A Level Further Mathematics 9231 Paper 33, May/June 2026

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Question paper

9231/33 · May/June 2026

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1

A particle P of mass m is moving in a horizontal circle with constant angular speed ω on the smooth inner surface of a hemispherical shell of radius a. The plane of the circular motion is at a height h=13a above the lowest point on the shell. The angle between the vertical and the normal reaction of the surface on P is θ. Find ω in terms of a and g.

[5]
2

A particle P of mass m is moving in a straight horizontal line towards a fixed point A on the line. The particle is subject to a resistive force that is inversely proportional to the square of the distance between P and A. Initially, the distance between P and A is d and the speed of P is u. When the distance between P and A is 12d, the particle has lost 34 of its kinetic energy. Find, in terms of d, the minimum distance of P from A.

[6]
3

A fixed smooth solid sphere has centre O and radius a. A particle of mass m is projected perpendicular to OA in a downward direction with speed u from the point A on the outer surface of the sphere, where OA makes an angle α with the upward vertical through O (see diagram). The particle moves in part of a vertical circle on the surface of the sphere. It loses contact with the sphere at the point B, where OB makes an angle β with the upward vertical through O. The speed of the particle at the point B is v. Given that the ratio of u to v is 2:3 find the ratio cos⁡α:cos⁡β.

[5]
Figure 1 of 1 for question 3
4

An object consists of a uniform solid cube of side length d, and two uniform solid square-based pyramids. The cube and the pyramids are made of the same material. One pyramid, of base length d and height h, is fixed to the top face of the cube so that the base of this pyramid and the face of the cube coincide. Point A is at the top vertex of this pyramid. Point B is in the centre of the bottom face of the cube, vertically below A. The second pyramid, of base length d and height kh, is fixed to a vertical face of the cube so that the base of this pyramid and the vertical face of the cube coincide (see diagram, which shows a cross-section, including the axes of symmetry).

Figure 1 of 1 for question 4
a

Find an expression for the distance of the centre of mass of the object from AB in terms of h, k, and d.

[4]
b

The object is placed on a horizontal plane so that the bottom face of the cube that contains point B is in contact with the plane. It is given that the object is on the point of toppling. Find the value of h in terms of k and d.

[3]
c

Given that h=1.5d, find the value of k.

[2]
5

Two uniform smooth spheres A and B of equal radii and equal masses m are moving with speeds uA and uB respectively on a smooth horizontal surface when they collide. Immediately before the collision, the direction of motion of A makes an angle α with the line of centres, and the direction of motion of B makes an angle β with the line of centres (see diagram). Immediately after the collision, the direction of motion of A makes an angle θ with the line of centres, and the direction of motion of B makes an angle ϕ with the line of centres. It is given that cos⁡α=cos⁡ϕ=45 and cos⁡β=2425 and cos⁡θ=35.

Figure 1 of 1 for question 5
a

Find the percentage loss in the kinetic energy of A.

[2]
b

Find the value of the coefficient of restitution between the spheres.

[4]
6

A smooth plane is inclined at an angle θ to the horizontal, where sin⁡θ=45. One end of a light elastic spring, with natural length 10a and modulus of elasticity 1.6mg, is attached to a fixed point O on the plane. The other end of the spring is attached to a particle P of mass 0.2m. The point O and the particle P lie on the same line of greatest slope of the plane, with P above O. The particle P is at rest.

a

Show that the distance OP is 9a.

[3]
b

A particle Q of mass 0.3m is held on the plane on the line of greatest slope through O, at a distance 10a above P. The particle Q is released from rest. The particle Q then collides and coalesces with particle P. Find, in terms of a and g, the speed of the combined particle immediately after the collision.

[4]
c

As a result of the collision, the spring is further compressed. Find, in terms of a, the minimum distance of the combined particle from O.

[4]
7

At time t=0 s, a particle P is projected from a point O on horizontal ground. The initial velocity of P is 25 ms−1 at an angle α above the horizontal, where sin⁡α=45. At time t=T s, a second particle Q is projected vertically upwards from the horizontal ground directly below the point where P will reach its greatest height. The initial velocity of Q is also 25 ms−1. Both particles move freely under gravity. The two particles collide when Q is travelling upwards and at the point where P reaches its greatest height.

a

Show that the two particles collide 2 seconds after P is projected.

[1]
b

Find the value of T.

[3]
c

As a result of the collision, the trajectory of P changes, but the horizontal component of its velocity remains unchanged. Immediately after the collision, the speed of P is v ms−1, at an angle θ above the horizontal. Find, in terms of tan⁡θ, the greatest height of P above the ground after the collision.

[4]

Mark scheme

9231/33 · May/June 2026

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1 · 5 marks

  • B1

    Rcos⁡θ=mg

  • B1

    Rsin⁡θ=m(asin⁡θ)ω2

  • M1

    g=(acos⁡θ)ω2 or ω2=gacos⁡θ or cos⁡θ=gaω2

  • B1

    13a=a−acos⁡θ, cos⁡θ=23

  • A1

    ω=3g2a

2 · 6 marks

  • B1

    N2L to form a differential equation in terms of vdvdx.

  • M1

    Separate variables correctly and integrate to obtain an equation of the form mv22=αx(+c) (or mv22=αd−x(+c)) where α≠0.

  • B1

    Use kinetic energy loss to find v when x=12d.

  • 1 mark

    Use both sets of boundary conditions correctly to obtain two equations in c, k and d using their integrated equation.

  • 1 mark

    Substitute v=0, dependent on all marks so far.

  • A1

    37d

3 · 5 marks

  • B1

    At B, mgcos⁡β=mv2a

  • M1

    Energy equation with 4 terms and correct dimensions (mass must be present). Allow sign errors and sin/cos mix only.

  • A1

    Fully correct energy equation: 12mv2−12mu2=mga(cos⁡α−cos⁡β)

  • 1 mark

    Combine their two equations and substitute u=23v to obtain an equation in, for example, cos⁡α and cos⁡β only.

  • A1

    cos⁡α:cos⁡β=23:18

4a · 4 marks

  • M1

    Dimensionally correct moments equation, which must be of the form (αd3+βd2h(k+1))xˉ=γkhd2(γd+δkh) where α,β,γ,δ are all non-zero constants.

  • B1

    Either side of equation correct.

  • A1

    Fully correct moments equation.

  • A1

    xˉ=kh(2d+kh)4(hk+h+3d)

4b · 3 marks

  • B1

    xˉ=12d

  • M1

    Rearrange to obtain a 3-term quadratic in h of the form [α]k2h2+βdh+γd2[=0].

  • A1

    h=d+d1+6k2k2[=dk2(1+1+6k2)]

4c · 2 marks

  • M1

    Equating their h to 1.5d or substituting h=1.5d into k2h2−2dh−6d2=0.

  • A1

    k=2 only

5a · 2 marks

  • B1

    35uA=45vA

  • 1 mark

    Calculate the percentage loss in KE of A: [12muA2−12m(34uA)212muA2×100=]43.75(%)

5b · 4 marks

  • B1

    725uB=35vB[7uB=15vB]

  • M1

    Dimensionally correct momentum equation (m required): m(45uA)−m(2425uB)=−mvA×35+mvB×45.

  • M1

    Use Newton's experimental law to form an equation: e(45uA+2425uB)=35vA+45vB

  • A1

    e=817[=0.471]

6a · 3 marks

  • B1

    T=1.6mgx10a

  • M1

    T=0.2mg×45[=0.16mg]

  • A1

    0.16mg=1.6mgx10a x=a OP=10a−a=9a.

6b · 4 marks

  • M1

    12×0.3mv2=10a×45×0.3mg OR v2=2×10a×g×45

  • A1

    v=4ag

  • 1 mark

    0.3m×′4ag′=(0.3m+0.2m)w

  • A1

    w=2.4ag

6c · 4 marks

  • B1

    At least one correct EPE.

  • M1

    Dimensionally correct energy equation with 2 EPE terms, one KE term and one/two GPE term(s) depending on how x is defined.

  • A1

    All terms correct. FT their w2.

  • A1

    [x=]3a

7a · 1 mark

  • B1

    Vy=uPsin⁡α−gt, and so Vy=25×sin⁡α−10t. At impact, Vy=0, hence, 25×45−10t=0, t=2.

7b · 3 marks

  • M1

    h=20×2−12g×22

  • 1 mark

    25tQ−12g×tQ2=’20’

  • A1

    tQ=1,T=2−tQ=1

7c · 4 marks

  • M1

    vcos⁡θ=uPcos⁡α[=15] leading to vsin⁡θ=15cos⁡θ×sin⁡θ

  • 1 mark

    [0=(vsin⁡θ)2−2g×ymax⁡] 0=(15cos⁡θsin⁡θ)2−2g×ymax⁡

  • A1

    ymax⁡=454tan⁡2θ

  • A1

    Greatest height above ground=20+454tan⁡2θ

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