| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Lamina with attached triangle |
| Difficulty | Standard +0.3 This is a standard M2 centre of mass question requiring decomposition of a trapezium into rectangles/triangles, application of the composite body formula, and basic equilibrium principles. The calculations are straightforward with no conceptual surprises, making it slightly easier than average for A-level. |
| Spec | 6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks |
|---|---|
| (i) \(168(10.5) + 94.5(7) = 262.5\bar{x}\) | M1 M1 A1 A1 |
| \(\bar{x} = 9.24\) | |
| (ii) \(168(4) + 94.5(11) = 262.5\bar{y}\) | M1 M1 A1 A1 |
| \(\bar{y} = 6.52\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tan \alpha = \frac{21 - 9.24}{6.52} = 1.804\) | M1 A1 A1 | |
| \(\alpha = 61.0°\) | Total: 11 marks |
### (a)
(i) $168(10.5) + 94.5(7) = 262.5\bar{x}$ | M1 M1 A1 A1 |
$\bar{x} = 9.24$ | |
(ii) $168(4) + 94.5(11) = 262.5\bar{y}$ | M1 M1 A1 A1 |
$\bar{y} = 6.52$ | |
### (b)
$\tan \alpha = \frac{21 - 9.24}{6.52} = 1.804$ | M1 A1 A1 |
$\alpha = 61.0°$ | | **Total: 11 marks**A uniform lamina is in the form of a trapezium $ABCD$, as shown. $AB$ and $DC$ are perpendicular to $BC$. $AB = 17$ cm, $BC = 21$ cm and $CD = 8$ cm.
\includegraphics{figure_7}
\begin{enumerate}[label=(\alph*)]
\item Find the distances of the centre of mass of the lamina from
\begin{enumerate}[label=(\roman*)]
\item $AB$,
\item $BC$. [8 marks]
\end{enumerate}
\end{enumerate}
The lamina is freely suspended from $C$ and rests in equilibrium.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the angle between $CD$ and the vertical. [3 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q7 [11]}}