A Level Further Mathematics 9231 Paper 32, Oct/Nov 2025

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Question paper

9231/32 · Oct/Nov 2025

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1

A particle P of mass m is attached to two light inextensible strings each of length l. The end of one string is attached to a fixed point A and the end of the other string is attached to a fixed point B, with A vertically above B. Angle APB is a right angle. The particle P rotates in a horizontal circle at a constant angular speed ω with both strings taut (see diagram). Find the tension in string AP in terms of m, g, l and ω.

[4]
Figure 1 of 1 for question 1
2

One end of a light elastic string of natural length a and modulus of elasticity 2mg is attached to a fixed point A on a rough horizontal surface. The other end of the string is attached to a particle P of mass m. The particle and string rest on the surface. The coefficient of friction between P and the surface is μ. The particle P is initially held in equilibrium at a distance 43a from A. The particle is then released from rest.

a

Given that the string never becomes slack, find the minimum value of μ.

[3]
b

It is now given that μ=12. Find the extension of the string when the particle comes to rest.

[3]
3

A uniform lamina OABCD is in the form of a rectangle, OBCD, joined along the edge OB to a quarter circle OAB. The length of DO is ka and the length of OB is a. The lamina rests in a vertical plane with its edge CB on a horizontal surface (see diagram).

Figure 1 of 1 for question 3
a

Find, in terms of k, a and π, an expression for the distance of the centre of mass above the horizontal surface. [You may use without proof the result for the centre of mass of a circular sector in the list of formulae (MF19).]

[4]
b

The lamina is on the point of toppling about B. Find the value of k.

[2]
4

A particle Q is initially positioned at a distance d vertically above a particle P. Particle P is projected with speed U at an angle α above the horizontal. At the same time, Q is projected at an angle β below the horizontal. Both particles move freely under gravity. The particles collide at time T after the projections.

a

Show that d=UT(sin⁡α+cos⁡αtan⁡β).

[4]
b

The particles collide when P is at its maximum height. Given that α=30∘ and β=60∘, find d in terms of U and g.

[3]
5

Two uniform smooth spheres, A and B, of equal radii are on a horizontal surface. They have masses of m and km respectively. Sphere A is moving with speed u at an angle α with the line of centres when it collides with sphere B which is stationary. Immediately after the collision, sphere A moves with speed v at an angle 2α with the line of centres (see diagram). It is given that tan⁡α=34.

Figure 1 of 1 for question 5
a

Find v in terms of u.

[2]
b

Find the coefficient of restitution between the spheres in terms of k.

[4]
c

Find the range of possible values of k.

[3]
6

A fixed smooth sphere with radius a and centre O rests on horizontal ground. A particle is projected horizontally from the highest point, A, of the sphere with speed u. The particle begins to move in a vertical circle along the surface of the sphere. The particle loses contact with the sphere at the point B, where the angle AOB is θ. After leaving the surface of the sphere, the particle moves freely under gravity before striking the horizontal ground with speed 3u at an angle β to the horizontal (see diagram). Find the value of β.

[8]
Figure 1 of 1 for question 6
7

A particle P of mass m kg moving along a rough horizontal table has displacement x m from a fixed point O on the table and velocity v ms−1 at time t s. The particle P is subject to a resistive force of magnitude mgkv N, where k is a positive constant, and a frictional force of magnitude μmg. The particle P is initially at O with speed U ms−1.

a

Show that t=1gkln⁡(kU+μkv+μ).

[4]
b

It is given that U=10, k=0.04 and μ=0.2. Find the distance P moves before coming to rest.

[4]
c

Find the average speed of P over the period it is moving.

[2]

Mark scheme

9231/32 · Oct/Nov 2025

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1 · 4 marks

  • B1

    TAcos⁡45∘+TBsin⁡45∘=mω2lsin⁡45∘

  • B1

    TAsin⁡45∘−TBcos⁡45∘=mg

  • M1

    2TA=mω2l+2mg

  • A1

    TA=12m(ω2l+2g)

2a · 3 marks

  • B1

    EPE=2mg2a(13a)2 Work done by friction=μmg(13a)

  • M1

    2mg2a(13a)2=μmg(13a)

  • A1

    μ=13

2b · 3 marks

  • M1

    Let x be the extension in the string when P comes to rest. 2mg2a(13a)2=2mg2ax2+12mg(13a−x)

  • A1

    Correct energy equation.

  • A1

    [18x2−9ax+a2=0,(6x−a)(3x−a)=0]x=16a only.

3a · 4 marks

  • B1

    Vertical distance of COM of sector from OA: (2acos⁡(14π))sin⁡(14π)3(14π)=4a3π

  • M1

    ka2(12a)+14πa2(a−4a3π)=(ka2+14πa2)yˉ

  • A1

    [6ka+3πa−4a=3(4k+π)yˉ]

  • A1

    yˉ=6k+3π−43(4k+π)a

3b · 2 marks

  • M1

    ka2(12ka)=14πa2(4a3π)

  • A1

    [k2=23]k=23=0.816

4a · 4 marks

  • B1

    Horizontal velocities are the same: Uqcos⁡β=Ucos⁡α

  • M1

    d=(Usin⁡α)T−12gT2+(Uqsin⁡β)T+12gT2

  • M1

    d=(Usin⁡α)T+(Ucos⁡αcos⁡βsin⁡β)T

  • A1

    d=UT(sin⁡α+cos⁡αtan⁡β)

4b · 3 marks

  • B1

    Usin⁡α=gT

  • M1

    d=U2sin⁡αg(sin⁡α+cos⁡αtan⁡β)

  • A1

    d=U2g

5a · 2 marks

  • B1

    usin⁡α=vsin⁡(2α)

  • B1

    v=58u

5b · 4 marks

  • B1

    mucos⁡α=mvcos⁡(2α)+kmw

  • B1

    eucos⁡α=w−vcos⁡(2α)

  • M1

    58k=e+45×58×725

  • A1

    e=2532k−732

5c · 3 marks

  • M1

    0⩽2532k−732⩽1

  • M1

    7k⩽25⩽39k

  • A1

    2539⩽k⩽257

6 · 8 marks

  • M1

    A to ground: 12mu2+2mga=12m(3u)2[u2=12ga]

  • M1

    A to B: 12mv2+mgacos⁡θ=12mu2+mga[v2=u2+2ag(1−cos⁡θ)] OR B to ground: 12mv2+mgacos⁡θ=12m(3u)2−mga[v2=9u2−2ag(1+cos⁡θ)]

  • A1

    At least one correct energy equation involving cos⁡θ.

  • B1

    mgcos⁡θ=mv2a[v2=agcos⁡θ]

  • 1 mark

    gacos⁡θ=12ga+2ag(1−cos⁡θ)

  • A1

    cos⁡θ=56

  • M1

    Horizontally at ground: vcos⁡θ=3ucos⁡β. cos⁡β=56ga(56)312ga=(55415) OR vx=vcos⁡θ=56ga(56),vy2=(vsin⁡θ)2+2ga(1+cos⁡θ)=847216ga tan⁡β=847216ga56ga(56)

  • A1

    β=69.0∘

7a · 4 marks

  • B1

    −mgkv−μmg=mdvdt

  • M1

    t=−1g∫1kv+μdv t=−1gkln⁡(kv+μ)[+C]

  • 1 mark

    When v=u, t=0: C=1gkln⁡(kU+μ)

  • A1

    t=−1gkln⁡(kv+μ)+1gkln⁡(kU+μ) t=1gkln⁡(kU+μkv+μ)

7b · 4 marks

  • M1

    −25v−2=vdvdx x=−52∫vv+5dv=−52∫(1−5v+5)dv

  • M1

    x=−52v+252ln⁡(v+5)[+C]

  • M1

    When x=0,v=10: C=25−252ln⁡(15)

  • A1

    When v=0, x=25−252ln⁡(15)+252ln⁡(5)=11.3 [m]

7c · 2 marks

  • M1

    Average speed="11.3"110×0.04ln⁡(0.04×10+0.20.2)="11.3"2.5ln⁡3

  • A1

    4.10 [ms−1]

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