A-LevelMathematicsMechanics9709/41 · Oct/Nov 2023 · Q69 Marks

A car of mass 1300kg is moving on a straight road.

(a)(i)On a horizontal section of the road, the car has a constant speed of 30 ms¯¹ and there is a constant force of 650N resisting the motion. Calculate, in kW, the power developed by the engine of the car.[2]
(a)(ii)Given that this power is suddenly increased by 9kW, find the instantaneous acceleration of the car.[3]
(b)On a section of the road inclined at sin¯¹ 0.08 to the horizontal, the resistance to the motion of the car is (1000 + 20v) N when the speed of the car is vms¯¹. The car travels downwards along this section of the road at constant speed with the engine working at 11.5kW. Find this constant speed.[4]

Mark scheme

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

[B2] (a)(i) Power = 19.5 kW. Or B1 for either 650×30 or 19 500.
[B1FT] (a)(ii) their 19500+9000=DF ×30. Oe FT their 19.5 in watts only.
[M1] (a)(ii) DF-650=1300a. Newton's second law horizontally; 3 relevant terms; dimensionally correct but allow sign errors; allow with their driving force or just DF.
[A1] (a)(ii) a = 3/13 = 0.231 ms⁻². 0.23076923...
[*B1] (a)(ii) Alternative scheme for 6(a)(ii). 9000=DF×30. oe e.g. DF = 9000/30.
[DM1] (a)(ii) DF=1300a. Resolving horizontally using Newton's second law; 2 relevant terms; dimensionally correct but allow sign errors.
[A1] (a)(ii) a = 3/13 = 0.231ms⁻². 0.23076923...
[B1] (b) DF = 11500/ν. oe e.g. DF×v=11500.
[M1] (b) Attempt at Newton's second law. 4 relevant terms, their DF or just DF; allow sign errors: allow sin/cos mix; allow g missing.
[A1] (b) 11500/ν +1300×g×0.08-(1000+20v) = 0. Correct equation.
[A1] (b) Speed = 25 m s⁻¹.

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