M5 June 2005 Q6
6. A rocket-driven car moves along a straight horizontal road. The car has total initial mass \(M\). It propels itself forwards by ejecting mass backwards at a constant rate \(\lambda\) per unit time at a constant speed \(U\) relative to the car. The car starts from rest at time \(t = 0\). At time \(t\) the speed of the car is \(v\). The total resistance to motion is modelled as having magnitude \(kv\), where \(k\) is a constant.
Given that \(t \lt \dfrac{M}{\lambda}\), show that
(a) \(\dfrac{\mathrm{d}v}{\mathrm{d}t} = \dfrac{\lambda U - kv}{M - \lambda t}\), (7)
(b) \(v = \dfrac{\lambda U}{k}\left\{1 - \left(1 - \dfrac{\lambda t}{M}\right)^{\frac{k}{\lambda}}\right\}\). (6)
| Scheme | Marks |
|---|---|
| Impulse-momentum: \(-kv\,\delta t = (m + \delta m)(v + \delta v) + (-\delta m)(v - U) - mv\) | M1 A3 |
| \(-kv\,\delta t = mv + m\,\delta v + \delta m\,v - \delta m\,v + \delta m\,U - mv\) | |
| \(-kv = m\dfrac{\delta v}{\delta t} + U\dfrac{\delta m}{\delta t}\) | |
| limit as \(\delta t \to 0\): \(-kv = m\dfrac{\mathrm{d}v}{\mathrm{d}t} + U\dfrac{\mathrm{d}m}{\mathrm{d}t}\) | |
| \(\dfrac{\mathrm{d}m}{\mathrm{d}t} = -\lambda \iff m = M - \lambda t\) | B1 |
| \(-kv = (M - \lambda t)\dfrac{\mathrm{d}v}{\mathrm{d}t} - \lambda U\) | M1 |
| \(\dfrac{\mathrm{d}v}{\mathrm{d}t} = \dfrac{\lambda U - kv}{M - \lambda t}\) * | A1 |
| (7) |
| Scheme | Marks |
|---|---|
| \(\displaystyle\int_0^t \frac{\mathrm{d}t}{M - \lambda t} = \int_0^v \frac{\mathrm{d}v}{\lambda U - kv}\) | M1 A1 |
| \(-\dfrac{1}{\lambda}\Big[\ln(M - \lambda t)\Big]_0^t = -\dfrac{1}{k}\Big[\ln(\lambda U - kv)\Big]_0^v\) | A1 |
| \(\dfrac{k}{\lambda}\big(\ln(M - \lambda t) - \ln M\big) = \ln(\lambda U - kv) - \ln\lambda U\) | M1 |
| \(\left(\dfrac{M - \lambda t}{M}\right)^{k/\lambda} = \dfrac{\lambda U - kv}{\lambda U}\) | M1 |
| \(v = \dfrac{\lambda U}{k}\left[1 - \left(1 - \dfrac{\lambda t}{M}\right)^{k/\lambda}\right]\) * | A1 |
| (6) | |
| (13 marks) |