A Level Further Mathematics 9231 Paper 33, Oct/Nov 2025

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Question paper

9231/33 · Oct/Nov 2025

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1

Two uniform smooth spheres A and B of equal radii have masses 4m and m respectively. Sphere B is at rest on a smooth horizontal surface. Sphere A is moving on the surface with speed u and collides directly with sphere B. After the collision, the momentum of A is three times the momentum of B. Find the value of the coefficient of restitution e.

[4]
2

A particle P of mass m is moving in a horizontal circle with angular speed ω1 on the smooth inner surface of a hemispherical shell of radius r. The angle between the upward vertical and the normal reaction of the surface on P is θ1, where tan⁡θ1=34. When the angular speed is increased to ω2, the angle between the upward vertical and the normal reaction of the surface on P becomes θ2, where tan⁡θ2=43. Find the ratio ω1ω2.

[4]
3

A particle P is moving in a straight horizontal line. At time t s, the displacement of P from a fixed point O on the line is x m and the velocity of P is v m s−1. The acceleration of P is 12(v2+4) m s−2 in the direction PO. Initially P is at O and is moving with velocity 2 m s−1.

a

Find an expression for x in terms of t.

[5]
b

Find the time when P next goes through O.

[2]
4

One end of a light elastic string of natural length a and modulus of elasticity 5mg is attached to a fixed point O. Two particles, P and Q, of masses m and 4m respectively are attached to the other end of the string and they hang vertically in equilibrium. Particle Q is then detached from the string, hence releasing particle P from rest. Find, in terms of a, the length of the string when the speed of particle P is first equal to 75ag.

[6]
Figure 1 of 1 for question 4
5

A uniform lamina OABCD consists of a rectangle OACD and a triangle ABC. The length of OA is ka, the length of OD is 2a, the height of triangle ABC is h and angle CAB is 45∘ (see diagram). Relative to axes through O, parallel and perpendicular to OA as shown, the centre of mass of triangle ABC is (xˉ,yˉ).

a

Show that xˉ is 13(3ka+h), and find an expression for yˉ.

[3]
b

The lamina OABCD is placed vertically on its edge OA on a horizontal plane. Find, in terms of a and k, the set of values of h for which the lamina is in equilibrium.

[4]
c

It is now given that k=33 and that the lamina is on the point of toppling. Find, in terms of a, the coordinates of the centre of mass of the triangle ABC.

[2]
6

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. Initially P is held with the string taut and making an angle of 60∘ with the upward vertical through O. The particle P is projected perpendicular to the string in a downwards direction with speed 17ag. It then starts to move along a circular path in a vertical plane with centre O (see diagram). At the lowest point of its path, vertically below O, the particle P collides with a stationary particle Q.

Figure 1 of 1 for question 6
a

Find, in terms of a and g, an expression for the speed of P immediately before the collision with Q.

[2]
b

As a result of the collision, P rebounds and moves back along a circular path with centre O. The string becomes slack when P reaches the point on the circle vertically above O. Find, in terms of a and g, an expression for the speed of P immediately after the collision with Q.

[3]
c

The mass of particle Q is km and the collision between P and Q is perfectly elastic. Find the value of k.

[3]
7

A particle P is projected from a point O on a horizontal plane and moves freely under gravity. The initial velocity of P is 25 m s−1 at an angle θ above the horizontal, where tan⁡θ=43. At point A, the direction of motion of P makes an angle of 45∘ with the downward vertical through A.

a

By differentiating the equation of the trajectory or otherwise, find the coordinates of A.

[5]
b

At point A, the particle strikes a fixed smooth barrier, rebounds, and lands on the horizontal plane. The barrier is inclined at an angle of 45∘ to the horizontal. Find the speed of P immediately before it collides with the barrier.

[3]
c

Given that the coefficient of restitution between the barrier and the particle is 19, find the horizontal distance travelled by P after it strikes the barrier.

[4]

Mark scheme

9231/33 · Oct/Nov 2025

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1 · 4 marks

  • M1

    Let vA and vB be the speeds of A and B respectively after the collision. 4mvA+mvB=4mu

  • M1

    −vA+vB=eu

  • A1

    vA=34u, vB=u

  • A1

    e=14

2 · 4 marks

  • B1

    ↑N1cos⁡θ1=mg →N1sin⁡θ1=m(rsin⁡θ1)ω12

  • B1

    ↑N2cos⁡θ2=mg →N2sin⁡θ2=m(rsin⁡θ2)ω22

  • M1

    54g53g=rω12rω22

  • A1

    ω1ω2=123

3a · 5 marks

  • M1

    dvdt=−12(v2+4) 12tan⁡−1(12v)=−12t[+A]

  • A1

    Correct integration 12tan⁡−1(12v)=−12t[+A]

  • 1 mark

    t=0,v=2:A=12tan⁡−1(1)=18π

  • 1 mark

    v=dxdt=2tan⁡(14π−t) x=2ln⁡(cos⁡(t−14π))[+B]

  • A1

    x=0,t=0:B=ln⁡2 x=2ln⁡(cos⁡(t−14π))+ln⁡2[=ln⁡(2cos⁡2(t−14π))]

3b · 2 marks

  • M1

    2ln⁡(cos⁡(t−14π))+ln⁡2=0

  • A1

    t=12π

4 · 6 marks

  • B1

    In equilibrium, 5mgea=5mg,e=a.

  • B1

    At least one correct EPE term seen.

  • M1

    5mga22a−5mg(x−a)22a=mg(e+a−x)+12m(75ag)

  • A1

    Fully correct energy equation.

  • M1

    25x2−60ax+27a2=0

  • A1

    x=95a only.

5a · 3 marks

  • M1

    xˉ=ka+13h

  • A1

    xˉ=13(3ka+h)

  • B1

    yˉ=a+13(h−a)=13(2a+h)

5b · 4 marks

  • M1

    Taking moments about the x-axis: xˉ(2ka2+ah)=2ka2×12ka+ah(ka+13h)

  • A1

    Fully correct moments equation: xˉ(2ka2+ah)=2ka2×12ka+ah(ka+13h)

  • M1

    For equilibrium, xˉ⩽ka, so a2k2+ahk+13h22ak+h⩽ka

  • A1

    (0⩽)h⩽3ka

5c · 2 marks

  • B1

    Point of toppling means h=3ka[=a]. xˉ=13a(3+1)

  • B1

    [h=a, meaning triangle ABC is isosceles.] yˉ=a (by symmetry)

6a · 2 marks

  • M1

    Let v be the speed of P before the collision. Energy conservation before the collision: 12m(17ag)+32mag=12mv2

  • A1

    v=20ag

6b · 3 marks

  • B1

    N2L: mV2a=(T)+mg[V=ag]

  • M1

    Energy equation: 12mwP2=12mV2+2mga

  • A1

    wP=5ag

6c · 3 marks

  • B1

    Let wQ the speed of Q after the collision. mv=kmwQ−mwP

  • B1

    wP+wQ=ev[=v] OR 12mv2=12mwP2+12kmwQ2

  • B1

    k=3

7a · 5 marks

  • M1

    dydx=tan⁡θ−2gx2u2cos⁡2θ

  • B1

    dydx=−1

  • M1

    −1=43−245xA

  • A1

    xA=1052

  • A1

    yA=354

7b · 3 marks

  • B1

    Let v be the velocity before the collision at A. ux=ucos⁡θ=15=vx

  • B1

    vy=−vx=−15

  • B1

    Speed: vx2+vy2=152 (m s−1)

7c · 4 marks

  • M1

    Let w be the velocity after the collision. w=ev=532

  • A1

    w=532

  • M1

    −354=xtan⁡45∘−gx22(532)2(cos⁡45∘)2,95x2−x−354=0. OR 354=−532×122t+12gt2,t=32 x=(wcos⁡45∘)t[=53×32]

  • A1

    x=52 (metres)

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