A particle A of mass 0.5 kg is projected vertically upwards from horizontal ground with speed 25 m s⁻¹.

Mark scheme

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

1. [M1] v²=25²+2(-g)×20 Use of v² =u² +2as with u=25, s=20 and a =±g. OR using change in KE = ± change in PE.
2. [A1] Speed = 15 ms⁻¹
3. [M1] Taking up as positive direction: 0.5×15+0.3×(-32.5)=0.5v+0 or Taking down as positive direction: 0.5×(-15)+0.3×32.5 = 0.5v + 0 For use of conservation of momentum, 3 non-zero terms, allow sign errors, using their speed 15 ms⁻¹. Must show how ±2.25 is obtained.
4. [A1] [Taking up as positive direction: velocity of A = −4.5 ms⁻¹] [Taking down as positive direction: velocity of A = 4.5 ms⁻¹] Speed = 4.5 m s⁻¹ direction downwards Any error seen in calculating v is A0. Must explicitly say 4.5 m s⁻¹ and downwards.
5. [M1] Downwards to be positive, for A 20 = 4.5t_A + ½ gt_A² and solve for t_A Upwards to be positive, for A −20 = −4.5t_A − ½ gt_A² and solve for t_A Using constant acceleration formula(e) to get a correct equation in t_A and solve for t_A. If using quadratic formula, must be the correct formula. If factorising, when brackets expanded, 2 terms correct.
6. [M1] For B 20=0+½ g t_B² [t = 2] and solve for t_B Using constant acceleration formula(e) to get a correct equation in t_B and solve for t_B.
7. [A1] t_A =1.6 or t_B = 2
8. [A1] Difference = 0.4s only