| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2013 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Variable power or two power scenarios |
| Difficulty | Challenging +1.2 This is a standard M4 variable force question requiring power-force-velocity relationships and separable differential equations. Part (a) involves routine manipulation of P=Fv and F=ma with resistance proportional to v to reach the given result. Part (b) requires separation of variables and partial fractions—all standard M4 techniques with no novel insight needed. Slightly above average difficulty due to the algebraic manipulation and integration required, but follows a well-established template for this topic. |
| Spec | 6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Resistance \(= kv\) (proportional to speed) | B1 | Stated |
| Driving force \(= \frac{P}{v} = \frac{40000}{v}\) | B1 | Power = force × velocity |
| Newton's second law: \(\frac{40000}{v} - kv = 1200a\) | M1 | Equation of motion |
| At \(v = 40\), \(a = 0.3\): \(\frac{40000}{40} - 40k = 1200 \times 0.3\) | M1 | Substituting given values |
| \(1000 - 40k = 360 \Rightarrow 40k = 640 \Rightarrow k = 16\) | A1 | |
| So: \(\frac{40000}{v} - 16v = 1200v\frac{dv}{dx}\cdot\)... \(= 1200\frac{dv}{dt}\) | ||
| Divide by 16: \(\frac{2500}{v} - v = 75\frac{dv}{dt}\), multiply by \(v\): \(75v\frac{dv}{dt} = 2500 - v^2\) | A1 | Completion — no errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{75v}{2500 - v^2}dv = dt\) | M1 | Separating variables |
| \(\int \frac{75v}{2500-v^2}dv = \int dt\) | ||
| \(-\frac{75}{2}\ln(2500 - v^2) = t + C\) | M1 A1 | Integration; correct form |
| At \(t=0\), \(v=0\): \(C = -\frac{75}{2}\ln 2500\) | M1 | Applying initial condition |
| \(-\frac{75}{2}\ln(2500-v^2) + \frac{75}{2}\ln 2500 = t\) | ||
| \(\frac{75}{2}\ln\!\left(\frac{2500}{2500-v^2}\right) = t\) | A1 | |
| \(\frac{2500}{2500-v^2} = e^{2t/75}\) | ||
| \(v^2 = 2500\!\left(1 - e^{-2t/75}\right)\) | A1 | \(v = 50\sqrt{1-e^{-2t/75}}\) |
# Question 5:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Resistance $= kv$ (proportional to speed) | B1 | Stated |
| Driving force $= \frac{P}{v} = \frac{40000}{v}$ | B1 | Power = force × velocity |
| Newton's second law: $\frac{40000}{v} - kv = 1200a$ | M1 | Equation of motion |
| At $v = 40$, $a = 0.3$: $\frac{40000}{40} - 40k = 1200 \times 0.3$ | M1 | Substituting given values |
| $1000 - 40k = 360 \Rightarrow 40k = 640 \Rightarrow k = 16$ | A1 | |
| So: $\frac{40000}{v} - 16v = 1200v\frac{dv}{dx}\cdot$... $= 1200\frac{dv}{dt}$ | | |
| Divide by 16: $\frac{2500}{v} - v = 75\frac{dv}{dt}$, multiply by $v$: $75v\frac{dv}{dt} = 2500 - v^2$ | A1 | Completion — no errors seen |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{75v}{2500 - v^2}dv = dt$ | M1 | Separating variables |
| $\int \frac{75v}{2500-v^2}dv = \int dt$ | | |
| $-\frac{75}{2}\ln(2500 - v^2) = t + C$ | M1 A1 | Integration; correct form |
| At $t=0$, $v=0$: $C = -\frac{75}{2}\ln 2500$ | M1 | Applying initial condition |
| $-\frac{75}{2}\ln(2500-v^2) + \frac{75}{2}\ln 2500 = t$ | | |
| $\frac{75}{2}\ln\!\left(\frac{2500}{2500-v^2}\right) = t$ | A1 | |
| $\frac{2500}{2500-v^2} = e^{2t/75}$ | | |
| $v^2 = 2500\!\left(1 - e^{-2t/75}\right)$ | A1 | $v = 50\sqrt{1-e^{-2t/75}}$ |
---5. A van of mass 1200 kg travels along a straight horizontal road against a resistance to motion which is proportional to the speed of the van. The engine of the van is working at a constant rate of 40 kW . The van starts from rest at time $t = 0$. At time $t$ seconds, the speed of the van is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. When the speed of the van is $40 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, the acceleration of the van is $0.3 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that
$$75 v \frac { \mathrm {~d} v } { \mathrm {~d} t } = 2500 - v ^ { 2 }$$
\item Find $v$ in terms of $t$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2013 Q5 [12]}}