Two particles A and B have masses 0.35 kg and 0.45 kg respectively. The particles are attached to the ends of a light inextensible string which passes over a small fixed smooth pulley which is 1 m above horizontal ground. Initially particle A is held at rest on the ground vertically below the pulley, with the string taut. Particle B hangs vertically below the pulley at a height of 0.64 m above the ground. Particle A is released.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2017). The mark or mark type is shown in brackets.

1. EITHER: (M1 Applies Newton's Second Law to one of the particles or forms system equation in a (mg - mg = (m + mb)a)
2. [T-0.35g = 0.35a or 0.45g - T = 0.45a or 0.45g -0.35g = 0.8a]
3. M1 Applies Newton's Second Law to form second equation in T and a and solves for a or solves system equation for a
4. [0.45g - T = 0.45a or T-0.35g = 0.35a] →a = ...
5. a = 1.25 m s⁻² A1
6. [v² = 2 × 1.25 × 0.64] (= 1.6) M1 Using v² = u² + 2as
7. Velocity = 1.26 ms⁻¹ A1)
8. OR: (M1 Attempts PE loss
9. [PE loss = 0.45g × 0.64 – 0.35g × 0.64]
10. M1 Attempts KE gain
11. [KE gain = ½ (0.35 + 0.45) v²]
12. A1 PE loss = 0.45g × 0.64 – 0.35g × 0.64 and KE gain = ½ (0.35 + 0.45) v²
13. [½ (0.8) v² = 0.1g × 0.64]] (v² = 1.6) M1 Using PE loss = KE gain
14. Velocity = 1.26 ms⁻¹ A1)
15. EITHER: (M1 Using v² = u² + 2as
16. [0 = 1.6-2 gs] (s = 0.08)
17. A1) Distance = 0.16 m
18. OR: (M1 Using PE gain = KE loss for particle A
19. [0.35gh = ½ (0.35) × 1.6] (h = 0.08)
20. A1) Distance = 0.16 m