| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2013 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | L-shaped or composite rectangular lamina |
| Difficulty | Moderate -0.3 This is a standard M2 centre of mass question involving composite shapes with clear symmetry. Part (a) requires recognizing symmetry (minimal calculation), part (b) is a routine application of the centre of mass formula for composite bodies, and part (c) applies basic equilibrium geometry with tan θ. While it requires multiple steps and careful coordinate work, it follows a well-practiced template with no novel problem-solving required, making it slightly easier than average. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
4 The diagram shows a uniform lamina which is in the shape of two identical rectangles $A X G H$ and $Y B C D$ and a square $X Y E F$, arranged as shown.
The length of $A X$ is 10 cm , the length of $X Y$ is 10 cm and the length of $A H$ is 30 cm .\\
\includegraphics[max width=\textwidth, alt={}, center]{85514b55-3f13-4746-a3ef-747239b64cca-3_1183_1278_513_374}
\begin{enumerate}[label=(\alph*)]
\item Explain why the centre of mass of the lamina is 15 cm from $A H$.
\item Find the distance of the centre of mass of the lamina from $A B$.
\item The lamina is freely suspended from the point $H$.
Find, to the nearest degree, the angle between $H G$ and the horizontal when the lamina is in equilibrium.
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2013 Q4 [8]}}