Cambridge A Level Mathematics - Further 9231 — 2022 May/June Paper 3 · Variant 1

1 A i 7.5 N A particle of weight 10 N is attached to one end of a light elastic string. The other end of the string is attached to a fixed point A on a horizontal ceiling. A horizontal force of 7.5 N acts on the particle. In the equilibrium position, the string makes an angle i with the ceiling (see diagram). The string has natural length 0.8 m and modulus of elasticity 50 N. (a) Find the tension in the string. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the vertical distance between the particle and the ceiling. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 1(a) B1   1 2 2 2 7.5 10 T   = 12.5 N B1 2 1(b) Hooke’s law: 50 0.8 x T  , 0.2 x  B1   10 0.8 sin 1 12.5 x    M1 4 Vertical distance 0.8 5   A1 3

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2 One end of a light inextensible string of length a is attached to a fixed point O. A particle of mass m is attached to the other end of the string. The particle is held at the point A with the string taut. The angle between OA and the downward vertical is equal to a, where cos a = 45 . The particle is projected from A, perpendicular to the string in an upwards direction, with a speed 3ga . It then moves along a circular path in a vertical plane. The string first goes slack when it makes an angle i with the upward vertical through O. Find the value of cosi. [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2   2 1 1 3 cos cos 2 2 mv m ga mga        M1 A1 Energy equation.   2 cos m T mg v a    B1 N2L 4 cos 3 2 cos 2 5 ag ga ga ga       M1 Combine to find cos. 7 cos 15  A1 5

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3 A particle P is moving in a horizontal straight line. Initially P is at the point O on the line and is moving -1 4000 -2 with velocity 25 m s . At time t s after passing through O, the acceleration of P is 3 m s in ( t5 + 4 ) the direction PO. The displacement of P from O at time t is x m. Find an expression for x in terms of t. [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 3   3 4000 5 4 dv dt t   ;     2 400 5 4 v A t    M1 A1 Integrate. Constant of integration needed for A1. 0, 25 25 25 0 t v A      M1 Find constant.   1 : 80 5 4 dx v x t B dt      0, 0 20 x t B    M1 Integrate and find constant. 80 100 20 5 4 5 4 t x t t             A1 5

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4 2a h a A B An object is composed of a hemispherical shell of radius 2a attached to a closed hollow circular cylinder of height h and base radius a. The hemispherical shell and the hollow cylinder are made of the same uniform material. The axes of symmetry of the shell and the cylinder coincide. AB is a diameter of the lower end of the cylinder (see diagram). (a) Find, in terms of a and h, an expression for the distance of the centre of mass of the object from AB. 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The object is placed on a rough plane which is inclined to the horizontal at an angle i, where tan i = 23 . The object is in equilibrium with AB in contact with the plane and lying along a line of greatest slope of the plane. (b) Find the set of possible values of h, in terms of a. 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Mark scheme: 4(a) Area Centre of mass from AB Cylinder 2 2π 2π ah a  1 2 h Shell   2 2π 2 a h a  M1 Moments equation, condone missing ends of cylinder. One expression on the RHS correct. Moments about AB          2 2 2 2 1 2π 2π 2π 2 2π 2 2π 2π 2 x ah a a a h a ah a h              A1 A1 One correct expression on RHS correct scores A1.   2 2 2 10 8 8 h a x h ah ah a        2 2 9 8 2 5 h ah a x h a     A1 4 4(b) tan a x  B1   2 2 9 8 3 2 5 2 h ah a x a h a      2 2 6 7 0 h ah a    M1 Form inequality and rearrange to quadratic, condone equation.    7 0 h a h a    M1 Attempt to solve, condone equation.   7a h a    A1 4

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5 C D i a i B 3a A A light inextensible string AB passes through two small holes C and D in a smooth horizontal table where AC = 3a and DB = a. A particle of mass m is attached at the end A and moves in a horizontal circle with angular velocity ~. A particle of mass 34 m is attached to the end B and moves in a horizontal circle with angular velocity k~. AC makes an angle i with the downward vertical and DB makes an angle i with the horizontal (see diagram). Find the value of k. 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Mark scheme: 5 For A: 2 sin T mr    M1 N2L horizontal. 3 sin r a   B1 Correct expression for radius. 2 3 T m a   A1 Similarly, for B: 2 2 3 cos 4 T m r k     M1 N2L horizontal 2 2 3 4 T mak   A1 2 2 2 3 3 4 m a mak     M1 Equate expressions for T. 2 4, 2 k k   A1 Question Answer Marks Guidance 5 Alternative method for question 5 For A: 2 cos , sin T mg T mr      M1 N2L horizontal and vertical. 3 sin r a   B1 Correct expression for radius. 2 2 5 5 3 , 4 12 g T m a mg a        A1 Combine to obtain expression for 2 . Similarly, for B: 2 2 3 cos 4 T m r k     M1 N2L horizontal. 2 2 3 4 T mak   A1 2 5 3 5 4 4 12 g mg mak a   M1 Substitute for T and . 2 4, 2 k k   A1 7

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6 Two uniform smooth spheres A and B of equal radii have masses m and km respectively. The two spheres are on a horizontal surface. Sphere A is travelling with speed u towards sphere B which is at rest. The spheres collide. Immediately before the collision, the direction of motion of A makes an angle a with the line of centres. The coefficient of restitution between the spheres is 1.2 3u cos a (a) Show that the speed of B after the collision is and find also an expression for the speed of 2 ( 1 + k) A along the line of centres after the collision, in terms of k, u and a. 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After the collision, the kinetic energy of A is equal to the kinetic energy of B. (b) Given that tan a = 23 , find the possible values of k. 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Mark scheme: 6(a) Let v and w be speeds after collision: cos mv kmw mu    M1 Momentum along line of centres. 1 cos 2 w v u    M1 NEL consistent signs. Add to give   3 cos 2 1 u k   A1 AG Convincing working. Substitute back or re-solve:     2 cos 2 1 k u v k     A1 Accept without modulus sign. 4 Question Answer Marks Guidance 6(b)     2 2 2 cos ( sin ) 2 1 k u u k             B1 For speed of A (SOI). Equal KE after collision:       2 2 2 2 cos 1 3 cos 1 ( sin ) 2 2 1 2 2 1 k u u km m u k k                                         2 2 2 2 2 9 cos 4 1 sin 2 cos k k k            M1 Equate KEs. Use 2 tan : 3      2 2 16 1 2 9 4 4 81 k k k k k       M1 2 25 85 52 0 k k    leading to    5 4 5 13 0 k k    M1 Obtain quadratic and attempt to solve. 4 13 or 5 5 k  A1 5

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7 Particles P and Q are projected in the same vertical plane from a point O at the top of a cliff. The height of the cliff exceeds 50 m. Both particles move freely under gravity. Particle P is projected with a above the horizontal, where tan a = 3 . Particle Q is projected with speed speed 352 ms -1 at an angle 4 u m s -1 at an angle b above the horizontal, where tan b = 12 . Particle Q is projected one second after the projection of particle P. The particles collide T s after the projection of particle Q. (a) Write down expressions, in terms of T, for the horizontal displacements of P and Q from O when they collide and hence show that 4uT = 21 5 ( T + 1) . 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(b) Find the value of T. 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(c) Find the horizontal and vertical displacements of the particles from O when they collide. 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Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................

Mark scheme: 7(a) For Q: cos x u T For P:   35 cos 1 2 x T    B1 Collision, so   35 cos 1 cos 2 T u T     M1 Equate and attempt to rearrange.   35 3 2 1 2 5 5 T u T       4 21 5 1 uT T   A1 AG Shown convincingly. 4 7(b) Vertical motion to collision: For Q: 2 1 sin 2 y u T gT    For P:   2 35 1 sin 1 ( 1) 2 2 y T g T      M1 A1 M1 for both expressions, one correct. Equate:   2 2 1 1 35 4 1 1 ( 1) 2 2 5 2 5 u T gT T g T              2 2 1 21 14 1 2 1 1 2 4 T g T T T T       M1 Equate and attempt to solve 16 36 21 21, 1 5 5 T T T     3 T  A1 4 Question Answer Marks Guidance 7(c) 42 x  B1 24 y  M1 24 y  (or 24 m below O ) A1 Correct sign or in words. 3

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