A-LevelMathematicsMechanics9709/42 · Oct/Nov 2018 · Q68 Marks

A car of mass 1200 kg is driving along a straight horizontal road at a constant speed of 15 m s¯¹. There is a constant resistance to motion of 350 N.

(i)Find the power of the car's engine.[1]
(ii)The car comes to a hill inclined at 1° to the horizontal, still travelling at 15 m s¯¹. The car starts to descend the hill with reduced power and with an acceleration of 0.12 m s⁻². Given that there is no change in the resistance force, find the new power of the car's engine at the instant when it starts to descend the hill.[3]
(iii)When the car is travelling at 20 m s¯¹ down the hill, the power is cut off and the car gradually slows down. Assuming that the resistance force remains 350 N, find the distance travelled from the moment when the power is cut off until the speed of the car is reduced to 18 m s⁻¹.[4]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2018). The mark or mark type is shown in brackets.

[B1] Power = 350 × 15 = 5250 W
[B1] Using Driving force DF = P/15
[M1] For using Newton's 2nd law down the slope
[A1] P = 4270 W (4268.56...)
[M1] [1200g sin 1- 350 = 1200a] Using Newton's 2nd law down the slope
[A1] Correct equation
[M1] [18² = 20² + 2as] Using constant acceleration formulae with a complete method to find distance, s, travelled.
[A1] Distance travelled s = 324 m (324.39)
[M1] Attempt either PE loss or KE loss
[A1] Both PE loss and KE loss correct
[M1] [1200g × s sin 1 + ½ × 1200 × (20² – 18²) = 350s] Apply work-energy equation to the car
[A1] Distance travelled s = 324 m (324.39)

On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.