M2 January 2013 Q2
2. A lorry of mass 1800 kg travels along a straight horizontal road. The lorry’s engine is working at a constant rate of 30 kW. When the lorry’s speed is 20 m s\(^{-1}\), its acceleration is 0.4 m s\(^{-2}\). The magnitude of the resistance to the motion of the lorry is \(R\) newtons.
The lorry now travels up a straight road which is inclined at an angle \(\alpha\) to the horizontal, where \(\sin\alpha = \dfrac{1}{12}\). The magnitude of the non-gravitational resistance to motion is \(R\) newtons. The lorry travels at a constant speed of 20 m s\(^{-1}\).
| Scheme | Marks |
|---|---|
 | |
| \(T = \dfrac{30000}{20}\ \ (= 1500)\) | B1 |
| \(T - R = 1800a\) | M1 |
| \(T - R = 1800 \times 0.4\) \(R = 1500 - 1800 \times 0.4\) | A1 |
| \(= 780\) | A1 |
| (4) |
Notes
B1 Use of \(P = Fv\)
M1 Equation of motion. Need all 3 terms. Condone sign errors
A1 Equation correct (their T)
A1 Only
| Scheme | Marks |
|---|---|
 | |
| \(T - 1800g\sin\alpha - R = 0\) | M1 A1 |
| \(T = 1800 \times \dfrac{1}{12}g + 780\) | |
| Power \(= \left(1800 \times \dfrac{1}{12}g + 780\right) \times 20\) | DM1 A1 |
| \(= 45000\) W or 45 kW | A1 |
| (5) | |
| (9 marks) |
Notes
M1 Equation of motion. Need all 3 terms. Weight must be resolved. Condone cos for sin. Condone sign errors
A1 Correct equation. Allow with \(R\) not substituted or with their \(R\).
DM1 Use of \(P = Tv\)
A1 Correctly substituted equation (for their \(R\))
A1 cao