 Finds T max from m × max(RA) T = 6 A1 R + 6cos60 = 0.3g M1 Resolves vertically with T max R = 0 and a higher value of ω makes A1 Additional justification needed of R negative which is impossible inequality (ii) (b) KE = 0.3(10 × 0.2cos30) 2 /2 M1 Attempt at KE with v = 10 × radius KE = 0.45 J A1 [2] [7]

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7 A B 15 N r m 23p O C q OABC is the cross-section through the centre of mass of a uniform prism of weight 20 N. The cross- section is in the shape of a sector of a circle with centre O, radius OA = r m and angle AOC = 230 radians. The prism lies on a plane inclined at an angle 1 radians to the horizontal, where 1 < 130. OC lies along a line of greatest slope with O higher than C. The prism is freely hinged to the plane at O. A force of magnitude 15 N acts at A, in a direction towards to the plane and at right angles to it (see diagram). Given that the prism remains in equilibrium, find the set of possible values of 1. [9]

Mark scheme: 7 OG = 2rsin(π /3)/(3π /3) B1 Centre of mass from O 15rcos(π – 2π /3) B1 Moment of 15 N about O 20 × OGcos(π /3 –θ ) B1ft Moment of weight about O, ft cv(OG) if used GCE AS/A LEVEL – May/June 2013 9709 51 M1 Uses moments, including 15 N and 20 N 15rcos(π /3) Y A1ft Accept ≺, =, ≻ as alternative to Y 20 × 2rsin(π /3)/π xcos(π /3 –θ ) cos(π /3 –θ ) [ 0.68(017..) A1 Accept ≻, =, ≺ as an alternative to [ π /3 – θ Y 0.82(279..) M1 Solves for θ , equation or inequality θ = 0.224 A1 Correct value θ ≥ 0.224 A1 [9] Correct sign, accept ≻ [9] SR deduct 1 mark for assuming r = 1

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