| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2002 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable mass problems |
| Type | Rocket ascending against gravity |
| Difficulty | Challenging +1.8 This is a challenging M5 variable mass rocket problem requiring derivation of the rocket equation from first principles using momentum conservation, then solving a differential equation. Part (a) demands careful application of Newton's second law to a variable mass system (8 marks indicates substantial working), and part (b) requires separating variables and integrating. This is significantly harder than standard mechanics questions, requiring both conceptual understanding of variable mass systems and advanced calculus techniques typical of Further Maths M5. |
| Spec | 6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(-mg\delta t = (m+\delta m)(v+\delta v) - \delta m(v-2000) - mv\) | M1 A2 | Accept \((-1\) each error) |
| \(-mg = m\frac{dv}{dt} + 2000\frac{dm}{dt}\) | A1 | |
| \(m = 1000 - 10t\) | B1 | |
| \(\frac{dm}{dt} = -10\) | B1 | |
| \(-9.8(1000-10t) = (1000-10t)\frac{dv}{dt} - 20000\) | M1 | |
| \(-9.8(100-t) = (100-t)\frac{dv}{dt} - 2000\) | A1 c.s.o. | (8) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(-9.8 + \frac{2000}{100-t} = \frac{dv}{dt}\) | M1 | |
| \(\int_0^{60} \left(-9.8 + \frac{2000}{100-t}\right)dt = \int_0^{v_{\max}} dv\) | M1 A1 | (limits) |
| \(\left[-9.8t - 2000\ln(100-t)\right]_0^{60} = V_{\max}\) | A1 | |
| \(-588 - 2000\ln 40 + 2000\ln 100 = V_{\max}\) | M1 | |
| \(1200 \approx 124 \text{ ms}^{-1} = V_{\max}\) | A1 | (6) |
## Question 5(a):
| Working | Marks | Notes |
|---------|-------|-------|
| $-mg\delta t = (m+\delta m)(v+\delta v) - \delta m(v-2000) - mv$ | M1 A2 | Accept $(-1$ each error) |
| $-mg = m\frac{dv}{dt} + 2000\frac{dm}{dt}$ | A1 | |
| $m = 1000 - 10t$ | B1 | |
| $\frac{dm}{dt} = -10$ | B1 | |
| $-9.8(1000-10t) = (1000-10t)\frac{dv}{dt} - 20000$ | M1 | |
| $-9.8(100-t) = (100-t)\frac{dv}{dt} - 2000$ | A1 c.s.o. | **(8)** |
---
## Question 5(b):
| Working | Marks | Notes |
|---------|-------|-------|
| $-9.8 + \frac{2000}{100-t} = \frac{dv}{dt}$ | M1 | |
| $\int_0^{60} \left(-9.8 + \frac{2000}{100-t}\right)dt = \int_0^{v_{\max}} dv$ | M1 A1 | (limits) |
| $\left[-9.8t - 2000\ln(100-t)\right]_0^{60} = V_{\max}$ | A1 | |
| $-588 - 2000\ln 40 + 2000\ln 100 = V_{\max}$ | M1 | |
| $1200 \approx 124 \text{ ms}^{-1} = V_{\max}$ | A1 | **(6)** |
---5. A rocket is launched vertically upwards from rest. Initially, the total mass of the rocket and its fuel is 1000 kg . The rocket burns fuel at a rate of $10 \mathrm {~kg} \mathrm {~s} ^ { - 1 }$. The burnt fuel is ejected vertically downwards with a speed of $2000 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ relative to the rocket, and burning stops after one minute. At time $t$ seconds, $t \leq 60$, after the launch, the speed of the rocket is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Air resistance is assumed to be negligible.
\begin{enumerate}[label=(\alph*)]
\item Show that
$$- 9.8 ( 100 - t ) = ( 100 - t ) \frac { \mathrm { d } v } { \mathrm {~d} t } - 2000$$
(8)
\item Find the speed of the rocket when burning stops.\\
(6)
\section*{6.}
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\includegraphics[alt={},max width=\textwidth]{4c585ec7-7b3e-4ff8-b7c2-83f58ad82ae9-4_316_929_391_573}
\end{center}
\end{figure}
A rough uniform rod, of mass $m$ and length $4 a$, is rod is held on a rough horizontal table. The rod is perpendicular to the edge of the table and a length $3 a$ projects horizontally over the edge, as shown in Fig. 1.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2002 Q5 [14]}}