A-LevelMathematicsMechanics9709/42 · Feb/Mar 2022 · Q35 Marks

A car of mass mkg is towing a trailer of mass 300kg down a straight hill inclined at 3° to the horizontal at a constant speed. There are resistance forces on the car and on the trailer, and the total work done against the resistance forces in a distance of 50 m is 40000 J. The engine of the car is doing no work and the tow-bar is light and rigid.

Mark scheme

Marking points from the official mark scheme (Feb/Mar 2022). The mark or mark type is shown in brackets.

[B1] PE lost in 50 m = (m +300) g × 50 sin 3
[M1] (m +300) g × 50 sin 3 – 40 000 = 0 (Use of the work-energy equation.)
[A1] m = 1230 to 3 sf (m = 1228.6)
[M1] T+ 300 g sin 3 – 200 = 0 (Trailer) or mg sin 3 = T + 600 (Car) (Apply Newton's 2nd law either to the trailer or to the car using a = 0, three terms in either case.)
[A1] T=43[.0] N to 3 sf

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Topic: Mechanics, A-Level Mathematics (syllabus 9709). Structured question, 5 marks.
Source: A-Level Mathematics, Feb/Mar 2022 Paper 4 Variant 2, question 3.

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