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

1 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 on a smooth horizontal plane. The particle P moves in horizontal circles about O. The tension in the string is 4mg. Find, in terms of a and g, the time that P takes to make one complete revolution. [2] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 1 2 4 ω = = T mg ma so 2 4 ω = g a B1 Time per revn = 2π ω = π a g B1 2

More questions on Circular motion

2 A particle Q of mass m kg falls from rest under gravity. The motion of Q is resisted by a force of magnitude mkv N, where v ms -1 is the speed of Q at time t s and k is a positive constant. Find an expression for v in terms of g, k and t. 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Mark scheme: 2 d d = − v t g kv M1 ( ) 1 ln − − = + g kv t A k M1 1 0, 0: ln = = = − t v A g k A1 1 ln − = − g kv t k g M1 ( ) 1 − = − kt g v e k M1A1 6

More questions on Linear motion under a variable force

3 A particle Q of mass m is attached to a fixed point O by a light inextensible string of length a. The particle moves in complete vertical circles about O. The points A and B are on the path of Q with AB a diameter of the circle. OA makes an angle of 60° with the downward vertical through O and OB makes an angle of 60° with the upward vertical through O. The speed of Q when it is at A is 2 ag . Given that TA and TB are the tensions in the string at A and B respectively, find the ratio TA : TB . 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Mark scheme: 3 4 cos60 . − = A ag T mg m a ( 9 2 = A T mg ) B1 2 cos60 . + = B v T mg m a B1 Energy: ( ) 2 1 4 .2 cos60 2 − = m ag v mg a 2 2 = v ag M1A1 3 2 = B mg T A1 Ratio is 3 : 1 A1 6

More questions on Circular motion

4 4r 2r kr 3r A uniform solid circular cone, of vertical height 4r and radius 2r, is attached to a uniform solid cylinder, of height 3r and radius kr, where k is a constant less than 2. 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). 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 vertex of the cone is ( 99 k 2 + 96) r 2 . 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The point C is on the circumference of the base of the cone. When the combined solid is freely suspended from C and hanging in equilibrium, the diameter through C makes an angle a with the downward vertical, where tan a = 18 . (b) Given that the centre of mass of the combined solid is within the cylinder, find the value of k. 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Mark scheme: 4(a) Mass Centre of mass from vertex of cone Cylinder ( ) 2 π .3 kr rρ 11 2 r Cone ( ) 2 1 π 2 .4 3 r rρ 3r Combined ( ) ( ) 2 2 1 π .3 π 2 .4 3 kr r r r ρ   +     x B1 Take moments about vertex: ( ) ( ) 2 2 1 11 π .3 π 2 .4 3 2 r x kr r r r   × + = ×     ( ) 2 π .3 3 kr r r + × ( ) 2 1 π 2 .4 3 r r leading to M1A1 ( ) 2 2 99 96 18 32 + = + k r x k A1 4 Question Answer Marks 4(b) 4 tan 2 α − = x r r ( = 1/8) M1 17 4 = x r A1 Equate to answer in (a) and attempt to find k M1 2 4 45 80: 3 = = k k A1 4

More questions on Equilibrium of a rigid body

5 A B a° (90 – a)° u m m 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 of a° with the line of centres, and B’s direction of motion is perpendicular to that of A (see diagram). The coefficient of restitution between the spheres is e. Immediately after the collision, B moves in a direction at right angles to the line of centres. 1 + e (a) Show that tana = . 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(b) Given that tan a = 2 , find the speed of A after the collision. 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Mark scheme: 5(a) Let w be speed of A along line of centres after collision cos sin α α ← = − + mw mu mu M1 0 ( cos sin α α − = + w e u u ) M1 Rearrange: ( ) ( ) sin cos α α − = + u eu u eu M1 1 tan 1 α + = − e e . AG A1 4 5(b) 1 tan 2 3 α =  = e B1 1 1 2 3 5 5 5   = + =     u w u M1 Speed = ( ) 2 2 sinα + w u M1 = 2 2 4 5 5 + = u u u A1 4

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6 A particle P is projected with speed u at an angle i above the horizontal from a point O on a horizontal plane and moves freely under gravity. The direction of motion of P makes an angle a above the horizontal when P first reaches three-quarters of its greatest height. i . [6] (a) Show that tan a = 12 tan ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Given that tan i = 43 , find the horizontal distance travelled by P when it first reaches three-quarters of its greatest height. Give your answer in terms of u and g. 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Mark scheme: 6(a) Greatest height = ( ) 2 sin 2 θ u g M1A1 At ¾ greatest height, cos cos α θ → = v u M1 At ¾ greatest height, ( ) ( ) ( ) 2 2 2 sin 3 sin sin 2 .4 2 θ α θ ↑ = − u v u g g M1 1 sin sin 2 α θ = v u A1 So 1 tan tan 2 α θ = AG A1 6 Question Answer Marks 6(b) 1 4 2 tan . 2 3 3 α = = 1 sin sin 2 θ θ ↑ = − u u gt M1 2 5 = u t g A1 / cosθ → = t d u M1 = 2 6 25 u g A1 4

More questions on Motion of a projectile

7 4 3 ga a A B O One end of a light spring of natural length a and modulus of elasticity 4mg is attached to a fixed point O. The other end of the spring is attached to a particle A of mass km, where k is a constant. Initially the spring lies at rest on a smooth horizontal surface and has length a. A second particle B, of mass m, is moving towards A with speed 43 ga along the line of the spring from the opposite direction to O (see diagram). The particles A and B collide and coalesce. At a point C in the subsequent motion, the length of the spring is 34 a and the speed of the combined particle is half of its initial speed. (a) Find the value of k. 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At the point C the horizontal surface becomes rough, with coefficient of friction n between the combined particle and the surface. The deceleration of the combined particle at C is 209 g . (b) Find the value of n. 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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) Collision: ( )1 + = k mv mu so 1 = + u v k B1 Loss in KE = ( ) 2 2 1 1 1 2 4   + −     m k v v B1 Gain in EPE = ½ .4mg/a. 2 4       a B1 So, ( ) ( ) 2 2 1 3 . . 1 2 4 8 1 + = + u mga m k k M1 Use value of u and solve M1 3 = k A1 6 Question Answer Marks 7(b) 9 4 . 20 + = g T F m M1 4 9 4 . 5 a mg mg F a + = M1 4 / 5 μ = R mg and compare with 4 = R mg B1 1 5 μ = A1 4

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