A car of mass 1600kg travels up a slope inclined at an angle of sin⁻¹0.08 to the horizontal. There is a constant resistance of magnitude 240N acting on the car.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2024). The mark or mark type is shown in brackets.

1. [M1] Must have correct number of relevant terms (weight component and 240 N resistance). Allow sign errors. Allow cos 4.58 or cos 4.6. Do not allow g missing. Using sin⁻¹ 0.08 or sin 0.08 scores M0B0A0. Must have either 0.08 or sin4.58... or sin 4.6, not just sinθ.
2. [B1] OE. E.g. Power/32 = their 1520. Allow any driving force provided it has a resistance and a weight component.
3. [A1] Allow 48 600 W or 48.64 kW or 48.6 kW. Must state units if given in kW.
4. [B1FT] DF = Power/v oe e.g. 0.95×their 48640=DF × 24. FT their power from part (a) Do not allow if not using power from part (a). Note: candidates who use sin 0.08 in part (a) should get a DF of 332.3 N, which can score B1FT and use of sin⁻¹ 0.08 should get a DF of 93299 N, can score B1FT. If candidate uses 48600 DF = 46170/24=1923.75. Candidates who omit the weight component in part (a) should get a DF of 304 N, and can score B1FT.
5. [M1] N2L Must have correct number of relevant terms (weight component and 240 N resistance). Allow sign errors. Must be dimensionally correct. Allow without using 95% or with using 5%. Must have either 0.08 or sin4.58... or sin 4.6, not just sinθ or sin⁻¹ 0.08 or sin 0.08.
6. [A1] Allow 19/75. Note: 0.25 scores A0. If candidate uses 48600 they must get 0.252(34...) rather than 0.253.