A car of mass 1200kg is travelling along a straight horizontal road AB. There is a constant resistance force of magnitude 500N. When the car passes point A, it has a speed of 15 ms⁻¹ and an acceleration of 0.8 ms⁻².

Mark scheme

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

1. [M1] Apply Newton's Second Law: Driving Force (D) - Resistance = mass × acceleration. D – 500 = 1200 × 0.8.
2. [B1] Calculate Driving Force, D = 1460 N. This mark can be implied by subsequent correct use or calculation of P=D*v.
3. [A1] Calculate Power = Driving Force × velocity = 1460 × 15 = 21900 W. Allow 21900 without units or 21.9 kW, but not simply 21.9 without units or with wrong units.
4. [B1] Calculate initial and final Kinetic Energies: KE_A = ½ × 1200 × 15² = 135000 J, KE_B = ½ × 1200 × 32² = 614400 J. Sight of both KEs or their difference (479400 J).
5. [B1ft] Calculate work done by engine: Work_engine = Power × time = (their 21900) × 53 = 1160700 J. Follow through on their power from (a).
6. [B1] Apply Work-Energy Theorem: Work_engine - Work_resistance = Change in KE. 1160700 – 500d = 479400, leading to d = 1362.6 m (AG).