Cambridge A Level Mathematics - Further 9231 — 2022 Oct/Nov Paper 3 · Variant 2

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. The string is held taut with OP making an angle a with the downward vertical, where cos a = 23 . The particle P is projected perpendicular to OP in an upwards direction with speed 3ag . It then starts to move along a circular path in a vertical plane. Find the cosine of the angle between the string and the upward vertical when the string first becomes slack. [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 m 2 2 B1 N2L When string goes slack, mg cos = v , v = ag cos  May include T, but B1 not awarded until T = 0. a 1 1 2 B1 Energy equation. m.3ag − mv = mg ( a cos+ a cos ) 2 2 2  2  M1 Combine. So u − ag cos  = 2 ag  cos +   3  2 4 A1 u − ag 3 5 cos  = = 3ag 9 4

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2 A 9a B D C x 6a A uniform lamina is in the form of a triangle ABC in which angle B is a right angle, AB = 9a and BC = 6a . The point D is on BC such that BD = x (see diagram). The region ABD is removed from the lamina. The resulting shape ADC is placed with the edge DC on a horizontal surface and the plane ADC is vertical. Find the set of values of x, in terms of a, for which the shape is in equilibrium. [6] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 Area Distance from AB ABC 27a 2 2a 9 1 ABD ax x 2 3 2 9 Shape ADC 27 a − ax x 2 Taking moments about AB M1 Moments equation with 3 terms.  2 9  2 9 1 x   27 a − ax  = 27a  2 a − ax  x A1 At least 2 terms correct.  2  2 3  3 3 2  A1 All correct. 54 a − ax   2  x =  2 9  27 a − ax   2  For equilibrium, x x , B1 Use correct condition: allow strict inequality. Can be implied by correct final answer x 3a . 3 3 2  2 9  54 a − ax x  27 a − ax  2  2  54 a 2 − 27 ax + 3 x 2 0 M1 Simplify and attempt to solve a quadratic inequality or equation. ( x − 3a )( x − 6 a ) 0 ( 0 ) x 3a [only] A1 CAO 2 Alternative method for question 2 Taking moments with B as origin. M1 1 1 A2 x = ( 0 + x + 6 a ) = 2 a + x 3 3 1 B1 Allow strict inequality. For equilibrium, x x , so x 2 a + x 3 ( 0 ) x 3a M1 A1 6

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3 One end of a light elastic string, of natural length a and modulus of elasticity 163 Mg , is attached to a fixed point O. A particle P of mass 4M is attached to the other end of the string and hangs vertically in equilibrium. Another particle of mass 2M is attached to P and the combined particle is then released from rest. The speed of the combined particle when it has descended a distance 14 a is v. Find an expression for v in terms of g and a. 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Mark scheme: 3 16 B1 Mge 3 3 In equilibrium, = 4 Mg , e = a a 4 In subsequent motion, M1 Energy equation with GPE and KE terms correct and at least one Loss in GPE = gain in EPE + gain in KE EPE term. Dimensionally correct. 6 Mga 1 16 Mg  2  3a  2  1 2 B1 EPE correct. = . . .  a −    + .6 Mv  4 2 3 a  2  4  A1 All correct.   3Mga 8 Mg 7 2 2 M1 Attempt to find v in terms of a and g. = . a + 3Mv etc 2 3a 16 ga 2 1 A1 = 3v , v = ga 3 3 6

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4 A particle P of mass 5 kg moves along a horizontal straight line. At time t s, the velocity of P is v m s -1 and its displacement from a fixed point O on the line is x m. The forces acting on P are a force of 500 1 2 magnitude N in the direction OP and a resistive force of magnitude 2 v N . When t = 0 , x = 0 and v v = 5 . (a) Find an expression for v in terms of x. 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(b) State the value that the speed approaches for large values of x. 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Mark scheme: 4(a) dv 500 1 2 B1 Sight of m or 5 is required. 5v = − v dx v 2 10 v 2 dv M1 Separate variables and attempt to integrate into a log term. = dx 1000 − v 3 3 A1 1000 − v = x ( + A ) − 10ln ( ) 3 10 M1 Evaluate constant: correct initial condition used. x = 0, v = 5, A = − ln875 3 10 875 M1 Make v the subject: correct use of logs. x = ln 3 1000 − v 3 1 A1 1 3 : A0 if eln terms. v =  (1000 − 875e −0.3 x  3 v = 5  (8 − 7 e −0.3 x      6 4(b) Maximum value of v is 10 B1 No FT: result can be found from initial equation. 1

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5 A particle P is projected with speed u m s -1 at an angle of i above the horizontal from a point O on a horizontal plane and moves freely under gravity. The horizontal and vertical displacements of P from O at a subsequent time t s are denoted by x m and y m respectively. (a) Show that the equation of the trajectory is given by gx 2 2 i) . [4] y = x tan i - 2 ( 1 + tan 2u ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ In the subsequent motion P passes through the point with coordinates (30, 20). (b) Given that one possible value of tani is 4,3 find the other possible value of tani. 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Mark scheme: 5(a) → x = u cost B1 Result quoted from MF19 scores 0/4. 1 2 B1  y = u sint − gt 2 2 M1 x 1  x  Eliminate t: y = u sin − g   cos 2  ucos  gx 2 A1 Must be an intermediate line of working. y = x tan− (1 + tan2 ) 2 AG 2u 4 5(b) 4 30 2   4  2  M1 Substituting values correctly. 20 = 30  − 10    1 +    2 3 2u   3     u 2 = 625, u = 25 A1 Substitute back into trajectory equation, M1 Obtain a 3-term quadratic. g 30 2 2 36 2 20 = 30tan− sec = 30tan− (1 + tan  ) 2.25 2 5 18tan 2− 75tan+ 68 = 0 One solution is 4, ( 3tan− 4 )( 6tan− 17 ) = 0 M1 3 17 A1 Giving tan= 6 5

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6 R B h A A light inextensible string is threaded through a fixed smooth ring R which is at a height h above a smooth horizontal surface. One end of the string is attached to a particle A of mass m. The other end of the string is attached to a particle B of mass 67 m . The particle A moves in a horizontal circle on the surface. The particle B hangs in equilibrium below the ring and above the surface (see diagram). When A has constant angular speed ~, the angle between AR and BR is i and the normal reaction between A and the surface is N. a and the normal reaction When A has constant angular speed 32 ~, the angle between AR and BR is between A and the surface is 12 N . a . [5] (a) Show that cos i = 49 cos ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find N in terms of m and g and find the value of cosa. 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Mark scheme: 6(a) 6 B1 May be implied. T = mg 7 T sin= mr2 = mh tan 2 B1 Allow r for radius. Radius of circle = h tan B1 2 6 g [So  = cos] 7 h 9 2 6 g M1 Second scenario, equivalent result . In second scenario,  = cos 4 7 h 6 g 4 6 g A1 Combine convincingly to obtain given result. Equate, cos=  cos giving 7 h 9 7 h 4 cos= cos AG 9 5 6(b) First scenario: N + T cos= mg 1 B1 Both. Second scenario, N + T cos= mg 2 6 12 M1 12cos− 6cos= 7 Equate: mg − mg cos= 2mg − mg cos 7 7 3 A1 cos= 4 5 A1 N = mg 7 4

More questions on Circular motion

7 u 2u A B α β 1 m m 2 Two uniform smooth spheres A and B of equal radii have masses m and 12 m respectively. The two spheres are moving on a horizontal surface when they collide. Immediately before the collision, sphere A is travelling with speed u and its direction of motion makes an angle a with the line of centres. Sphere B is travelling with speed 2u and its direction of motion makes an angle b with the line of a + b = 90° . centres (see diagram). The coefficient of restitution between the spheres is 58 and (a) Find the component of the velocity of B parallel to the line of centres after the collision, giving your answer in terms of u and a. 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The direction of motion of B after the collision is parallel to the direction of motion of A before the collision. (b) Find the value of tana. 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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: 7(a) Let v , w be speeds of A and B along line of centres after collision M1 1 1 mv + mw = mu cos− m.2u cos  Momentum: masses correct, opposite signs on RHS. 2 2 w −=v e(2u cos + u cos) M1 NEL: LHS signs must be consistent with momentum equation, same sign for both terms on RHS. + = 90, so cos = sin M1 Solve to find an expression of the correct form. Use this fact and solve to find w 2  1 13  A1 w = u  sin+ cos 3  4 8  4 7(b) Perpendicular to line of centres, speed of B is B1 2u sin = 2u cos After, velocity of B makes angle  with line of centres, so B1 2u cos tan= w sin 2u cos M1* Obtain homogeneous equation in cos and sin or an equation in tan = giving cos 2  1 13  u  sin+ cos 3  4 8  2 1 2 13 DM1 Obtain quadratic and solve to find values of tan 3 ( cos) = ( sin) + sincos 4 8 2 ( tan) 2 + 13tan− 24 = 0 , ( 2tan − 3 )( tan + 8 ) = 0 3 A1 tan= 2 5

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