A-LevelMathematicsMechanics9709/43 · Oct/Nov 2019 · Q25 Marks

A train of mass 150000 kg ascends a straight slope inclined at α° to the horizontal with a constant driving force of 16000 N. At a point A on the slope the speed of the train is 45 m s⁻¹. Point B on the slope is 500 m beyond A. At B the speed of the train is 42 m s⁻¹. There is a resistance force acting on the train and the train does 4 × 10⁶ J of work against this resistance force between A and B. Find the value of α.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2019). The mark or mark type is shown in brackets.

[B1] PE gain = 150000g × 500sinα
[B1] KE loss = (1/2 × 150000 × 45²) - (1/2 × 150000 × 42²)
[M1] For 5 term work energy equation (or 4 terms if using loss in KE as 1 term): 150000g × 500sinα = 19575000 + 16000×500 – 4×10⁶
[A1]
[A1] α = 1.8

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Topic: Mechanics, A-Level Mathematics (syllabus 9709). Structured question, 5 marks.
Source: A-Level Mathematics, Oct/Nov 2019 Paper 4 Variant 3, question 2.

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