| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2006 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable mass problems |
| Type | Body collecting atmospheric moisture |
| Difficulty | Challenging +1.8 This is a challenging M5 variable mass problem requiring derivation of the rocket equation for an unusual scenario (mass increasing rather than decreasing), followed by differential equation manipulation and integration. While the steps are guided, it demands strong understanding of momentum principles, product rule application, and recognizing the integrating factor technique—significantly harder than standard mechanics but scaffolded enough to be accessible to well-prepared FM students. |
| Spec | 3.04a Calculate moments: about a point6.02i Conservation of energy: mechanical energy principle6.04e Rigid body equilibrium: coplanar forces6.05e Radial/tangential acceleration |
| Answer | Marks |
|---|---|
| (a) \(O\uparrow 05m\), \(u\uparrow (m)\), \((m + \delta m)(u + \delta u) - mu = -mg\delta t\) | M1, A2, L0 |
| \(mu\frac{du}{dt} + u\frac{dm}{dt} = -mg\) | M1, A1 (5) |
| (b) \(m = Me^{kt} \Rightarrow \frac{dm}{dt} = kMe^{kt}\) | B1 |
| Hence \(Me^{kt}\frac{du}{dt} + ku.kMe^{kt} = -Me^{kt}g\) | M1 |
| \(\Rightarrow \frac{d}{dt}(ue^{kt}) = -ge^{kt}\) | M1, A1 (3) |
| (c) \(ue^{kt} = -g\int e^{kt}dt = -\frac{g}{k}e^{kt}(+C)\) | M1, A1 |
| \(t = 0, u = \frac{3}{2k} \Rightarrow C = \frac{3g}{2k}\) | B1 |
| At height yr \(u = 0\): \(-\frac{g}{k}e^{kt} + \frac{3g}{2k} = 0\) | M1 |
| \(\Rightarrow e^{kt} = \frac{3}{2}\) | A1 |
| Hence \(m = Me^{kt} = \frac{3M}{2}\) | M1, A1 (7) |
(a) $O\uparrow 05m$, $u\uparrow (m)$, $(m + \delta m)(u + \delta u) - mu = -mg\delta t$ | M1, A2, L0 |
$mu\frac{du}{dt} + u\frac{dm}{dt} = -mg$ | M1, A1 (5) |
(b) $m = Me^{kt} \Rightarrow \frac{dm}{dt} = kMe^{kt}$ | B1 |
Hence $Me^{kt}\frac{du}{dt} + ku.kMe^{kt} = -Me^{kt}g$ | M1 |
$\Rightarrow \frac{d}{dt}(ue^{kt}) = -ge^{kt}$ | M1, A1 (3) |
(c) $ue^{kt} = -g\int e^{kt}dt = -\frac{g}{k}e^{kt}(+C)$ | M1, A1 |
$t = 0, u = \frac{3}{2k} \Rightarrow C = \frac{3g}{2k}$ | B1 |
At height yr $u = 0$: $-\frac{g}{k}e^{kt} + \frac{3g}{2k} = 0$ | M1 |
$\Rightarrow e^{kt} = \frac{3}{2}$ | A1 |
Hence $m = Me^{kt} = \frac{3M}{2}$ | M1, A1 (7) |7. At time $t = 0$, a small body is projected vertically upwards. While ascending it picks up small drops of moisture from the atmosphere. The drops of moisture are at rest before they are picked up. At time $t$, the combined body $P$ has mass $m$ and speed $v$.
\begin{enumerate}[label=(\alph*)]
\item Show that, while $P$ is moving upwards, $m \frac { \mathrm {~d} v } { \mathrm {~d} t } + v \frac { \mathrm {~d} m } { \mathrm {~d} t } = - m g$.
The initial mass of $P$ is $M$, and $m = M \mathrm { e } ^ { k t }$, where $k$ is a positive constant.
\item Show that, while $P$ is moving upwards, $\frac { \mathrm { d } } { \mathrm { d } t } \left( v \mathrm { e } ^ { k t } \right) = - g \mathrm { e } ^ { k t }$.
Given that the initial projection speed of $P$ is $\frac { g } { 2 k }$,
\item find, in terms of $M$, the mass of $P$ when it reaches its highest point.\\
(Total 15 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2006 Q7 [15]}}