A Level Further Mathematics 9231 Paper 33, May/June 2025

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Question paper

9231/33 · May/June 2025

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1

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. The particle moves in a horizontal circle with constant angular speed ω and with the string inclined at an angle of θ to the downward vertical. Given that tan⁡θ=43, find ω in terms of a and g.

[3]
2

A particle P of mass m is attached to one end of a light elastic string of natural length a and modulus of elasticity mg. The other end of the string is attached to a fixed point O on a rough plane inclined at an angle of 30∘ to the horizontal. The particle P is held at rest at point O before being released. The frictional force acting on P as it slides down the plane is 1130mg.

a

Find, in terms of a, the distance that P moves down the plane before coming to rest.

[5]
b

It is given that P remains at rest in this new position. Find, in terms of m and g, the magnitude of the frictional force in this position.

[3]
3

A ball of mass m kg is projected vertically upwards with initial speed U ms−1 and moves under gravity. At time t s after projection, the ball has travelled a distance x m and its speed is v ms−1. There is a resistive force of magnitude mkv2 N, where k is a positive constant.

a

Show that the distance travelled by the ball when it is moving upwards is x=12kln⁡(g+kU2g+kv2).

[4]
b

It is given that k=0.025 and that U=20. Find the time taken for the ball to reach its maximum height.

[4]
4

An object consists of a uniform lamina with a particle attached. The uniform lamina ABCEFD of mass m is formed from a rectangle ABCD and an isosceles triangle CEF, where F is the midpoint of CD. The rectangle has sides AB=2a and AD=a. The triangle CEF has base a and height 2a. The particle of mass km is attached to the lamina at E. The object rests in a vertical plane with its edge AD on horizontal ground (see diagram). Given that the object is on the point of toppling in its vertical plane about the vertex D, find the value of k.

[4]
Figure 1 of 1 for question 4
5

A hollow cylinder of radius r is fixed with its axis horizontal. Points A, B and O are in the same vertical plane perpendicular to the axis of the cylinder, with A and B on the smooth inner surface and O on the axis. OA and OB make angles 90∘ and α respectively with the upward vertical through O, with A and B on opposite sides of the vertical. A particle of mass m is projected vertically downwards from point A with speed 32rg and moves in a vertical circle inside the cylinder (see diagram). The particle loses contact with the cylinder at point B.

Figure 1 of 2 for question 5
Figure 2 of 2 for question 5
a

Find the value of α.

[4]
b

In the subsequent motion find, in terms of r, the greatest height above O reached by the particle.

[4]
6

Two identical uniform smooth spheres A and B, each with mass m, are moving on a horizontal surface with speeds 2u and u respectively when they collide. Immediately before the collision, the spheres are moving parallel to each other in opposite directions such that their directions of motion each make an angle θ with the line of centres (see diagram). As a result of the collision, B moves in a direction which is perpendicular to its initial direction of motion. The coefficient of restitution between the spheres is e.

a

Find an expression for tan⁡θ in terms of e.

[6]
b

As a result of the collision, A moves in a direction which is perpendicular to the line of centres. Find the value of θ.

[2]
7

A particle P is projected from a point O with speed U at an angle 45∘ above the horizontal and moves freely under gravity.

a

State the vertical and horizontal components of velocity at time t.

[1]
b

At time T, particle P is moving at an angle of 60∘ below the horizontal. Show that T=U2g(2+6).

[3]
c

At time T, the particle strikes a smooth horizontal plane at a point which is a horizontal distance D from O and a vertical distance H below O. Find the ratio H:D.

[4]
d

After striking the horizontal plane, P rebounds with speed w. The coefficient of restitution between P and the plane is 23. Find w in terms of U.

[3]

Mark scheme

9231/33 · May/June 2025

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1 · 3 marks

  • B1

    Tcos⁡θ=mg Tsin⁡θ=mω2r

  • M1

    '53mg' sin⁡θ=mω2asin⁡θ

  • A1

    ω=5g3a

2a · 5 marks

  • M1

    Energy equation with all three terms. Dimensionally correct.

  • B1

    Two terms correct (need not be in equation).

  • A1

    mg2a(x−a)2=mgxsin⁡30−1130mgx

  • M1

    15x2−34ax+15a2=0 (3x−5a)(5x−3a)=0

  • A1

    x=53a

2b · 3 marks

  • B1

    T=23mg

  • M1

    F=T−mgsin⁡30[=23mg−12mg]

  • A1

    16mg

3a · 4 marks

  • B1

    −mg−mkv2=mvdvdx

  • M1

    ∫dx=−∫vg+kv2dv leading to x=−12kln⁡(g+kv2)[+c].

  • M1

    v=U when x=0, so c=12kln⁡(g+kU2).

  • A1

    x=12kln⁡(g+kU2g+kv2) (metres)

3b · 4 marks

  • M1

    10+140v2=−dvdt ∫dt=−∫40400+v2dv

  • A1

    t=−4020tan⁡−1(120v)(+C)

  • 1 mark

    When t=0,v=20, so C=12π t=−2tan⁡−1(0)+12π

  • A1

    t=1.57 (seconds)

4 · 4 marks

  • B1

    Areas and distances from D table: 2a2,12a(2a),2a2+12a(2a), with distances −12a,23a,xˉ

  • M1

    −2a2(12a)+12(2a2)(23a)=(3a2)xˉ

  • 1 mark

    m(19a)=km(2a)

  • A1

    k=118

5a · 4 marks

  • M1

    12mv2+mgrcos⁡α=12m(32gr)[v2=32gr−2grcos⁡α]

  • B1

    mgcos⁡α=mv2r

  • M1

    mgcos⁡α=m(32g−2gcos⁡α)

  • A1

    cos⁡α=12,α=60∘

5b · 4 marks

  • B1

    v2=12gr

  • M1

    0 = (\sin'60^\circ'\sqrt{'v^2'})^2 - 2gh

  • A1

    h=316r

  • A1

    Total height =316r+rcos⁡60=1116r

6a · 6 marks

  • M1

    Let v and w denote the horizontal components of the velocities for A and B respectively after the collision. 2mucos⁡θ−mucos⁡θ=mw+mv

  • M1

    w−v=e(3ucos⁡θ)

  • A1

    w=12(3eucos⁡θ+ucos⁡θ)[=12(3e+1)ucos⁡θ]

  • M1

    12(3e+1)ucos⁡θusin⁡θ=tan⁡θ

  • M1

    tan⁡2θ=12(3e+1)

  • A1

    tan⁡θ=12(3e+1)

6b · 2 marks

  • M1

    w=3eucos⁡θ=ucos⁡θ e=13

  • A1

    tan⁡θ=1,θ=45∘

7a · 1 mark

  • B1

    vy=Usin⁡45−gT vx=Ucos⁡45

7b · 3 marks

  • B1

    vyvx=−tan⁡60∘

  • M1

    [U2−gTU2=−3]U−2gT=−3U

  • A1

    T=U2g(1+3)=2U2g(1+3)=U2g(2+6)

7c · 4 marks

  • B1

    D=U2×T

  • B1

    H=∣U2×T−12gT2∣

  • M1

    ∣U2×T−12gT2∣U2×T=∣122−12(12U(2+6))∣122=∣1−32∣

  • A1

    H:D=1:1+3

7d · 3 marks

  • B1

    wy=±2332U[=±23U]

  • M1

    w2=wy2+(U2)2[=23U2+12U2]

  • A1

    w=U23+12=U76

Questions

  1. 1
    Question 13 marks
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  2. 2
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  3. 3
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  4. 4
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  5. 5
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  6. 6
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  7. 7
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