Cambridge A Level Mathematics - Further 9231 — 2022 Oct/Nov Paper 3 · Variant 1

1 A particle of mass 2 kg is attached to one end of a light inextensible string of length 0.6 m. The other end of the string is attached to a fixed point on a smooth horizontal surface. The particle is moving in a circular path on the surface. The tension in the string is 20 N. Find how many revolutions the particle makes per minute. [3] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 2  v 2 2 M1 Use F = ma: 20 = OR 20 = 2  0.6 0.6 2 2 50 A1 v = 6 OR = 3 60v 60 A1 FT 38.9848…. Number of revolutions per min = OR 0.6  2π 2π so 39(.0) revolutions 3

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2 A light elastic string has natural length a and modulus of elasticity 4 mg. One end of the string is fixed to a point O on a smooth horizontal surface. A particle P of mass m is attached to the other end of the string. The particle P is projected along the surface in the direction OP. When the length of the string is 5 4 a , the speed of P is v. When the length of the string is 32 a , the speed of P is 12 v . (a) Find an expression for v in terms of a and g. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find, in terms of g, the acceleration of P when the stretched length of the string is 32 a . [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 2(a) Loss in KE = Gain in EPE, so B1 EPE terms correct. 1 2 1  v  2 1 4 mg   1  2  1  2  mv − m   =    a  −  a   M1 All 4 terms and no extras.   2 2  2  2 a  2   4    3 2 4mg 3 2 M1 Simplify. mv =  a 4 a 16 v 2 = ag , v = ag A1 4 2(b) 4 mg 1 M1 Hooke’s law: tension =  a (= 2 mg ) a 2 2mg A1 Accept -2g. Acceleration = = 2 g m 2

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3 B C A i A smooth cylinder is fixed to a rough horizontal surface with its axis of symmetry horizontal. A uniform rod AB, of length 4a and weight W, rests against the surface of the cylinder. The end A of the rod is in contact with the horizontal surface. The vertical plane containing the rod AB is perpendicular to the axis of the cylinder. The point of contact between the rod and the cylinder is C, where AC = 3a . The angle between the rod and the horizontal surface is i where tan i = 34 (see diagram). The coefficient of friction between the rod and the horizontal surface is 6.7 A particle of weight kW is attached to the rod at B. The rod is about to slip. The normal reaction between the rod and the cylinder is N. (a) Show that N = 8 W ( 1 + 2 k) . 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(b) Find the value of k. 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Mark scheme: 3(a) Let F and R be friction and normal reaction at A M1 Correct terms, allow sign errors and cos/sin mix. Take moments about A, for rod N  3a = W  2a cos+ kW  4a cos 4 A1 At least one intermediate line of working. 3 N = ( 2 + 4k ) W  5 8 N = W (1 + 2 k ) AG 15 2 3(b)  N cos+ R = W + kW B1 Resolve (to include R) for rod. 6 B1 Both. → F = N sin and F = R 7 28 21 M1 Find R or N . so R = W (1 + 2k ) or R = W (1 + k ) 75 45 Eliminate to find k M1 Complete method. 1 A1 k = 3 5

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4 A particle of mass 0.5 kg moves along a horizontal straight line. Its velocity is v m s -1 at time t s. The forces acting on the particle are a driving force of magnitude 50 N and a resistance of magnitude 2v 2 N . The initial velocity of the particle is 3 m s -1 . (a) Find an expression for v in terms of t. 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(b) Deduce the limiting value of v. 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Mark scheme: 4(a) dv 2 dv 2 B1 N2L m = 50 − 2v = 4 25 − v ( ) dt dt 1 1 1 M1 Separate variables and use partial fractions.  + dv =  4 dt 10 5 − v 5 + v 1 ( − ln ( 5 − v ) + ln ( 5 + v ) ) = 4t + A M1 A1 Integrate(Note: formulainto logonterms.MF19). 10 1 M1 Use initial condition. Use t = 0, v = 3 to give A = ln 4 10 1 5 + v 5 + v 40 t M1 Rearrange to make v the subject. 4t = ln leading to = e 10 4 ( 5 − v ) 20 − 4v −40 t A1 5 4 − e ( ) v = 4 + e −40 t 7 4(b) As t →, v → 5 B1 1

More questions on Linear motion under a variable force

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 string is held taut with OP horizontal. The particle P is projected vertically downwards with speed 13 ag and starts to move in a vertical circle. P passes through the lowest point of the circle and reaches the point Q where OQ makes an angle i with the downward vertical. At Q the speed of P is kag and the tension in the string is 116 mg . (a) Find the value of k and the value of cosi. 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At Q the particle P becomes detached from the string. (b) In the subsequent motion, find the greatest height reached by P above the level of the lowest point of the circle. 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Mark scheme: 5(a) 1 2 1 2 B1 Energy equation. mv − mu = mga cos 2 2  1  kag = ag + 2 ag cos    3  m 2 B1 N2L at B. T − mg cos= v a 11 m 11 So mg − mg cos= .kag , − cos= k 6 a 6 Solve simultaneously. M1 4 1 A1 Both. k = , cos= 3 2 4 5(b) Initial speed = kag sin B1 2 M1 2 2 Use v = u + 2 as : 0 = kag sin − 2 gs ( ) 1 A1 s = a 2 1 1 A1 FT Height above lowest point = s + a − a cos = a + a − a = a 2 2 4

More questions on Circular motion

6 A B u a 5 8 u m km Two uniform smooth spheres A and B of equal radii have masses m and km respectively. The two spheres are moving on a horizontal surface with speeds u and 58 u respectively. Immediately before the spheres collide, A is travelling along the line of centres, and B’s direction of motion makes an angle a with the line of centres (see diagram). The coefficient of restitution between the spheres is 23 and tan a = 34 . After the collision, the direction of motion of B is perpendicular to the line of centres. (a) Find the value of k. 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(b) Find the loss in the total kinetic energy as a result of the collision. 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Mark scheme: 6(a) Let speed of A after collision be → v A and speed of B M1 perpendicular to line of centres be v 5 Along line of centres: mu − km u cos= mv A Momentum. 8  5  M1 NEL NEL: 0 − v A = e  u cos+ u   8  5 2  5  M1 Solve. So u − ku cos= −  u cos+ u  8 3  8  Substitute for cos, to give k = 4 A1 4 6(b) 5 3 B1 vB = u sin = Velocity perpendicular to line of centres 8 8u v A = −u B1 FT 2 M1 NOTE: KE before and after for A is unchanged. 1 2 1  5  1 2 25 2 41 2 KE before = mu + km u = mu + mu = mu   2 2  8  2 32 32 Both. 1 2 1 2 1 2 9 2 25 2 KE after = mv A + kmvB = mu + 2m u = mu 2 2 2 64 32 2  41 25  1 2 A1 Loss = mu  −  = mu  32 32  2 4

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7 A particle P is projected with speed V ms -1 at an angle 75° above the horizontal from a point O on a horizontal plane. It then moves freely under gravity. 2V (a) Show that the total time of flight, in seconds, is sin 75°. 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A smooth vertical barrier is now inserted with its lower end on the plane at a distance 15 m from O. The particle is projected as before but now strikes the barrier, rebounds and returns to O. The coefficient of restitution between the barrier and the particle is 3.5 (b) Explain why the total time of flight is unchanged. 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(c) Find an expression for V in terms of 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: 7(a) 1 2 M1  0 = V sin75t − gt 2 2V A1 AG t = sin75 g 2 7(b) Vertical component of velocity is unchanged. B1 1 7(c)  15  B1 Horizontally to wall, → 15 = V cos75t ;  t =   Vcos 75  eVcos75 B1 Speed after rebound. 15 M1 Time back to O T = eVcos 75  3   t = T   5  Vertically for whole flight: M1 2V t + T = sin75  g 15 15 2V A1 + = sin75 Vcos 75 eVcos 75 g V 2 cos75sin75 = 20 g M1 Multiply by 2: V 2 sin150 = 40 g , V 2 = 80 g A1 V = 4 5 g = 8.94 g ( ) 7

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