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Learn/ A Level Mathematics/ Past Papers/ 2022/ Feb/March 2022 /Paper 4 (Variant 2)

Mechanics — Variant 2

A Level Mathematics · 9709/42 · Feb/March 2022

Question Paper

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Mark Scheme

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Questions

5 of 7 questions shown here, checked against the paper. The others are in the PDF above — they contain diagrams or maths we could not yet verify, and we would rather show you the original than a transcription we are not sure of.

1
A crane is used to raise a block of mass 600600 kg vertically upwards at a constant speed through a height of 1515 m. There is a resistance to the motion of the block, which the crane does 1000010000 J of work to overcome.
1(a) [2]
Find the total work done by the crane.
1(b) [2]
Given that the average power exerted by the crane is 12.512.5 kW, find the total time for which the block is in motion.
2
A particle PP is projected vertically upwards from horizontal ground with speed uu m s−1\text{m s}^{-1}. PP reaches a maximum height of 2020 m above the ground.
2(a) [2]
Find the value of uu.
2(b) [3]
Find the total time for which PP is at least 1515 m above the ground.
4
The total mass of a cyclist and her bicycle is 7070 kg. The cyclist is riding with constant power of 180180 W up a straight hill inclined at an angle α\alpha to the horizontal, where sin⁡α=0.05\sin \alpha = 0.05. At an instant when the cyclist's speed is 66 m s−1\text{m s}^{-1}, her acceleration is −0.2-0.2 m s−2\text{m s}^{-2}. There is a constant resistance to motion of magnitude FF N.
4(a) [4]
Find the value of FF.
4(b) [2]
Find the steady speed that the cyclist could maintain up the hill when working at this power.
5
Four coplanar forces act at a point. The magnitudes of the forces are 1010 N, FF N, GG N and 2F2F N. The directions of the forces are as shown in the diagram.
5(a) [5]
Given that the forces are in equilibrium, find the values of FF and GG.
5(b) [2]
Given instead that F=3F = 3, find the value of GG for which the resultant of the forces is perpendicular to the 1010 N force.
6
A cyclist starts from rest at a fixed point OO and moves in a straight line, before coming to rest kk seconds later. The acceleration of the cyclist at time tt s after leaving OO is aa m s−2\text{m s}^{-2}, where a=2t−12−35t12a = 2t^{-\frac{1}{2}} - \frac{3}{5}t^{\frac{1}{2}} for 0<t⩽k0 < t \leqslant k.
6(a) [4]
Find the value of kk.
6(b) [3]
Find the maximum speed of the cyclist.
6(c) [4]
Find an expression for the displacement from OO in terms of tt. Hence find the total distance travelled by the cyclist from the time at which she reaches her maximum speed until she comes to rest.

Published by Cambridge Assessment International Education. Reproduced for study use.