A Level Further Mathematics 9231 Paper 33, May/June 2024

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Question paper

9231/33 · May/June 2024

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1

Two smooth uniform spheres A and B of equal radii have masses m and 2m respectively. The two spheres are moving on a smooth horizontal surface when they collide with speeds u and 12u respectively. Immediately before the collision, A's direction of motion is along the line of centres, and B's direction of motion makes an angle θ with the line of centres (see diagram). As a result of the collision, the direction of motion of A is reversed and its speed is reduced to 14u. The direction of motion of B again makes an angle θ with the line of centres, but on the opposite side of the line of centres. The speed of B is unchanged. Find the value of the coefficient of restitution between the spheres.

[4]
Figure 1 of 2 for question 1
Figure 2 of 2 for question 1
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 2mg. A particle Q of mass km is attached to the other end of the string. Particle P lies on a smooth horizontal table. The string has part of its length in contact with the table and then passes through a small smooth hole H in the table. Particle P moves in a horizontal circle on the surface of the table with constant speed 12ga. Particle Q hangs in equilibrium vertically below the hole with HQ=14a.

a

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

[4]
b

Find the value of k.

[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. When the particle is hanging vertically below O, it is projected horizontally with speed u so that it begins to move along a circular path. When P is at the lowest point of its motion, the tension in the string is T. When OP makes an angle θ with the upward vertical, the tension in the string is S.

a

Show that S=T−3mg(1+cos⁡θ).

[5]
b

Given that u=4ag, find the value of cos⁡θ when the string goes slack.

[2]
4

A light spring of natural length a and modulus of elasticity kmg is attached to a fixed point O on a smooth plane inclined to the horizontal at an angle θ, where sin⁡θ=34. A particle of mass m is attached to the lower end of the spring and is held at the point A on the plane, where OA=2a and OA is along a line of greatest slope of the plane (see diagram). The particle is released from rest and is moving with speed V when it passes through the point B on the plane, where OB=32a. The speed of the particle is 12V when it passes through the point C on the plane, where OC=34a. Find the value of k.

[7]
Figure 1 of 1 for question 4
5

A uniform lamina is in the form of a triangle OBC, with OC=18a, OB=24a and angle COB=90∘. The point A on OB is such that OA=x (see diagram). The triangle OAC is removed from the lamina.

Figure 1 of 1 for question 5
a

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

[3]
b

The object ABC is suspended from C. In its equilibrium position, the side AB makes an angle θ with the vertical, where tan⁡θ=65. Find x in terms of a.

[4]
6

A particle P is projected with speed u ms−1 at an angle θ above the horizontal from a point O and moves freely under gravity. After 5 seconds the speed of P is 34u.

a

Show that 716u2−100usin⁡θ+2500=0.

[3]
b

It is given that the velocity of P after 5 seconds is perpendicular to the initial velocity. Find, in either order, the value of u and the value of sin⁡θ.

[5]
7

A parachutist of mass m kg opens his parachute when he is moving vertically downwards with a speed of 50 ms−1. At time t s after opening his parachute, he has fallen a distance x m from the point where he opened his parachute, and his speed is v ms−1. The forces acting on him are his weight and a resistive force of magnitude mv N.

a

Find an expression for v in terms of t.

[6]
b

Find an expression for x in terms of t.

[3]
c

Find the distance that the parachutist has fallen, since opening his parachute, when his speed is 15 ms−1.

[2]

Mark scheme

9231/33 · May/June 2024

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1 · 4 marks

  • M1

    Along line of centres, PCLM: −2m×12ucos⁡θ+mu=2m×12ucos⁡θ−m×14u

  • A1

    cos⁡θ=58

  • M1

    NEL: 12ucos⁡θ+14u=e(12ucos⁡θ+u)

  • A1

    e=37

2a · 4 marks

  • B1

    Hooke’s law: T=2mgxa

  • B1

    →T=mga2a−a4+x

  • M1

    Equate: 2mgxa=mga23a4+x, 4x2+3ax−a2=0

  • A1

    x=a4

2b · 2 marks

  • B1

    ↑T=kmg

  • B1

    T=2mgxa, k=12

3a · 5 marks

  • B1

    At lowest point, T−mg=mu2a

  • B1

    When string makes angle θ with upward vertical, S+mgcos⁡θ=mv2a

  • M1

    Energy: 12mu2−12mv2=mga(1+cos⁡θ)

  • M1

    Eliminate u2 and v2

  • A1

    S=T−3mg(1+cos⁡θ)

3b · 2 marks

  • M1

    When string goes slack, S=0 so T=3mg(1+cos⁡θ)

  • A1

    But T=mg+mu2a=mg+4mg=5mg, so cos⁡θ=23

4 · 7 marks

  • B1

    A to B: Loss in EPE =kmg2a×(a2−(12a)2)=38kmga

  • M1

    Energy: 12mV2+mga2sin⁡θ=38kmga

  • A1

    (V2=34ga(k−1))

  • B1

    A to C: Loss in EPE =kmg2a×(a2−(−a4)2)=1532kmga

  • M1

    Energy: 12m(12V)2+mg5a4sin⁡θ=1532kmga(V2=154ag(k−2))

  • M1

    Eliminate V2 from two energy equations to obtain expression involving only k, a and possibly sin⁡θ

  • A1

    k=94

5a · 3 marks

  • M1

    Moments about OC: (216a2−9ax)xˉ=216a2×8a−9ax×13x

  • A1

    All correct in moments equation

  • A1

    xˉ=576a2−x272a−3x or x+24a3

5b · 4 marks

  • M1

    tan⁡θ=18a−6axˉ or 18a−yˉxˉ or yˉ15a−xˉ or 18a12a+12x

  • A1

    Correct expression for tan⁡θ

  • M1

    x2−30ax+144a2=0 or, with simplified form, 30a=x+24a

  • A1

    x=6a

6a · 3 marks

  • B1

    Components of velocity are ucos⁡θ, usin⁡θ−gt

  • M1

    (ucos⁡θ)2+(usin⁡θ−gt)2=(34u)2

  • A1

    716u2−100usin⁡θ+2500=0

6b · 5 marks

  • B1

    Let α be angle of direction of motion with horizontal at t=5, then (tan⁡α=)usin⁡θ−5gucos⁡θ

  • M1

    tan⁡αtan⁡θ=−1, so tan⁡θ(usin⁡θ−5gucos⁡θ)=−1

  • A1

    u=50sin⁡θ

  • M1

    Use in result from part (a) to form equation in u or sin⁡θ

  • A1

    u2=1600, u=40 and sin⁡θ=45

7a · 6 marks

  • B1

    mdvdt=m(10−v)

  • 1 mark

    Separate variables and integrate to obtain a ln term

  • A1

    −ln⁡∣10−v∣=t+A or −ln⁡(v−10)=t+A

  • 1 mark

    Use t=0,v=50:A=−ln⁡∣−40∣

  • M1

    0.1vet=4+et

  • A1

    v=10+40e−t

7b · 3 marks

  • M1

    x=10t−40e−t+B

  • M1

    Use t=0,x=0:B=40

  • A1

    x=10t−40e−t+40

7c · 2 marks

  • M1

    When v=15,e−t=18,t=2.08 or ln⁡8

  • A1

    x=55.8 (metres)

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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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