Cambridge A Level Mathematics - Further 9231 — 2023 Oct/Nov Paper 3 · Variant 1

1 u u A B 60° i m 2m Two uniform smooth spheres A and B of equal radii have masses m and 2m respectively. The two spheres are moving with equal speeds u on a smooth horizontal surface when they collide. Immediately before the collision, A’s direction of motion makes an angle of 60° with the line of centres, and B’s direction of motion makes an angle i with the line of centres (see diagram). The coefficient of restitution between the spheres is e. After the collision, the component of the velocity of A along the line of centres is v and B moves perpendicular to the line of centres. Sphere A now has twice as much kinetic energy as sphere B. i - 1) . [1] (a) Show that v = 12 u ( 4 cos ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the value of cosi. 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(c) Find the value of e. 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Mark scheme: Question Answer Marks Guidance 1(a) PCLM: mv = −mu cos60+ 2mu cos B1 First line must be seen. 1 1 v = − u + 2u cos= v = u ( 4cos− 1) AG 2 2 1 1(b) 1 2 2 B1 KE of A = m v + ( u sin60 ) ( ) 2 KE of A = 2  KE of B, so M1 1 2 2 1 2 m v + ( u sin60 ) = 2   2m ( u sin) ( ) 2 2 2 M1 Use result of (a) and rearrange.  1  3 2 2 2 u = 4u ( sin) Obtain 3-term quadratic.  u ( 4cos− 1)  +  2  4 8cos 2 − 2cos− 3 = 0 3 1 3 A1 cos= , − but angle is acute, so cos= 4 2 4 4 1(c) NEL: v = e ( u cos60 + u cos) M1  1 3  A1 v = u , u = e u + u    2 4  4 e = 5 2

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2 A ball of mass 2 kg is projected vertically downwards with speed 5 ms -1 through a liquid. At time t s after projection, the velocity of the ball is v ms -1 and its displacement from its starting point is x m. The forces acting on the ball are its weight and a resistive force of magnitude 0.2v 2 N . (a) Find an expression for v in terms of t. 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(b) Deduce what happens to v for large values of t. 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Mark scheme: 2(a) dv 2 B1 2 = 2 g − 0.2v dt dv M1 Integrate to a ln term of the correct form. Separate variables and attempt to integrate = dt 0.1 100 − v 2 ( ) 1  10 + v A1 ln = 0.1t + c   20  10 − v  1 M1 Use initial condition. t = 0, v = 5, c = ln3 20 10 + v 2 t 10 + v M1 Rearrange, removing ln. 2t = ln , e = 3 (10 − v ) 3 (10 − v ) 30 − 10e −2 t A1 AEF v = 3 + e −2 t 6 2(b) v →10 B1FT FT from expression of correct form. 1

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3 P A i 2a E B D C A uniform square lamina of side 2a and weight W is suspended from a light inextensible string attached to the midpoint E of the side AB. The other end of the string is attached to a fixed point P on a rough vertical wall. The vertex B of the lamina is in contact with the wall. The string EP is perpendicular to the side AB and makes an angle i with the wall (see diagram). The string and the lamina are in a vertical plane perpendicular to the wall. The coefficient of friction between the wall and the lamina is 1.2 Given that the vertex B is about to slip up the wall, find the value of tani. 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Mark scheme: 3 Let N be normal reaction at B and F the frictional force acting downwards B1  T cos= F + W → T sin= N B1 Moments about B: Ta = W sin+a W cos a M1A1 A moments equation with all relevant forces. 1 M1 F = N used 2  1  M1 Combine to obtain equation in . cos− sin = 1 oe ( cos+ sin)   Equation in trigonometric functions only.  2  1 3 2 M1 Solve trigonometric equation. cossin= ( sin) 2 2 sin( cos− 3sin) = 0 1 A1 tan= 3 8

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4 A L 12a P L B A light elastic string has natural length 8a and modulus of elasticity 5 mg. A particle P of mass m is attached to the midpoint of the string. The ends of the string are attached to points A and B which are a distance 12a apart on a smooth horizontal table. The particle P is held on the table so that AP = BP = L (see diagram). The particle P is released from rest. When P is at the midpoint of AB it has speed 80ag . (a) Find L in terms of a. 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(b) Find the initial acceleration of P in terms of g. 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Mark scheme: 4(a) Energy equation: gain in KE = loss in EPE B1 At least one correct EPE. M1 3-term energy equation. 1 1 5mg 2 2 A1 All correct. m  80ag =  ( 2 L − 8a ) − ( 4a ) ( ) 2 2 8a 4 L2 − 8aL − 20 a 2 = 0 M1 Simplify to quadratic in L. L = 10a A1 Correct single answer. 5 4(b) At P: 2T cos= m  acceleration M1 5mg 15 M1 Hooke’s law: T =  12 a = mg 8a 2 Solve, acceleration = 12g A1 3

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5 A particle P is projected with speed u ms -1 at an angle i above the horizontal from a point O on a horizontal plane and moves freely under gravity. During its flight P passes through the point which is a horizontal distance 3a from O and a vertical distance 38 a above the horizontal plane. It is given that tan i = 13 . (a) Show that u 2 = 8ag . 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A particle Q is projected with speed V ms -1 at an angle a above the horizontal from O at the instant when P is at its highest point. Particles P and Q both land at the same point on the horizontal plane at the same time. (b) Find V in terms of a and g. 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Mark scheme: 5(a) 2 M1 Use equation of trajectory: y = x tan −gx sec 2: 2u 2 3 1 ( 3a ) 2  1  a = 3a  − g  1 + 2   8 3 2u  9  5 5 ga 2 A1 At least one step of working. a = , u = 8 ga 2 AG 8 u 2 5(b) 2u sin B1 For P, time of flight T = g For P, range = Tu cos B1 1 1 B1 For Q, time of flight = 2T , so range = 2 TV cos Equate: V cos= 2u cos (1) 1 2 M1 gT 1 2 4V sin For Q vertically: 0 = V sin  T − , so T = 2 4 g Equate with result for P: A1 4V sin 2u sin 1 = so V sin= u sin (2) g g 2 2 2  2 1 2  M1 From (1) and (2): V = u  4 ( cos) + ( sin)   4  2 2  9 1 1  145 2 A1 V = u  4  +  = u    10 4 10  40 V = 29ag 7

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6 A particle P of mass m is attached to one end of a light inextensible rod of length 3a. An identical particle Q is attached to the other end of the rod. The rod is smoothly pivoted at a point O on the rod, where OQ = x . The system, of rod and particles, rotates about O in a vertical plane. At an instant when the rod is vertical, with P above Q, the particle P is moving horizontally with speed u. When the rod has turned through an angle of 60° from the vertical, the speed of P is 2 ag , and the tensions in the two parts of the rod, OP and OQ, have equal magnitudes. (a) Show that the speed of Q when the rod has turned through an angle of 60° from the vertical is 2x ag . 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(b) Find x in terms of a. 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(c) Find u in terms of a and g. 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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: 6(a) Angular speeds of P and Q are equal, M1 vQ vP so = x 3a − x A1 Shown convincingly: angular speeds equal stated. 2 x ag vQ = AG 3a − x 2 6(b) m  4ag B1 For P: T + mg cos60= 3a − x mvQ2 B1 For Q: T − mg cos60= x m.4 ag mvQ2 M1 Eliminate T: − mg cos60+ = mg cos60+ 3a − x x m  4ag mx 4ag M1 Solve to find x . = 1 + 2 Obtain 3-term quadratic equation. 3a − x ( 3a − x ) 4a ( 3a − x ) = ( 3a − x ) 2 + 4ax , x 2 + 2 ax − 3a 2 = 0 ( x − a )( x + 3a ) = 0, x = a A1 5 6(c) Energy changes from initial position: B1 KEs correct. 1 2 Gain in KE of P: m 4 ag − u ( ) B1FT GPEs correct. 2 1   u 2 2  Loss in KE of Q: m   − vQ  2   2     Loss in GPE of P = mg ( 3a − x )(1 − cos60 ) ( = mga )  1  Gain in GPE of Q = mgx (1 − cos60 )  = mga   2   1 2 1  u  2 2  M1 Energy equation. m 4 ag − u − m  = − mgx (1 − cos60+) mg ( 3a − x )(1 − cos60 ) − vQ    ( )   2 2  2    5 2 A1 AEF Simplify: 4 ag − u + ag = ag 4 2 16 4 u = ag , u = 5ag 5 5 4

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