A car of mass 900kg is moving up a hill inclined at sin⁻¹ 0.12 to the horizontal. The initial speed of the car is 11 ms⁻¹. After 12s, the car has travelled 150m up the hill and has speed 16ms⁻¹. The engine of the car is working at a constant rate of 24 kW.

Mark scheme

Marking points from the official mark scheme (May/June 2022). The mark or mark type is shown in brackets.

1. [B1] KE change = ±(0.5×900×16² -0.5×900×11²) [= ±(115200-54450) = ±60750]
2. [B1] PE = 900g×150×0.12 [=162000]; Allow 900g×150×sin 6.89° or 900g×150×sin 6.9°; Not from use of constant acceleration/Newton's second law.
3. [B1] [Work done by car's engine =] 24000×12 [= 288000]; OE e.g. 24000 = WD/12
4. [M1] Use of work-energy 5 terms; dimensionally correct. Work done by car's engine not from using one of the given speeds. Allow sign errors.; 24000×12-900g×150×0.12-0.5×900×16² +0.5×900×11² = 288000-162000-115200+54450
5. [A1] Work done = 65250 J; or 65.25 kJ; Allow AWRT 65300 J or 65.3 kJ from correct work
6. [B1] Driving Force = 32000/ν; OE e.g. 32000 = DF × v
7. [M1] Apply N2L to the car with a = 0 (3 terms) and attempt to solve a 3-term quadratic in v. For reference 4v² +1520v-32000 = 0; Allow if no working seen and have correct real solution(s) to their 3-term quadratic. If working shown and if using the formula, it must be using the correct formula. If factorising must have 3 of the 4 terms correct of (v – 20) (ν+400)
8. [A1] Speed = 20ms⁻¹; Only.