A-LevelMathematicsMechanics9709/43 · Oct/Nov 2021 · Q49 Marks
A car of mass 1400kg is moving on a straight road against a constant force of 1250 N resisting the motion.
- (a)(i)The car moves along a horizontal section of the road at a constant speed of 36 m s¯¹. Calculate the work done against the resisting force during the first 8 seconds.[2]
- (a)(ii)Calculate, in kW, the power developed by the engine of the car.[2]
- (a)(iii)Given that this power is suddenly increased by 12kW, find the instantaneous acceleration of the car.[3]
- (b)The car now travels at a constant speed of 32 ms¯¹ up a section of the road inclined at θ° to the horizontal, with the engine working at 64 kW. Find the value of θ.[2]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2021). The mark or mark type is shown in brackets.
- [M1] For using Work Done = Force × Distance. ([WD = 1250 × 36 × 8])
- [A1] WD = 360000 J or 360 kJ
- [B1 FT] Work Done from a(i) / 8 (Power = 1250 × 36 or P = 360000/8 [= 45000 J])
- [B1] = 45 kW
- [M1] Use changed Power in P = DF × v. (DF = 57000/36 [= 1583.3..])
- [M1] For using Newton's 2nd law applied to the car. (57000/36 – 1250 = 1400a)
- [A1] a = 0.238 ms⁻²
- [M1] For using DF = resistance + component of the weight of the car. (64000/32 = 1250 + 1400g sin θ)
- [A1] θ = 3.1 [3.0708...]
Write your answer and get it marked
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.
Sign up free →