| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Given velocity function find force |
| Difficulty | Standard +0.3 This is a straightforward M3 variable force question requiring standard techniques: using F=ma with the chain rule (a = v dv/dx) for part (a), then separating variables and integrating for part (b). The given constant dv/dx=3 makes the integration trivial, requiring only routine application of formulas without problem-solving insight. |
| Spec | 6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dv}{dx} = 3 \Rightarrow v = 3x - 3\) | M1 A1 | Integration |
| \(a = 3(3x - 3)\) | DM1 | Using \(a = v\frac{dv}{dx}\) with their \(v\) |
| When \(x = 5, F = 0.25 \times 3(15 - 3) = 9\) N | A1 | Correct integration |
| (4) |
| Answer | Marks |
|---|---|
| M1 | |
| \(\frac{dx}{dt} = 3(x-1)\) | A1 |
| \(\int \frac{dx}{(x-1)} = \int 3dt\) | DM1 |
| \([\ln(x-1)]_2^t = 3t\) | A1 |
| \(t = \frac{1}{3}\ln 4 = 0.4620...\) | (4) |
## 1(a)
$\frac{dv}{dx} = 3 \Rightarrow v = 3x - 3$ | M1 A1 | Integration
$a = 3(3x - 3)$ | DM1 | Using $a = v\frac{dv}{dx}$ with their $v$
When $x = 5, F = 0.25 \times 3(15 - 3) = 9$ N | A1 | Correct integration
| (4) |
## 1(b)
| M1 |
$\frac{dx}{dt} = 3(x-1)$ | A1 |
$\int \frac{dx}{(x-1)} = \int 3dt$ | DM1 |
$[\ln(x-1)]_2^t = 3t$ | A1 |
$t = \frac{1}{3}\ln 4 = 0.4620...$ | (4) |
---A particle $P$ of mass $0.25$ kg is moving along the positive $x$-axis under the action of a single force. At time $t$ seconds $P$ is $x$ metres from the origin $O$ and is moving away from $O$ with speed $v$ m s$^{-1}$ where $\frac{\mathrm{d}v}{\mathrm{d}x} = 3$. It is given that $x = 2$ and $v = 3$ when $t = 0$
\begin{enumerate}[label=(\alph*)]
\item Find the magnitude of the force acting on $P$ when $x = 5$ [4]
\item Find the value of $t$ when $x = 5$ [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2014 Q1 [8]}}