A-LevelMathematicsMechanics9709/43 · Oct/Nov 2018 · Q59 Marks

Two particles P and Q, of masses 0.3 kg and 0.5 kg respectively, are attached to the ends of a light inextensible string. The string passes over a fixed smooth pulley with the particles hanging freely below it. Q is held at rest with the string taut at a height of h m above a horizontal floor (see diagram). Q is now released and both particles start to move. The pulley is sufficiently high so that P does not reach it at any stage. The time taken for Q to reach the floor is 0.6 s.

(i)Find the acceleration of Q before it reaches the floor and hence find the value of h.[6]
(ii)Q remains at rest when it reaches the floor, and P continues to move upwards. Find the velocity of P at the instant when Q reaches the floor and the total time taken from the instant at which Q is released until the string becomes taut again.[3]

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Mark scheme

Marking points from the official mark scheme (Oct/Nov 2018). The mark or mark type is shown in brackets.

[M1] Use of Newton's second law for P or Q or use of a = (mQ - mP)g / (mQ + mP) [T−0.3g = 0.3a or 0.5g – T = 0.5a]
[A1] T−0.3g = 0.3a and 0.5g – T = 0.5a or a = (0.5g – 0.3g) / (0.5 + 0.3)
[M1] Solve for a [0.5g – 0.3g = 0.8a]
[A1] a = 2.5
[M1] For use of s = ut + ½at² [h = 0 + ½ × 2.5 × 0.6²]
[A1] h = 0.45
[B1ft] ft a from (i) × 0.6 Velocity of P when Q reaches floor = 0 + 0.6 × 2.5 = 1.5 m s¯¹
[M1] Use of suvat to find time to highest point [0 = 1.5 – gt → t = ...] (t = 0.15)
[A1] Total time = 2 × 0.15 + 0.6 = 0.9 s

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