| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2004 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Force depends on velocity v |
| Difficulty | Challenging +1.8 This M4 variable force question requires understanding power-work relationships, chain rule manipulation (v dv/dx), and integration of a non-standard form. Part (a) needs insight to connect constant power to force, parts (b-c) involve careful integration and algebraic manipulation. More demanding than typical M4 questions but follows established patterns for this advanced module. |
| Spec | 6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(F = \frac{kmg}{v}\) | B1 | |
| \(R(\uparrow),\ F - mg = ma\) | M1 | |
| \(\frac{kmg}{v} - mg = mv\frac{dv}{dx}\) | M1 | |
| \(g(k-v) = v^2\frac{dv}{dx}\) | A1 | (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(g\,dx = \frac{v^2}{k-v}\,dv\) | M1 | |
| \(\int g\,dx = \int -v - k + \frac{k^2}{k-v}\,dv\) | M1 A1 | |
| \(gx = -\frac{v^2}{2} - kv - k^2\ln(k-v) + c\) | M1 A1 | |
| \(x=0,\ v=0\): \(\quad 0 = 0 - 0 - k^2\ln k + c\) | M1 | |
| \(c = k^2\ln k\) | ||
| \(gx = -\frac{v^2}{2} - kv - k^2\ln\!\left(\frac{k}{k-v}\right)\) | A1 | (7 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| Work done by engine = Energy gain | ||
| \(kmgt = \frac{1}{2}mv^2 + mgx\) | M1 A1 | |
| \(kmgt = mk^2\ln\!\left(\frac{k}{k-v}\right) - mkv\) | M1 | |
| \(\Rightarrow t = \frac{k}{g}\ln\!\left(\frac{k}{k-v}\right) - \frac{v}{g}\) | A1 | (5 marks) |
# Question 6:
## Part (a)
| Answer/Working | Marks | Notes |
|---|---|---|
| $F = \frac{kmg}{v}$ | B1 | |
| $R(\uparrow),\ F - mg = ma$ | M1 | |
| $\frac{kmg}{v} - mg = mv\frac{dv}{dx}$ | M1 | |
| $g(k-v) = v^2\frac{dv}{dx}$ | A1 | (4 marks) |
## Part (b)
| Answer/Working | Marks | Notes |
|---|---|---|
| $g\,dx = \frac{v^2}{k-v}\,dv$ | M1 | |
| $\int g\,dx = \int -v - k + \frac{k^2}{k-v}\,dv$ | M1 A1 | |
| $gx = -\frac{v^2}{2} - kv - k^2\ln(k-v) + c$ | M1 A1 | |
| $x=0,\ v=0$: $\quad 0 = 0 - 0 - k^2\ln k + c$ | M1 | |
| $c = k^2\ln k$ | | |
| $gx = -\frac{v^2}{2} - kv - k^2\ln\!\left(\frac{k}{k-v}\right)$ | A1 | (7 marks) |
## Part (c)
| Answer/Working | Marks | Notes |
|---|---|---|
| Work done by engine = Energy gain | | |
| $kmgt = \frac{1}{2}mv^2 + mgx$ | M1 A1 | |
| $kmgt = mk^2\ln\!\left(\frac{k}{k-v}\right) - mkv$ | M1 | |
| $\Rightarrow t = \frac{k}{g}\ln\!\left(\frac{k}{k-v}\right) - \frac{v}{g}$ | A1 | (5 marks) |
**Total: 16 marks**6. A particle $P$ of mass $m$ is attached to one end of a light inextensible string and hangs at rest at time $t = 0$. The other end of the string is then raised vertically by an engine which is working at a constant rate $k m g$, where $k > 0$. At time $t$, the distance of $P$ above its initial position is $x$, and $P$ is moving upwards with speed $v$.
\begin{enumerate}[label=(\alph*)]
\item Show that $v ^ { 2 } \frac { \mathrm {~d} v } { \mathrm {~d} x } = ( k - v ) g$.
\item Show that $g x = k ^ { 2 } \ln \left( \frac { k } { k - v } \right) - k v - \frac { 1 } { 2 } v ^ { 2 }$.
\item Hence, or otherwise, find $t$ in terms of $k , v$ and $g$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2004 Q6 [16]}}