A-LevelMathematicsMechanics9709/43 · Oct/Nov 2018 · Q610 Marks
A van of mass 3200 kg travels along a horizontal road. The power of the van's engine is constant and equal to 36 kW, and there is a constant resistance to motion acting on the van.
- (i)When the speed of the van is 20 m s¯¹, its acceleration is 0.2 m s¯². Find the resistance force.[3]
- (ii)When the van is travelling at 30m s¯¹, it begins to ascend a hill inclined at 1.5° to the horizontal. The power is increased and the resistance force is still equal to the value found in part (i). Find the power required to maintain this speed of 30 m s¯¹.[3]
- (iii)The engine is now stopped, with the van still travelling at 30m s¯¹, and the van decelerates to rest. Find the distance the van moves up the hill from the point at which the engine is stopped until it comes to rest.[4]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2018). The mark or mark type is shown in brackets.
- [B1] For use of power = Fv [Driving force = 36000 / 20]
- [M1] Use of Newton's Second Law [36000/20 - R = 3200 × 0.2]
- [A1] R = 1160 N
- [M1] Resolving along plane [Driving force F = 3200gsin1.5 + 1160]
- [M1] Use of P = Fv [Power = (3200gsin1.5 + 1160) × 30]
- [A1] Power = 59900W (59929.87...)
- [M1] Use of Newton's Second Law [-(3200gsin1.5 + 1160) = 3200a]
- [A1] (a = -0.62426...)
- [M1] Use of v² = u² + 2as to find s [0² = 30² + 2as]
- [A1] Distance s = 721 m (720.84...)
- [M1] For PE gain or KE loss [3200gsin1.5s] or [½ × 3200 × 900]
- [A1] For PE gain and KE loss [3200gsin1.5s and ½ × 3200 × 900]
- [M1] For work / energy equation [½ × 3200 × 900 = 1160s + 3200gsin1.5s]
- [A1] Distance s = 721 m (720.84...)
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