Cambridge A Level Mathematics - Further 9231 — 2024 May/June Paper 3 · Variant 2

9231/32/M/J/24 · 7 questions · 50 marks · ≈56 min

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Mark scheme13 pages

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Questions as text

Q1 · U A B i m 5m Two smooth uniform spheres A and B of equal radii have masses m and 5m…

1 u A B i m 5m Two smooth uniform spheres A and B of equal radii have masses m and 5m respectively. Sphere A is moving on a smooth horizontal surface with speed u when it collides with sphere B which is at rest on the surface. Immediately before the collision, A’s direction of motion makes an angle of i with the line of centres. After the collision, the kinetic energies of A and B are equal. The coefficient of restitution between the spheres is 1.2 Find the value of tani. 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Mark scheme: 1 Along line of centres, PCLM: 5 B A mv mv mu NEL: 1 2 cos B A v v u    M1 Signs consistent with PCLM equation. cos , cos 4 4     B A u u v v A1 Perpendicular to line of centres: speed of A is sin u B1   2 2 2 1 1 cos sin 5 cos 2 4 2 4                           u u m u m M1 Equate final kinetic energies, 3 terms, correct masses.   2 4 5 cos   , 2 1 2 5 cos , tan     A1 6

More questions on Momentum

Q2 · The points A and B are at the same horizontal level a distance 4a apart

2 The points A and B are at the same horizontal level a distance 4a apart. The ends of a light elastic string, of natural length 4a and modulus of elasticity m, are attached to A and B. A particle P of mass m is attached to the midpoint of the string. The system is in equilibrium with P at a distance 3a2 below M, the midpoint of AB. (a) Find m in terms of m and g. 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The particle P is pulled down vertically and released from rest at a distance 8a3 below M. (b) Find, in terms of a and g, the speed of P as it passes through M in the subsequent motion. 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Mark scheme: 2(a) T mg Hooke’s law: 5 2 2 2 4             a T a a B1 Equate and use 3 5 cos :   10 3 mg   A1 3 2(b) EPE loss = 2 1 20 4 2 4 3        a a a B1 80 27 mga Energy equation: 2 1 8 80 ' ' 2 3 27   a mv mg mga M1 A1 All 3 terms required, dimensionally correct, their . 4 3 3  ga v A1 Any equivalent form. 4

More questions on Hooke's law

Q3 · At time t = 0 seconds, a particle P is projected with speed u m s -1 at an angle 60°…

3 At time t = 0 seconds, a particle P is projected with speed u m s -1 at an angle 60° above the horizontal from a point O. In the subsequent motion P moves freely under gravity. The direction of motion of P when t = 5 is perpendicular to its direction of motion when t = 15 . Find the value of u. 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Mark scheme: 3 u u B1 If  is direction of velocity at  sin60 5 5, tan cos60      u g t u M1* Accept equivalent for 15  t . For perpendicular directions, sin60 5 sin60 15 1 cos60 cos60       u g u g u u M1dep Multiply two expressions involving relevant velocities and equate to – 1. Simplify: 3 1 2 2 2 2 4 4 75 10 3 0, 100 3 7500 0 u g ug u u u        M1 Simplify to quadratic in u (may see g ). 5 3  u g A1 OE. Accept 50 3 or 86.6. 5

More questions on Motion of a projectile

Q4 · P Q a O a A ring of weight W, with radius a and centre O, is at rest on a rough surface…

4 P Q a O a A ring of weight W, with radius a and centre O, is at rest on a rough surface that is inclined to the horizontal at an angle a where tan a = 12 . The plane of the ring is perpendicular to the inclined surface and parallel to a line of greatest slope of the surface. The point P on the circumference of the ring is such that OP is parallel to the surface. A light inextensible string is attached to P and to the point Q, which is on the surface, such that PQ is horizontal (see diagram). The points O, P and Q are in the same vertical plane. The system is in limiting equilibrium and the coefficient of friction between the ring and the surface is n. (a) Find, in terms of W, the tension in the string PQ. 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(b) Find the value of n. 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Mark scheme: 4(a) Frictional force F and normal reaction R at point of contact of ring with plane. Resolve parallel to plane: cos sin     F T W M1 Only allow cos/sin errors for T and W components, sign errors. Accept equations for vertical and horizontal (both needed). Moments about O: sin  Fa Ta B1 Combine and substitute for : M1 Expression for T in terms of W . 1 3 T W  A1 CAO Alternative solution for question 4(a) Moments about point where ring touches plane: sin cos sin      Ta Ta Wa M1 A1 Only allow cos/sin errors, sign errors. Must be dimensionally correct. Rearrange and substitute for : M1 Expression for T in terms of W . 1 3 T W  A1 CAO 4 4(b) Resolve perpendicular to plane: sin cos     R T W M1 Only allow cos/sin errors for T and W components, sign errors. Use   F R and combine to reach an equation in  only. M1 From part (a), cos sin F T W    or sin  F T . 1 7   A1 3

More questions on Equilibrium of a rigid body

Q5 · Two particles A and B of masses m and km respectively are connected by a light…

5 Two particles A and B of masses m and km respectively are connected by a light inextensible string of length a. The particles are placed on a rough horizontal circular turntable with the string taut and lying along a radius of the turntable. Particle A is at a distance a from the centre of the turntable and particle B is at a distance 2a from the centre of the turntable. The coefficient of friction between each particle and the turntable is 1.5 2 g When the turntable is made to rotate with angular speed , the system is in limiting equilibrium. 5 a (a) Find the tension in the string, in terms of m and g. 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(b) Find the value of k. 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Mark scheme: 5(a) For A: 2     A F T m a M1 Only allow sign errors. 1 5 A F mg mg    B1 Accept with g replaced by 10. Combine: 1 4 5 25 T mg mg   M1 To reach an equation in T and mg only. Accept with g replaced by 10. 1 25 T mg  A1 CAO 4 5(b) For B: 2 2     B F T km a M1 Only allow sign errors. 1 5 B F kmg kmg    and combine to find k M1 To reach an equation in k only. 1 3 k  A1 3

More questions on Circular motion

Q6 · A particle P of mass 2 kg moving on a horizontal straight line has displacement x m from…

6 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 has magnitude 1 2 -t 10 ( 2v - 1) e N and acts towards O. When t = 0 , x = 1 and v = 3 . (a) Find an expression for v in terms of t. 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(b) Find an expression for x in terms of t. 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Mark scheme: 6(a)   2 d 1 2 2 1 e d 10    t v v t so   2 d 1 e d 20 2 1    t v t v   e 2 1     t p q A v *M1 Separate variables and attempt to integrate both sides. Where p and q are constants.   1 1 e 2 2 1 20      t A v A1 AEF 3 0, 3, 20          t v A DM1 Substituting the boundary condition and obtain a value. 1 5e 2 3e 1    t t v *M1 A1 Find v in terms of t . AEF. 5 6(b) Integrate:   ln(re )      t x pt q s B *M1   1 5ln(3e 1) 2 3     t x t B A1 AEF 5 0, 1, 1 ln2 3    t x B DM1 Substituting the boundary condition and obtain a value. 1 5 (3e 1) 1 ln 2 3 2    t x t A1 AEF 4

More questions on Differential equations

Q7 · A smooth sphere with centre O and of radius a is fixed to a horizontal plane

7 A smooth sphere with centre O and of radius a is fixed to a horizontal plane. A particle P of mass m is projected horizontally from the highest point of the sphere with speed u, so that it begins to move along the surface of the sphere. The particle P loses contact with the sphere at the point Q on the sphere, where OQ makes an angle i with the upward vertical through O. u 2 + 2ag (a) Show that cos i = . 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It is given that cos i = 56 . (b) Find, in terms of a and g, an expression for the vertical component of the velocity of P just before it hits the horizontal plane to which the sphere is fixed. 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(c) Find an expression for the time taken by P to fall from Q to the plane. Give your answer in the a form k , stating the value of k correct to 3 significant figures. 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Mark scheme: 7(a) Energy: 1 1 2 2 2 2 1 cos mu mv mga     *M1 m must be present, dimensionally correct, no missing terms. Allow sin instead of cos . Allow sign errors. N2L: 2 cosmv mg a B1 No reaction when P loses contact. Eliminate 2 v DM1 2 2 cos 3 u ag ag    A1 AG 4 7(b) Vertical component of velocity of P when it leaves the sphere: sin  v 55 216          ag *B1 Must not come from u .     2 2 sin 2 1 cos       V v g a DM1 Use of 2 2 ' 2 .   v u as Allow sin for cos. Allow sign errors. 847 216  ag V A1 AEF 3 Question Answer Marks Guidance 7(c) 1 847 55 216 216           ag ag t g M1   1 847 55 1.48 6 6   a a g g A1 2

More questions on Motion of a projectile

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