A-LevelPhysicsDynamics9702/22 · Feb/Mar 2017 · Q310 Marks
A car travels in a straight line up a slope, as shown in [Figure 3.1]. The car has mass 850 kg and travels with a constant speed of 9.0 m s⁻¹. The car's engine exerts a force on the car of 2.0 kN up the slope. A resistive force FD, due to friction and air resistance, opposes the motion of the car. The variation of FD with the speed v of the car is shown in [Figure 3.2].
- (a)Define velocity.[1]
- (b)(i)State and explain whether the car is in equilibrium as it moves up the slope.[2]
- (b)(ii)Consider the forces that act along the slope. Use data from [Figure 3.2] to determine the component of the weight of the car that acts down the slope.[2]
- (b)(iii)Show that the power output of the car is 1.8 × 10⁴ W.[2]
- (b)(iv)The car now travels along horizontal ground. The output power of the car is maintained at 1.8 × 10⁴ W. The variation of the resistive force FD acting on the car is given in [Figure 3.2]. Calculate the acceleration of the car when its speed is 15 m s⁻¹.[3]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (Feb/Mar 2017). The mark or mark type is shown in brackets.
- [B1] (a) change of displacement/time (taken)
- [M1] (b)(i) constant velocity, so resultant force is zero
- [A1] (b)(i) (so car is) in (dynamic) equilibrium
- [C1] (b)(ii) FD = 0.40 (kN) or 0.40 × 10³ (N)
- [A1] (b)(ii) component of weight = 2.0 × 10³ - 0.40 × 10³ = 1.6 × 10³ N
- [C1] (b)(iii) P = Fv
- [A1] (b)(iii) = 2.0 × 10³ × 9.0 = 1.8 × 10⁴ W
- [C1] (b)(iv) (driving) force = 1.8 × 10⁴/15 (= 1.2 × 10³)
- [C1] (b)(iv) FD = 0.66 (kN) or 0.66 × 10³(N)
- [A1] (b)(iv) acceleration = (1.2 × 10³ – 0.66 × 10³)/850 = 0.64 (0.635) m s⁻²
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