Two particles P and Q, of masses 2kg and 0.25 kg respectively, are connected by a light inextensible string that passes over a fixed smooth pulley. Particle P is on an inclined plane at an angle of 30° to the horizontal. Particle Q hangs below the pulley. Three points A, B and C lie on a line of greatest slope of the plane with AB = 0.8 m and BC = 1.2 m (see diagram). Particle P is released from rest at A with the string taut and slides down the plane. During the motion of P from A to C, Q does not reach the pulley. The part of the plane from A to B is rough, with coefficient of friction 0.3 between the plane and P. The part of the plane from B to C is smooth. [Figure 7.1]

Mark scheme

Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.

1. 7(a)(i)
2. Particle P: 2g sin 30 – F – T = 2a [10 - F - T = 2a]
3. [M1] Newton's second law on either particle or for the system with correct masses; correct number of terms, allow sin/cos mix, allow sign errors. Allow with their F.
4. Particle Q: T - 0.25g = 0.25a
5. [A1] Both particle equations correct (with the same T) or system equation correct. Allow with their F. If their a direction is different to ours, allow if their a is consistently used e.g. -2g sin 30 + F + T = 2a' and 0.25g - T = 0.25a'.
6. System: 2g sin 30 – F – 0.25g = (2 + 0.25)a
7. F = 0.3R = 0.3×2g cos 30 [= 3√3 = 5.196...]
8. [M1] Use of F = 0.3R, where R is a component of weight.
9. Acceleration from A to B = 1.02 m s⁻²
10. [A1] Solving for the acceleration from A to B. Allow (10 - 4√3)/3; AWRT 1.02. May see T = (10 - √3)/3 = 2.7559...
11. 7(a)(ii)
12. Use of suvat from A to B to get an equation in v² or v.
13. [M1] Using u = 0 and their |a| from (a)(i) to get a positive v². E.g. v² = 2×0.8×(their 1.02) or v² = 2×(-0.8)×(-their 1.02).
14. v² = (80 - 32√3)/15 = 1.64 OR v = 1.28
15. [A1] Not v = 1.29. Allow 2sf or better without wrong work, i.e. v² = 1.6 or v = 1.3.
16. Find the acceleration from B to C: 2g sin 30 - 0.25g = (2+0.25)a
17. [M1] Resolving on both particles and eliminate T (if their a direction is different to ours, allow if their a is consistently used) OR for the system to get an equation in a only. Correct number of relevant terms, allow sin/cos mix, allow sign errors. For reference a = 10/3 or 3.33. May see T = 10/3.
18. v² = (their 1.28)² + 2×(their 10/3)×1.2
19. [DM1] Use of suvat from B to C, allow their positive a ≠ g from (a)(ii) not their a from (a)(i) and their 1.28 . Dependent on previous two marks.
20. Velocity = 3.1(0) ms⁻¹
21. [A1] AWRT 3.1(0) to 3sf.
22. Alternative Method for Question 7(a)(ii): using suvat in first stage and energy in second stage
23. Use of suvat from A to B to get an equation in v² or v.
24. [M1] Using u = 0 and their positive a from (a)(i). E.g. v² = 2 × 0.8×(their 1.02) .
25. v² = (80-32√3)/15 = 1.64 OR v = 1.28
26. [A1] Allow 2sf or better, i.e. v² = 1.6 or v = 1.3 .
27. Change in PE = ±(2g×1.2sin30-0.25g×1.2)
28. [B1]
29. ½(2+0.25)v² - ½(2+0.25)(their 1.28)² = 2g×1.2sin30-0.25g×1.2
30. [DM1] Use of work-energy 6 terms; dimensionally correct. Allow sign errors. Allow sin/cos mix on PE. Dependent on previous M.
31. Velocity = 3.1(0) m s⁻¹
32. [A1] AWRT 3.1(0) to 3sf.
33. Alternative Method for Question 7(a)(ii): using energy for complete motion
34. Change in PE = ±(2g×2sin30-0.25g×2)
35. [B1]
36. Work done against friction = 0.3×2g cos30×0.8
37. [B1]
38. Change in KE = ½(2+0.25)v²
39. [B1]
40. ½(2+0.25)v² + 0.3×2g cos30×0.8=(2g×2sin30-0.25g×2)
41. [M1] Use of work-energy 5 terms; dimensionally correct. Must be considering both particles. Allow sign errors. Allow sin/cos mix on PE and/or WD against friction.
42. Velocity = 3.1(0) m s⁻¹
43. [A1] AWRT 3.1(0) to 3sf.
44. 7(b)
45. 0.8 = 0 + ½ × (their positive answer to (a)(i)) × t₁² and solve for t₁
46. [M1] Use of suvat from A to B to find t₁, using s = 0.8 and their positive a ≠ g from (a)(i). Must get to t₁ =... OR using their positive 1.28. OR using their positive a ≠ g from (a)(i) and positive 1.28.
47. 1.2 = (their 1.28 from (a)(ii)) t₂ + ½ × (their 10/3) × t₂² and solve for t₂
48. [M1] Use of suvat from B to C to find t₂, using s = 1.2 and their positive a ≠ g from (a)(ii) (not their a from (a)(i)) and their 1.28 which would lead to a positive t₂ value. OR using their 1.28 and/or their answer to (a)(ii) which would lead to a positive t₂ value. OR using their 1.28 and/or their answer to (a)(ii) and their a ≠ g from (a)(ii) (not their a from (a)(i)) which would lead to a positive t₂ value.
49. t₁ = 1.25 or t₂ = 0.547
50. [A1] These can be seen as an expression. Allow t₁ = 0.8/0.64 = 1.25 OE. t₂ = 1.2/2.19 = 0.547 OE. Allow 2sf or better, e.g. t₁ = 1.3 or t₁ = 0.55.
51. Total time = 1.8(0) s
52. [A1] WWW. AWRT 1.8(0) to 3sf.