M5 June 2008 Q4
4. At time \(t = 0\) a rocket is launched from rest vertically upwards. The rocket propels itself upwards by expelling burnt fuel vertically downwards with constant speed \(U\) m s\(^{-1}\) relative to the rocket. The initial mass of the rocket is \(M_0\) kg. At time \(t\) seconds, where \(t \lt 2\), its mass is \(M_0\left(1 - \tfrac{1}{2}t\right)\) kg, and it is moving upwards with speed \(v\) m s\(^{-1}\).
(a) Show that \[\frac{\mathrm{d}v}{\mathrm{d}t} = \frac{U}{(2 - t)} - 9.8.\] (7)
(b) Hence show that \(U \gt 19.6\). (2)
(c) Find, in terms of \(U\), the speed of the rocket one second after its launch. (5)
| Scheme | Marks |
|---|---|
| \(-mg\,\delta t = (m + \delta m)(v + \delta v) + \delta m(U - v) - mv\) | |
| \(-mg\,\delta t = mv + m\,\delta v + v\,\delta m + U\,\delta m - v\,\delta m - mv\) | M1 A2 |
| \(-mg = m\dfrac{\mathrm{d}v}{\mathrm{d}t} + U\dfrac{\mathrm{d}m}{\mathrm{d}t}\) | A1 |
| \(m = M_0\left(1 - \tfrac{1}{2}t\right) \Rightarrow \dfrac{\mathrm{d}m}{\mathrm{d}t} = -\tfrac{1}{2}M_0\) | B1 |
| \(-M_0g\left(1 - \tfrac{1}{2}t\right) = M_0\left(1 - \tfrac{1}{2}t\right)\dfrac{\mathrm{d}v}{\mathrm{d}t} - \tfrac{1}{2}M_0U\) | M1 |
| \(U - g(2 - t) = (2 - t)\dfrac{\mathrm{d}v}{\mathrm{d}t}\) | |
| \(\dfrac{U}{(2 - t)} - 9.8 = \dfrac{\mathrm{d}v}{\mathrm{d}t}\) * | A1 |
| (7) |
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}v}{\mathrm{d}t} \gt 0\) when \(t = 0 \Rightarrow \dfrac{U}{2} - 9.8 \gt 0\) | M1 |
| \(\Rightarrow\ U \gt 19.6\) * | A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(v = \displaystyle\int\frac{U}{(2 - t)} - 9.8\,\mathrm{d}t\) | |
| \(= -U\ln(2 - t) - 9.8t + C\) | M1 |
| \(t = 0,\ v = 0\): \(0 = -U\ln 2 + C \Rightarrow C = U\ln 2\) | A1 |
| so, \(v = U\ln\dfrac{2}{(2 - t)} - 9.8t\) | M1 |
| \(t = 1\): \(v = U\ln 2 - 9.8\) | M1 A1 |
| (5) | |
| (14 marks) |