| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Inverse power force - non-gravitational context |
| Difficulty | Challenging +1.8 This M3 question requires applying F=ma with variable force, using energy methods or integration to derive a velocity-distance relationship, then solving a non-trivial time integral with a provided substitution. Part (a) is a standard 'show that' requiring work-energy principles (6 marks suggests multiple steps). Part (b) requires setting up dt = dx/v and integrating with the given result, which is procedurally demanding but the hard integral is provided. The combination of inverse-square force law, deriving velocity formula, and executing the time calculation with substitution makes this harder than average but not exceptionally difficult for M3 students who have practiced similar problems. |
| Spec | 6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(mv\frac{dv}{dx} = -\frac{kv}{x^2}\) | \(\frac{v}{2} = \frac{k}{x} + c\) | \(x=a\), \(v=0\); \(c = -\frac{k}{a}\) |
| \(\frac{v}{2} = k\left(\frac{1}{x} - \frac{1}{a}\right)\) | \(v = \sqrt{\frac{2k}{a}\left(\frac{a-x}{x}\right)}\) | M1 A1 |
| (b) \(v = \frac{dr}{dt}\), so \(\frac{dr}{dt} = \sqrt{\frac{2k(a-x)}{ax}} - \sqrt{\frac{1-x}{x}}\) | \(\int dr = \int\frac{x}{1-x}dx\) | M1 A1 A1 |
| \(T = [\arcsin(\sqrt{x}) - \sqrt{(x-x^2)}]_{1/2}^{\pi/2} = \left(\frac{\pi}{2} - \left(\frac{\pi}{4} - \frac{1}{2}\right)\right) = \frac{\pi+2}{4}\) | M1 A1 |
(a) $mv\frac{dv}{dx} = -\frac{kv}{x^2}$ | $\frac{v}{2} = \frac{k}{x} + c$ | $x=a$, $v=0$; $c = -\frac{k}{a}$ | M1 A1 M1 A1
$\frac{v}{2} = k\left(\frac{1}{x} - \frac{1}{a}\right)$ | $v = \sqrt{\frac{2k}{a}\left(\frac{a-x}{x}\right)}$ | M1 A1
(b) $v = \frac{dr}{dt}$, so $\frac{dr}{dt} = \sqrt{\frac{2k(a-x)}{ax}} - \sqrt{\frac{1-x}{x}}$ | $\int dr = \int\frac{x}{1-x}dx$ | M1 A1 A1
$T = [\arcsin(\sqrt{x}) - \sqrt{(x-x^2)}]_{1/2}^{\pi/2} = \left(\frac{\pi}{2} - \left(\frac{\pi}{4} - \frac{1}{2}\right)\right) = \frac{\pi+2}{4}$ | M1 A1
**Total: 11 marks**A particle $P$ of mass $m$ kg moves along a straight line under the action of a force of magnitude $\frac{km}{x^2}$ N, where $k$ is a constant, directed towards a fixed point $O$ on the line, where $OP = x$ m. $P$ starts from rest at $A$, at a distance $a$ m from $O$. When $OP = x$ m, the speed of $P$ is $v$ ms$^{-1}$.
\begin{enumerate}[label=(\alph*)]
\item Show that $v = \sqrt{\frac{2k(a-x)}{ax}}$. [6 marks]
\end{enumerate}
$B$ is the point half-way between $O$ and $A$. When $k = \frac{1}{2}$ and $a = 1$, the time taken by $P$ to travel from $A$ to $B$ is $T$ seconds
Assuming the result that, for $0 \leq x \leq 1$, $\int \sqrt{\frac{x}{1-x}} dx = \arcsin(\sqrt{x}) - \sqrt{x(1-x^2)} + \text{constant}$,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item find the value of $T$. [5 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q4 [11]}}