A Level Mathematics 9709 Paper 43, May/June 2023

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Question paper

9709/43 · May/June 2023

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1

Two particles P and Q, of masses 0.1 kg and 0.4 kg respectively, are free to move on a smooth horizontal plane. Particle P is projected with speed 4 m s−1 towards Q which is stationary. After P and Q collide, the speeds of P and Q are equal. Find the two possible values of the speed of P after the collision. [3]

[3]
2

A car of mass 1500 kg is towing a trailer of mass m kg along a straight horizontal road. The car and the trailer are connected by a tow-bar which is horizontal, light and rigid. There is a resistance force of F N on the car and a resistance force of 200 N on the trailer. The driving force of the car’s engine is 3200 N, the acceleration of the car is 1.25 m s−2 and the tension in the tow-bar is 300 N. Find the value of m and the value of F. [4]

[4]
3

A smooth ring R of mass 0.2 kg is threaded on a light string ARB. The ends of the string are attached to fixed points A and B with A vertically above B. The string is taut and angle ABR=90∘. The angle between the part AR of the string and the vertical is 60∘. The ring is held in equilibrium by a force of magnitude X N, acting on the ring in a direction perpendicular to AR (see diagram). Calculate the tension in the string and the value of X. [5]

[5]
Figure 1 of 1 for question 3
4

A lorry of mass 15 000 kg moves on a straight horizontal road in the direction from A to B. It passes A and B with speeds 20 m s−1 and 25 m s−1 respectively. The power of the lorry’s engine is constant and there is a constant resistance to motion of magnitude 6000 N. The acceleration of the lorry at B is 0.5 times the acceleration of the lorry at A.

a

Show that the power of the lorry’s engine is 200 kW, and hence find the acceleration of the lorry when it is travelling at 20 m s−1. [5]

[5]
b

The lorry begins to ascend a straight hill inclined at 1∘ to the horizontal. It is given that the power of the lorry’s engine and the resistance force do not change. Find the steady speed up the hill that the lorry could maintain. [2]

[2]
5

A particle starts from rest from a point O and moves in a straight line. The acceleration of the particle at time t s after leaving O is a m s−2, where a=kt12 for 0⩽t⩽9 and where k is a constant. The velocity of the particle at t=9 is 1.8 m s−1.

a

Show that k=0.1. [3]

[3]
b

For t>9, the velocity v m s−1 of the particle is given by v=0.2(t−9)2+1.8. Show that the distance travelled in the first 9 seconds is one tenth of the distance travelled between t=9 and t=18. [4]

[4]
c

Find the greatest acceleration of the particle during the first 10 seconds of its motion. [3]

[3]
6

An elevator is pulled vertically upwards by a cable. The elevator accelerates at 0.4 m s−2 for 5 s, then travels at constant speed for 25 s. The elevator then decelerates at 0.2 m s−2 until it comes to rest.

a

Find the greatest speed of the elevator and hence draw a velocity-time graph for the motion of the elevator. [3]

[3]
b

Find the total distance travelled by the elevator. [2]

[2]
c

The mass of the elevator is 1200 kg and there is a crate of mass m kg resting on the floor of the elevator. Given that the tension in the cable when the elevator is decelerating is 12 250 N, find the value of m. [3]

[3]
d

Find the greatest magnitude of the force exerted on the crate by the floor of the elevator, and state its direction. [3]

[3]
7

The diagram shows the vertical cross-section XYZ of a rough slide. The section YZ is a straight line of length 2 m inclined at an angle of α to the horizontal, where sin⁡α=0.28. The section YZ is tangential to the curved section XY at Y, and X is 1.8 m above the level of Y. A child of mass 25 kg slides down the slide, starting from rest at X. The work done by the child against the resistance force in moving from X to Y is 50 J.

Figure 1 of 1 for question 7
a

Find the speed of the child at Y. [4]

[4]
b

It is given that the child comes to rest at Z. Use an energy method to find the coefficient of friction between the child and YZ, giving your answer as a fraction in its simplest form. [6]

[6]

Mark scheme

9709/43 · May/June 2023

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1 · 3 marks

  • M1

    For attempt at use of conservation of momentum in one case

  • A1

    Speed = 0.8[ms−1] or 45

  • A1

    Speed = 43 [ms−1] Allow 1.33

2 · 4 marks

  • M1

    Attempt to use Newton's second law

  • A1

    Trailer 300−200=m×1.25 or Car 3200−F−300=1500×1.25 System 3200−F−200=(1500+m)×1.25

  • M1

    Solve for m or F

  • A1

    m=80 and F=1025

3 · 5 marks

  • M1

    For attempt to resolve in one direction

  • A1

    Xsin⁡60+Tsin⁡30−0.2g=0

  • A1

    Xcos⁡60−T−Tcos⁡30=0

  • M1

    For attempt to solve for tension or X

  • A1

    X=2, tension in string = 0.536[N]

4a · 5 marks

  • B1

    For use of P=Fv

  • M1

    Attempt to use Newton's second law in at least one case

  • A1

    P20−6000=15000a and P25−6000=15000(12a)

  • M1

    For solving simultaneously

  • A1

    Power [= 200 000W] = 200 kW, a=415 [ms−2]

4b · 2 marks

  • M1

    For attempt at resolving up hill 200000v−6000−15000gsin⁡1=0

  • A1

    Steady speed = 23.2[m s−1]

5a · 3 marks

  • M1

    For attempt at integration

  • A1

    v=23kt32[+c]

  • B1

    1.8=23k×932⇒k=[32×1.8÷27=]0.1

5b · 4 marks

  • M1

    For attempt at integration of either ∫(23kt32)dt or ∫(0.2(t−9)2+1.8)dt or ∫(0.2t2−3.6t+18)dt

  • A1

    [4150t52]09 and [(0.23(t−9)3+1.8t)]918 or [0.23t3−3.62t2+18t]918

  • M1

    =4150×952[=6.48 or 16225] or =0.23(18−9)3+1.8×18−1.8×9=[64.8 or 3245] or (0.23×183−3.62×182+18×18)−(0.23×93−3.62×92+18×9) [=6485−3245=3245]

  • A1

    6.48=110×64.8 or 64.8=10×6.48 or 3245=10×16225

5c · 3 marks

  • M1

    For differentiation Should get a=0.4(t−9) or a=0.4t−3.6

  • A1

    0.4[ms−2] [at t=10]

  • B1

    0.3 seen (from the first phase) and state that 0.4 is final answer

6a · 3 marks

  • B1

    Greatest speed = 2[m s−1] [0.4×5]

  • B1

    Trapezium shape

  • B1

    [Graph of v against t showing trapezium sitting on t-axis, starting at origin, with height 2 and t-values 5, 30, 40]

6b · 2 marks

  • M1

    Distance = 12(25+5+25+their 10)×their 2 or 12×5×their 2+25×their 2+12×their 10×their 2

  • A1

    Distance = 65[m]

6c · 3 marks

  • M1

    Attempt at Newton's second law

  • A1

    12250−1200g−mg=(1200+m)×(−0.2) Or 1200g+mg−12250=(1200+m)×0.2

  • A1

    m=50

6d · 3 marks

  • M1

    Realise that this is when accelerating and attempt Newton's second law for the crate only

  • 1 mark

    R−50g=50×0.4 or 50g−R=50×(−0.4)

  • A1

    Force R=520[N], upwards

7a · 4 marks

  • B1

    PE lost = mgh=25g×1.8[=450]

  • M1

    For work energy equation

  • A1

    25g×1.8−50=12×25v2

  • A1

    v=42 [ms−1] or 5.66 [5.6568...]

7b · 6 marks

  • 1 mark

    PE gained/lost = ±25g×2×0.28[=±140] or KE gained/lost = ±12×25(their 42)2 [KE = ±400]

  • M1

    For work energy equation

  • 1 mark

    F×2=25g×2×0.28+12×25(42)2 [⇒F=270]

  • B1

    R=25g×0.96 [= 240]

  • M1

    Use of F=μR to form an equation in μ only

  • A1

    μ=98

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