M5 June 2012 Q2
2. A rocket, with initial mass 1500 kg, including 600 kg of fuel, is launched vertically upwards from rest. The rocket burns fuel at a rate of 15 kg s\(^{-1}\) and the burnt fuel is ejected vertically downwards with a speed of 1000 m s\(^{-1}\) relative to the rocket. At time \(t\) seconds after launch \((t \leqslant 40)\) the rocket has mass \(m\) kg and velocity \(v\) m s\(^{-1}\).
(a) Show that \[\frac{\mathrm{d}v}{\mathrm{d}t} + \frac{1000}{m}\frac{\mathrm{d}m}{\mathrm{d}t} = -9.8\] (5)
(b) Find \(v\) at time \(t\), \(0 \leqslant t \leqslant 40\) (5)
| Scheme | Marks |
|---|---|
| \((m + \delta m)(v + \delta v) - (-\delta m)(1000 - v) - mv = -mg\delta t\) | M1 A2 |
| \(\delta v + \dfrac{1000}{m}\delta m = -g\delta t\) | |
| \(\dfrac{\mathrm{d}v}{\mathrm{d}t} + \dfrac{1000}{m}\dfrac{\mathrm{d}m}{\mathrm{d}t} = -9.8\) PRINTED ANSWER | DM1 A1 |
| (5) |
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}v}{\mathrm{d}t} - \dfrac{15000}{1500 - 15t} = -9.8\) \(\dfrac{\mathrm{d}v}{\mathrm{d}t} - \dfrac{1000}{100 - t} = -9.8\) | M1 |
| \(v = \displaystyle\int_0^t \dfrac{1000}{100 - t} - 9.8\,\mathrm{d}t\) | M1 |
| \(= \big[-1000\ln(100 - t) - 9.8t\big]_0^t\) | A1 |
| \(v = 1000\ln\dfrac{100}{(100 - t)} - 9.8t\) | DM1 A1 |
| (5) | |
| (10 marks) |