| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2012 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable mass problems |
| Type | Rocket ascending against gravity |
| Difficulty | Challenging +1.2 This is a standard variable mass/rocket equation problem from M5. Part (a) requires applying the rocket equation with gravity, which is a bookwork derivation covered in the syllabus. Part (b) involves separating variables and integrating with a logarithm, which is routine for students who have practiced this topic. While it requires careful algebra and understanding of variable mass systems, it follows a well-established template without requiring novel insight. |
| Spec | 4.10a General/particular solutions: of differential equations4.10b Model with differential equations: kinematics and other contexts6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| \((m + \delta m)(v + \delta v) - (-\delta m)(1000 - v) - mv = -mg\delta t\) | M1 A2 | |
| \(\delta v + \frac{1000}{m}\delta m = -g\delta t\) | PRINTED ANSWER | |
| \(\frac{dv}{dt} + \frac{1000}{m}\frac{dm}{dt} = -9.8\) | DM1 A1 |
| Answer | Marks |
|---|---|
| \(\frac{dv}{dt} = \frac{15000}{1500 - 15t} - 9.8\) | M1 |
| \(\frac{dv}{dt} = \frac{1000}{100 - t} - 9.8\) | |
| \(v = \int \frac{1000}{100 - t} - 9.8 \, dt\) | M1 |
| \(= [-1000\ln(100 - t) - 9.8t]_0^t\) | A1 |
| \(v = 1000\ln\frac{100}{(100 - t)} - 9.8t\) | DM1 A1 |
## (a)
$(m + \delta m)(v + \delta v) - (-\delta m)(1000 - v) - mv = -mg\delta t$ | M1 A2 |
$\delta v + \frac{1000}{m}\delta m = -g\delta t$ | | PRINTED ANSWER |
$\frac{dv}{dt} + \frac{1000}{m}\frac{dm}{dt} = -9.8$ | DM1 A1 |
## (b)
$\frac{dv}{dt} = \frac{15000}{1500 - 15t} - 9.8$ | M1 |
$\frac{dv}{dt} = \frac{1000}{100 - t} - 9.8$ | |
$v = \int \frac{1000}{100 - t} - 9.8 \, dt$ | M1 |
$= [-1000\ln(100 - t) - 9.8t]_0^t$ | A1 |
$v = 1000\ln\frac{100}{(100 - t)} - 9.8t$ | DM1 A1 |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$^{-1}$.
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\frac{dv}{dt} + \frac{1000}{m}\frac{dm}{dt} = -9.8$$
[5]
\item Find $v$ at time $t$, $0 \leqslant t \leqslant 40$
[5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2012 Q2 [10]}}