M5 June 2009 Q3
3. A spaceship is moving in a straight line in deep space and needs to increase its speed. This is done by ejecting fuel backwards from the spaceship at a constant speed \(c\) relative to the spaceship. When the speed of the spaceship is \(v\), its mass is \(m\).
(a) Show that, while the spaceship is ejecting fuel, \[\frac{\mathrm{d}v}{\mathrm{d}m} = -\frac{c}{m}.\] (5)
The initial mass of the spaceship is \(m_0\) and at time \(t\) the mass of the spaceship is given by \(m = m_0(1 - kt)\), where \(k\) is a positive constant.
(b) Find the acceleration of the spaceship at time \(t\). (4)
| Scheme | Marks |
|---|---|
| \(mv = (m + \delta m)(v + \delta v) - (-\delta m)(c - v)\) \(mv = mv + m\delta v + v\delta m + c\delta m - v\delta m\) | M1 A2 |
| \(-m\delta v = c\delta m\) \(\dfrac{\mathrm{d}v}{\mathrm{d}m} = -\dfrac{c}{m}\) * | DM1 A1 |
| (5) |
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}m}{\mathrm{d}t} = -m_0k\) | B1 |
| \(\dfrac{\mathrm{d}v}{\mathrm{d}t} = \dfrac{\mathrm{d}v}{\mathrm{d}m} \times \dfrac{\mathrm{d}m}{\mathrm{d}t}\) | M1 |
| \(= -\dfrac{c}{m} \times -m_0k\) \(= \dfrac{cm_0k}{m_0(1 - kt)}\) | DM1 |
| \(= \dfrac{ck}{(1 - kt)}\) | A1 |
| (4) | |
| (9 marks) |