| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2007 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Find acceleration given power |
| Difficulty | Standard +0.3 This is a standard M4 variable acceleration problem requiring the power-resistance equation and separable differential equations. Part (a) is straightforward derivation from P=Fv. Part (b) involves routine separation of variables and integration using partial fractions or logarithms—techniques well-practiced at this level. The setup is clear with no geometric complexity or novel insight required, making it slightly easier than average. |
| Spec | 4.10b Model with differential equations: kinematics and other contexts6.02l Power and velocity: P = Fv |
| Answer | Marks |
|---|---|
| \(F = \frac{Ru}{v}\) | B1 |
| \(R(\rightarrow), \frac{Ru}{v} - R = M\frac{dv}{dt}\) | M1 |
| \(R(u-v) = Mv\frac{dv}{dt}\) * | A1 |
| Total: 3 marks |
| Answer | Marks |
|---|---|
| \(\int_0^t dt = \frac{M}{R} \int_4^v \frac{v dv}{u - v}\) | M1A1 |
| \(\Rightarrow T = \frac{M}{R} \int_4^u \left[-1 + \frac{u}{u-v}\right] dv\) | DM1 |
| \(= \frac{M}{R}[-v - u\ln(u-v)]_4^u\) | A1 |
| \(= \frac{M}{R}\left[\frac{u}{3} - u\ln\left(\frac{2u}{3}\right) + \frac{u}{4} + u\ln\left(\frac{3u}{4}\right)\right]\) | M1 |
| \(= \frac{Mu}{R}\left[-\frac{1}{12} + \ln\frac{9}{8}\right]\) | M1 |
| \(\text{Hence } k = \ln\frac{9}{8} - \frac{1}{12}\) | A1 |
| \(\left(C = -\frac{Mu}{R}\left(\ln\frac{3u}{4} + \frac{1}{4}\right)\right)\) | M1 |
| Total: 7 marks |
## Part (a)
| $F = \frac{Ru}{v}$ | B1 |
| $R(\rightarrow), \frac{Ru}{v} - R = M\frac{dv}{dt}$ | M1 |
| $R(u-v) = Mv\frac{dv}{dt}$ * | A1 |
| **Total: 3 marks** | |
## Part (b)
| $\int_0^t dt = \frac{M}{R} \int_4^v \frac{v dv}{u - v}$ | M1A1 |
| $\Rightarrow T = \frac{M}{R} \int_4^u \left[-1 + \frac{u}{u-v}\right] dv$ | DM1 |
| $= \frac{M}{R}[-v - u\ln(u-v)]_4^u$ | A1 |
| $= \frac{M}{R}\left[\frac{u}{3} - u\ln\left(\frac{2u}{3}\right) + \frac{u}{4} + u\ln\left(\frac{3u}{4}\right)\right]$ | M1 |
| $= \frac{Mu}{R}\left[-\frac{1}{12} + \ln\frac{9}{8}\right]$ | M1 |
| $\text{Hence } k = \ln\frac{9}{8} - \frac{1}{12}$ | A1 |
| $\left(C = -\frac{Mu}{R}\left(\ln\frac{3u}{4} + \frac{1}{4}\right)\right)$ | M1 |
| **Total: 7 marks** | |
### Guidance:
**a)** B1: Correct expression involving the driving force. M1: Use of $F = ma$ to form a differential equation. Condone sign errors. A1: Rearrange to given form.
**b)** M1: Separate the variables. A1: Separation correct (limits not necessarily seen at this stage). DM1: Attempt a complete integration process. A1: Integration correct. M1: Correct use of both limits – substitute and subtract. Condone wrong order. M1: Simplify to find $k$ from an expression involving a logarithm. A1: Answer as given, or exact equivalent. Need to see $k = \ln A + B$.
---A lorry of mass $M$ moves along a straight horizontal road against a constant resistance of magnitude $R$. The engine of the lorry works at a constant rate $RU$, where $U$ is a constant. At time $t$, the lorry is moving with speed $v$.
\begin{enumerate}[label=(\alph*)]
\item Show that $Mv\frac{dv}{dt} = R(U - v)$. [3]
\end{enumerate}
At time $t = 0$, the lorry has speed $\frac{1}{4}U$ and the time taken by the lorry to attain a speed of $\frac{3}{4}U$ is $\frac{kMU}{R}$, where $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the exact value of $k$. [7]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2007 Q2 [10]}}