AS June 2022 Q1
1. A car of mass 1200 kg moves up a straight road that is inclined to the horizontal at an angle \(\alpha\), where \(\sin\alpha = \dfrac{1}{15}\)
The total resistance to the motion of the car from non-gravitational forces is modelled as a constant force of magnitude \(R\) newtons.
At the instant when the engine of the car is working at a rate of 32 kW and the speed of the car is \(20\ \text{m s}^{-1}\), the acceleration of the car is \(0.5\ \text{m s}^{-2}\)
Find the value of \(R\) (5)
| Scheme | Marks | AO |
|---|---|---|
| \(F = \dfrac{32000}{20}\) | M1 | 3.3 |
| Equation of motion | M1 | 3.1b |
| \(F - 1200g\sin\alpha - R = 1200 \times 0.5\) | A1 | 1.1b |
| Substitute for \(g\), trig and \(F\) and solve for \(R\) | DM1 | 1.1b |
| \(R = 216\) or 220 (N) | A1 | 1.1b |
| (5) | ||
| (5 marks) |
Notes
M1: Use of \(P = Fv\). Allow \(\dfrac{32}{20}\).
Allow \(32000 = 20F\) or \(32 = 20F\), followed by an error when dividing
M0 for \(32000 = 20(F - R)\) or similar
M1: Correct no. of terms, condone sign errors and sin/cos confusion
M0 if they use power in equation of motion
A1: Correct equation
DM1: Dependent on second M1 (allow if \(g\) missing)
A1: Cao (\(R = 215.2\) if they use \(g = 9.81\))