Particles P and Q have masses m kg and 2m kg respectively. The particles are initially held at rest 6.4 m apart on the same line of greatest slope of a rough plane inclined at an angle α to the horizontal, where sin α = 0.8 (see diagram). Particle P is released from rest and slides down the line of greatest slope. Simultaneously, particle Q is projected up the same line of greatest slope at a speed of 10 m s⁻¹. The coefficient of friction between each particle and the plane is 0.6. [Diagram]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2021). The mark or mark type is shown in brackets.

1. M1 Apply Newton's 2nd law along or perpendicular to the plane to particle Q. Must use values for α or sin α or cos α.
2. A1 Both correct.
3. M1 Using F = 0.6R where R is a component of 2mg only
4. A1 AG Acceleration of Q up the plane while moving up the plane is a = -11.6 ms⁻²
5. M1 Apply Newton's 2nd law to attempt to find the acceleration of particle P. Must use values for α or sin α.
6. M1 For using constant acceleration equations to attempt to determine when v_Q = 0 .
7. M1 Use constant acceleration equations to attempt to find either S_P(down) or S_Q(up) at time T_1.
8. M1 For attempting to find the extra distance d[=0.455] needed to reach 6.4 m and using u_P =4.4T_1 at T_1 to find T_2 as d=(4.4T_1)T_2+1/2×4.4T_2²-1/2x4.4T_2².
9. A1 t=0.98194357... Time before collision = [t = T_1 + T_2 = 0.862 +0.12 =]0.982
10. Alternative method for Question 7(b)
11. M1 Apply Newton's 2nd law to attempt to find the acceleration of particle P. Must use values for α or sin α.
12. M1 For using constant acceleration equations to attempt to determine when v_Q = 0
13. M1 Use constant acceleration equations to attempt to find either S_P(down) or S_Q(up) at time t where t is the total time before collision.
14. M1 For applying (±) s_P + (±) s_Q = 6.4 using their expressions for s_P and s_Q to set up and solve a 3-term quadratic equation in t to obtain at least 1 solution.
15. A1 t=0.98194357... Time before collision is t = 0.982 s
16. Special case for those who do not take into account the fact that Q comes to rest and then changes its direction
17. M1 Apply Newton's 2nd law to attempt to find the acceleration of particle P. Must use values for α or sin α.
18. M1 For using constant acceleration equations to attempt to find either S_P(down) or S_Q(up).
19. M1 For applying (±) s_P + (±) s_Q = 6.4 using their expressions for s_P and s_Q to set up and solve a 3-term quadratic equation in t to obtain at least 1 solution.
20. A1 Must reject t = 16/9. Maximum mark 4 out of 5. Time that particles are in motion before collision t = 1 s
21. B1 FT Allow ±4.4. FT on their 4.4 and their 0.982. U_P(down) = 0 + 4.4 × 0.982 [= 4.3208]
22. B1 FT Allow ±4.4. FT on their 4.4 and their 0.12. U_Q(down) = 4.4 × 0.12 [= 0.528]
23. M1 Apply conservation of momentum, 4 terms, using their u_P and u_Q values with m and 2m respectively. Velocity of P and Q after impact must be equal.
24. A1 Must be positive. Speed of combined particle immediately after impact = v = 1.79 ms⁻¹
25. Special case for those who do not take into account the fact that Q comes to rest and then changes its direction
26. B1 FT Allow ±4.4, FT on their 1 and their 4.4. U_P(down) = 0 + 4.4 × 1 [= 4.4]
27. B1 FT Allow ± (10 – 11.6 × 1), FT on their 1. U_Q(up) = 10 - 11.6 × 1 [= –1.6] so u_Q(down) = 1.6
28. M1 Apply conservation of momentum, 4 terms, using their u_P and u_Q values with m and 2m respectively. Velocity of P and Q after impact must be equal.
29. A1 Allow v = 38/15. Speed of combined particle immediately after impact = v = 2.53 ms⁻¹