A-LevelMathematicsMechanics9709/43 · Oct/Nov 2019 · Q713 Marks

Two particles A and B have masses m kg and km kg respectively, where k > 1. The particles are attached to the ends of a light inextensible string. The string passes over a fixed smooth pulley and the particles hang vertically below it. Both particles are at a height of 0.81 m above horizontal ground (see diagram). The system is released from rest and particle B reaches the ground 0.9 s later. The particle A does not reach the pulley in its subsequent motion. [Figure 7.1]

(i)Find the value of k and show that the tension in the string before B reaches the ground is equal to 12m N.[7]
(ii)At the instant when B reaches the ground, the string breaks. Show that the speed of A when it reaches the ground is 5.97 m s⁻¹, correct to 3 significant figures, and find the time taken, after the string breaks, for A to reach the ground.[4]
(iii)Sketch a velocity-time graph for the motion of particle A from the instant when the system is released until A reaches the ground.[2]

This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2019). The mark or mark type is shown in brackets.

[M1] For use of s = ut + (1/2)at² (0.81 = 0 + (1/2)a × 0.9²)
[A1] a = 2
[M1] Use of Newton's Second Law for A or B or use of a = (m_B - m_A)g / (m_B + m_A) (T-mg = ma or kmg - T= kma)
[A1] T-mg = ma and kmg - T = kma or a = (km-m)g / (km+m)
[M1] Solves to find k (a = (kg-g)/(k+1) = 2 → k = ...)
[A1] k = 1.5
[B1] AG (T = 10m + 2m = 12m N)
[B1FT] For use of v=u+at ft a from (i) (Velocity of A when string breaks = 2 × 0.9 = 1.8 ms⁻¹ upwards)
[M1] For use of suvat to find v_A at ground (v²=1.8²+2g×1.62 → v=...)
[A1] AG (Speed is 5.97 ms⁻¹)
[B1] Time taken = (1.8+5.97)/g = 0.777s (0.7769...)
[B1] Straight line from (0, 0) to (0.9, 1.8)
[B1FT] Straight line from (0.9, 1.8) to approx. (1.7, −6)

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