A Level Further Mathematics 9231 Paper 32, May/June 2024

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9231/32 · May/June 2024

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1

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 θ 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 12. Find the value of tan⁡θ.

[6]
Figure 1 of 1 for question 1
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 λ, 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 32a below M, the midpoint of AB.

a

Find λ in terms of m and g.

[3]
b

The particle P is pulled down vertically and released from rest at a distance 83a below M. Find, in terms of a and g, the speed of P as it passes through M in the subsequent motion.

[4]
3

At time t=0 seconds, a particle P is projected with speed u ms−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.

[5]
4

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 α where tan⁡α=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 μ.

Figure 1 of 1 for question 4
a

Find, in terms of W, the tension in the string PQ.

[4]
b

Find the value of μ.

[3]
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 15. When the turntable is made to rotate with angular speed 25ga, the system is in limiting equilibrium.

a

Find the tension in the string, in terms of m and g.

[4]
b

Find the value of k.

[3]
6

A particle P of mass 2kg moving on a horizontal straight line has displacement xm from a fixed point O on the line and velocity v ms−1 at time t s. The only horizontal force acting on P has magnitude 110(2v−1)2e−t N and acts towards O. When t=0, x=1 and v=3.

a

Find an expression for v in terms of t.

[5]
b

Find an expression for x in terms of t.

[4]
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 θ with the upward vertical through O.

a

Show that cos⁡θ=u2+2ag3ag.

[4]
b

It is given that cos⁡θ=56. 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.

[3]
c

Find an expression for the time taken by P to fall from Q to the plane. Give your answer in the form kag, stating the value of k correct to 3 significant figures.

[2]

Mark scheme

9231/32 · May/June 2024

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1 · 6 marks

  • M1

    Along line of centres, PCLM: 5mvB+mvA=mucos⁡θ

  • M1

    NEL: vB−vA=12ucos⁡θ

  • A1

    vB=u4cos⁡θ,  vA=−u4cos⁡θ

  • B1

    Perpendicular to line of centres: speed of A is usin⁡θ

  • M1

    12m((−u4cos⁡θ)2+(usin⁡θ)2)=125m(u4cos⁡θ)2

  • A1

    (cos⁡θ)2=45,  cos⁡θ=25,  tan⁡θ=12

2a · 3 marks

  • M1

    In equilibrium: 2Tcos⁡θ=mg

  • B1

    Hooke’s law: T=λ2a×(5a2−2a)=λ4

  • A1

    Equate and use cos⁡θ=35: λ=103mg

2b · 4 marks

  • B1

    EPE loss = 12λ4a(203a−4a)2

  • M1

    Energy equation: 12mv2+mg8a3=‘8027mga’

  • A1

    Energy equation: 12mv2+mg8a3=‘8027mga’

  • A1

    v=43ga3

3 · 5 marks

  • B1

    usin⁡60∘−5g, and ucos⁡60∘, or usin⁡60∘−15g, and ucos⁡60∘

  • 1 mark

    If θ is direction of velocity at t=5,[tan⁡θ=]usin⁡60∘−5gucos⁡60∘

  • 1 mark

    For perpendicular directions, usin⁡60∘−5gucos⁡60∘×usin⁡60∘−15gucos⁡60∘=−1

  • M1

    Simplify: 34u2+75g2−103ug+14u2=0,u2−1003u+7500=0

  • A1

    u=53g

4a · 4 marks

  • M1

    Frictional force F and normal reaction R at point of contact of ring with plane. Resolve parallel to plane: F+Tcos⁡α=Wsin⁡α

  • B1

    Moments about O: Fa=Tasin⁡α

  • M1

    Combine and substitute for α:

  • A1

    T=13W

4b · 3 marks

  • M1

    Resolve perpendicular to plane: R=Tsin⁡α+Wcos⁡α

  • M1

    Use F=μR and combine to reach an equation in μ only.

  • A1

    μ=17

5a · 4 marks

  • M1

    For A: FA−T=m×aω2

  • B1

    FA=μmg=15mg

  • M1

    Combine: T=15mg−425mg

  • A1

    T=125mg

5b · 3 marks

  • M1

    For B: FB+T=km×2aω2

  • M1

    FB=μkmg=15kmg and combine to find k

  • A1

    k=13

6a · 5 marks

  • 1 mark

    2dvdt=−110(2v−1)2e−t so dv(2v−1)2=−120e−tdt p(2v−1)=qe−t+A

  • A1

    −12(2v−1)=+120e−t+A

  • 1 mark

    t=0,v=3,[A=−320]

  • 1 mark

    v=12+5et3et−1

  • A1

    v=12+5et3et−1

6b · 4 marks

  • 1 mark

    Integrate: x=pt+qln⁡(re±t−s)[+B]

  • A1

    x=12t+53ln⁡(3et−1)[+B]

  • 1 mark

    t=0,x=1,B=1−53ln⁡2

  • A1

    x=1+12t+53ln⁡(3et−1)2

7a · 4 marks

  • 1 mark

    Energy: 12mu2=12mv2−mga(1−cos⁡θ)

  • B1

    N2L: mgcos⁡θ=mv2a

  • 1 mark

    Eliminate v2

  • A1

    cos⁡θ=u2+2ag3ag

7b · 3 marks

  • 1 mark

    Vertical component of velocity of P when it leaves the sphere: vsin⁡θ [=55ag216]

  • 1 mark

    V2=(vsin⁡θ)2+2g×a(1+cos⁡θ)

  • A1

    V=847ag216

7c · 2 marks

  • M1

    t=1g(847ag216−55ag216)

  • A1

    16(847−55)a6g=1.48ag

Questions

  1. 1
    Question 16 marks
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  2. 2
    Question 27 marks
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  3. 3
    Question 35 marks
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  4. 4
    Question 47 marks
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  5. 5
    Question 57 marks
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  6. 6
    Question 69 marks
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  7. 7
    Question 79 marks
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