Cambridge A Level Mathematics - Further 9231 — 2024 Oct/Nov Paper 3 · Variant 1

1 A particle P is projected with speed u m s -1 at an angle tan -1 2 above the horizontal from a point O on a horizontal plane and moves freely under gravity. When P has travelled a distance 56 m horizontally from O, it is at a vertical height H m above the plane. When P has travelled a distance 84 m horizontally from O, it is at a vertical height 1H m above the plane. 2 Find, in either order, the value of u and the value of H. [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1  1  M1 For one equation with one error. Use equation of trajectory with point (56, H) or  84, H   2  A1 Both correct. 5 g 2 1 5 g 2 H = 112 −  56 or H = 168 −  84 2u 2 2 2u 2 Eliminate to find u or H M1 u = 35 A1 H = 48 A1 5

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2 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 particle P is held at the point A with the string taut. It is given that OA makes an angle i with the downward vertical through O, where tan i = 3 . The particle P 4 is projected perpendicular to OA in an upwards direction with speed 5ag , and it starts to move along a circular path in a vertical plane. When P is at the point B, where angle AOB is a right angle, the tension in the string is T. Find T in terms of m and g. [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 mv 2 B1 At B, T + mg sin= a 1 2 1 2 M1A1 Energy A to B: mu − mv = mga ( cos+ sin) 2 2 Substitute for u and  to find T: M1  4 3  T = mg  5 −2 −3   5 5  8 A1 T = mg 5 5

More questions on Circular motion

3 A particle P of mass m kg is attached to one end of a light elastic string of natural length 2 m and modulus of elasticity 2mg N. The other end of the string is attached to a fixed point O. The particle P hangs in equilibrium vertically below O. The particle P is pulled down vertically a distance d m below its equilibrium position and released from rest. (a) Given that the particle just reaches O in the subsequent motion, find the value of d. 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(b) Hence find the speed of P when it is 2 m below O. 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Mark scheme: 3(a) 2 mg M1 Equilibrium position. Hooke’s law: T =  extension and T = mg 2 Extension = 1 m A1 1 2mg 2 B1 EPE loss =   (1 + d ) 2 2 Gain in GPE = mg ( 2 + 1 + d ) B1 1 2 M1 Equate: mg (1 + d ) = mg ( 3 + d ) 2 d = 5 A1 SC: 3 marks for final answer of 5 + 1 . SC: 2 marks for final answer of 5 + k , k  1 . 6 3(b) 1 2 1 2 mg 2 M1 GPE, KE, EPE terms. Energy equation: mV + mg (1 + d ) =   (1 + d ) 2 2 2 2 2 A1 V = g d − 1 V = 40 = 2 10 ( ) Alternatively: M1 Using KE and GPE from 2 m below O to point O 1 2 mV = 2mg 2 V 2 = 4 g V = 40 = 2 10 A1 2

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4 A 2 a B 3 kh h a An object is formed by removing a cylinder of radius 2 a and height kh ( k 1 1 ) from a uniform solid 3 cylinder of radius a and height h. The vertical axes of symmetry of the two cylinders coincide. The upper faces of the two cylinders are in the same plane as each other. The points A and B are the opposite ends of a diameter of the upper face of the object (see diagram). (a) Find, in terms of h and k, the distance of the centre of mass of the object from AB. 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When the object is suspended from A, the angle between AB and the vertical is i, where tan i = 3 . 2 (b) Given that h = 8 a , find the possible values of k. 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Mark scheme: 4(a) B1 Correct volumes and distances for large and small. Large Small Object M1 Moments equation with 3 terms, dimensionally 2 correct. 2 2  2  4  Volume a h  a  kh a h  1 − k   3   9  A1 Correct, unsimplified. Centre of 1 1 mass from h kh x AB 2 2 Moments about AB: 2 2  4  2 1 2  1 a h  1 − k   y = a h  h −  a  kh  kh  9  2  3  2 2 A1 9 − 4 k h ( ) y = 2 ( 9 − 4 k ) 4 4(b) 2 B1 FT FT their part (a) 9 − 4 k h 3 y ( ) tan= : = a 2 ( 9 − 4 k ) a 2 8 2 M1 Use h = a and simplify to quadratic in k: 32k − 36k + 9 = 0 3 3 3 A1 k = , 8 4 3

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5 A particle P of mass 2 kg moving on a horizontal straight line has displacement x m from a fixed point O on the line and velocity v m s -1 at time t s. The only horizontal force acting on P is a variable force F N which can be expressed as a function of t. It is given that v 3 - t = x 1 + t and when t = 0, x = 5 . (a) Find an expression for x in terms of t. 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(b) Find the magnitude of F when t = 3. 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Mark scheme: 5(a) d x  4  M1 Separate variables, obtain RHS in integrable form. =  − 1  d t x  t + 1  ln x = 4ln t + 1 −+t A A1 t = 0, x = 5: A = ln5 M1 4 t A1 x = 5 ( t + 1) e− 4 5(b) 3 t M1 v = ( 3 − t )  5 ( t + 1) e− dv − t 3 2 3 Acceleration = = 5e − ( t + 1) + ( 3 − t ) 3 ( t + 1) − ( 3 − t )( t + 1) ( ) dt − t 2 AEF Acceleration = 5e ( t + 1) ( 5 − t )(1 − t ) − t 2 M1 F = 2  acceleration, so at F = 10e ( t + 1) ( 5 − t )(1 − t ) At t = 3, magnitude of force is 640 e− N3 A1 31.9 N 3

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6 O a 1 m P 0.8 m b Q 1.4 m A particle P of mass 0.05 kg is attached to one end of a light inextensible string of length 1 m. The other end of the string is attached to a fixed point O. A particle Q of mass 0.04 kg is attached to one end of a second light inextensible string. The other end of this string is attached to P. The particle P moves in a horizontal circle of radius 0.8 m with angular speed ~ rad s -1 . The particle Q moves in a horizontal circle of radius 1.4 m also with angular speed ~ rad s -1 . The centres of the circles are vertically below O, and O, P and Q are always in the same vertical plane. The strings OP and PQ remain at constant angles a and b respectively to the vertical (see diagram). (a) Find the tension in the string OP. 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(b) Find the value of ~. 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(c) Find the value of b. 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Mark scheme: 6(a) At P:  T1 cos= T2 cos + 0.05 g B1 At Q:  T2 cos = 0.04 g B1 OR: whole system: T1 cos= 0.09 g T1 = 0.15 g = 1.5 N B1 3 6(b) T1 sin − T2 sin  = 0.05  0.82 M1 Allow sin/cos mix T2 sin  = 0.04  1.42 M1 T1sin = 0.05  0.82 + 0.04  1.42 A1 2 5  = 12.5, = 2 2 3 6(c) T2 cos = 0.04 g and T2 sin  = 0.04  1.42 M1 From part (a) and part (b) 7 Divide: tan = 4 = 60.3 A1 2

More questions on Circular motion

7 A particle P is projected with speed u at an angle tan -1 b 4 l above the horizontal from a point O on a 3 horizontal plane and moves freely under gravity. When P is moving horizontally, it strikes a smooth inclined plane at the point A. This plane is inclined to the horizontal at an angle a, and the line of greatest slope through A lies in the vertical plane through O and A. As a result of the impact, P moves vertically upwards. The coefficient of restitution between P and the inclined plane is e. (a) Show that e tan 2a = 1. 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In its subsequent motion, the greatest height reached by P above A is 3 of the vertical height of A 16 above the horizontal plane. (b) Find the value of e. 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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) When P strikes plane, velocity is →u cos, M1 3 5u 3 3 Before impact: parallel to inclined plane u cos, perpendicular to plane u sin 5 5 3 3 A1 After impact: components u cos (parallel) and eu sin (perpendicular) 5 5 3 3 M1 Since velocity is vertical after impact, tan= u cos/ eu sin 5 5 tan= 1/ e tan, e tan2 = 1 A1 AG 4 7(b) ( u sin) 2 8u 2 M1A1 Note: alternative methods. Greatest height of P before impact: H = = 2 g 25 g 3 2 2 2 M1 After impact, vertical speed of P is u ( cos) + e ( sin) 5 2 2 3 M1 Use V = U + 2 as to greatest height, equal to H 16 9 2 2 2 2 3 u ( cos) + e ( sin) = 2 g  H ( ) 25 16 1 e 1 M1 Use part (a): tan = , cos= , sin = e 1 + e 1 + e Substitute to find e 2 1 A1 3e + 2e −=1 0, e = 3 6

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