| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Lamina with removed triangle/rectangle/square |
| Difficulty | Standard +0.3 This is a standard M2 centre of mass question using composite bodies (rectangle minus triangle) and equilibrium under suspension. Part (a) requires routine application of the composite body formula with symmetry simplifying calculations. Part (b) involves taking moments about the suspension point, which is a standard technique. The symmetry and vertical alignment conditions make this slightly easier than average, requiring methodical application of learned techniques rather than problem-solving insight. |
| Spec | 6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks |
|---|---|
| (a) \(750(12.5) = 180(21) + 570\bar{y}\), \(\bar{y} = 9.82 \text{ cm}\) | M1 A1 M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(9.816m + 13M = 12.5(m + M)\), \(0.5M = 2.684m\), \(M = 5.37m\) | B1 | M1 A1 M1 A1 |
(a) $750(12.5) = 180(21) + 570\bar{y}$, $\bar{y} = 9.82 \text{ cm}$ | M1 A1 M1 A1 |
(b) Must have centre of mass 12.5 cm from $ED$
$9.816m + 13M = 12.5(m + M)$, $0.5M = 2.684m$, $M = 5.37m$ | B1 | M1 A1 M1 A1 | Total: 9 marks\includegraphics{figure_4}
The diagram shows a uniform lamina $ABCDE$ formed by removing a symmetrical triangular section from a rectangular sheet of metal measuring 30 cm by 25 cm.
\begin{enumerate}[label=(\alph*)]
\item Find the distance of the centre of mass of the lamina from $ED$. [4 marks]
\end{enumerate}
The lamina has mass $m$. A particle $P$ is attached to the lamina at $B$. The lamina is then suspended freely from $A$ and hangs in equilibrium with $AD$ vertical.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find, in terms of $m$, the mass of $P$. [5 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q4 [9]}}