A lorry of mass 15000kg moves on a straight horizontal road in the direction from A to B. It passes A and B with speeds 20ms⁻¹ and 25ms⁻¹ respectively. The power of the lorry's engine is constant and there is a constant resistance to motion of magnitude 6000N. The acceleration of the lorry at B is 0.5 times the acceleration of the lorry at A.

Mark scheme

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

1. [B1] For use of P = Fv. P=20F or P=25F OE (e.g. F = P/20 or F = P/25). But not with wrong F substituted (e.g. 6000).
2. [M1] Attempt to use Newton's second law in at least one case. Must have 3 terms. Allow sign errors. Allow F.
3. [A1] P/20 - 6000 = 15000a and P/25 - 6000 = 15000(1/2)a. OE for both. Allow 2a' and a'. Must be the same P for both.
4. [M1] For solving simultaneously. Dependent on 2 equations of the correct form with the correct number of relevant terms. Must get to 'P =' or ‘a =', but P = 200kW or 200000W with no attempt at a gets M0. Must be the same P for both.
5. [A1] Power [= 200 000W] = 200kW, a = 4/15 [ms⁻²]. AG. OE awrt 0.267. Do not allow 200000 [W] as final answer. Must show some working when they find P.
6. [M1] For attempt at resolving up hill. P/v - 6000 - 15000g sin1 = 0. Or 200000/v - 6000 - 2618 = 0. May see 200000/8618. Must have correct number of terms. Allow sin/cos mix. Allow sign errors. Allow g missing, but not a different acceleration. Do not allow F.
7. [A1] Steady speed = 23.2[ms⁻¹].