A-LevelMathematicsMechanics9709/43 · Oct/Nov 2023 · Q47 Marks

A car has mass 1600kg.

(a)The car is moving along a straight horizontal road at a constant speed of 24 ms⁻¹ and is subject to a constant resistance of magnitude 480N. Find, in kW, the rate at which the engine of the car is working.[2]
(b)The car now moves down a hill inclined at an angle of θ to the horizontal, where sin θ = 0.09. The engine of the car is working at a constant rate of 12kW. The speed of the car is 24 ms⁻¹ at the top of the hill. Ten seconds later the car has travelled 280m down the hill and has speed 32ms⁻¹. Given that the resistance is not constant, use an energy method to find the total work done against the resistance during the ten seconds.[5]

Mark scheme

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

[M1] P=480×24 or, e.g. P/24 - 480 = 0 For P/v - F = 0 or P=Fv oe.
[A1] P=11.52 [kW] Allow 11.5 M1A0 for 11 520 or 11 500.
[B1] [KE before]= (1/2)×1600×24² [= 460800] [KE after]= (1/2)×1600×32² [=819200] For either correct. Do not allow (1/2)×1600×(32-24)².
[B1] [PE loss]=1600g×280×0.09[=1600g × 25.2 = 403200] Allow 1600g×280×sin 5.16° or 1600g×280×sin 5.2° but not simply 1600g×280×sinθ (unless implied by correct final answer).
[B1] Total WD = 12000×10 [=120000] oe, e.g. 12000 = WD/10
[M1] Attempt at work energy equation with 5 relevant terms (4 relevant terms plus work done against resistance); dimensionally correct. Allow sign errors. M0 for use of constant acceleration. Do not allow (1/2)×1600×(32-24)². (e.g., 12000×10+1600g×280×0.09 - (1/2)×1600×32² + (1/2)×1200×24² = WD_resistance which implies 120000+403200-819200+460800 = WD_resistance)
[A1] WD = 164 800 [J] Or 164.8 kJ CAO but condone 165 kJ or 165 000 [J] Not from use of constant acceleration or Newton's second law. ISW attempt to find force after correct WD found.

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