A Level Further Mathematics 9231 Paper 33, May/June 2023

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Question paper

9231/33 · May/June 2023

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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 P is held at the point A, where OA makes an angle α 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 3ag. It then starts to move along a circular path in a vertical plane. The string goes slack when P is at B, where OB makes an angle θ with the upward vertical. Given that cos⁡α=45, find the value of cos⁡θ.

[4]
2

One end of a light elastic string, of natural length a and modulus of elasticity λmg, is attached to a fixed point O. The string lies on a smooth horizontal surface. A particle P of mass m is attached to the other end of the string. The particle P is projected in the direction OP. When the length of the string is 43a, the speed of P is 2ag. When the length of the string is 53a, the speed of P is 122ag. Find the value of λ.

[4]
3

A uniform lamina is in the form of a triangle ABC, with AC=8a, BC=6a and angle ACB=90∘. The point D on AC is such that AD=3a. The point E on CB is such that CE=x (see diagram). The triangle CDE is removed from the lamina.

Figure 1 of 1 for question 3
a

Find, in terms of a and x, the distance of the centre of mass of the remaining object ADEB from AC.

[4]
b

The object ADEB is on the point of toppling about the point E when the object is in the vertical plane with its edge EB on a smooth horizontal surface. Find x in terms of a.

[3]
4

Two identical smooth uniform spheres A and B each have mass m. The two spheres are moving on a smooth horizontal surface when they collide with speeds u and 2u respectively. Immediately before the collision, A’s direction of motion makes an angle of 30∘ with the line of centres, and B’s direction of motion is perpendicular to the line of centres (see diagram). After the collision, A and B are moving in the same direction. The coefficient of restitution between the spheres is e.

Figure 1 of 1 for question 4
a

Find the value of e.

[5]
b

Find the loss in the total kinetic energy of the spheres as a result of the collision.

[3]
5

One end of a light elastic string, of natural length 12a and modulus of elasticity kmg, is attached to a fixed point O. The other end of the string is attached to a particle of mass m. The particle moves with constant speed 323ag in a horizontal circle with centre at a distance 12a below O. The string is inclined at an angle θ to the downward vertical through O.

a

Find, in terms of a, the extension of the string.

[5]
b

Find the value of k.

[3]
6

A particle of mass m kg falls vertically under gravity, from rest. At time t s, P has fallen x m and has velocity v m s−1. The only forces acting on P are its weight and a resistance of magnitude kmgv N, where k is a constant.

a

Find an expression for v in terms of t, g and k.

[5]
b

Given that k=0.05, find, in metres, how far P has fallen when its speed is 12 m s−1.

[5]
7

The points O and P are on a horizontal plane, a distance 8 m apart. A ball is thrown from O with speed u m s−1 at an angle θ above the horizontal, where tan⁡θ=43. At the same instant, a model aircraft is launched with speed 5 m s−1 parallel to the horizontal plane from a point 4 m vertically above P. The model aircraft moves in the same vertical plane as the ball and in the same horizontal direction as the ball. The model aircraft moves horizontally with a constant speed of 5 m s−1. After T s, the ball and the model aircraft collide.

a

Find the value of T.

[6]
b

Find the direction in which the ball is moving immediately before the collision.

[3]

Mark scheme

9231/33 · May/June 2023

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1 · 4 marks

  • M1

    12m.3ag−12mv2=mga(cos⁡α+cos⁡θ)

  • B1

    mgcos⁡θ=mv2a

  • M1

    32mag−12m.agcos⁡θ=mga(45+cos⁡θ) 32cos⁡θ=710

  • A1

    cos⁡θ=715

2 · 4 marks

  • M1

    12m(v2−v24)=12λmga((23a)2−(13a)2)

  • A1

    12m(2ag−12ag)=12λmga((23a)2−(13a)2)

  • M1

    Solve [34v2=λg(39a)]

  • A1

    λ=92

3a · 4 marks

  • B1

    All correct for ABC and DEC.

  • M1

    Moments [about AC] xˉ(24a2−52xa)=24a2×2a−13x×52ax

  • A1

    All correct moments about AC.

  • A1

    xˉ=288a2−5x23(48a−5x)

3b · 3 marks

  • B1

    On the point of toppling about E: xˉ=x, 288a2−5x23(48a−5x)=x

  • M1

    Rearrange to 3-term quadratic: 10x2−144ax+288a2=0

  • A1

    2(5x−12a)(x−12a)=0, x=125a

4a · 5 marks

  • M1

    VA+VB=ucos⁡30∘(1)

  • M1

    −VA+VB=eucos⁡30∘(2)

  • B1

    Speeds perpendicular to line of centres after collision are usin⁡30∘ and 2u Moving in same direction, so VAusin⁡30∘=VB2u(3)

  • M1

    Use VB=4VA in (1): 5VA=ucos⁡30∘ From (2): 3VA=eucos⁡30∘ then Combine to find equation in e only.

  • A1

    e=35

4b · 3 marks

  • B1

    KE after=12m(VA2+(u2)2)+12m((2u)2+VB2)

  • B1

    KE for A after=750mu2 or KE for B after=5625mu2 or KE loss for A=925mu2 or KE gain for B=625mu2

  • B1

    Total loss in KE=325mu2

5a · 5 marks

  • B1

    ↑Tcos⁡θ=mg

  • B1

    →Tsin⁡θ=[mv2r]=m×27ag4r

  • M1

    r=12atan⁡θ used

  • M1

    Divide: tan⁡θ=274×12tan⁡θ, so tan⁡θ=34

  • A1

    r=9a, extension of string=3a

5b · 3 marks

  • B1

    Hooke’s law: T=kmg(L−12a)12a

  • M1

    Eliminate T:kmg(L−12a)12a=mgL12a

  • A1

    k=LL−12a=5

6a · 5 marks

  • B1

    mdvdt=mg(1−kv)

  • M1

    −1kln⁡(1−kv)=gt+A

  • A1

    Correct, with constant of integration.

  • M1

    t=0,v=0 [A=0]

  • A1

    v=1k(1−e−kgt)

6b · 5 marks

  • M1

    k=0.05 and so dxdt=20(1−e−0.5t)

  • A1

    Integrate: x=20(t+2e−0.5t)+B

  • M1

    t=0,x=0 [B=−40]

  • M1

    When v=12, from part (a), e−0.5t=1−0.05×12=0.4,t=−2ln⁡0.4

  • A1

    x=−40ln⁡0.4+40×0.4−40=12.7

7a · 6 marks

  • B1

    For aircraft, d=5T

  • B1

    For ball, ↑4=usin⁡θ T−12×10×T2

  • B1

    For ball, →ucos⁡θ T=d+8=5T+8

  • M1

    Eliminate u: 4=u45T−12×10×T2,u=5(4+5T2)4Tand u=5(5T+8)3T 3(4+5T2)=4(5T+8)

  • M1

    3T2−4T−4=0

  • A1

    T=2

7b · 3 marks

  • M1

    tan⁡−1(usin⁡θ−10T)ucos⁡θ

  • A1

    tan⁡−189

  • A1

    Direction is 41.6∘ below the horizontal

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