Cambridge A Level Mathematics - Further 9231 — 2024 Oct/Nov Paper 3 · Variant 3
9231/33/O/N/24 · 7 questions · 50 marks · ≈56 min
The question paper and its mark scheme, free to read here and free to download. This is Cambridge’s own paper, exactly as it was sat.
Question paper16 pages
















Mark scheme12 pages
Answers below. Sit the paper first if you are practising.












Questions as text
Q1 · A particle P is projected with speed u m s -1 at an angle tan -1 2 above the horizontal…
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
Q2 · A particle P of mass m is attached to one end of a light inextensible string of length a
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
Q3 · A particle P of mass m kg is attached to one end of a light elastic string of natural…
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
Q4 · A 2 a B 3 kh h a An object is formed by removing a cylinder of radius 2 a and height kh (…
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 4 2 1 2 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
Q5 · A particle P of mass 2 kg moving on a horizontal straight line has displacement x m from…
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
Q6 · 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…
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 0.82 M1 Allow sin/cos mix T2 sin = 0.04 1.42 M1 T1sin = 0.05 0.82 + 0.04 1.42 A1 2 5 = 12.5, = 2 2 3 6(c) T2 cos = 0.04 g and T2 sin = 0.04 1.42 M1 From part (a) and part (b) 7 Divide: tan = 4 = 60.3 A1 2
Q7 · A particle P is projected with speed u at an angle tan -1 b 4 l above the horizontal from…
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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