M4 June 2007 Q2
2. A lorry of mass \(M\) moves along a straight horizontal road against a constant resistance of magnitude \(R\). The engine of the lorry works at a constant rate \(RU\), where \(U\) is a constant. At time \(t\), the lorry is moving with speed \(v\).
At time \(t = 0\), the lorry has speed \(\tfrac{1}{4}U\) and the time taken by the lorry to attain a speed of \(\tfrac{1}{3}U\) is \(\dfrac{kMU}{R}\), where \(k\) is a constant.
| Scheme | Marks |
|---|---|
| \(F = \dfrac{Ru}{v}\) | B1 |
| R\((\rightarrow)\), \(\ \dfrac{Ru}{v} - R = M\dfrac{dv}{dt}\) | M1 |
| \(R(u - v) = Mv\dfrac{dv}{dt}\ *\) | A1 |
| (3) |
Notes
The mark scheme writes \(u\) for the constant \(U\).
B1 Correct expression involving the driving force.
M1 Use of F = ma to form a differential equation. Condone sign errors. a must be expressed as a derivative, but could be any valid form.
A1 Rearrange to given form.
| Scheme | Marks |
|---|---|
| \(\displaystyle\int_0^T dt = \frac{M}{R}\int_{\frac{1}{4}U}^{\frac{1}{3}U} \frac{v\,dv}{u - v}\) | M1A1 |
| \(\displaystyle\Rightarrow T = \frac{M}{R}\int_{\frac{1}{4}U}^{\frac{1}{3}U} -1 + \frac{u}{u - v}\,dv\) | DM1 |
| \(= \dfrac{M}{R}\Big[-v - u\ln(u - v)\Big]_{\frac{1}{4}U}^{\frac{1}{3}U}\) | A1 |
| \(= \dfrac{M}{R}\left[-\dfrac{u}{3} - u\ln\left(\dfrac{2u}{3}\right) + \dfrac{u}{4} + u\ln\left(\dfrac{3u}{4}\right)\right]\qquad \left(C = -\dfrac{Mu}{R}\left(\ln\dfrac{3u}{4} + \dfrac{1}{4}\right)\right)\) | M1 |
| \(= \dfrac{Mu}{R}\left(-\dfrac{1}{12} + \ln\dfrac{9}{8}\right)\) | M1 |
| Hence \(\ k = \ln\dfrac{9}{8} - \dfrac{1}{12}\) | A1 |
| (7) | |
| (10 marks) |
Notes
M1 Separate the variables
A1 Separation correct (limits not necessarily seen at this stage)
DM1 Attempt a complete integration process
A1 Integration correct
M1 Correct use of both limits – substitute and subtract. Condone wrong order.
M1 Simplify to find k from an expression involving a logarithm
A1 Answer as given, or exact equivalent. Need to see k = lnA + B