| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Elastic string – conical pendulum (string inclined to vertical) |
| Difficulty | Standard +0.8 This M3 conical pendulum problem requires resolving forces in two directions, applying Hooke's law for elastic strings, and using circular motion equations. While the setup is standard for M3, students must correctly handle the geometry, tension calculation, and centripetal force simultaneously across multiple steps, making it moderately challenging but within typical M3 scope. |
| Spec | 6.02g Hooke's law: T = k*x or T = lambda*x/l6.02h Elastic PE: 1/2 k x^26.05b Circular motion: v=r*omega and a=v^2/r |
| Answer | Marks |
|---|---|
| \(T \cos 60° = mg\), so \(T = 2mg\) | M1 A1 M1 A1 |
| \(T \sin 60° = m(L \sin 60°)\omega^2\) | M1 A1 |
| Thus \(\frac{mg}{l}(L - l) = 2mg\) | M1 A1 |
| \(L - l = 2l\) | |
| \(L = 3l\) | |
| \(\omega^2 = \frac{2g}{3l}\) | M1 A1 |
| (b) \(2mg = m(3l)\omega^2\) | M1 A1 |
| \(\omega = \sqrt{\frac{2g}{3l}}\) | M1 A1 |
$T \cos 60° = mg$, so $T = 2mg$ | M1 A1 M1 A1 |
$T \sin 60° = m(L \sin 60°)\omega^2$ | M1 A1 |
Thus $\frac{mg}{l}(L - l) = 2mg$ | M1 A1 |
$L - l = 2l$ | |
$L = 3l$ | |
$\omega^2 = \frac{2g}{3l}$ | M1 A1 |
(b) $2mg = m(3l)\omega^2$ | M1 A1 |
$\omega = \sqrt{\frac{2g}{3l}}$ | M1 A1 |
**Total: 8 marks**
---A particle $P$ of mass $m$ kg moves in a horizontal circle at one end of a light elastic string of natural length $l$ m and modulus of elasticity $mg$ N. The other end of the string is attached to a fixed point $O$. Given that the string makes an angle of $60°$ with the vertical,
\begin{enumerate}[label=(\alph*)]
\item show that $OP = 3l$ m. [4 marks]
\item Find, in terms of $l$ and $g$, the angular speed of $P$. [4 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q1 [8]}}