| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Vertical circle – surface contact (sphere/track, leaving surface) |
| Difficulty | Challenging +1.8 This M3 question requires geometric constraint analysis, energy conservation with two particles, and circular motion dynamics. Part (a) demands careful geometric reasoning about string constraints. Part (b) combines energy methods with resolving forces in circular motion, requiring multiple sophisticated steps and careful algebraic manipulation. The multi-stage reasoning and integration of several mechanics topics elevates this significantly above average difficulty. |
| Spec | 3.03k Connected particles: pulleys and equilibrium6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks |
|---|---|
| (a) \(P\) moves up \(r\sin\theta\) while \(Q\) moves down by arc length \(r\theta\) | M1 A1 A1 |
| Ratio of vertical distances moved \(= \frac{r\theta}{r\sin\theta} = \frac{\theta}{\sin\theta}\) | A1 |
| (b) \(mg\sin\theta - R = \frac{m\dot{v}^2}{r}\) | M1 A1 A1 M1 A1 A1 |
| Energy: \(mgr\theta - mgr\sin\theta = \frac{1}{2}mv^2 + \frac{1}{2}mv^2\) | M1 A1 A1 M1 A1 A1 |
| \(R = mg(2\sin\theta - \theta) = mg(1 - \frac{e}{3})\) when \(\theta = \frac{\pi}{6}\) | M1 A1 A1 |
| Total: 13 marks |
(a) $P$ moves up $r\sin\theta$ while $Q$ moves down by arc length $r\theta$ | M1 A1 A1 |
Ratio of vertical distances moved $= \frac{r\theta}{r\sin\theta} = \frac{\theta}{\sin\theta}$ | A1 |
(b) $mg\sin\theta - R = \frac{m\dot{v}^2}{r}$ | M1 A1 A1 M1 A1 A1 |
Energy: $mgr\theta - mgr\sin\theta = \frac{1}{2}mv^2 + \frac{1}{2}mv^2$ | M1 A1 A1 M1 A1 A1 |
$R = mg(2\sin\theta - \theta) = mg(1 - \frac{e}{3})$ when $\theta = \frac{\pi}{6}$ | M1 A1 A1 |
| **Total: 13 marks** |The diagram shows two identical particles, each of mass $m$ kg, connected by a thin, light inextensible string. $P$ slides on the surface of a smooth right circular cylinder fixed with its axis, through $O$, horizontal. $Q$ moves vertically. $OP$ makes an angle $\theta$ radians with the horizontal.
\includegraphics{figure_6}
The system is released from rest in the position where $\theta = 0$.
\begin{enumerate}[label=(\alph*)]
\item Show that the vertical distance moved by $Q$ is $\frac{\theta}{\sin \theta}$ times the vertical distance moved by $P$. [4 marks]
\item In the position where $\theta = \frac{\pi}{6}$, prove that the reaction of the cylinder on $P$ has magnitude $\left(1-\frac{\pi}{6}\right)mg$ N. [9 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q6 [13]}}