A2 June 2019 Q4
4. A car of mass 600 kg pulls a trailer of mass 150 kg along a straight horizontal road. The trailer is connected to the car by a light inextensible towbar, which is parallel to the direction of motion of the car. The resistance to the motion of the trailer is modelled as a constant force of magnitude 200 N. At the instant when the speed of the car is \(v\ \text{m s}^{-1}\), the resistance to the motion of the car is modelled as a force of magnitude \((200 + \lambda v)\) N, where \(\lambda\) is a constant.
When the engine of the car is working at a constant rate of 15 kW, the car is moving at a constant speed of \(25\ \text{m s}^{-1}\)
Later on, the car is pulling the trailer up a straight road inclined at an angle \(\theta\) to the horizontal, where \(\sin\theta = \dfrac{1}{15}\)
The resistance to the motion of the trailer from non-gravitational forces is modelled as a constant force of magnitude 200 N at all times. At the instant when the speed of the car is \(v\ \text{m s}^{-1}\), the resistance to the motion of the car from non-gravitational forces is modelled as a force of magnitude \((200 + 8v)\) N.
The engine of the car is again working at a constant rate of 15 kW.
When \(v = 10\), the towbar breaks. The trailer comes to instantaneous rest after moving a distance \(d\) metres up the road from the point where the towbar broke.
| Scheme | Marks | AO |
|---|---|---|
| Use of \(P = Fv\): \(F = \dfrac{15000}{25}\ (= 600)\) | B1 | 3.3 |
| Equation of motion: | M1 | 3.4 |
| \(F - (200 + 200 + 25\lambda) = 0\) | A1 | 1.1b |
| \(\lambda = 8\) * | A1* | 2.2a |
| (4) |
Notes
B1: 600 or equivalent
M1: Use the model to form the equation of motion
If they start with two separate equations each one must be correct.
A1: Correct unsimplified equation
A1*: Deduce given answer from correct working.
| Scheme | Marks | AO |
|---|---|---|
| Equation of motion: | M1 | 3.4 |
| \(\dfrac{15000}{10} - 280 - 600g\sin\theta = 600a\) | A1 A1 | 1.1b 1.1b |
| \(a = 1.38\ \text{m s}^{-2}\ \ (1.4)\) | A1 | 1.1b |
| (4) |
Notes
M1: Use the model to form the equation of motion for the car (with \(v = 10\) used). All terms required. Dimensionally correct. Condone sign error and sin/cos confusion
A1 A1: Unsimplified equation with at most one error.
Correct unsimplified equation
A1: 2 or 3 sf only – follows use of 9.8
| Scheme | Marks | AO |
|---|---|---|
| Work energy equation | M1 | 3.1b |
| \(\dfrac{1}{2} \times 150 \times 100 = 200d + 150gd\sin\theta\) | A1 A1 | 1.1b 1.1b |
| \(d = 25.2\ (\text{m})\ \ (25)\) | A1 | 1.1b |
| (4) | ||
| (12 marks) |
Notes
M1: Complete strategy to form the work-energy equation. Condone sin/cos confusion and sign errors
A1 A1: Unsimplified equation with at most one error
Correct unsimplified equation for \(d\)
A1: Max 3 sf – follows use of 9.8