Cambridge A Level Mathematics - Further 9231 — 2020 May/June Paper 3 · Variant 2

1 A particle P is projected with speed u at an angle of 30° above the horizontal from a point O on a horizontal plane and moves freely under gravity. The particle reaches its greatest height at time T after projection. Find, in terms of u, the speed of P at time 23 T after projection. [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 1 For greatest height, 2 = u T g At 2 2 , 3 2 3 6 = ↑ = − = v T u Tg u t v M1 3 2 → = h u v A1 Speed = 2 2 + v h v v = 2 2 3 36 4 + u u M1 = 7 3 u A1 5

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2 R x i A B A light inextensible string of length a is threaded through a fixed smooth ring R. One end of the string is attached to a particle A of mass 3m. The other end of the string is attached to a particle B of mass m. The particle A hangs in equilibrium at a distance x vertically below the ring. The angle between AR and g BR is i (see diagram). The particle B moves in a horizontal circle with constant angular speed 2 . a Show that cos i = 13 and find x in terms of a. 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Mark scheme: 2 For A: 3 = T mg For B: cosθ ↑ = T mg Equate: 3 cosθ = mg mg 1 cos 3 θ = A1 2 sinθ ω → = T mr with ( )sinθ = − r a x M1 Equate: ( ) 2 3 ω = − mg m a x A1 4 = a x A1 5

More questions on Circular motion

3 One end of a light elastic spring, of natural length a and modulus of elasticity 5mg, is attached to a fixed point A. The other end of the spring is attached to a particle P of mass m. The spring hangs with P vertically below A. The particle P is released from rest in the position where the extension of the spring is 12 a . (a) Show that the initial acceleration of P is 32 g upwards. 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(b) Find the speed of P when the spring first returns to its natural length. 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Mark scheme: 3(a) − = 1 5 . / 2 = T mg a a = 5 2 mg M1 3 2 = a g (upwards) AG A1 3 Question Answer Marks 3(b) Gain in KE = 2 1 2 mv Gain in GPE = 1 2 mga B1 Loss in EPE = 2 1 5 . 1 2 2       mg a a B1 2 2 1 5 . 1 1 1 2 2 2 2       + = mg a mv mga a [  2 1 1 5 2 2 8 + = mv mga mga ] M1 1 2 = v ga A1 4

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4 B A E 7.5 cm C x cm F D A uniform square lamina ABCD has sides of length 10 cm. The point E is on BC with EC = 7.5cm , and the point F is on DC with CF = xcm . The triangle EFC is removed from ABCD (see diagram). The centre of mass of the resulting shape ABEFD is a distance x cm from CB and a distance y cm from CD. 400 - x 2 (a) Show that x = and find a corresponding expression for y. 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The shape ABEFD is in equilibrium in a vertical plane with the edge DF resting on a smooth horizontal surface. (b) Find the greatest possible value of x, giving your answer in the form a + b 2 , where a and b are constants to be determined. 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Mark scheme: 4(a) Area Centre of mass from BC Centre of mass from DC Square 100 5 5 Triangle ½ x. 15/2 1 3 x 5 2 Shape ABEFD 15 100 4 − x x y Take moments about BC: 15 100 4 σ   −     x . 15 1 500 . 4 3 σ σ = − x x x (M1 for all terms present) M1 2 400 80 3 − = − x x x AG A1 Take moments about DC: 15 100 4   −     x . y = 15 5 100 5 . 4 2 × − x M1 y = 800 15 160 6 − − x x A1 4 4(b) Use condition: x x  B1 2 2 80 400 0 − + x x  M1 20 10 2 = − x A1 3

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5 A particle P is moving along a straight line with acceleration 3ku - kv where v is its velocity at time t, u is its initial velocity and k is a constant. The velocity and acceleration of P are both in the direction of increasing displacement from the initial position. (a) Find the time taken for P to achieve a velocity of 2u. 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(b) Find an expression for the displacement of P from its initial position when its velocity is 2u. 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Mark scheme: 5(a) d d 3 = − v k t u v M1 ( ) ln 3 − − = + u v kt d 0, : ln 2 = = = − t v u d u M1 1 2 : ln 2 = = v u t k A1 3 5(b) d 3 d = − v v ku kv x [  d d 3 = − v v k x u v ] B1 ( ) ( ) 3 3 d d 3 − − + = − u v u v k x u v so ( ) 3 ln 3 −− − = + v u u v kx c M1A1 0, : 3 ln 2 = = = −− x v u c u u u M1 ( ) 2 : 3ln 2 1 = = − u v u x k A1 5

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6 A particle P of mass m is moving with speed u on a fixed smooth horizontal surface. The particle strikes a fixed vertical barrier. At the instant of impact the direction of motion of P makes an angle a with the barrier. The coefficient of restitution between P and the barrier is e. As a result of the impact, the direction of motion of P is turned through 90°. 1 (a) Show that tan 2a = . 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The particle P loses two-thirds of its kinetic energy in the impact. (b) Find the value of a and the value of e. 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Mark scheme: 6(a) Let components of velocity (parallel to plane and perpendicular) after impact be (x, y) cos sin α α = = y v eu B1 sin cos α α = = x v u B1 Divide: 1 tan tan α α = e : tan2 α = 1.e B1 3 Question Answer Marks 6(b) 2 2 1 3 = v u B1 2 2 cos 1 sin 3 α α  =     u u M1 ( ) 2 tan 3 α = M1 60 α = ° A1 1 3 = e A1 Alternative method for 6(b) KE after impact = ( ) 2 2 1 2 + m x y ( ) ( ) ( ) 2 2 2 1 cos sin 2 α α = + m u e u M1 From (a) ( ) sin 1/ 1 α = + e and ( ) cos / 1 α = + e e B1 KE = 2 2 1 2 1 1   +   + +   e e mu e e = 2 1 2 mu e A1 This is equal to 2 1 1 3 2 × mu so 1 3 = e M1 tan 3, 60 α α = = ° A1 5

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7 A hollow cylinder of radius a is fixed with its axis horizontal. A particle P, of mass m, moves in part of a vertical circle of radius a and centre O on the smooth inner surface of the cylinder. The speed of P when it is at the lowest point A of its motion is 72 ga . The particle P loses contact with the surface of the cylinder when OP makes an angle i with the upward vertical through O. (a) Show that i = 60° . 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(b) Show that in its subsequent motion P strikes the cylinder at the point A. 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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) ( ) 2 cosθ + = mv N mg a B1 ( ) 2 1 1 7 cos 2 2 2 θ − = − + ag mv m mg a a M1A1 Loses contact when N = 0, so combine and simplify M1 1 cos : 60 2 θ θ = = ° AG A1 5 Question Answer Marks 7(b) When P is vertically below O, its horizontal displacement is sin 60 a , so time T = sin60 cos60 a v = 3 / a v 6 = a g M1 From (a), 2 1 2 = v ag A1 Vert: 2 3 1 . 2 2 = − v h T g T M1 3 3 3 2 2 − = −a a a A1 This corresponds to the point A A1 Alternative method for question 7(b) 2 4 3 = − x y x a M1A1 Coordinates of A: 1 3 3, 2 2 = = − x a y a B1 Substitute coordinates into 2 4 3 = − x y x a and show that these satisfy this equation M1A1 5

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