A Level Further Mathematics 9231 Paper 31, Oct/Nov 2024

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Question paper

9231/31 · Oct/Nov 2024

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1

A particle P is projected with speed u m s−1 at an angle tan⁡−12 above the horizontal from a point O on a horizontal plane and moves freely under gravity. When P has travelled a distance 56 m horizontally from O, it is at a vertical height H m above the plane. When P has travelled a distance 84 m horizontally from O, it is at a vertical height 12H m above the plane. Find, in either order, the value of u and the value of H.

[5]
2

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 P is held at the point A with the string taut. It is given that OA makes an angle θ with the downward vertical through O, where tan⁡θ=34. The particle P is projected perpendicular to OA in an upwards direction with speed 5ag, and it starts to move along a circular path in a vertical plane. When P is at the point B, where angle AOB is a right angle, the tension in the string is T. Find T in terms of m and g.

[5]
3

A particle P of mass m kg is attached to one end of a light elastic string of natural length 2 m and modulus of elasticity 2mg N. The other end of the string is attached to a fixed point O. The particle P hangs in equilibrium vertically below O. The particle P is pulled down vertically a distance d m below its equilibrium position and released from rest.

a

Given that the particle just reaches O in the subsequent motion, find the value of d.

[6]
b

Hence find the speed of P when it is 2 m below O.

[2]
4

An object is formed by removing a cylinder of radius 23a and height kh (k<1) from a uniform solid cylinder of radius a and height h. The vertical axes of symmetry of the two cylinders coincide. The upper faces of the two cylinders are in the same plane as each other. The points A and B are the opposite ends of a diameter of the upper face of the object (see diagram).

Figure 1 of 2 for question 4
Figure 2 of 2 for question 4
a

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

[4]
b

When the object is suspended from A, the angle between AB and the vertical is θ, where tan⁡θ=32. Given that h=83a, find the possible values of k.

[3]
5

A particle P of mass 2 kg moving on a horizontal straight line has displacement x m from a fixed point O on the line and velocity v m s−1 at time t s. The only horizontal force acting on P is a variable force F N which can be expressed as a function of t. It is given that vx=3−t1+t and when t=0, x=5.

a

Find an expression for x in terms of t.

[4]
b

Find the magnitude of F when t=3.

[3]
6

A particle P of mass 0.05 kg is attached to one end of a light inextensible string of length 1 m. The other end of the string is attached to a fixed point O. A particle Q of mass 0.04 kg is attached to one end of a second light inextensible string. The other end of this string is attached to P. The particle P moves in a horizontal circle of radius 0.8 m with angular speed ω rad s−1. The particle Q moves in a horizontal circle of radius 1.4 m also with angular speed ω rad s−1. The centres of the circles are vertically below O, and O, P and Q are always in the same vertical plane. The strings OP and PQ remain at constant angles α and β respectively to the vertical (see diagram).

Figure 1 of 1 for question 6
a

Find the tension in the string OP.

[3]
b

Find the value of ω.

[3]
c

Find the value of β.

[2]
7

A particle P is projected with speed u at an angle tan⁡−1(43) above the horizontal from a point O on a horizontal plane and moves freely under gravity. When P is moving horizontally, it strikes a smooth inclined plane at the point A. This plane is inclined to the horizontal at an angle α, and the line of greatest slope through A lies in the vertical plane through O and A. As a result of the impact, P moves vertically upwards. The coefficient of restitution between P and the inclined plane is e.

a

Show that etan⁡2α=1.

[4]
b

In its subsequent motion, the greatest height reached by P above A is 316 of the vertical height of A above the horizontal plane. Find the value of e.

[6]

Mark scheme

9231/31 · Oct/Nov 2024

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1 · 5 marks

  • M1

    Use equation of trajectory with point (56,H) or (84,12H)

  • A1

    H=112−5g2u2×562 or 12H=168−5g2u2×842

  • M1

    Eliminate to find u or H

  • A1

    u=35

  • A1

    H=48

2 · 5 marks

  • B1

    At B, T+mgsin⁡θ=mv2a

  • M1

    Energy A to B: 12mu2−12mv2=mga(cos⁡θ+sin⁡θ)

  • A1

    12mu2−12mv2=mga(cos⁡θ+sin⁡θ)

  • M1

    Substitute for u and θ to find T: T=mg(5−2×45−3×35)

  • A1

    T=85mg

3a · 6 marks

  • M1

    Hooke’s law: T=2mg2×extension and T=mg

  • A1

    Extension =1 m

  • B1

    EPE loss =12×2mg2×(1+d)2

  • B1

    Gain in GPE =mg(2+1+d)

  • M1

    Equate: 12mg(1+d)2=mg(3+d)

  • A1

    d=5

3b · 2 marks

  • M1

    Energy equation: 12mV2+mg(1+d)=12×2mg2×(1+d)2 Alternatively: Using KE and GPE from 2 m below O to point O 12mV2=2mg

  • A1

    V2=g(d2−1)V=40=210 Alternatively: V2=4gV=40=210

4a · 4 marks

  • B1

    Correct volumes and distances for large and small.

  • M1

    Moments about AB: πa2h(1−49k)yˉ=πa2h×12h−π(23a)2kh×12kh

  • A1

    Correct, unsimplified.

  • A1

    yˉ=(9−4k2)h2(9−4k)

4b · 3 marks

  • B1

    tan⁡θ=yˉa:(9−4k2)h2(9−4k)a=32

  • M1

    Use h=83a and simplify to quadratic in k: 32k2−36k+9=0

  • A1

    k=38,34

5a · 4 marks

  • M1

    dxx=(4t+1−1)dt

  • A1

    ln⁡∣x∣=4ln⁡∣t+1∣−t+A

  • M1

    t=0,x=5:A=ln⁡5

  • A1

    x=5(t+1)4e−t

5b · 3 marks

  • M1

    v=(3−t)×5(t+1)3e−t Acceleration =dvdt=5e−t(−(t+1)3+(3−t)3(t+1)2−(3−t)(t+1)3) Acceleration =5e−t(t+1)2(5−t)(1−t)

  • M1

    F=2×acceleration, so at F=10e−t(t+1)2(5−t)(1−t)

  • A1

    At t=3, magnitude of force is 640e−3 N

6a · 3 marks

  • B1

    At P: ↑T1cos⁡α=T2cos⁡β+0.05g

  • B1

    At Q: ↑T2cos⁡β=0.04g

  • B1

    T1=0.15g=1.5 N

6b · 3 marks

  • M1

    T1sin⁡α−T2sin⁡β=0.05×0.8ω2

  • M1

    T2sin⁡β=0.04×1.4ω2

  • A1

    T1sin⁡α=0.05×0.8ω2+0.04×1.4ω2 ω2=12.5,ω=522

6c · 2 marks

  • M1

    T2cos⁡β=0.04g and T2sin⁡β=0.04×1.4ω2 Divide: tan⁡β=74

  • A1

    β=60.3∘

7a · 4 marks

  • M1

    When P strikes plane, velocity is →ucos⁡θ, Before impact: parallel to inclined plane 35ucos⁡α, perpendicular to plane 35usin⁡α

  • A1

    After impact: components 35ucos⁡α (parallel) and 35eusin⁡α (perpendicular)

  • M1

    Since velocity is vertical after impact, tan⁡α=35ucos⁡α/35eusin⁡α

  • A1

    tan⁡α=1/etan⁡α,etan⁡2α=1

7b · 6 marks

  • M1

    Greatest height of P before impact: H=(usin⁡θ)22g=8u225g

  • A1

    H=8u225g

  • M1

    After impact, vertical speed of P is 35u(cos⁡α)2+e2(sin⁡α)2

  • M1

    Use V2=U2+2as to greatest height, equal to 316H 925u2((cos⁡α)2+e2(sin⁡α)2)=2g×316H

  • M1

    Use part (a): tan⁡α=1e, cos⁡α=e1+e, sin⁡α=11+e Substitute to find e

  • A1

    3e2+2e−1=0,e=13

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