| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2013 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Conical pendulum – horizontal circle in free space (no surface) |
| Difficulty | Moderate -0.8 This is a standard conical pendulum problem requiring straightforward application of resolving forces (vertical: T cos θ = mg, horizontal: T sin θ = mv²/r) with all values given. The three parts are routine calculations with no conceptual challenges beyond basic circular motion mechanics, making it easier than average for A-level. |
| Spec | 6.05c Horizontal circles: conical pendulum, banked tracks |
6 A light inextensible string has one end attached to a particle, $P$, of mass 2 kg . The other end of the string is attached to the fixed point $A$. The point $A$ is vertically above the point $B$. The particle moves at a constant speed in a horizontal circle of radius 0.8 m and centre $B$. The tension in the string is 34 N .
The string is inclined at an angle $\theta$ to the vertical, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{85514b55-3f13-4746-a3ef-747239b64cca-4_760_816_1436_596}
\begin{enumerate}[label=(\alph*)]
\item Find the angle $\theta$.
\item Find the speed of the particle.
\item Find the time taken for the particle to make one complete revolution.
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2013 Q6 [8]}}