| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2005 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Air resistance kv² - horizontal motion or engine power |
| Difficulty | Standard +0.8 Part (a) is trivial equilibrium (2 marks). Part (b) requires setting up F=ma with v²-resistance, separating variables in the form v dv/(F-kv²), integrating to find distance, then substituting V²=F/k and simplifying logarithms—this is a standard M4 differential equation but demands careful algebraic manipulation across 9 marks, placing it moderately above average difficulty. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y)6.02k Power: rate of doing work6.02l Power and velocity: P = Fv6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks |
|---|---|
| For constant speed: \(F - kv^2 = 0\) \(\Rightarrow\) \(v = \sqrt{\frac{F}{k}}\) | M1 A1 (2) |
| Answer | Marks |
|---|---|
| \(x = -\frac{M}{2k} \ln(F - kv^2)\) (+ C) | M1 A1, M1, A1, M1 A1 |
| Answer | Marks |
|---|---|
| \(= -\frac{M}{2k} \ln\frac{3}{4}\) | M1, A1 (7), M1 A1 (9), (11) |
## Part (a)
For constant speed: $F - kv^2 = 0$ $\Rightarrow$ $v = \sqrt{\frac{F}{k}}$ | M1 A1 (2) |
## Part (b)
$F - kv^2 = Ma$
$\Rightarrow$ $F - kv^2 = M\frac{dv}{dx}$
$\int dx = M \int \frac{dv}{F - kv^2}$
$x = -\frac{M}{2k} \ln(F - kv^2)$ (+ C) | M1 A1, M1, A1, M1 A1 |
$x = 0, v = 0$ $\Rightarrow$ $C = \frac{M}{2k} \ln F$
$x = \frac{M}{2k} \ln\left[\frac{F}{F - kv^2}\right]$
$x = \frac{M}{2k} \ln\left(\frac{F}{F - k \cdot \frac{E}{4k}}\right)$
$= -\frac{M}{2k} \ln\frac{3}{4}$ | M1, A1 (7), M1 A1 (9), (11) |
---A lorry of mass $M$ is moving along a straight horizontal road. The engine produces a constant driving force of magnitude $F$. The total resistance to motion is modelled as having magnitude $kv^2$, where $k$ is a constant, and $v$ is the speed of the lorry.
Given the lorry moves with constant speed $V$,
\begin{enumerate}[label=(\alph*)]
\item show that $V = \sqrt{\frac{F}{k}}$. [2]
\end{enumerate}
Given instead that the lorry starts from rest,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item show that the distance travelled by the lorry in attaining a speed of $\frac{1}{2}V$ is
$$\frac{M}{2k}\ln\left(\frac{4}{3}\right).$$ [9]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2005 Q4 [11]}}