Cambridge A Level Mathematics 9709 — 2015 May/June Paper 5 · Variant 1

1 One end of a light elastic string of natural length 0.7 m is attached to a fixed point A on a smooth horizontal surface. The other end of the string is attached to a particle P of mass 0.3 kg which is held at a point B on the horizontal surface, where AB = 1.2 m. It is given that P is released from rest at B and that when AP = 0.9 m, the particle has speed 4 m s−1. Calculate the modulus of elasticity of the string. [3]

Mark scheme: 1 λ × 0.5 2 5λ  EE(B) = = B1 Correct EE when AP = 1.2 m 2 × 0.7  28  Correct EE when AP = 0.9 m λ × 0.2 2 λ  OR EE = = 2 × 0.7  35  Λ× 0.5 2 λ × 0.2 2 0.3 × 4 2 – = M1 Using EE loss = KE gain 2 × 0.7 2 × 0.7 2 λ = 16 N A1 [3]

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2 A stone is projected from a point O on horizontal ground. The equation of the trajectory of the stone is y = 1.2x −0.15x2, where x m and y m are respectively the horizontal and vertically upwards displacements of the stone from O. Find (i) the greatest height of the stone, [2] (ii) the distance from O of the point where the stone strikes the ground. [2]

Mark scheme: 2 (i) d y M1 Solving trajectory derivative = 0 = 1.2 – 2(0.15x) = 0 d x y = 2.4 A1 [2] From x = 4 OR Greatest height at half range (ii) M1 Uses y = 1.2x – 0.15 x2 with x = distance in (ii) ÷ 2 y = 2.4 A1 (ii) Twice 'x' for greatest height M1 Using 'x' from (i) Distance = 8 m A1 [2] No ft OR 1.2x – 0.15 x2 = 0 M1 Solves quadratic with y = 0 Distance = 8 m A1

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3 A 1 5 rad s−1 O P One end of a light inextensible string is attached to a fixed point A and the other end of the string is attached to a particle P. The particle P moves with constant angular speed 5 rad s−1 in a horizontal circle which has its centre O vertically below A. The string makes an angle 1 with the vertical (see diagram). The tension in the string is three times the weight of P. (i) Show that the length of the string is 1.2 m. [3] (ii) Find the speed of P. [4]

Mark scheme: 3 (i) Tsinθ = m ω 2 r M1 Newton’s 2nd law, acceleration = 52r and component of T  ω  2 3ωsinθ =  5 (Lsinθ) A1 3mgsinθ = m 52 (Lsinθ)  g  L = 1.2 m AG A1 [3] (ii) 3ω cosθ = ω M1 Resolves vertically for P –1  1  θ = 70.53° A1 OR θ = cos   ,  3  8 θ = sin–1 9 etc. v = 5×1.2sinθ M1 v = ωr v = 5.66 ms –1 A1 [4]

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4 15 m s−1 O 30Å B A small ball B is projected from a point O above horizontal ground, with initial speed 15 m s−1 at an angle of projection of 30Å above the horizontal (see diagram). The ball strikes the ground 3 s after projection. (i) Calculate the speed and direction of motion of the ball immediately before it strikes the ground. [5] (ii) Find the height of O above the ground. [2]

Mark scheme: 4 (i) x' = 15cos30 ( = 12.990..) B1 y' = 15sin30 – 3g ( = – 22.5 ) B1 Or with signs reversed v2 = (15cos30)2 + (15sin30 – 3g)2 or M1 3 g – 15sin30 tanθ = 15cos30 v = 26(.0) ms –1 A1 θ = 60° to the horizontal A1 [5] Or 30° to the vertical (ii) 23 g y = 3(15sin30) – M1 Uses s = ut + 1at2 2 2 Height = 22.5 m A1 [2] N.B. this is also y' OR ( – 22.5)2 = (15sin30)2 – 2 × 10y M1 Uses v2 = u2 + 2as Height = 22.5 m A1

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5 A particle P of mass 0.3 kg is attached to one end of a light elastic string of natural length 0.9 m and modulus of elasticity 18 N. The other end of the string is attached to a fixed point O which is 3 m above the ground. (i) Find the extension of the string when P is in the equilibrium position. [2] P is projected vertically downwards from the equilibrium position with initial speed 6 m s−1. At the instant when the tension in the string is 12 N the string breaks. P continues to descend vertically. (ii) (a) Calculate the height of P above the ground at the instant when the string breaks. [2] (b) Find the speed of P immediately before it strikes the ground. [4]

Mark scheme: 5 (i) 18e λ x 0.3g = M1 Uses T = 0.9 l e = 0.15 m A1 [2] (ii) (a) 18ext 12 = and ht = 3 – 0.9 – ext M1 Both ideas needed, ext = 0.6 0.9 ht = 1.5 m A1 [2] (ii) (b) 0.3 × 6 2 0.3u 2 – + 0.3g(0.6 – 0.15) M1 KE/PE/EE balance up to string 2 2 A1 breaking 18 × 0.6 2 18 × 0.15 2 = – 2 × 0.9 2 × 0.9  3.0u 2   = .3 375  u2 = 22.5  2  0.3v2 = 0.3u2 + 0.3g(3 – 0.6 – 0.9) M1 KE/PE balance after string breaks or OR v2 = u2 + 2g(3 – 0.6 – 0.9) v2 = u2 + 2g(ht) using ht from (ii)(a) v = 7.25 ms–1 A1 4 7.2456

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6 A particle P of mass 0.1 kg moves with decreasing speed in a straight line on a smooth horizontal surface. A horizontal resisting force of magnitude 0.2e−x N acts on P, where x m is the displacement of P from a fixed point O on the line. The velocity of P is v m s−1 when its displacement from O is x m. (i) Show that vdv = ke−x, dx where k is a constant to be found. [2] P passes through O with velocity 2.2 m s−1. (ii) Calculate the value of x at the instant when the velocity of P is 2 m s−1. [4] (iii) Show that the speed of P does not fall below 0.917 m s−1, correct to 3 significant figures. [2] [Question 7 is printed on the next page.]

Mark scheme: 6 (i) d v 0.1v = – 0.2 e–x M1 Newton’s 2nd law, 1 force d x Must have negative coefficient d v v = – 2e–x d x k = –2 A1* [2] (ii) ∫ vd v = ∫ − 2 e− x dx M1 Integrates v 2 = 2e–x ( + c ) D*A1 Needs first A1 in (i) 2  2.2 2 0  c =  − 2 e  = .0 42 M1 Or uses limits of 2 and 2.2 for v and x  2  and 0 for x 2 2 − x = 2 e + 0.42 A1 [4] 2 x = 0.236 (iii) v 2 − ∞ = 2 e + 0.42 M1 OR finds x when v = 0.9165 2 v = 0.917 ms–1 AG A1 [2] No solution when v = 0.917 2

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7 D 0.8 m E 0.6 m C A O B 45Å The diagram shows the cross-section OABCDE through the centre of mass of a uniform prism on a rough inclined plane. The portion ADEO is a rectangle in which AD = OE = 0.6 m and DE = AO = 0.8 m; the portion BCD is an isosceles triangle in which angle BCD is a right angle, and A is the mid-point of BD. The plane is inclined at 45Å to the horizontal, BC lies along a line of greatest slope of the plane and DE is horizontal. (i) Calculate the distance of the centre of mass of the prism from BD. [3] The weight of the prism is 21 N, and it is held in equilibrium by a horizontal force of magnitude P N acting along ED. (ii) (a) Find the smallest value of P for which the prism does not topple. [2] (b) It is given that the prism is about to slip for this smallest value of P. Calculate the coefficient of friction between the prism and the plane. [3] The value of P is gradually increased until the prism ceases to be in equilibrium. (iii) Show that the prism topples before it begins to slide, stating the value of P at which equilibrium is broken. [5]

Mark scheme: 7 (i) d(0.6 × 0.8 + 0.62) = M1 Moments about BAD 0.4(0.6 × 0.8) – (0.6/3) × 0.62 A1 d = 0.143 m A1 [3] Exact 1/7 (ii) (a) 21 × 0.143 = 1.2 P M1 Moments about B P = 2.5(0) A1 [2] (ii) (b) Fr = 21sin45 – 2.5cos45 and B1 R = 21cos45 + 2.5sin45 M1 For using Fr = µR µ = 0.787 A1 [3] (iii) P × 0.6 = 21 × (0.143 + 0.6) M1 Moments about C P = 26(.0) A1 Required Fr = 26sin45 – 21sin45 M1 3.5355.. Max Fr = 26(.155) A1 0.787 × (26cos45 + 21cos45) As actual Fr < max Fr , no sliding A1 [5]

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