The diagram shows two particles P and Q which lie on a line of greatest slope of a plane ABC. Particles P and Q are each of mass m kg. The plane is inclined at an angle θ to the horizontal, where sin θ = 0.6. The length of AB is 0.75 m and the length of BC is 3.25 m. The section AB of the plane is smooth and the section BC is rough. The coefficient of friction between each particle and the section BC is 0.25. Particle P is released from rest at A. At the same instant, particle Q is released from rest at B. [Figure 7]

Mark scheme

Marking points from the official mark scheme (Feb/Mar 2024). The mark or mark type is shown in brackets.

1. [*M1] (m)a=(m)gsinθ = (m)g×0.6 [⇒a=6]. For attempt at Newton's second law; 2 terms; allow sign errors; allow sin/cos mix, allow θ=36.9 or better. Allow a=6 seen/used.
2. [A1] v²=[0²]+2×6×0.75⇒v=3. AG.
3. [A1] 3=[0]+6t⇒ t = 0.5. AG. Allow A2 for use of a=6 with any 2 of v=3, t=0.5 and s = 0.75 to get the third value.
4. [*M1] For BC (m)a=(m)gsinθ-(m)g×0.8×0.25= (m)g×0.6-(m)g×0.8×0.25 [⇒a = 4]. For attempt at Newton's second law; 3 terms; allow sign errors; allow g missing; allow sin/cos mix, allow θ=36.9 or better.
5. [A1] For either S_P or S_Q. t is the time after P arrives at B. T is the time from when both particles released.
6. [DM1] For use of s_P = s_Q; s_P and s_Q of correct form. Using their a ≠ their 6 from part (a), a ≠±g. Using same a in both s_P and s_Q.
7. [A1] t = 0.5, so total time = 1 s or time T=1s
8. [B1] Immediately before the collision speed of P = 3+4x[0.5] = 5 ms⁻¹. For either. Speed of Q [=0+4×1] = 4ms⁻¹.
9. [*M1] 5(m)+4(m)=[(m)+(m)]×v [⇒v=4.5]. Use of conservation of momentum; 4 non-zero terms; allow sign errors; allow their 4ms⁻¹ (≠0or2) and 5ms⁻¹ (≠3). Use of mg then withhold final A mark.
10. [B1] Distance from B at collision = ½×4×1² = 2m. May be implied by 1.25 m. OR =3×0.5+½×4×0.5² = 2 m.
11. [DM1] 1.25=4.5t + ½x4t². Use of constant acceleration with u =their 4.5, a = their 4(≠±g) and s =3.25-their 2, s≠3.25, s≠0.75, s ≠ 4, s ≠ 2.75, s ≠ 0.5. If using two formulae, must be a complete method to get an equation in t only.
12. [A1] t = 0.25s only. A0 if from use of mg in conservation of momentum.