| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2009 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Two strings, two fixed points |
| Difficulty | Standard +0.3 This is a standard M2 circular motion problem with two strings at different angles. Part (a) requires resolving forces vertically (given that tensions are equal) to find tension - straightforward application of equilibrium. Part (b) requires horizontal resolution to find centripetal force, then using v²/r. The geometry is given, making this a routine 2-part question slightly easier than average due to the scaffolding in part (a). |
| Spec | 3.03m Equilibrium: sum of resolved forces = 06.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks |
4 Two light inextensible strings each have one end attached to a particle, $P$, of mass 6 kg . The other ends of the strings are attached to the fixed points $B$ and $C$. The point $C$ is vertically above the point $B$. The particle moves, at constant speed, in a horizontal circle, with centre 0.6 m below point $B$, with the strings inclined at $40 ^ { \circ }$ and $60 ^ { \circ }$ to the vertical, as shown in the diagram. Both strings are taut.\\
\includegraphics[max width=\textwidth, alt={}, center]{9cfa110c-ee11-447a-b21a-3f436432e27d-4_761_542_539_751}
\begin{enumerate}[label=(\alph*)]
\item As the particle moves in the horizontal circle, the tensions in the two strings are equal.
Show that the tension in the strings is 46.4 N , correct to three significant figures.
\item Find the speed of the particle.
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2009 Q4 [8]}}