M2 June 2012 Q6
6. A car of mass 1200 kg pulls a trailer of mass 400 kg up a straight road which is inclined to the horizontal at an angle \(\alpha\), where \(\sin\alpha = \dfrac{1}{14}\). The trailer is attached to the car by a light inextensible towbar which is parallel to the road. The car’s engine works at a constant rate of 60 kW. The non-gravitational resistances to motion are constant and of magnitude 1000 N on the car and 200 N on the trailer.
At a given instant, the car is moving at 10 m s\(^{-1}\). Find
The towbar breaks when the car is moving at 12 m s\(^{-1}\).
| Scheme | Marks |
|---|---|
| \(F = \dfrac{60000}{10} = 6000\) | B1 |
| \(F - 1200g\sin\alpha - 400g\sin\alpha - 1000 - 200 = 1600a\) | M1 A1 A1 |
| \(a = 2.3\) (m s\(^{-2}\)) | A1 |
| (5) |
Notes
B1 Correct application of \(P = Fv\) seen or implied.
M1 Use of \(F = ma\) parallel to the slope for the car and trailer. Must have all the terms, but condone sign errors.
A1 At most one error (with \(F\) or their \(F\))
A1 Correct equation (with \(F\) or their \(F\))
A1 only
OR (a)
| \(F = 6000\) | B1 |
| \(T - 400g\sin\alpha - 200 = 400 \times a\) \(6000 - 1200g\sin\alpha - 1000 - T = 1200 \times a\) | |
| \(6000 - 1600g\sin\alpha - 1200 = 1600a\) | M1A1A1 |
| \(a = 2.3\) (m s\(^{-2}\)) | A1 |
Simultaneous equations in \(T\) and \(a\)
M1A1A1 Add to eliminate \(T\)
| Scheme | Marks |
|---|---|
| \(T - 400g\sin\alpha - 200 = 400 \times 2.3\) | M1 A1 ft A1 ft |
| \(T = 1400\) | A1 |
| (4) |
Notes
M1 Use of \(F = ma\) parallel to the slope for the trailer
A1 ft At most one error (their \(a\))
A1 ft All correct (their \(a\))
A1 only
OR
| \(6000 - 1200g\sin\alpha - 1000 - T = 1200 \times 2.3\) | M1 A1 ft A1 ft |
| \(T = 1400\) | A1 |
M1 Use of \(F = ma\) parallel to the slope for the car
A1 ft At most one error (their \(a\))
A1 ft All correct (their \(a\))
A1 only
OR (b) (following OR (a))
| \(-800a = 2T + 800g\sin\alpha + 800 - 6000\) | M1A1A1 |
| \(2T = 5200 - 800g\sin\alpha - 800 \times 2.3\) | |
| \(T = 1400\) | A1 |
M1A1A1 Subtract and / or substitute to eliminate \(a\)
| Scheme | Marks |
|---|---|
| \(200d = \dfrac{1}{2}400.12^2 - 400gd\sin\alpha\) | M1 A1 A1 |
| \(d = 60\) (m) | DM1 A1 |
| (5) | |
| (14 marks) |
Notes
M1 Use of work-energy. Must have all three terms. Do not accept duplication of terms, but condone sign errors. Equation in only one unknown, but could be vertical distance.
A1 At most one error in the equation
A1 All correct in one unknown
DM1 Solve for \(d\) – dependent on M for work-energy equation.
A1 only
For vertical distance \(\left(= \dfrac{60}{14} = 4.29\right)\) allow 3/5