| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2009 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Rocket/thrust problems (mass decreasing) |
| Difficulty | Challenging +1.8 This is a variable mass rocket equation problem requiring derivation of the fundamental rocket equation using momentum conservation, followed by application of the chain rule. While M5/Further Mechanics topics are inherently advanced, this is a relatively standard textbook derivation with clear guidance ('show that') and straightforward differentiation in part (b). The conceptual leap needed is moderate but well-scaffolded. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Momentum of system at time \(t\): \(mv\) | M1 | Setting up momentum equation |
| At time \(t + \delta t\): mass ejected \(= -\delta m\) at velocity \(v - c\) (backwards); ship has mass \(m + \delta m\), velocity \(v + \delta v\) | M1 | |
| Conservation of momentum: \(mv = (m+\delta m)(v+\delta v) + (-\delta m)(v-c)\) | M1 | |
| \(mv = mv + m\delta v + v\delta m + \delta m \delta v - v\delta m + c\delta m\) | A1 | |
| Neglect \(\delta m \delta v\): \(0 = m\delta v + c\delta m\) | M1 | |
| \(\frac{dv}{dm} = -\frac{c}{m}\) | A1 | Given answer — must show working |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(m = m_0(1-kt)\), so \(\frac{dm}{dt} = -m_0 k\) | M1 | Differentiating m |
| \(a = \frac{dv}{dt} = \frac{dv}{dm}\cdot\frac{dm}{dt} = \left(-\frac{c}{m}\right)(-m_0 k)\) | M1 | Chain rule |
| \(a = \frac{ckm_0}{m_0(1-kt)}\) | A1 | |
| \(a = \frac{ck}{1-kt}\) | A1 | cao |
# Question 3:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Momentum of system at time $t$: $mv$ | M1 | Setting up momentum equation |
| At time $t + \delta t$: mass ejected $= -\delta m$ at velocity $v - c$ (backwards); ship has mass $m + \delta m$, velocity $v + \delta v$ | M1 | |
| Conservation of momentum: $mv = (m+\delta m)(v+\delta v) + (-\delta m)(v-c)$ | M1 | |
| $mv = mv + m\delta v + v\delta m + \delta m \delta v - v\delta m + c\delta m$ | A1 | |
| Neglect $\delta m \delta v$: $0 = m\delta v + c\delta m$ | M1 | |
| $\frac{dv}{dm} = -\frac{c}{m}$ | A1 | **Given answer — must show working** |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $m = m_0(1-kt)$, so $\frac{dm}{dt} = -m_0 k$ | M1 | Differentiating m |
| $a = \frac{dv}{dt} = \frac{dv}{dm}\cdot\frac{dm}{dt} = \left(-\frac{c}{m}\right)(-m_0 k)$ | M1 | Chain rule |
| $a = \frac{ckm_0}{m_0(1-kt)}$ | A1 | |
| $a = \frac{ck}{1-kt}$ | A1 | cao |\begin{enumerate}
\item A spaceship is moving in a straight line in deep space and needs to increase its speed. This is done by ejecting fuel backwards from the spaceship at a constant speed $c$ relative to the spaceship. When the speed of the spaceship is $v$, its mass is $m$.\\
(a) Show that, while the spaceship is ejecting fuel,
\end{enumerate}
$$\frac { \mathrm { d } v } { \mathrm {~d} m } = - \frac { c } { m } .$$
The initial mass of the spaceship is $m _ { 0 }$ and at time $t$ the mass of the spaceship is given by $m = m _ { 0 } ( 1 - k t )$, where $k$ is a positive constant.\\
(b) Find the acceleration of the spaceship at time $t$.
\hfill \mbox{\textit{Edexcel M5 2009 Q3 [9]}}