| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: string becomes slack |
| Difficulty | Challenging +1.2 This is a standard M3 circular motion problem requiring energy conservation and tension conditions. Part (a) uses the standard technique that tension becomes zero when the string goes slack, combined with energy conservation—a routine application for M3 students. Part (b) requires projectile motion after the string slacks, involving energy methods again. While multi-step with 17 marks total, it follows predictable M3 patterns without requiring novel insight or particularly complex algebra. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings6.05d Variable speed circles: energy methods6.05e Radial/tangential acceleration |
| Answer | Marks |
|---|---|
| (a) K.E. lost = P.E. gained: \(\frac{1}{2}mu^2 - \frac{1}{2}mv^2 = mgl - mgl \cos \theta\) | M1 A1 |
| \(v^2 = u^2 - 2gl + 2gl \cos \theta\) | |
| \(T - mg \cos \theta = \frac{mv^2}{l}\) | M1 A1 M1 A1 |
| \(T = \frac{m}{l}(u^2 - 2gl + 2gl \cos \theta) + mg \cos \theta\) | |
| \(T = 0\) when \(\theta = 120°\): \(\frac{u^2}{2} - 2g - g - \frac{g}{8} = 0\) | M1 A1 I1 |
| \(u^2 = 7gl\) | |
| (b) Now moves as projectile with initial speed \(v\), where \(v^2 = u^2 - 3gl\) | M1 |
| so \(v = \sqrt{\frac{gl}{2}}\), and angle of projection \(60°\) | A1 B1 |
| At highest point above \(O\), \(h = \frac{l}{2} + \frac{v^2\sin 60°}{2g} = \frac{l}{2} + \frac{gl}{2} \cdot \frac{3}{4g^2} = \frac{l}{2} + \frac{3l}{8g} - \frac{11l}{16} \, \text{m}\) | M1 A1 M1 A1 |
| Total: 17 marks |
**(a)** K.E. lost = P.E. gained: $\frac{1}{2}mu^2 - \frac{1}{2}mv^2 = mgl - mgl \cos \theta$ | M1 A1 |
$v^2 = u^2 - 2gl + 2gl \cos \theta$ | |
$T - mg \cos \theta = \frac{mv^2}{l}$ | M1 A1 M1 A1 |
$T = \frac{m}{l}(u^2 - 2gl + 2gl \cos \theta) + mg \cos \theta$ | |
$T = 0$ when $\theta = 120°$: $\frac{u^2}{2} - 2g - g - \frac{g}{8} = 0$ | M1 A1 I1 |
$u^2 = 7gl$ | |
**(b)** Now moves as projectile with initial speed $v$, where $v^2 = u^2 - 3gl$ | M1 |
so $v = \sqrt{\frac{gl}{2}}$, and angle of projection $60°$ | A1 B1 |
At highest point above $O$, $h = \frac{l}{2} + \frac{v^2\sin 60°}{2g} = \frac{l}{2} + \frac{gl}{2} \cdot \frac{3}{4g^2} = \frac{l}{2} + \frac{3l}{8g} - \frac{11l}{16} \, \text{m}$ | M1 A1 M1 A1 |
| | **Total: 17 marks** |A particle $P$ is attached to one end of a light inextensible string of length $l$ m. The other end of the string is attached to a fixed point $O$. When $P$ is hanging at rest vertically below $O$, it is given a horizontal speed $u$ ms$^{-1}$ and starts to move in a vertical circle.
Given that the string becomes slack when it makes an angle of 120° with the downward vertical through $O$,
\begin{enumerate}[label=(\alph*)]
\item show that $u^2 = \frac{7gl}{2}$. [10 marks]
\item Find, in terms of $l$, the greatest height above $O$ reached by $P$ in the subsequent motion. [7 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q7 [17]}}