| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2015 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Variable mass problems (mass increasing) |
| Difficulty | Challenging +1.8 This is a variable mass mechanics problem requiring application of the rocket equation F = d(mv)/dt with mass increasing exponentially. Part (a) requires careful differentiation of momentum and understanding that picked-up droplets have zero velocity, yielding a non-standard differential equation. Part (b) requires solving this DE with given initial conditions. While the topic (M5 variable mass) is advanced, the mathematical steps are systematic once the correct framework is applied, making it challenging but not exceptional for Further Maths students. |
| Spec | 4.10a General/particular solutions: of differential equations6.03a Linear momentum: p = mv6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Mass at time \(t\) is \(m_0 e^{kt}\), so \(\frac{dm}{dt} = km_0 e^{kt}\) | B1 | Correct derivative of mass |
| The droplets are at rest, so momentum principle: \(\frac{d}{dt}(mv) = -mg\) | M1 | Applying Newton's second law in the form \(\frac{d(mv)}{dt}\) |
| \(\frac{d}{dt}(m_0 e^{kt} \cdot v) = -m_0 e^{kt} g\) | A1 | Correct equation |
| \(m_0 e^{kt}\frac{dv}{dt} + km_0 e^{kt} v = -m_0 e^{kt} g\) | M1 | Expanding derivative using product rule |
| Dividing through by \(m_0 e^{kt}\): | M1 | Dividing to simplify |
| \(kv + \frac{dv}{dt} = -g\) | A1 | Correct result shown \(\square\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Integrating factor: \(e^{\int k\, dt} = e^{kt}\) | M1 | Correct integrating factor |
| \(\frac{d}{dt}(ve^{kt}) = -ge^{kt}\) | M1 | Multiplying through and recognising exact derivative |
| \(ve^{kt} = -\frac{g}{k}e^{kt} + C\) | A1 | Correct integration |
| \(v = -\frac{g}{k} + Ce^{-kt}\) | ||
| At \(t=0\), \(v = \frac{g}{k}\): \(\frac{g}{k} = -\frac{g}{k} + C \Rightarrow C = \frac{2g}{k}\) | M1 | Applying initial condition |
| \(v = -\frac{g}{k} + \frac{2g}{k}e^{-kt}\) | A1 | Correct expression for \(v\) |
| At greatest height, \(v = 0\): \(\frac{g}{k} = \frac{2g}{k}e^{-kt} \Rightarrow e^{-kt} = \frac{1}{2}\) | M1 | Setting \(v=0\) |
| \(e^{kt} = 2\), so mass \(= m_0 e^{kt} = 2m_0\) | A1 | Correct final answer |
# Question 4:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Mass at time $t$ is $m_0 e^{kt}$, so $\frac{dm}{dt} = km_0 e^{kt}$ | B1 | Correct derivative of mass |
| The droplets are at rest, so momentum principle: $\frac{d}{dt}(mv) = -mg$ | M1 | Applying Newton's second law in the form $\frac{d(mv)}{dt}$ |
| $\frac{d}{dt}(m_0 e^{kt} \cdot v) = -m_0 e^{kt} g$ | A1 | Correct equation |
| $m_0 e^{kt}\frac{dv}{dt} + km_0 e^{kt} v = -m_0 e^{kt} g$ | M1 | Expanding derivative using product rule |
| Dividing through by $m_0 e^{kt}$: | M1 | Dividing to simplify |
| $kv + \frac{dv}{dt} = -g$ | A1 | Correct result shown $\square$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Integrating factor: $e^{\int k\, dt} = e^{kt}$ | M1 | Correct integrating factor |
| $\frac{d}{dt}(ve^{kt}) = -ge^{kt}$ | M1 | Multiplying through and recognising exact derivative |
| $ve^{kt} = -\frac{g}{k}e^{kt} + C$ | A1 | Correct integration |
| $v = -\frac{g}{k} + Ce^{-kt}$ | | |
| At $t=0$, $v = \frac{g}{k}$: $\frac{g}{k} = -\frac{g}{k} + C \Rightarrow C = \frac{2g}{k}$ | M1 | Applying initial condition |
| $v = -\frac{g}{k} + \frac{2g}{k}e^{-kt}$ | A1 | Correct expression for $v$ |
| At greatest height, $v = 0$: $\frac{g}{k} = \frac{2g}{k}e^{-kt} \Rightarrow e^{-kt} = \frac{1}{2}$ | M1 | Setting $v=0$ |
| $e^{kt} = 2$, so mass $= m_0 e^{kt} = 2m_0$ | A1 | Correct final answer |
---4. A particle $P$, whose initial mass is $m _ { 0 }$, is projected vertically upwards from the ground at time $t = 0$ with speed $\frac { g } { k }$, where $k$ is a constant. As the particle moves upwards it gains mass by picking up small droplets of moisture from the atmosphere. The droplets are at rest before they are picked up. At time $t$ the speed of $P$ is $v$ and its mass has increased to $m _ { 0 } \mathrm { e } ^ { k t }$. Assuming that, during the motion, the acceleration due to gravity is constant,
\begin{enumerate}[label=(\alph*)]
\item show that, while $P$ is moving upwards,
$$k v + \frac { \mathrm { d } v } { \mathrm {~d} t } = - g$$
\item find, in terms of $m _ { 0 }$, the mass of $P$ when it reaches its greatest height above the ground.\\
(6)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2015 Q4 [12]}}