| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Given velocity function find force |
| Difficulty | Challenging +1.2 This M3 question requires applying F=ma with variable resistance, using the chain rule (v dv/dx = a), and integrating to find work done. While it involves multiple steps and careful manipulation of the given velocity equation, the techniques are standard for M3 and the question provides significant scaffolding through the given v² expression and work integral formula. The algebraic manipulation is moderately demanding but follows predictable patterns for this module. |
| Spec | 6.02c Work by variable force: using integration6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks |
|---|---|
| (a) \(mv\frac{dv}{dx} = -(mg + mf(x))\) | M1 A1 |
| \(v\frac{dv}{dx} = -g - f(x)\) | M1 A1 |
| \(v^2 = 2e^{-2gx} - 1\), so \(2v\frac{dv}{dx} = -4ge^{-2gx}\) | M1 A1 |
| \(-2ge^{-2gx} = -g - f(x)\) | M1 A1 |
| \(f(x) = g(2e^{-2gx} - 1)\) | |
| (b) \(v = 0\) when \(2e^{-2gx} = 1\) | A1; M1 A1 |
| \(x = \frac{1}{2g}\ln 2\) | |
| (c) W.D. \(= m[-e^{-2gx} - gx]_0^{\ln 2/2g} = m(-e^{-\ln 2} - \frac{1}{2}\ln 2 + 1) = \frac{1}{2}m(1 - \ln 2)\) | M1 A1 A1 |
(a) $mv\frac{dv}{dx} = -(mg + mf(x))$ | M1 A1 |
$v\frac{dv}{dx} = -g - f(x)$ | M1 A1 |
$v^2 = 2e^{-2gx} - 1$, so $2v\frac{dv}{dx} = -4ge^{-2gx}$ | M1 A1 |
$-2ge^{-2gx} = -g - f(x)$ | M1 A1 |
$f(x) = g(2e^{-2gx} - 1)$ | |
(b) $v = 0$ when $2e^{-2gx} = 1$ | A1; M1 A1 |
$x = \frac{1}{2g}\ln 2$ | |
(c) W.D. $= m[-e^{-2gx} - gx]_0^{\ln 2/2g} = m(-e^{-\ln 2} - \frac{1}{2}\ln 2 + 1) = \frac{1}{2}m(1 - \ln 2)$ | M1 A1 A1 |
**Total: 10 marks**
---A particle $P$ of mass $m$ kg moves vertically upwards under gravity, starting from ground level. It is acted on by a resistive force of magnitude $m f(x)$ N, where $f(x)$ is a function of the height $x$ m of $P$ above the ground. When $P$ is at this height, its upward speed $v$ ms$^{-1}$ is given by
$$v^2 = 2e^{-2gx} - 1.$$
\begin{enumerate}[label=(\alph*)]
\item Write down a differential equation for the motion of $P$ and hence determine $f(x)$ in terms of $g$ and $x$. [5 marks]
\item Show that the greatest height reached by $P$ above the ground is $\frac{1}{2g} \ln 2$ m. [2 marks]
\end{enumerate}
Given that the work, in J, done by $P$ against the resisting force as it moves from ground level to a point $H$ m above the ground is equal to $\int_0^H m f(x) dx$,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item show that the total work done by $P$ against the resistance during its upward motion is $\frac{1}{2}m(1 - \ln 2)$ J. [3 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q2 [10]}}