A-LevelMathematicsMechanics9709/41 · Oct/Nov 2021 · Q511 Marks

A car of mass 1600 kg travels at constant speed 20 m s⁻¹ up a straight road inclined at an angle of sin⁻¹ 0.12 to the horizontal.

(a)Find the change in potential energy of the car in 30 s.[3]
(b)Given that the total work done by the engine of the car in this time is 1960 kJ, find the constant force resisting the motion.[3]
(c)Calculate, in kW, the power developed by the engine of the car.[2]
(d)Given that this power is suddenly decreased by 15%, find the instantaneous deceleration of the car.[3]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2021). The mark or mark type is shown in brackets.

[B1] s = 30×20
[M1] Attempt change in PE. May use angle = 6.9°. Allow sin/cos error only. PE change = 1600 × g × s × 0.12
[A1] Change in PE = 1152000 J
[M1] Using work-energy, allow sign error. 1960 000 = WD_res + their PE
[B1] Using WD_res = R×600. R=WD_res ÷600
[A1] Force resisting motion = R = 1350 N to 3sf. Allow R = 4040/3 N. Allow R negative.
[M1] For using P = DF × v. Allow use of their R. P = (4040/3 + 1600×g×0.12)×20
[A1] P = 65.3 kW
[B1 FT] Using the reduction in power as the cause of the deceleration. 0.85 × 1960000/3 = DF × 20. P=DFxv. [DF = 8330/3]. FT on their P.
[M1] Newton's 2nd law, four terms, allow sin/cos error, their R and their DF. DF - R - 1600g × 0.12 = 1600a
[A1] a = [-]0.306 ms⁻². a = [-]490/1600 = [-]49/160

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