A-LevelMathematicsMechanics9709/41 · May/June 2023 · Q711 Marks

A car of mass 1200kg is travelling along a straight horizontal road. The power of the car's engine is constant and is equal to 16kW. There is a constant resistance to motion of magnitude 500N.

(a)Find the acceleration of the car at an instant when its speed is 20ms¯¹.[3]
(b)Assuming that the power and the resistance forces remain unchanged, find the steady speed at which the car can travel.[2]
(c)The car comes to the bottom of a straight hill of length 316m, inclined at an angle to the horizontal of sin¯¹(¹/₆₀). The power remains constant at 16kW, but the magnitude of the resistance force is no longer constant and changes such that the work done against the resistance force in ascending the hill is 128 400 J. The time taken to ascend the hill is 15 s. Given that the car is travelling at a speed of 20 m s¯¹ at the bottom of the hill, find its speed at the top of the hill.[6]

Mark scheme

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

[B1] OE e.g. 16000 = 20 × F
[M1] Use of Newton's second law; allow sign errors but must be 3 terms. Allow F or any non-zero value for the driving force (allow 0.8 from using 16 rather than 16000) but not 16000, 16, 20 or 500 for F.
[A1]
[M1] Allow sign errors but must be 2 terms. Condone 16/ν - 500 = 0 for M1.
[A1]
[B1] Or 16000 = WD/15
[B1] ±(600v² – 240000)
[B1] Allow 1200×g×316×sin 0.955 or 1200×g×316×sin 0.95 or 1200×g × 79/15 or 1200x gx 5.266....
[M1] Use of work-energy principle with 5 terms; dimensionally correct. Allow sign errors and sin/cos mix on PE term
[A1] Allow a value in the interval [62870,63600] for the PE term from using non-exact values for the given angle (but not if from incorrect working).
[A1] 21.924111...

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