| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | January |
| Marks | 10 |
| 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 coordinate geometry setup, area calculation, and equilibrium analysis. Part (a) involves routine application of the centroid formula for a trapezium (though students must first find the height using Pythagoras), and part (b) uses the standard principle that the centre of mass hangs directly below the suspension point. While multi-step, all techniques are textbook-standard with no novel insight 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 |
| Answer | Marks |
|---|---|
| For an appropriate division of the trapezium into standard shapes with: correct ratio of masses and correct distances of c.o.m. from AB | B1 B1 |
| e.g. three equilateral triangles of height \(\sqrt{3}\), mass \(m\) kg, com \(\frac{\sqrt{3}}{3}\) from bases of each | |
| \(3md = \left(m \times \frac{2}{3} \times \sqrt{3}\right) + \left(2 \times m \times \frac{1}{3}\sqrt{3}\right) = \frac{4\sqrt{3}}{3}m\) | M1 A1 |
| \(d = \frac{4\sqrt{3}}{9}\) AG | A1 |
| Answer | Marks |
|---|---|
| Horizontal distance of c of m from D = 1m | B1 |
| Vertical distance \(\sqrt{3} - \frac{4\sqrt{3}}{9} = \frac{5\sqrt{3}}{9}\) (0.962....) | B1 |
| \(\tan^{-1}\frac{0.962...}{1}\) | M1 A1ft |
| Angle = 43.9° | A1 |
## (a)
| For an appropriate division of the trapezium into standard shapes with: correct ratio of masses and correct distances of c.o.m. from AB | B1 B1 |
| e.g. three equilateral triangles of height $\sqrt{3}$, mass $m$ kg, com $\frac{\sqrt{3}}{3}$ from bases of each | |
| $3md = \left(m \times \frac{2}{3} \times \sqrt{3}\right) + \left(2 \times m \times \frac{1}{3}\sqrt{3}\right) = \frac{4\sqrt{3}}{3}m$ | M1 A1 |
| $d = \frac{4\sqrt{3}}{9}$ AG | A1 |
**Total: 5 marks**
## (b)
| Horizontal distance of c of m from D = 1m | B1 |
| Vertical distance $\sqrt{3} - \frac{4\sqrt{3}}{9} = \frac{5\sqrt{3}}{9}$ (0.962....) | B1 |
| $\tan^{-1}\frac{0.962...}{1}$ | M1 A1ft |
| Angle = 43.9° | A1 |
**Total: 5 marks**
**Question 4 Total: 10 marks**
---\includegraphics{figure_1}
The trapezium $ABCD$ is a uniform lamina with $AB = 4$ m and $BC = CD = DA = 2$ m, as shown in Figure 1.
\begin{enumerate}[label=(\alph*)]
\item Show that the centre of mass of the lamina is $\frac{4\sqrt{3}}{9}$ m from $AB$.
[5]
\end{enumerate}
The lamina is freely suspended from $D$ and hangs in equilibrium.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the angle between $DC$ and the vertical through $D$.
[5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2012 Q4 [10]}}