Two particles A and B of masses m kg and 4 kg respectively are connected by a light inextensible string that passes over a fixed smooth pulley. Particle A is on a rough fixed slope which is at an angle of 30° to the horizontal ground. Particle B hangs vertically below the pulley and is 0.5 m above the ground (see diagram). The coefficient of friction between the slope and particle A is 0.2. [Figure X.X]

Mark scheme

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

1. B1 Resolves normally
2. M1 Uses F = μR
3. B1 Particle B
4. M1 Resolves parallel to plane for particle A
5. A1 Equation in m
6. A1 AG All correct and no errors seen
7. (B1 [R = 3g cos30] F = 0.2 × 3g cos30 (3√3 = 5.196))
8. M1 Applies Newton's Second Law to one of the particles or forms system equation in a
9. M1 Applies Newton's Second Law to form second equation in T and a and solves for a or solves system equation for a
10. A1 a = (25-3√3)/7 = 2.83.
11. B1 FT v as T becomes zero FT on a
12. M1 Applies Newton's Second Law and solves for a
13. M1 Uses v² = u² + 2as and solves for s
14. A1 Total distance = 0.710 m
15. (B1 [R = 3g cos30] F = 0.2 × 3g cos30 (3√3 = 5.196))
16. M1 For 4kg mass, uses PE loss – WDfr = KE gain
17. M1 For 3kg mass, uses WDT = KE gain + PE gain + WDFr
18. A1 v² = (25 – 3√3)/7 or v = 1.68
19. B1
20. M1 For 3kg mass, uses KE loss = PE gain + WDFr
21. M1 Solves for s
22. A1 Total distance = 0.710 m