A-LevelMathematicsMechanics9709/41 · Oct/Nov 2019 · Q69 Marks

A particle of mass 0.4 kg is released from rest at a height of 1.8 m above the surface of the water in a tank. There is no instantaneous change of speed when the particle enters the water. The water exerts an upward force of 5.6 N on the particle when it is in the water.

(i)Find the velocity of the particle at the instant when it reaches the surface of the water.[2]
(ii)Find the time that it takes from the instant when the particle enters the water until it comes to instantaneous rest in the water. You may assume that the tank is deep enough so that the particle does not reach the bottom of the tank.[4]
(iii)Sketch a velocity-time graph for the motion of the particle from the instant at which it is released until it comes to instantaneous rest in the water.[3]

Mark scheme

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

[M1] KE gain = PE lost (½ × 0.4 × v² = 0.4g × 1.8)
[A1] v = 6 m s⁻¹
[M1] Use Newton's second law for the particle in the vertical (3 terms) (0.4g - 5.6 = 0.4a)
[A1] a = -4 m s⁻²
[M1] Use of constant acceleration equation(s) such as v = u + at to find t (0 = 6 - 4t)
[A1] t = 1.5 s
[B1] Straight line starting at (0,0) with positive gradient
[B1] Second straight line starting at end of the first line with negative gradient and ending with v = 0
[B1FT] All correct, start at (0, 0) with max velocity v = 6 at t = 0.6 i.e. (0.6, 6) and finishing at (2.1, 0) (FT on their v from (i) and/or their t from (ii))

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