| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2016 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable mass problems |
| Type | Rocket ascending against gravity |
| Difficulty | Challenging +1.8 This is a variable mass rocket equation problem requiring application of the rocket equation F = v(dm/dt) - mg, differentiation of the given mass function, and integration to find final velocity. While conceptually demanding for A-level (variable mass is an advanced M5 topic), the mathematical steps are guided by part (a) showing the differential equation, making it a structured multi-step problem rather than requiring novel insight. |
| Spec | 4.10a General/particular solutions: of differential equations6.03a Linear momentum: p = mv6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Mass at time \(t\): \(M = m_0(5-2t)\), rate of mass loss: \(\frac{dM}{dt} = -2m_0\) | B1 | |
| Thrust force \(= u \cdot \left | \frac{dM}{dt}\right | = 2m_0 u\) (upward) |
| Newton's second law (rocket equation): \(M\frac{dv}{dt} = \text{Thrust} - Mg\) | M1 | |
| \(m_0(5-2t)\frac{dv}{dt} = 2m_0 u - m_0(5-2t)g\) | A1 | |
| \(\frac{dv}{dt} = \frac{2u}{5-2t} - 9.8\) | A1 | Completion (given answer) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Fuel all burnt when \(M = m_0\), i.e. \(m_0(5-2t) = m_0\), so \(t = 2\) | B1 | |
| Integrate: \(v = -u\ln(5-2t) - 9.8t + C\) | M1 A1 | |
| At \(t=0\), \(v=0\): \(0 = -u\ln 5 + C\), so \(C = u\ln 5\) | M1 | |
| \(v = u\ln 5 - u\ln(5-2t) - 9.8t = u\ln\frac{5}{5-2t} - 9.8t\) | A1 | |
| At \(t=2\): \(v = u\ln\frac{5}{1} - 19.6 = u\ln 5 - 19.6\) | A1 | Accept numerical forms |
# Question 6:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Mass at time $t$: $M = m_0(5-2t)$, rate of mass loss: $\frac{dM}{dt} = -2m_0$ | B1 | |
| Thrust force $= u \cdot \left|\frac{dM}{dt}\right| = 2m_0 u$ (upward) | M1 A1 | |
| Newton's second law (rocket equation): $M\frac{dv}{dt} = \text{Thrust} - Mg$ | M1 | |
| $m_0(5-2t)\frac{dv}{dt} = 2m_0 u - m_0(5-2t)g$ | A1 | |
| $\frac{dv}{dt} = \frac{2u}{5-2t} - 9.8$ | A1 | Completion (given answer) |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Fuel all burnt when $M = m_0$, i.e. $m_0(5-2t) = m_0$, so $t = 2$ | B1 | |
| Integrate: $v = -u\ln(5-2t) - 9.8t + C$ | M1 A1 | |
| At $t=0$, $v=0$: $0 = -u\ln 5 + C$, so $C = u\ln 5$ | M1 | |
| $v = u\ln 5 - u\ln(5-2t) - 9.8t = u\ln\frac{5}{5-2t} - 9.8t$ | A1 | |
| At $t=2$: $v = u\ln\frac{5}{1} - 19.6 = u\ln 5 - 19.6$ | A1 | Accept numerical forms |6. A firework rocket, excluding its fuel, has mass $m _ { 0 } \mathrm {~kg}$. The rocket moves vertically upwards by ejecting burnt fuel vertically downwards with constant speed $u \mathrm {~m} \mathrm {~s} ^ { - 1 } , u > 24.5$, relative to the rocket. The rocket starts from rest on the ground at time $t = 0$. At time $t$ seconds, $t \leqslant 2$, the speed of the rocket is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and the mass of the rocket including its fuel is $m _ { 0 } ( 5 - 2 t ) \mathrm { kg }$. It is assumed that air resistance is negligible and the acceleration due to gravity is constant.
\begin{enumerate}[label=(\alph*)]
\item Show that, for $t \leqslant 2$
$$\frac { \mathrm { d } v } { \mathrm {~d} t } = \frac { 2 u } { 5 - 2 t } - 9.8$$
\item Find the speed of the rocket at the instant when all of its fuel has been burnt.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2016 Q6 [12]}}