AS June 2024 Q2
2. A lorry has mass 5000 kg.
In all circumstances, when the speed of the lorry is \(v\ \text{m s}^{-1}\), the resistance to motion of the lorry from non-gravitational forces is modelled as having magnitude \(490v\) newtons.
The lorry moves along a straight horizontal road at \(12\ \text{m s}^{-1}\), with its engine working at a constant rate of 84 kW.
Using the model,
Another straight road is inclined to the horizontal at an angle \(\alpha\) where \(\sin\alpha = \dfrac{1}{14}\)
With its engine again working at a constant rate of 84 kW, the lorry can maintain a constant speed of \(V\ \text{m s}^{-1}\) up the road.
Using the model,
| Scheme | Marks | AO |
|---|---|---|
| Use of \(F = \dfrac{84000}{12}\) | M1 | 3.4 |
| Equation of motion horizontally | M1 | 3.1b |
| \(\dfrac{84000}{12} - 490 \times 12 = 5000a\) | A1 | 1.1b |
| \(\dfrac{28}{125}\) or 0.224 or 0.22 \((\text{m s}^{-2})\) | A1 | 1.1b |
| (4) |
Notes
M1: Allow use of 84
M1: Correct no. of terms, condone sign errors.
Allow if they use 84 or 84000 as the driving force
A1: Correct equation
A1: Accept 0.22
| Scheme | Marks | AO |
|---|---|---|
| Use of \(D = \dfrac{84000}{V}\) | M1 | 3.4 |
| Equation of motion parallel to the road: \(D - 490V - 5000g\sin\alpha = 0\) | M1 | 2.1 |
| \(\dfrac{84000}{V} - 490V - 5000g\sin\alpha = 0\) | A1 | 1.1b |
| \(V = 10\) only | A1 | 1.1b |
| (4) | ||
| (8 marks) |
Notes
M1: Allow use of 84
M1: Equation in \(V\) only with correct no. of terms, condone sign errors and sin/cos confusion and omitted \(g\) with \(D\) in terms of \(V\).
A1: Correct equation in \(V\) only
A1: cao