Cambridge A Level Mathematics - Further 9231 — 2021 Oct/Nov Paper 3 · Variant 2

1 A particle is projected with speed u at an angle a above the horizontal from a point O on a horizontal plane. The particle moves freely under gravity. (a) Write down the horizontal and vertical components of the velocity of the particle at time T after projection. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ At time T after projection, the direction of motion of the particle is perpendicular to the direction of projection. (b) Express T in terms of u, g and a. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ u (c) Deduce that T 2 . [1] g ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 1(a) Velocity: α sinα ↑ − u gT B1 Allow 10 for g. Must be T. 2 1(b) cos sin sin cos α α α α = − − u u gT oe M1 FT Allow missing minus sign on RHS for M1. FT from (a). sinα = u T g A1 2 1(c) sin 1 α < giving > u T g B1 AG 1

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2 A light spring AB has natural length a and modulus of elasticity 5mg. The end A of the spring is attached to a fixed point on a smooth horizontal surface. A particle P of mass m is attached to the end B of the spring. The spring and particle P are at rest on the surface. Another particle Q of mass km is moving with speed 4ga along the horizontal surface towards P in the direction BA. The particles P and Q collide directly and coalesce. In the subsequent motion the greatest amount by which the spring is compressed is 15 a . Find the value of k. [6] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 At the collision of P and Q: ( ) + = m km v kmu M1 Momentum conserved, allow missing k on RHS. So ( ) 4 1 = + k ga v k A1 EPE = 2 1 5 2 5 mg a a   × ×    10 mga   =     B1 Loss in KE = Gain in EPE: ( ) 2 2 1 1 5 1 2 2 5 mg a m k v a   + = × ×    M1 Energy equation, LHS correct, EPE dimensionally correct. Substitute for v and rearrange to form quadratic equation in k 2 20 1 = + k k M1 1 4 = k A1 6

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3 R i 3m B m A Particles A and B, of masses m and 3m respectively, are connected by a light inextensible string of length a that passes through a fixed smooth ring R. Particle B hangs in equilibrium vertically below the ring. Particle A moves in horizontal circles with speed v. Particles A and B are at the same horizontal level. The angle between AR and BR is i (see diagram). (a) Show that cos i = 13 . 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(b) Find an expression for v in terms of a and g. 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Mark scheme: 3(a) = M1 Must see both of these separately. Combining, 1 cos 3 θ = A1 At least one step of working, AG. 2 3(b) (cos , θ − = a x x where x = AR) 3 1 or 4 4 AR a BR a = = or radius = 2 a B1 8 sin 3 θ   =       2 sinθ = mv T r M1 Combining to find an equation in v2, a and g only. DM1 2 2 = v ga , 2 = v ga A1 4

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4 A 1 2 a B ka a An object is formed by removing a solid cylinder, of height ka and radius 12 a , from a uniform solid hemisphere of radius a. The axes of symmetry of the hemisphere and the cylinder coincide and one circular face of the cylinder coincides with the plane face of the hemisphere. AB is a diameter of the circular face of the hemisphere (see diagram). 3a ( 2 - k 2 ) (a) Show that the distance of the centre of mass of the object from AB is . 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When the object is freely suspended from the point A, the line AB makes an angle i with the downward vertical, where tan i = 187 . (b) Find the possible values of k. 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Mark scheme: 4(a) Volume Centre of mass from AB Hemisphere 3 2 π 3 a 3 8 a Cylinder 2 π ( ) 2 a ka 2 ka Remainder 2 3 2 π π 3 2 a a ka  −     x M1 Attempt at moments, 3 terms. Taking moments about AB: 2 3 2 π π 3 2 a a ka         −       × 2 3 2 3 π π 3 8 2 2 a ka x a a ka  × ×       = −             A1 A1 Any 2 terms correct. All correct. ( ) ( ) 2 3 2 2 8 3 a k x k − = − A1 Shown convincingly, AG. 4 4(b) tanθ = x a ( ) ( ) 2 3 2 7 2 8 3 18 − = − k k B1 2 27 21 2 0 − + = k k M1 Rearrange to form quadratic. 2 3 k = and 1 9 k = A1 Both answers correct. 3

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5 A B u 60° m 32 m u Two uniform smooth spheres A and B of equal radii have masses m and 32 m respectively. The two spheres are each moving with speed u on a horizontal surface when they collide. Immediately before the collision A’s direction of motion is along the line of centres, and B’s direction of motion makes an angle of 60° with the line of centres (see diagram). The coefficient of restitution between the spheres is 2.3 (a) Find the angle through which the direction of motion of B is deflected by the collision. 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(b) Find the loss in the total kinetic energy of the system as a result of the collision. 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Mark scheme: 5(a) Let speeds of A and B along line of centres after collision be 1v and 2v 1 2 3 3 cos60 2 2 4   + = − ° + =     u mv mv mu mu M1 Momentum with masses correct. ( ) 2 1 2 cos60 ( 3 − = − − ° − = v v u u u ) M1 Restitution, with consistent signs on LHS. 1 2 1 1 2 2   = − =     v u v u A1 Perpendicular to line of centres, speed of B is 3 sin60 2 ° = u u B1 Direction of B is now 60° above line of centres. M1 Angle of deflection is 60°. A1 FT FT (120° – their direction of B angle) 6 5(b) KE before = 2 2 2 1 1 3 5 . 2 2 2 4 + = m mu u mu B1 KE after = 2 2 2 1 1 3 3 . 2 2 2 2 2 2           + +                   u m u u m 2 7 8 mu       = B1 FT FT only their speeds from (a) Loss in KE = 2 3 8 mu B1 3

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6 A particle P of mass 2 kg moves along a horizontal straight line. The point O is a fixed point on this line. At time t s the velocity of P is v ms -1 and the displacement of P from O is x m. 128 A force of magnitude 8x - 3 N acts on P in the direction OP. When t = 0, x = 8 and v = - 15 . e x o 2 2 (a) Show that v = - ( x - 4) . 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(b) Find an expression for x in terms of t. 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Mark scheme: 6(a) 3 128 2 8 = − dv v x dx x 2 2 2 4 64 − = + + v x x c A1 OE. x = 8, v = ‒15 and c = ‒32 M1 Use initial condition. ( ) 2 4 2 2 4 8 16 = − + v x x x or 2 2 64 4 32 − + x x A1 Correct expression for 2v , AEF. ( ) 2 2 2 2 4 4 v x x = − giving ( ) 2 2 4 = − − v x x A1 Convincingly shown, e.g. v is negative initially, AG. 5 6(b) ( ) 2 1 ln( 4) 2 2 − = − + x t A M1 Use = dx v dt and integrate. 1 0, 8, ln60 2 t x A = = = DM1 Use initial condition. 2 1 4 ln 2 2 60   − = −     x t giving 2 4 4 60 − − = t x e M1 Remove log. 4 4 60 − = + t x e A1 CAO 4

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7 One end of a light inextensible string of length a is attached to a fixed point O. The other end of the string is attached to a particle P of mass m. The particle P is held vertically below O with the string taut and then projected horizontally. When the string makes an angle of 60° with the upward vertical, P becomes detached from the string. In its subsequent motion, P passes through the point A which is a distance a vertically above O. (a) The speed of P when it becomes detached from the string is V. Use the equation of the trajectory of a projectile to find V in terms of a and g. 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(b) Find, in terms of m and g, the tension in the string immediately after P is initially projected horizontally. 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Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. ..................................................................................................................................................................

Mark scheme: 7(a) Coordinates of A: = = − B1 ( ) 2 2 2 3 3 2 3 1 2 2 2 .4 = − a g a a V M1 Substitute their (x, y) into correct trajectory equation. Rearrange to find 2 V . M1 2 3 3 , 2 2 = = V ag V ag A1 4 7(b) ( ) 2 2 1 1 1 cos60 2 2 − = + mu mV mga M1 Energy equation. 2 9 2 = u ag A1 u is the speed at P. 2 − = m T mg u a M1 N2L 11 2 = T mg A1 4

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