M5 June 2006 Q5
5. A space-ship is moving in a straight line in deep space and needs to reduce its speed from \(U\) to \(V\). This is done by ejecting fuel from the front of the space-ship at a constant speed \(k\) relative to the space-ship. When the speed of the space-ship is \(v\), its mass is \(m\).
(a) Show that, while the space-ship is ejecting fuel, \(\dfrac{\mathrm{d}m}{\mathrm{d}v} = \dfrac{m}{k}\). (6)
The initial mass of the space-ship is \(M\).
(b) Find, in terms of \(U\), \(V\), \(k\) and \(M\), the amount of fuel which needs to be used to reduce the speed of the space-ship from \(U\) to \(V\). (6)
| Scheme | Marks |
|---|---|
| \(mv \simeq (m + \delta m)(v + \delta v) + (-\delta m)(k + v + \delta v)\) | M1 A3 |
| \(mv \simeq mv + m\,\delta v + v\,\delta m - k\,\delta m - v\,\delta m\) | M1 |
| \(k\,\delta m \simeq m\,\delta v\) | |
| In the limit, as \(\delta t \to 0\), \(\dfrac{\mathrm{d}m}{\mathrm{d}v} = \dfrac{m}{k}\) * | A1 |
| (6) |
| Scheme | Marks |
|---|---|
| \(\displaystyle\int_M^{m_1}\frac{\mathrm{d}m}{m} = \int_U^V\frac{\mathrm{d}v}{k}\) | M1 |
| \(\ln m_1 - \ln M = \dfrac{1}{k}(V - U)\) | A1 |
| \(\ln\dfrac{m_1}{M} = \dfrac{1}{k}(V - U)\) | M1 |
| \(m_1 = M\mathrm{e}^{\left(\frac{V - U}{k}\right)}\) | A1 |
| Amount of fuel \(= M - m_1 = M\left(1 - \mathrm{e}^{\frac{V - U}{k}}\right)\) | M1 A1 |
| (6) | |
| (12 marks) |