A-LevelMathematicsMechanics9709/41 · May/June 2018 · Q610 Marks
A car has mass 1250 kg.
- (i)The car is moving along a straight level road at a constant speed of 36 m s¯¹ and is subject to a constant resistance of magnitude 850 N. Find, in kW, the rate at which the engine of the car is working.[2]
- (ii)The car travels at a constant speed up a hill and is subject to the same resistance as in part (i). The hill is inclined at an angle of θ° to the horizontal, where sin θ° = 0.1, and the engine is working at 63 kW. Find the speed of the car.[3]
- (iii)The car descends the same hill with the engine of the car working at a constant rate of 20 kW. The resistance is not constant. The initial speed of the car is 20 m s¯¹. Eight seconds later the car has speed 24 m s¯¹ and has moved 176 m down the hill. Use an energy method to find the total work done against the resistance during the eight seconds.[5]
Mark scheme
Marking points from the official mark scheme (May/June 2018). The mark or mark type is shown in brackets.
- [M1] Apply P = DF × v with DF = Resistance force
- [A1] Power = rate of working = 30.6 kW
- [M1] Driving force comprising of resistance plus a weight component
- [M1] DF = P/v
- [A1] v = 30 so speed of car is 30 m s⁻¹
- [B1] Gain in KE = ½ x 1250 x (24² – 20²) [= 110 000]
- [B1] Loss in PE = 1250 g × 176 × 0.1 [= 220 000]
- [B1] WD by car's engine = 20 000 × 8 [= 160 000]
- [M1] 4 term work energy equation
- [A1] WD = 270 000 J = 270 kJ
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