A particle of mass 0.3 kg is released from rest above a tank containing water. The particle falls vertically, taking 0.8 s to reach the water surface. There is no instantaneous change of speed when the particle enters the water. The depth of water in the tank is 1.25 m. The water exerts a force on the particle resisting its motion. The work done against this resistance force from the instant that the particle enters the water until it reaches the bottom of the tank is 1.2 J.

Mark scheme

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

1. [B1] At liquid surface, speed = 0 + g × 0.8 [= 8] Using constant acceleration equation v = u + at or PE loss = KE gain
2. [B1] PE lost in water = 0.3g × 1.25 [ = 3.75]
3. [M1] [½ × 0.3 × (8² – v²) + 0.3g × 1.25 = 1.2] Using work-energy for downward motion in the tank PE loss + KE loss = Work done against resistance
4. [A1] v = 9 m s⁻¹
5. [B1] Height above tank = ½ × g × 0.8² [= 3.2]
6. [B1] Total PE loss = 0.3g × (3.2 + 1.25) [= 13.35]
7. [M1] [0.3g × (3.2 + 1.25) = ½ × 0.3 × v² + 1.2] Work-energy equation for the total downward motion
8. [A1] v = 9 m s⁻¹
9. [M1] [-0.3g - 1.8 = 0.3a] Using Newton's 2nd law for the upward motion in the tank
10. [A1] a = -16
11. [M1] [1.25 = 7T + ½ × (−16) × T²] Using constant acceleration equations to find the time, T, for the particle to travel from the bottom to the surface of the liquid
12. [A1] T = 0.25 (or 0.625, on the way down)
13. [B1] [v at surface = 7 + (−16) × 0.25 = 3] Using v = u + aT or equivalent to find v at surface
14. [M1] [0 = 3 - gt → t = 0.3] Attempt to find the time, t, taken for the particle to travel from the surface to reach maximum height using their v ≠ 7
15. [A1] Total time = T + t = 0.55 s
16. [B1] v² = 7² + 2 × (−16) × 1.25 = 9 → v = 3 Using constant acceleration equations to find v at the surface
17. [M1] [1.25 = ½ (7 + 3) × T or 3 = 7 + (-16) × T] Using s = ½ (u + v) × T or v = u + aT to find the time, T, for the particle to travel from the bottom to the surface of the liquid
18. [M1] [½ x 0.3 × (7² – v²) = 0.3g × 1.25 + 1.8 × 1.25] Work-energy equation for motion from bottom to surface
19. [B1] v = 3 Find v at surface from rearrangement of work-energy