| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2009 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Variable resistance or force differential equation |
| Difficulty | Challenging +1.2 This is a multi-step M3 work-energy problem requiring derivation of a differential equation from power/resistance principles, separation of variables with integration (including partial fractions), and numerical methods. While technically demanding with several components, it follows standard M3 patterns: establishing the resistance constant from maximum speed, applying P=Fv, and using F=ma to get the differential equation. The integration in part (b) is routine for M3 students familiar with partial fractions, and the trapezium rule application is straightforward. More challenging than average due to length and algebraic manipulation, but represents expected M3 material without requiring novel insight. |
| Spec | 1.09f Trapezium rule: numerical integration6.02l Power and velocity: P = Fv6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks |
|---|---|
| At max v, driving force = resistance | B1 |
| Driving force \(= \frac{80}{v}\) | |
| \(\Rightarrow \frac{80}{20} = k \times 20^2 \Rightarrow k = \frac{1}{100}\) | M1A1 |
| \(F = ma \Rightarrow 100a = \frac{80}{v} - kv^2 \left(= \frac{8000 - v^3}{100v}\right)\) | M1 |
| \(\Rightarrow v\frac{dv}{dx} = \frac{8000 - v^3}{10000v}\) | A1 |
| Answer | Marks |
|---|---|
| \(\int_4^8 \frac{10000v^2}{8000 - v^3} dv = \int_0^D 1dx\) | M1A1 |
| \(D = \left[-\frac{10000}{3}\ln(8000 - v^3)\right]_4^8\) | A1 |
| \(= \left(-\frac{10000}{3}\ln\frac{7488}{7936}\right) = 193.7.... \approx 194\) m (accept 190) | M1A1 |
| Answer | Marks |
|---|---|
| \(\frac{dv}{dt} = \frac{8000 - v^3}{10000v} \Rightarrow \int_0^t 1dt = \int_4^8 \frac{10000v}{8000 - v^3} dv\) | M1A1 |
| \(\Rightarrow T \approx \frac{1}{2} \times 2 \times 10000 \times \left(\frac{4}{7936} + \frac{2 \times 6}{7784} + \frac{8}{7488}\right)\) | M1 |
| \(\Rightarrow T(= 31.1409....) \approx 31\) | A1 |
## Part (a)
At max v, driving force = resistance | B1 |
Driving force $= \frac{80}{v}$ | |
$\Rightarrow \frac{80}{20} = k \times 20^2 \Rightarrow k = \frac{1}{100}$ | M1A1 |
$F = ma \Rightarrow 100a = \frac{80}{v} - kv^2 \left(= \frac{8000 - v^3}{100v}\right)$ | M1 |
$\Rightarrow v\frac{dv}{dx} = \frac{8000 - v^3}{10000v}$ | A1 |
## Part (b)
$\int_4^8 \frac{10000v^2}{8000 - v^3} dv = \int_0^D 1dx$ | M1A1 |
$D = \left[-\frac{10000}{3}\ln(8000 - v^3)\right]_4^8$ | A1 |
$= \left(-\frac{10000}{3}\ln\frac{7488}{7936}\right) = 193.7.... \approx 194$ m (accept 190) | M1A1 |
## Part (c)
$\frac{dv}{dt} = \frac{8000 - v^3}{10000v} \Rightarrow \int_0^t 1dt = \int_4^8 \frac{10000v}{8000 - v^3} dv$ | M1A1 |
$\Rightarrow T \approx \frac{1}{2} \times 2 \times 10000 \times \left(\frac{4}{7936} + \frac{2 \times 6}{7784} + \frac{8}{7488}\right)$ | M1 |
$\Rightarrow T(= 31.1409....) \approx 31$ | A1 |
---A cyclist and her bicycle have a combined mass of $100$ kg. She is working at a constant rate of $80$ W and is moving in a straight line on a horizontal road. The resistance to motion is proportional to the square of her speed. Her initial speed is $4$ m s$^{-1}$ and her maximum possible speed under these conditions is $20$ m s$^{-1}$. When she is at a distance $x$ m from a fixed point $O$ on the road, she is moving with speed $v$ m s$^{-1}$ away from $O$.
\begin{enumerate}[label=(\alph*)]
\item Show that
$$v \frac{dv}{dx} = \frac{8000 - v^3}{10000v}.$$ [5]
\item Find the distance she travels as her speed increases from $4$ m s$^{-1}$ to $8$ m s$^{-1}$. [5]
\item Use the trapezium rule, with 2 intervals, to estimate how long it takes for her speed to increase from $4$ m s$^{-1}$ to $8$ m s$^{-1}$. [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2009 Q6 [14]}}