| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | L-shaped or composite rectangular lamina |
| Difficulty | Standard +0.3 This is a standard M2 centre of mass question involving a composite rectangular lamina with straightforward decomposition into two rectangles, followed by a routine equilibrium calculation using tan θ = horizontal distance / vertical distance. The method is well-practiced and requires no novel insight, making it slightly easier than average. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass |
| Answer | Marks | Guidance |
|---|---|---|
| (a) (i) \(132(11) + 84(3) = 216\bar{x}\) \(\quad \bar{x} = 7.89\) | M1 A1 A1 | |
| (ii) \(132(3) + 84(13) = 216\bar{y}\) \(\quad \bar{y} = 6.89\) | M1 A1 A1 | |
| (b) \(\tan\alpha = 0.89 \div 1.89 = 0.471\) \(\quad \alpha = 25.2°\) | M1 A1 M1 A1 | Total: 10 marks |
**(a)** (i) $132(11) + 84(3) = 216\bar{x}$ $\quad \bar{x} = 7.89$ | M1 A1 A1 |
(ii) $132(3) + 84(13) = 216\bar{y}$ $\quad \bar{y} = 6.89$ | M1 A1 A1 |
**(b)** $\tan\alpha = 0.89 \div 1.89 = 0.471$ $\quad \alpha = 25.2°$ | M1 A1 M1 A1 | Total: 10 marks6.\\
\includegraphics[max width=\textwidth, alt={}, center]{3c084e42-d304-4b77-afee-7e4bd801a03c-2_424_492_813_379}
The diagram shows a uniform lamina $A B C D E F$.
\begin{enumerate}[label=(\alph*)]
\item Calculate the distance of the centre of mass of the lamina from (i) $A F$, (ii) $A B$.
The lamina is hung over a smooth peg at $D$ and rests in equilibrium in a vertical plane.
\item Find the angle between $C D$ and the vertical.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q6 [10]}}