A-LevelMathematicsMechanics9709/42 · May/June 2023 · Q47 Marks

An athlete of mass 84kg is running along a straight road.

(a)Initially the road is horizontal and he runs at a constant speed of 3 ms¯¹. The athlete produces a constant power of 60 W. Find the resistive force which acts on the athlete.[1]
(b)The athlete then runs up a 150m section of the road which is inclined at 0.8° to the horizontal. The speed of the athlete at the start of this section of road is 3ms¯¹ and he now produces a constant driving force of 24N. The total resistive force which acts on the athlete along this section of road has constant magnitude 13 N. Use an energy method to find the speed of the athlete at the end of the 150 m section of road.[6]

Mark scheme

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

4(a)
Resistive force = DF = (60/3 =) 20 N
[B1]
4(b)
PE = ±84g×150 sin 0.8
[B1] ±1759.23...
KE change = ±(½×84v² - ½×84×3²)
[B1] (½×84v² - 378)
Work Done = ±(24×150 - 13×150)
[B1] ±(3600 - 1950) = ±1650
Attempt at work-energy equation.
[M1] 5 terms, dimensionally correct, allow sign errors, sin/cos mix on PE term, PE must include sin 0.8 or cos 0.8.
84g×150sin 0.8 + ½×84v² - ½×84×3² = 24×150 - 13×150
[A1]
[1759.23... + 42v² - 378 = 3600 - 1950→42v² = 268.765...]
[v =] 2.53 ms⁻¹
[A1] AWRT 2.53; 2.5296...
Special case for use of constant acceleration: Maximum 4 marks
Resolve parallel to slope and use Newton's second law
24 - 13 - 84g sin 0.8 = ±84a
Use constant acceleration formula to get an equation in v or v²
[DM1] E.g. v² =3² +2×(their a)×150.
[v =] 2.53 ms⁻¹
[A1] AWRT 2.53; 2.5296...

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