AS June 2019 Q1
1. A lorry of mass 16 000 kg moves along a straight horizontal road.
The lorry moves at a constant speed of \(25\ \text{m s}^{-1}\)
In an initial model for the motion of the lorry, the resistance to the motion of the lorry is modelled as having constant magnitude 16 000 N.
The model for the motion of the lorry along the same road is now refined so that when the speed of the lorry along the same road is \(V\ \text{m s}^{-1}\), the resistance to the motion of the lorry is modelled as having magnitude \(640V\) newtons.
Assuming that the engine of the lorry is working at the same rate of 400 kW
| Scheme | Marks | AO |
|---|---|---|
| Equation of motion parallel to the road with \(a = 0\) and using the model | M1 | 3.3 |
| \(F - 16000 = 0\) | A1 | 1.1b |
| \(P = 16\,000 \times 25\) | M1 | 3.4 |
| \(= 400\,000 = 400\ \text{kW}\) * | A1* | 1.1b |
| (4) |
Notes
M1: Correct no. of terms with \(a = 0\), condone sign errors
Given answer, so step must be seen, but allow if in verbal form or on a diagram.
A1: Correct equation
M1: Use of \(P = Fv\)
Independent mark - could be the first mark seen
A1*: Obtain given answer from correct working
| Scheme | Marks | AO |
|---|---|---|
| Use of \(\dfrac{400\,000}{V}\) | M1 | 3.3 |
| Equation of motion parallel to the road and using the refined model | M1 | 3.4 |
| \(\dfrac{400\,000}{V} - 640V = 16000 \times 2.1\) | A1 | 1.1b |
| \(2V^2 + 105V - 1250 = 0 \qquad \left(640V^2 + 33600V - 400000 = 0\right)\) | A1 | 1.1b |
| Solve for \(V\) | M1 | 1.1b |
| \(V = 10\) (i.e. speed is \(10\ \text{m s}^{-1}\)) | A1 | 1.1b |
| (6) | ||
| (10 marks) |
Notes
M1: Use of \(P = Fv\)
M1: Correct no. of terms, condone sign errors.
Dimensionally correct
A1: Correct unsimplified equation
A1: Correct 3 term quadratic
M1: For solving a 3 term quadratic – this mark can be implied by a correct value of \(V\) but otherwise can only be earned for evidence of an explicit method being used.
A1: \(V = 10\) only