Cambridge A Level Mathematics 9709 — 2014 Oct/Nov Paper 5 · Variant 2

1 A particle P is projected with speed V m s−1 at an angle of 30Å above the horizontal from a point O on horizontal ground. At the instant 2 s after projection, OP makes an angle of 15Å above the horizontal. Calculate V. [4]

Mark scheme: 1 X = 2Vcos30 B1 1.731V 22 Y = 2V sin 30 − g B1 V–20 2  22   2V sin 30 − g   2  tan15 = M1 2V cos30 V = 37.3 A1 [4] 3

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2 F N V 30Å 0.8 m P A uniform solid cone with height 0.8 m and semi-vertical angle 30Å has weight 20 N. The cone rests in equilibrium with a single point P of its base in contact with a rough horizontal surface, and its vertex V vertically above P. Equilibrium is maintained by a force of magnitude F N acting along the axis of symmetry of the cone and applied to V (see diagram). (i) Show that the moment of the weight of the cone about P is 6 N m. [2] (ii) Hence find F. [2]

Mark scheme: 3 2 (i) Horizontal distance = 0.8 × × sin30 P to centre of mass (= 0.3 m) 4 0.8 OR 0.8tan30cos30 – sin30 M1 4 Mom. = (0.6sin30 × 20 =) 6 Nm AG A1 OR [2] 0.8 Mom = 20cos30 × 0.8tan30 – 20sin30 × M1 Resolves Wt // and perp axis and finds 4 moments of both components Mom = 6 Nm AG A1 (ii) 6 = F × 0.8tan30 M1 Takes moments about P F = 13(.0) A1 [2]

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3 One end of a light elastic string of natural length 1.6 m and modulus of elasticity 28 N is attached to a fixed point O. The other end of the string is attached to a particle P of mass 0.35 kg which hangs in equilibrium vertically below O. The particle P is projected vertically upwards from the equilibrium position with speed 1.8 m s−1. Calculate the speed of P at the instant the string first becomes slack. [5]

Mark scheme: 3 28 e = 0.35g M1 Equates λext/l and weight 1.6 e = 0.2 A1 OP = 1.8 m 0.35 v 2 0.2 2 1.8 2 = 28 × × 1.6 + 0.35 × − 0.35 g × 0.2 M1 EE/KE/PE balance 2 2 2 A1 All correct terms with candidate’s value of e v = 1.11 m s–1 A1 [5]

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4 B C 1.8 m D T N 0.4 m E 1.6 m A F 0.4 m ABCDEF is the cross-section through the centre of mass of a uniform solid prism. ABCF is a rectangle in which AB = CF = 1.6 m, and BC = AF = 0.4 m. CDE is a triangle in which CD = 1.8 m, CE = 0.4 m, and angle DCE = 90Å. The prism stands on a rough horizontal surface. A horizontal force of magnitude T N acts at B in the direction CB (see diagram). The prism is in equilibrium. (i) Show that the distance of the centre of mass of the prism from AB is 0.488 m. [4] (ii) Given that the weight of the prism is 100 N, find the greatest and least possible values of T. [3]

Mark scheme: 4 (i) ABCF area = 0.64 and CDE = 0.36 B1 Both areas correct 0.4 1.8 (0.64 + 0.36)d = 0.64× + 0.36×(0.4 + ) M1 Table of moments idea 2 3 A1 All terms correct d = 0.488 m AG A1 [4] (ii) 0.488 × 100 = 1.6T M1 Either limiting case T = 30.5 N A1 (no turning about A) (0.488 – 0.4) × 100 = 1.6T T = 5.5 A1 [3] (no turning about F)

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5 The equation of the trajectory of a small ball B projected from a fixed point O is y = −0.05x2, where x and y are, respectively, the displacements in metres of B from O in the horizontal and vertically upwards directions. (i) Show that B is projected horizontally, and find its speed of projection. [3] (ii) Find the value of y when the direction of motion of B is 60Å below the horizontal, and find the corresponding speed of B. [6]

Mark scheme: 5 (i) xtanα = 0 so α = 0 B1 Justification needed gx 2 2 2 2 = 0.05 x M1 Comparison with standard eqn 2V cos 0 V = 10 m s–1 A1 [3] dy (ii) = –0.1x M1 dx –0.1x = –tan60 M1 y (= –0.05(10tan60)2) = –15 A1 v2 = M1 Uses Pythagoras 102 + 2g15 A1 ft candidate’s value (V(i), y) v = 20 m s–1 A1 OR [6] y' = 10tan 60 M1 y' = B’s downward velocity =10√3 (10√3)2 =2gh M1 y = –15 A1 Negative, y = –h v2 = M1 Uses Pythagoras 102 + (10√3)2 A1 ft candidate’s value (V(i)) v = 20 m s–1 A1 OR vcos60 = 10 M1 v = 20 m s–1 A1 10 3 = 10t M1 t = 3 A1 3 y = 10 3 × M1 2 y = 15 (below) or –15 A1 dv 1/2 dv

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6 O, A and B are three points in a straight line on a smooth horizontal surface. A particle P of mass 0.6 kg moves along the line. At time t s the particle has displacement x m from O and speed v m s−1. 1 The only horizontal force acting on P has magnitude 0.4v 2 N and acts in the direction OA. Initially the particle is at A, where x = 1 and v = 1. 1 dv (i) Show that 3v 2 = 2. [2] dx (ii) Express v in terms of x. [4] (iii) Given that AB = 7 m, find the value of t when P passes through B. [3] [Question 7 is printed on the next page.]

Mark scheme: dv dv 6 (i) 0.6v = 0.4v1/2 M1 Newton’s 2nd law, a = v dx dx dv 3v1/2 = 2 AG A1 dx [2] 1 (ii) 3 ∫ v 2 dv = 2 ∫ dx M1 Integrates 3 3v 2 = 2 x (+c) A1 Accept omission of +c 3 Evaluates c (=0) 2 3 2 3 × 21 × = 2 +c M1 3 2 v = x 3 A1 [4] − 2 d x (iii) ∫ x 3 dx = ∫ dt M1 Integrates using v = d t 8 1    x 3  = t A1   1    3 1 t = 3 A1 [3]  0.4  15    cosθ  λ ext

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7 A 2 m R 0.4 m 1 P 7 rad s−1 One end of a light elastic string with modulus of elasticity 15 N is attached to a fixed point A which is 2 m vertically above a fixed small smooth ring R. The string has natural length 2 m and it passes through R. The other end of the string is attached to a particle P of mass m kg which moves with constant angular speed 7 rad s−1 in a horizontal circle which has its centre 0.4 m vertically below the ring. PR makes an acute angle 1 with the vertical (see diagram). 3 (i) Show that the tension in the string is N and hence find the value of m. [4] cos 1 (ii) Show that the value of 7 does not depend on 1. [4] It is given that for one value of 1 the elastic potential energy stored in the string is twice the kinetic energy of P. (iii) Find this value of 1. [4]

Mark scheme:  cos θ  λ ext 7 (i) T = M1 Uses T = 2 2 3 T = AG A1 cos θ Tcosθ = mg M1 Resolves vertically for P m = 0.3 A1 [4] (ii) r = 0.4tanθ B1 0.3v 2 = T sin θ OR 0.3ω2r = Tsinθ M1 Newton’s 2nd law with correct r expression for radial accn, ft cv(m(i)) 3 0.3ω2(0.4tanθ) = × sinθ A1 cosθ ω = 5 A1 SC [4] Candidates who choose at least two specific values of θ: Calculation of r twice B1 Both calculations give ω = 5 B1  0.4  2 15    cos θ  (iii) EPE = B1 2 × 2 0.3 ( 5 × 0.4 tan θ ) 2 KE = B1 ft candidate’s value of ω 2 Award if × 2 is with wrong term  0.4  2 15   2  cos θ   0.3(2 tan θ )  =   × 2 M1 2 × 2  2  cos2θ tan2θ = 0.5 OR sin2θ = 0.5 θ = 45 A1 www [4]

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