| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2007 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Variable mass problems (mass increasing) |
| Difficulty | Challenging +1.8 This M5 variable mass mechanics problem requires applying momentum principles to a non-standard situation (leaking boat), deriving and solving a differential equation with variable coefficients, and interpreting the physical constraints. While systematic, it demands careful handling of the momentum equation for variable mass systems, integration techniques beyond basic A-level, and multi-step reasoning across three connected parts—significantly harder than typical mechanics questions but follows established M5 patterns. |
| Spec | 4.10a General/particular solutions: of differential equations4.10b Model with differential equations: kinematics and other contexts6.02a Work done: concept and definition6.03a Linear momentum: p = mv6.03b Conservation of momentum: 1D two particles |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \((m+\delta m)(v+\delta v) - mv = -2\lambda v\,\delta t\) | M1 A1 | |
| \(m\frac{dv}{dt} + v\frac{dm}{dt} = -2\lambda v\) | ||
| \(\frac{dm}{dt} = \lambda\); \(m = M + \lambda t\) | B1; B1 | |
| \((M+\lambda t)\frac{dv}{dt} + 3\lambda v = 0\) | D M1 A1 | (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(-\int\frac{dv}{3\lambda v} = \int\frac{dt}{M+\lambda t}\) | M1 | |
| \(-\frac{1}{3\lambda}[\ln v]_u^v = \frac{1}{\lambda}[\ln(M+\lambda t)]_0^T\) | DM1 A1 | |
| \(\frac{1}{3}\ln 2 = \ln\frac{(M+\lambda T)}{M}\) | DM1 | |
| \(T = \frac{M}{\lambda}(2^{\frac{1}{3}}-1)\) | DM1 A1 | (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Sinks at \(T_s = \frac{M}{\lambda}\); reaches speed \(\frac{1}{2}U\) at \(T = \frac{M}{\lambda}(2^{\frac{1}{3}}-1)\) | M1 | |
| Since \((2^{\frac{1}{3}}-1) < 1\), \(T < T_s\), i.e. reaches speed \(\frac{1}{2}U\) before it sinks | A1 c.s.o. | (2),(14) |
## Question 7:
### Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $(m+\delta m)(v+\delta v) - mv = -2\lambda v\,\delta t$ | M1 A1 | |
| $m\frac{dv}{dt} + v\frac{dm}{dt} = -2\lambda v$ | | |
| $\frac{dm}{dt} = \lambda$; $m = M + \lambda t$ | B1; B1 | |
| $(M+\lambda t)\frac{dv}{dt} + 3\lambda v = 0$ | D M1 A1 | **(6)** |
### Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $-\int\frac{dv}{3\lambda v} = \int\frac{dt}{M+\lambda t}$ | M1 | |
| $-\frac{1}{3\lambda}[\ln v]_u^v = \frac{1}{\lambda}[\ln(M+\lambda t)]_0^T$ | DM1 A1 | |
| $\frac{1}{3}\ln 2 = \ln\frac{(M+\lambda T)}{M}$ | DM1 | |
| $T = \frac{M}{\lambda}(2^{\frac{1}{3}}-1)$ | DM1 A1 | **(6)** |
### Part (c):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Sinks at $T_s = \frac{M}{\lambda}$; reaches speed $\frac{1}{2}U$ at $T = \frac{M}{\lambda}(2^{\frac{1}{3}}-1)$ | M1 | |
| Since $(2^{\frac{1}{3}}-1) < 1$, $T < T_s$, i.e. reaches speed $\frac{1}{2}U$ before it sinks | A1 c.s.o. | **(2),(14)** |
---7. A motor boat of mass $M$ is moving in a straight line, with its engine switched off, across a stretch of still water. The boat is moving with speed $U$ when, at time $t = 0$, it develops a leak. The water comes in at a constant rate so that at time $t$, the mass of water in the boat is $\lambda t$. At time $t$ the speed of the boat is $v$ and it experiences a total resistance to motion of magnitude $2 \lambda v$.
\begin{enumerate}[label=(\alph*)]
\item Show that $( M + \lambda t ) \frac { \mathrm { d } v } { \mathrm {~d} t } + 3 \lambda v = 0$.\\
(6)
\item Show that the time taken for the speed of the boat to reduce to $\frac { 1 } { 2 } U$ is $\frac { M } { \lambda } \left( 2 ^ { \frac { 1 } { 3 } } - 1 \right)$.\\
(6)
The boat sinks when the mass of water inside the boat is $M$.
\item Show that the boat does not sink before the speed of the boat is $\frac { 1 } { 2 } U$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2007 Q7 [14]}}