Two particles P and Q, of masses 6kg and 2kg respectively, lie at rest 12.5m apart on a rough horizontal plane. The coefficient of friction between each particle and the plane is 0.4. Particle P is projected towards Q with speed 20ms¯¹.

Mark scheme

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

1. [*B1] (a) Resolve horizontally using Newton's second law; 2 relevant terms; must be either -0.4×6g=6a or 0.4×6g=6a.
2. [DM1] (a) Use complete suvat method to get an equation in v or v² – must be using u = 20, s = 12.5 and their a.
3. [A1] (a) v² = 300 ⇒ v=10√3. AG. Condone correct expression for v or v² followed by correct answer.
4. [*B1] (a) Alternative method for Question 4(a). RF =0.4×6g. Correct application of F = µR for P.
5. [DM1] (a) 0.5×6×20² -0.5×6× v² = 12.5×(0.4×6g). 3 relevant terms; dimensionally correct; allow sign errors only.
6. [A1] (a) v² = 300 ⇒ v=10√3. AG. Condone correct expression for v or v² followed by correct answer.
7. [M1] (b) 6×10√3=(6+2)ν'. For use of conservation of momentum, 3 non-zero terms, allow sign errors. Use of 20 is M0.
8. [A1] (b) v' = 7.5√3. 12.99038...
9. [B1] (b) Initial KE = 0.5×6×(10√3)² [=900] or Final KE = 0.5×8×(7.5√3)² [=675]. Either initial kinetic energy or final kinetic energy correct. Allow unsimplified.
10. [A1] (b) Loss of KE = 225 J.
11. [M1] (c) 0=(their 7.5√3)²+2x(their-4)×s. Use complete suvat method to find distance. This must be using their v' from part (b), so it is dependent on scoring the first M mark in part (b) and either their a from part (a), or from ±0.4×8g=8a.
12. [A1] (c) [Distance =] 21.1m. 21.1 or better (21.09375).