A-LevelMathematicsMechanics9709/42 · May/June 2021 · Q511 Marks

A car of mass 1250 kg is pulling a caravan of mass 800 kg along a straight road. The resistances to the motion of the car and caravan are 440 N and 280 N respectively. The car and caravan are connected by a light rigid tow-bar. (a) The car and caravan move along a horizontal part of the road at a constant speed of 30 m s⁻¹. (b) The car and caravan now travel along a part of the road inclined at sin⁻¹ 0.06 to the horizontal. The car and caravan travel up the incline at constant speed with the engine of the car working at 28 kW.

(a)(i)Calculate, in kW, the power developed by the engine of the car.[2]
(a)(ii)Given that this power is suddenly decreased by 8 kW, find the instantaneous deceleration of the car and caravan and the tension in the tow-bar.[4]
(b)(i)Find this constant speed.[3]
(b)(ii)Find the increase in the potential energy of the caravan in one minute.[2]

Mark scheme

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

[M1] Using P = Fv with F as total resistance
P = (440+280)×30
[A1] P=720×30=21.6 kW Answer must be in kW
[B1] FT Follow through on their power from 5(a)(i) Allow Driving Force (DF) = (21600-8000)/30 = 13600/30 = 453.333... as the force due to solely to the change in power provided correct equation(s) used.
[M1] Apply Newton's 2nd law to either the car or to the caravan or to the system. Must be correct number of relevant terms.
Car: DF-440-T=1250a
Caravan: T-280=800a
System: DF-(440 + 280) = 2050a
If DF = 8000/30 is used then the equations must be either -DF = 2050a or T-280=800a
[M1] Solve for either a or T Using equation(s) with no missing/extra terms, DF ≠ 720. Solving for a either from the system equation or from the car AND caravan equation. OR solving for T from the car AND caravan equation.
[A1] a=-0.13 ms⁻² and T=176 N
[M1] Apply Newton's 2nd law with a = 0, either to the system OR by eliminating T between the equations for the car and the caravan, no extra or missing relevant terms, dimensionally correct, to find DF
System: DF = 720+2050g×0.06 [=1950]
Car: DF-440-T-1250g×0.06=0
Caravan: T-280-800g×0.06 = 0
[B1] P=DFxv. 28000/v SOI.
[A1] v=14.4 ms⁻¹
[M1] Using PE = mgh with h being height gained in 60 s, using their v
PE = 800g×d×0.06=800g×14.4×60×0.06
[A1] PE = 414 000 (J) or PE = 414 kJ Using v = 560/39 = 14.359

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