| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2018 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Rocket/thrust problems (mass decreasing) |
| Difficulty | Challenging +1.8 This is a challenging M5 variable mass rocket problem requiring derivation from first principles of the rocket equation, solving a differential equation with exponential solutions, and applying energy considerations. While the steps are guided, it demands strong understanding of momentum conservation in variable mass systems and confident manipulation of exponentials and calculus—significantly harder than typical A-level questions but standard for Further Maths M5. |
| Spec | 6.03f Impulse-momentum: relation6.03g Impulse in 2D: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \((m+\delta m)(v+\delta v)+(-\delta m)(v-U)-mv=-mg\delta t\) | M1 A2 | M1 for use of Impulse-Momentum principle, dimensionally correct, with correct no. of terms; A1A1 for correct equation |
| \(mv+v\delta m+m\delta v+U\delta m-v\delta m-mv=-mg\delta t\) | Simplification step | |
| \(m\frac{dv}{dt}+U\frac{dm}{dt}=-mg\) | A1 | Third A1 for correct given answer correctly obtained |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(mg+U\frac{dm}{dt}=-mg\) | Setting \(\frac{dv}{dt}=g\) | |
| \(U\frac{dm}{dt}=-2mg\) | M1 | M1 for putting \(\frac{dv}{dt}=g\) into the DE and collecting terms |
| \(\int_M^m \frac{dm}{m}=\frac{-2g}{U}\int_0^t dt\) | M1 | M1 for separating variables and attempting to integrate both sides |
| \([\ln m]_M^m=\frac{-2gt}{U}\) | A1 | A1 for correct integral on both sides with correct limits |
| \(m=Me^{\frac{-2gt}{U}}\) | A1 | A1 for GIVEN ANSWER correctly obtained |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \((1-\lambda)M=Me^{\frac{-2gT}{U}}\) | M1 | M1 for putting \(m=(1-\lambda)M\) and solving for \(T\) |
| \(T=\frac{U}{2g}\ln\!\left(\frac{1}{1-\lambda}\right)\) | A1 | A1 for correct expression for \(T\) |
| \(v=gT=\frac{U}{2}\ln\!\left(\frac{1}{1-\lambda}\right)\) | M1 A1 | M1 for using \(v=u+at\) with \(u=0,\, a=g,\, t=T\); A1 for correct \(v\) |
| \(\text{KE}=\frac{1}{2}(1-\lambda)M\left[\frac{U}{2}\ln\!\left(\frac{1}{1-\lambda}\right)\right]^2=\frac{1}{8}MU^2(1-\lambda)\left[\ln\!\left(\frac{1}{1-\lambda}\right)\right]^2\) | M1 A1 | M1 for use of KE formula (M0 if \(M\) used for mass); A1 for correct expression in any form |
# Question 5:
## Part 5(a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $(m+\delta m)(v+\delta v)+(-\delta m)(v-U)-mv=-mg\delta t$ | M1 A2 | M1 for use of Impulse-Momentum principle, dimensionally correct, with correct no. of terms; A1A1 for correct equation |
| $mv+v\delta m+m\delta v+U\delta m-v\delta m-mv=-mg\delta t$ | | Simplification step |
| $m\frac{dv}{dt}+U\frac{dm}{dt}=-mg$ | A1 | Third A1 for correct given answer correctly obtained |
## Part 5(b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $mg+U\frac{dm}{dt}=-mg$ | | Setting $\frac{dv}{dt}=g$ |
| $U\frac{dm}{dt}=-2mg$ | M1 | M1 for putting $\frac{dv}{dt}=g$ into the DE and collecting terms |
| $\int_M^m \frac{dm}{m}=\frac{-2g}{U}\int_0^t dt$ | M1 | M1 for separating variables and attempting to integrate both sides |
| $[\ln m]_M^m=\frac{-2gt}{U}$ | A1 | A1 for correct integral on both sides with correct limits |
| $m=Me^{\frac{-2gt}{U}}$ | A1 | A1 for GIVEN ANSWER correctly obtained |
## Part 5(c):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $(1-\lambda)M=Me^{\frac{-2gT}{U}}$ | M1 | M1 for putting $m=(1-\lambda)M$ and solving for $T$ |
| $T=\frac{U}{2g}\ln\!\left(\frac{1}{1-\lambda}\right)$ | A1 | A1 for correct expression for $T$ |
| $v=gT=\frac{U}{2}\ln\!\left(\frac{1}{1-\lambda}\right)$ | M1 A1 | M1 for using $v=u+at$ with $u=0,\, a=g,\, t=T$; A1 for correct $v$ |
| $\text{KE}=\frac{1}{2}(1-\lambda)M\left[\frac{U}{2}\ln\!\left(\frac{1}{1-\lambda}\right)\right]^2=\frac{1}{8}MU^2(1-\lambda)\left[\ln\!\left(\frac{1}{1-\lambda}\right)\right]^2$ | M1 A1 | M1 for use of KE formula (M0 if $M$ used for mass); A1 for correct expression in any form |
---5. At time $t = 0$ a rocket is launched. The rocket has initial mass $M$, of which mass $\lambda M$, $0 < \lambda < 1$, is fuel. The rocket is launched vertically upwards, from rest, from the surface of the Earth. The rocket burns fuel and the burnt fuel is ejected vertically downwards with constant speed $U$ relative to the rocket. At time $t$, the rocket has mass $m$ and velocity $v$. Ignoring air resistance and any variation in $g$,
\begin{enumerate}[label=(\alph*)]
\item show, from first principles, that until all the fuel is used,
$$m \frac { \mathrm {~d} v } { \mathrm {~d} t } + U \frac { \mathrm {~d} m } { \mathrm {~d} t } = - m g$$
The rocket accelerates vertically upwards with constant acceleration $g$.
\item Show that $m = M \mathrm { e } ^ { \frac { - 2 g t } { U } }$
\item Find, in terms of $M , U$ and $\lambda$, an expression for the kinetic energy of the rocket at the instant when all of the fuel has been used.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2018 Q5 [14]}}