Cambridge A Level Mathematics - Further 9231 — 2021 May/June Paper 3 · Variant 2

1 A particle P of mass 1 kg is moving along a straight line against a resistive force of magnitude 10 v -1 2 N, where v ms is the speed of P at time t s. When t = 0, v = 25. ( t + 1 ) Find an expression for v in terms of t. [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 1 ( ) 2 d 10d 1 v t v t = − + 10 2 1 = + + v A t M1 A1 Attempt to integrate. 0, 25, 0 t v A = = = M1 Use correct initial condition. ( ) 2 25 1 = + v t A1 CAO 5

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2 A hollow hemispherical bowl of radius a has a smooth inner surface and is fixed with its axis vertical. A particle P of mass m moves in horizontal circles on the inner surface of the bowl, at a height x above the lowest point of the bowl. The speed of P is 83 ga . Find x in terms of a. [6] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 cosθ ↑ = R 2 sinθ → = mv R r B1 sinθ = r a B1 ( ) ( ) 2 8cos 3 1 cos θ θ = − M1 Quadratic equation in cosθ . 1 cos 3 θ = A1 2 3 = x a A1 6

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3 One end of a light elastic string, of natural length a and modulus of elasticity kmg, is attached to a fixed point A. The other end of the string is attached to a particle P of mass 4m. The particle P hangs in equilibrium a distance x vertically below A. 4a (a) Show that k = . [1] x - a ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ An additional particle, of mass 2m, is now attached to P and the combined particle is released from rest at the original equilibrium position of P. When the combined particle has descended a distance 13 a , its speed is 13 ga . (b) Find x in terms of a. [6] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 3(a) Use Hooke’s Law: ( ) 4 kmg x a mg a − = leading to 4a k x a = − B1 AG. Shown convincingly. 1 3(b) Gain in KE + gain in EPE = loss in GPE B1 One correct EPE term seen. ( ) 2 2 1 1 6 6 2 9 2 3 3 ga kmg a a m x a x a mg a     × × + + − − − = ×           M1 A1 All 3 types of energy included in energy equation. All terms correct. Simplify and substitute for k from part (a) M1 Obtain linear equation in x and a M1 5 3 = x a A1 (k = 6) 6

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4 kh h r A uniform solid circular cone has vertical height kh and radius r. A uniform solid cylinder has height h and radius r. The base of the cone is joined to one of the circular faces of the cylinder so that the axes of symmetry of the two solids coincide (see diagram, which shows a cross-section). The cone and the cylinder are made of the same material. (a) Show that the distance of the centre of mass of the combined solid from the base of the cylinder h ( k 2 + 4k + 6) is . 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The solid is placed on a plane that is inclined to the horizontal at an angle i. The base of the cylinder is in contact with the plane. The plane is sufficiently rough to prevent sliding. It is given that 3h = 2r and that the solid is on the point of toppling when tan i = 43 . (b) Find the value of k. 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Mark scheme: 4(a) Volume Centre of mass from base of cylinder Cone 2 1 π 3 r kh 4 + kh h Cylinder 2 πr h 2 h Combined 2 πr h 1 1 3 k  +      x Take moments about base: 2 πr h 1 1 3 k   +     x = 2 1 π 3 r kh 4 kh h  +      + 2 πr h 2 h A1 A1 2 terms correct. All terms correct. x = ( ) ( ) 2 4 6 4 3 h k k k + + + A1 AG. Shown convincingly. 4 4(b) tanθ = r x M1 ( ) ( ) 2 6 3 4 3 4 6 + = + + h k h k k 2 2 15 0 − − = k k M1 Equate to 4 3 and simplify to quadratic. 3 = k A1 CAO. No other solutions. 3

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5 A particle P of mass m is attached to one end of a light inextensible string of length a. The other end of the string is attached to a fixed point O. The particle completes vertical circles with centre O. The points A and B are on the path of P, both on the same side of the vertical through O. OA makes an angle i with the downward vertical through O and OB makes an angle i with the upward vertical through O. The speed of P when it is at A is u and the speed of P when it is at B is ag. The tensions in the string at A and B are TA and TB respectively. It is given that TA = 7TB. Find the value of i and find an expression for u in terms of a and g. 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Mark scheme: 5 2 cosθ − = A mu T mg a B1 cosθ + = B mag T mg a B1 TA = 7TB so 2 cos 7 cos θ θ   + = − +     mu mag mg mg a a ( ) 2 7 8cosθ = − u ag M1 Use given relationship and combine. Energy: ( ) 2 1 1 cos cos 2 2 θ θ − = + mu mag mg a a So ( ) 2 4cos 1 θ = + u ag M1 A1 Energy equation. Equate expressions for 2 u M1 1 cos , 60 2 θ θ = = ° A1 CAO 3 = u ga A1 CAO 8

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6 A B a b m m u u Two uniform smooth spheres A and B of equal radii each have mass m. 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 makes an angle a with the line of centres, and B’s direction of motion makes an angle b with the line of centres (see diagram). The coefficient of restitution between the spheres is 13 and 2 cos b = cos a . (a) Show that the direction of motion of A after the collision is perpendicular to the line of centres. 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The total kinetic energy of the spheres after the collision is 34 mu 2 . (b) Find the value of a. 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Mark scheme: 6(a) Along line of centres, speeds 1 v and 2 v 1 2 cos cos α β + = − mv mv mu mu M1 Momentum (condone missing masses). ( ) 2 1 cos cos β α − = + v v eu M1 Restitution. Both correct, masses seen. A1 1 0 = v so A has no speed along line of centres: moves perpendicular to line of centres A1 AG. 4 6(b) 2 1 ( cos cos 2 α β = = v u u ) KE of B after collision is ( ) ( ) 2 2 2 1 sin 2 β + m v u KE of A after collision = ( ) 2 1 sin 2 α m u M1 Both components. Add both KEs and equate to 2 3 4 mu M1 Simplify to equation in sinα M1 1 sin 2 α = , 45 α = ° A1 4

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7 A particle P is projected from a point O on a horizontal plane and moves freely under gravity. The initial velocity of P is 100ms -1 at an angle i above the horizontal, where tan i = 43 . The two times at which P’s height above the plane is H m differ by 10 s. (a) Find the value of H. 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(b) Find the magnitude and direction of the velocity of P one second before it strikes the plane. 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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) At greatest height 0 100sinθ = −gt t = 8 A1 Therefore times at height H are 3 = t (and 13 = t ) B1 Substitute into 2 1 100sin 2 θ = − H t gt M1 195 = H A1 Alternative method to question 7(a) 2 1 100sin 2 θ ↑ = − H t gt M1 And ( ) ( ) 2 1 100sin 10 10 2 θ = + − + H t g t A1 Subtract: ( ) 1 1000sin 20 100 2 θ = + g t M1 3 = t B1 195 = H A1 Question Answer Marks Guidance 7(a) Alternative method to question 7(a) 2 1 100sin 2 θ ↑ = − H t gt B1 Difference between roots = 2 (100sin ) 2 1 2 θ −gH g M1 A1 Equate to 10 and rearrange to find H M1 195 = H A1 5 7(b) Time to required point = 15 s B1 ( ) 100sin 10 15 70 θ ↑= − × =− v 100cos 60 θ →= = v B1 Both components. Magnitude = 92.2 B1 Angle below horizontal = tan -1 (70/60) = 49.4° B1 4

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