| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2008 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: complete revolution conditions |
| Difficulty | Standard +0.3 This is a standard vertical circle problem requiring energy conservation and the critical condition (T=0 at top). Part (a) is a bookwork derivation, part (b) applies the result with straightforward calculation, and part (c) asks for a modeling assumption. While it requires multiple steps, the approach is well-rehearsed in M2 courses with no novel insight needed. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.05d Variable speed circles: energy methods |
| Answer | Marks | Guidance |
|---|---|---|
| At top, for complete revolutions: \(\frac{mv^2}{a} = mg\) where \(v\) is speed at top, so \(v^2 = ag\). Conservation of energy from B to top: \(\frac{1}{2}mv^2 + mg(2a) = \frac{1}{2}mu^2\), \(u^2 = 4ag + v^2 = 5ag\), \(u = \sqrt{5ag}\) | M1 A1 M1 A1 A1 | 5 marks |
| Answer | Marks | Guidance |
|---|---|---|
| At C, speed of particle is \(\sqrt{3ag}\). Resolving horizontally at C: \(T = \frac{mv^2}{a} = \frac{m\sqrt{3ag}}{a} = 3mg\) | B1 M1 A1 | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| No air resistance. Bead is a particle | B1 | 1 mark |
**Part (a)**
At top, for complete revolutions: $\frac{mv^2}{a} = mg$ where $v$ is speed at top, so $v^2 = ag$. Conservation of energy from B to top: $\frac{1}{2}mv^2 + mg(2a) = \frac{1}{2}mu^2$, $u^2 = 4ag + v^2 = 5ag$, $u = \sqrt{5ag}$ | M1 A1 M1 A1 A1 | 5 marks | 3 terms, 2 KE and PE. AG
**Part (b)**
At C, speed of particle is $\sqrt{3ag}$. Resolving horizontally at C: $T = \frac{mv^2}{a} = \frac{m\sqrt{3ag}}{a} = 3mg$ | B1 M1 A1 | 3 marks | Needs 2 correct terms
**Part (c)**
No air resistance. Bead is a particle | B1 | 1 mark
**Total: 9 marks**
---7 A small bead, of mass $m$, is suspended from a fixed point $O$ by a light inextensible string, of length $a$. The bead is then set into circular motion with the string taut at $B$, where $B$ is vertically below $O$, with a horizontal speed $u$.\\
\includegraphics[max width=\textwidth, alt={}, center]{03994596-21ad-4201-8d64-ba2d7b7e0a77-5_451_458_461_760}
\begin{enumerate}[label=(\alph*)]
\item Given that the string does not become slack, show that the least value of $u$ required for the bead to make complete revolutions about $O$ is $\sqrt { 5 a g }$.
\item In the case where $u = \sqrt { 5 a g }$, find, in terms of $g$ and $m$, the tension in the string when the bead is at the point $C$, which is at the same horizontal level as $O$, as shown in the diagram.
\item State one modelling assumption that you have made in your solution.
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2008 Q7 [9]}}