| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Session | Specimen |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable mass problems |
| Type | Body collecting atmospheric moisture |
| Difficulty | Challenging +1.2 This is a variable mass mechanics problem requiring chain rule differentiation, application of Newton's second law with variable mass, and solving a first-order linear ODE. While it involves multiple steps (12 marks total) and combines calculus with mechanics, the techniques are standard for M5: the sphere volume formula leads directly to part (a), momentum considerations give part (b), and part (c) is a routine separable/integrating factor ODE. The limiting behavior in (d) is immediate. More challenging than basic mechanics but follows established M5 patterns without requiring novel insight. |
| Spec | 4.10a General/particular solutions: of differential equations4.10c Integrating factor: first order equations6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| \(m = \frac{4}{3}\pi r^3\rho\) (\(\rho\) constant) | ||
| \(\frac{dm}{dt} = \frac{4}{3}\pi \times 3r^2\frac{dr}{dt} = 4\pi r^2 \cdot kr = 3km\) | M1 A1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(mg\delta t = (m + \delta m)(v + \delta v) - mv\) | M1 | |
| \(mg = v\frac{dm}{dt} + m\frac{dv}{dt}\) | A1 | |
| \(mg = v \times 3km + m\frac{dv}{dt}\) | M1 | |
| \(g - 3kv = \frac{dv}{dt}\) | A1 | (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int dt = \int\frac{dv}{g - 3kv}\) | M1 | |
| \(t = -\frac{1}{3k}\ln(g - 3kv) + c\) | A1 | |
| At \(t = 0, v = u\): \(c = \frac{1}{3k}\ln(g - 3ku)\) | A1 | |
| \(\frac{g - 3ku}{g - 3kv} = e^{3kt}\) (or equivalent) | M1 | |
| \(v = \frac{g}{3k} - \left(\frac{g}{3k} - u\right)e^{-3kt}\) | A1 | (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| As \(t \to \infty\), \(e^{-3kt} \to 0\) | ||
| \(v \to \frac{g}{3k}\) | B1 | (1 mark) |
Total: 12 marks
## Part (a)
$m = \frac{4}{3}\pi r^3\rho$ ($\rho$ constant) | |
$\frac{dm}{dt} = \frac{4}{3}\pi \times 3r^2\frac{dr}{dt} = 4\pi r^2 \cdot kr = 3km$ | M1 A1 | (2 marks)
## Part (b)
$mg\delta t = (m + \delta m)(v + \delta v) - mv$ | M1 |
$mg = v\frac{dm}{dt} + m\frac{dv}{dt}$ | A1 |
$mg = v \times 3km + m\frac{dv}{dt}$ | M1 |
$g - 3kv = \frac{dv}{dt}$ | A1 | (4 marks)
## Part (c)
$\int dt = \int\frac{dv}{g - 3kv}$ | M1 |
$t = -\frac{1}{3k}\ln(g - 3kv) + c$ | A1 |
At $t = 0, v = u$: $c = \frac{1}{3k}\ln(g - 3ku)$ | A1 |
$\frac{g - 3ku}{g - 3kv} = e^{3kt}$ (or equivalent) | M1 |
$v = \frac{g}{3k} - \left(\frac{g}{3k} - u\right)e^{-3kt}$ | A1 | (5 marks)
## Part (d)
As $t \to \infty$, $e^{-3kt} \to 0$ | |
$v \to \frac{g}{3k}$ | B1 | (1 mark)
Total: 12 marks
---As a hailstone falls under gravity in still air, its mass increases. At time $t$ the mass of the hailstone is $m$. The hailstone is modelled as a uniform sphere of radius $r$ such that
$$\frac{dr}{dt} = kr,$$
where $k$ is a positive constant.
\begin{enumerate}[label=(\alph*)]
\item Show that $\frac{dm}{dt} = 3km$. [2]
\end{enumerate}
Assuming that there is no air resistance,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item show that the speed $v$ of the hailstone at time $t$ satisfies
$$\frac{dv}{dt} = g - 3kv.$$ [4]
\end{enumerate}
Given that the speed of the hailstone at time $t = 0$ is $u$,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item find an expression for $v$ in terms of $t$. [5]
\item Hence show that the speed of the hailstone approaches the limiting value $\frac{g}{3k}$. [1]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 Q7 [12]}}