A Level Further Mathematics 9231 Paper 31, May/June 2023

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Question paper

9231/31 · May/June 2023

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1

One end of a light elastic string, of natural length a and modulus of elasticity 3mg, is attached to a fixed point O. The other end of the string is attached to a particle P of mass m. The string hangs with P vertically below O. The particle P is pulled vertically downwards so that the extension of the string is 2a. The particle P is then released from rest.

a

Find the speed of P when it is at a distance 34a below O.

[3]
b

Find the initial acceleration of P when it is released from rest.

[2]
2

A particle P of mass m is moving with speed u on a fixed smooth horizontal surface. It collides at an angle α with a fixed smooth vertical barrier. After the collision, P moves at an angle θ with the barrier, where tan⁡θ=12 (see diagram). The coefficient of restitution between P and the barrier is e. The particle P loses 20% of its kinetic energy as a result of the collision. Find the value of e.

[5]
Figure 1 of 1 for question 2
3

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, where OA makes an angle θ with the downward vertical through O, and with the string taut. The particle P is projected perpendicular to OA in an upwards direction with speed u. It then starts to move along a circular path in a vertical plane. The string goes slack when P is at B, where angle AOB is 90∘ and the speed of P is 45ag.

a

Find the value of sin⁡θ.

[2]
b

Find, in terms of m and g, the tension in the string when P is at A.

[5]
4

An object is formed from a solid hemisphere, of radius 2a, and a solid cylinder, of radius a and height d. The hemisphere and the cylinder are made of the same material. The cylinder is attached to the plane face of the hemisphere. The line OC forms a diameter of the base of the cylinder, where C is the centre of the plane face of the hemisphere and O is common to both circumferences (see diagram). Relative to axes through O, parallel and perpendicular to OC as shown, the centre of mass of the object is (xˉ,yˉ).

Figure 1 of 1 for question 4
a

Show that xˉ=32a2+3ad16a+3d and find an expression, in terms of a and d, for yˉ.

[5]
b

The object is placed on a rough plane which is inclined to the horizontal at an angle θ where sin⁡θ=16. The object is in equilibrium with CO horizontal, where CO lies in a vertical plane through a line of greatest slope. Find d in terms of a.

[3]
5

A light elastic string of natural length a and modulus of elasticity λmg has one end attached to a fixed point O on a smooth horizontal surface. When a particle of mass m is attached to the free end of the string, it moves with speed v in a horizontal circle with centre O and radius x. When, instead, a particle of mass 2m is attached to the free end of the string, this particle moves with speed 12v in a horizontal circle with centre O and radius 34x.

a

Find x in terms of a.

[5]
b

Given that v=12ag, find the value of λ.

[2]
6

A particle P moving in a straight line has displacement x m from a fixed point O on the line and velocity v ms−1 at time t s. The acceleration of P, in ms−2, is given by 6vv+9. When t=0, x=2 and v=72.

a

Find an expression for v in terms of x.

[4]
b

Find an expression for x in terms of t.

[5]
7

At time t s, a particle P is projected with speed 40 ms−1 at an angle θ above the horizontal from a point O on a horizontal plane and moves freely under gravity. The greatest height achieved by P during its flight is H m and the corresponding time is T s.

a

Obtain expressions for H and T in terms of θ.

[2]
b

During the time between t=T and t=3, P descends a distance 14H. Find the value of θ.

[4]
c

Find the speed of P when t=3.

[3]

Mark scheme

9231/31 · May/June 2023

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1a · 3 marks

  • B1

    Correct EPE term seen

  • M1

    Dimensionally correct energy equation. Must have one KE, one EPE term and at least one GPE. Allow sign errors.

  • A1

    v=152ag[2.74ag]

1b · 2 marks

  • M1

    N2L and Hooke’s law

  • A1

    Acceleration = 5g [upwards]

2 · 5 marks

  • M1

    Parallel to wall vcos⁡θ=ucos⁡α Perpendicular to wall vsin⁡θ=eusin⁡α Both

  • A1

    Dividing, e=12tan⁡α

  • M1

    KE reduced by 20%, so 12mu2(cos⁡2α+e2sin⁡2α)=45×12mu2

  • A1

    Eliminate e: cos⁡α=45

  • A1

    e=23

3a · 2 marks

  • M1

    At B, mgsin⁡θ=m4ag5a

  • A1

    sin⁡θ=45

3b · 5 marks

  • B1

    At A, T−mgcos⁡θ=mu2a

  • M1

    Energy equation with 4 terms, dimensionally correct. Mass must be present, allow sign errors. Must see 12 in the KE terms.

  • A1

    Energy 12mu2−12m×4ag5=mga(cos⁡θ+sin⁡θ)

  • M1

    Solve to find T

  • A1

    T=215mg

4a · 5 marks

  • M1

    Moments equation, dimensionally correct, correct number of terms. Allow sign errors.

  • A1

    (23π(2a)3+πa2d)xˉ=163πa3×2a+πa2d×a

  • A1

    Simplify to xˉ=32a2+3ad16a+3d

  • M1

    (23π(2a)3+πa2d)yˉ=163πa3×(−38×2a)+πa2d×12d

  • A1

    yˉ=3(d2−8a2)2(16a+3d)

4b · 3 marks

  • B1

    sin⁡θ=2a−xˉ2a

  • M1

    Remove fractions

  • A1

    d=83a

5a · 5 marks

  • B1

    Hooke’s law, T1=λmga(x−a) or T2=λmga(34x−a)

  • M1

    Also, T1=mv2x and equate mv2x=λmga(x−a)

  • M1

    Similarly: λmga(3x4−a)=2m(12v)234x

  • M1

    Equate expressions for v2 and solve for x in terms of a.

  • A1

    x=4a

5b · 2 marks

  • M1

    λ=axg(x−a)v2 or λ=2a3xg(34x−a)v2 and substitute x=4a,v=12ag

  • A1

    1

6a · 4 marks

  • M1

    vdvdx=6vv+9 and attempt to separate variables and integrate

  • A1

    2v+9=6x+A

  • M1

    Use initial condition to find constant.

  • A1

    v=9(x+1)2−9

6b · 5 marks

  • M1

    Separate variables and write in the form (ax−bx+c)dx=dt

  • A1

    12ln⁡(xx+2)=9t+B

  • M1

    Use initial condition to find constant.

  • M1

    Take logarithms

  • A1

    x=2e18t2−e18t or x=22e−18t−1

7a · 2 marks

  • B1

    H=80sin⁡2θ or 800sin⁡2θg

  • B1

    T=4sin⁡θ or 40sin⁡θg

7b · 4 marks

  • M1

    Between t=T and t=3↑14H=12×10×(3−T)2

  • A1

    Correct equation with no extra terms.

  • M1

    Substitute their expressions for H and T from part (a) and obtain a quadratic equation in sin⁡θ with no more than three terms.

  • A1

    sin⁡θ=12,θ=30∘

7c · 3 marks

  • B1

    When t=3 speeds →40cos⁡θ and ↑40sin⁡θ−10×3

  • M1

    Square and add to find square of speed: v2=(203)2+(−10)2

  • A1

    v2=1300,v=1013 [= 36.1]

Questions

  1. 1
    Question 15 marks
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  2. 2
    Question 25 marks
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  3. 3
    Question 37 marks
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  4. 4
    Question 48 marks
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  5. 5
    Question 57 marks
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  6. 6
    Question 69 marks
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  7. 7
    Question 79 marks
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