| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2015 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Elastic string equilibrium |
| Difficulty | Challenging +1.2 This is a multi-part mechanics problem requiring Hooke's law, geometric reasoning about string angles, and taking moments about a hinge. While it involves several concepts (elasticity, equilibrium, moments), each part is relatively straightforward: (i) is a show-that using standard Hooke's law with given geometry, (ii) requires understanding that a smooth ring means no friction so the string must be perpendicular, and (iii) involves taking moments about A with standard techniques. The problem is more involved than a basic moments question but doesn't require novel insight—it's a standard textbook-style equilibrium problem that's slightly above average difficulty due to the elastic string component and multi-step nature. |
| Spec | 3.04b Equilibrium: zero resultant moment and force6.02h Elastic PE: 1/2 k x^2 |
\includegraphics{figure_2}
A uniform rigid rod $AB$ of length $1.2\text{ m}$ and weight $8\text{ N}$ has a particle of weight $2\text{ N}$ attached at the end $B$. The end $A$ of the rod is freely hinged to a fixed point. One end of a light elastic string of natural length $0.8\text{ m}$ and modulus of elasticity $20\text{ N}$ is attached to the hinge. The string passes over a small smooth pulley $P$ fixed $0.8\text{ m}$ vertically above the hinge. The other end of the string is attached to a small light smooth ring $R$ which can slide on the rod. The system is in equilibrium with the rod inclined at an angle $\theta°$ to the vertical (see diagram).
\begin{enumerate}[label=(\roman*)]
\item Show that the tension in the string is $20\sin\theta\text{ N}$. [1]
\item Explain why the part of the string attached to the ring is perpendicular to the rod. [1]
\item Find $\theta$. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2015 Q2 [5]}}