| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2006 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: speed at specific point |
| Difficulty | Standard +0.3 This is a standard vertical circle problem requiring energy conservation and circular motion equations. Part (a) uses straightforward energy conservation with a given angle, (b) applies Newton's second law for circular motion, and (c) uses symmetry. All are routine M2 techniques with no novel insight required, making it slightly easier than average. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.05d Variable speed circles: energy methods |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\frac{1}{2}mU^2 = \frac{1}{2}mv^2 + mgl(1 - \cos 60°)\) and \(U^2 = v^2 + gl\) with \(v = \sqrt{U^2 - gl}\) | M1, A1, dM1, A1 | 4 marks |
| (b) \(T - mg\cos 60° = m\frac{v^2}{l}\) and \(T = m\left(\frac{U^2 - gl}{l} - \frac{g}{2}\right) = m\left(\frac{U^2}{l} - \frac{g}{2}\right)\) | M1, dM1, A1, dM1, A1 | 5 marks |
| (c) \(T - mg = m\frac{U^2}{l}\) and \(T = m\left(\frac{U^2}{l} + g\right)\) | M1, A1 | 2 marks |
**(a)** $\frac{1}{2}mU^2 = \frac{1}{2}mv^2 + mgl(1 - \cos 60°)$ and $U^2 = v^2 + gl$ with $v = \sqrt{U^2 - gl}$ | M1, A1, dM1, A1 | 4 marks | Three/four term energy equation with a trig term; correct equation; solving for $v$ or $v^2$; correct $v$ in a simplified form
**(b)** $T - mg\cos 60° = m\frac{v^2}{l}$ and $T = m\left(\frac{U^2 - gl}{l} - \frac{g}{2}\right) = m\left(\frac{U^2}{l} - \frac{g}{2}\right)$ | M1, dM1, A1, dM1, A1 | 5 marks | Resolving towards the centre of the circle with three terms; substituting for $v^2$; correct equation; making $T$ the subject; correct expression for $T$. Simplification not necessary.
**(c)** $T - mg = m\frac{U^2}{l}$ and $T = m\left(\frac{U^2}{l} + g\right)$ | M1, A1 | 2 marks | Considering the vertical forces and using Newton's second law with $\frac{U^2}{l}$; correct $T$
**Total for Question 4: 11 marks**
---4 A particle of mass $m$ is suspended from a fixed point $O$ by a light inextensible string of length $l$. The particle hangs in equilibrium at the point $P$ vertically below $O$. The particle is then set into motion with a horizontal velocity $U$ so that it moves in a complete vertical circle with centre $O$. The point $Q$ on the circle is such that $\angle P O Q = 60 ^ { \circ }$, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{c02cf013-365b-44e2-8c16-aa8209cbe250-3_566_540_1797_751}
\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $g , l$ and $U$, the speed of the particle at $Q$.
\item Find, in terms of $g , l , m$ and $U$, the tension in the string when the particle is at $Q$.
\item Find, in terms of $g , l , m$ and $U$, the tension in the string when the particle returns to $P$.\\
(2 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2006 Q4 [11]}}