| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | June |
| Marks | 7 |
| 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 into simple shapes (rectangle + triangle), finding individual centroids, using the composite formula, then applying equilibrium conditions. All techniques are routine textbook methods with straightforward geometry and no novel insight required. Slightly easier than average due to clear setup and standard approach. |
| Spec | 6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Mass ratio: ABDE \(\frac{8a^3\rho}{8}\), BCD \(\frac{a^3\rho}{1}\), Lamina \(\frac{9a^2\rho}{9}\) | B1 | |
| Dist of C of M From AE: \(2a\), \(4\frac{1}{3}a\), \(\bar{x}\) | B1 | |
| \(8 \times 2a + 1 \times 4\frac{1}{3}a = 9\bar{x}\) | M1 | |
| \(\bar{x} = \frac{61}{27}a\) (2.26a) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\tan \phi = \frac{a}{61} = \frac{27}{61} \cdot \frac{a}{\frac{27}{a}}\) | M1 A1 ft | |
| \(\phi = 23.87\ldots = 24°\) (accept 23.9), 0.417 radians | A1 |
## Part (a):
| **Answer/Working** | **Marks** | **Guidance** |
|---|---|---|
| Mass ratio: ABDE $\frac{8a^3\rho}{8}$, BCD $\frac{a^3\rho}{1}$, Lamina $\frac{9a^2\rho}{9}$ | B1 | |
| Dist of C of M From AE: $2a$, $4\frac{1}{3}a$, $\bar{x}$ | B1 | |
| $8 \times 2a + 1 \times 4\frac{1}{3}a = 9\bar{x}$ | M1 | |
| $\bar{x} = \frac{61}{27}a$ (2.26a) | A1 | |
## Part (b):
| **Answer/Working** | **Marks** | **Guidance** |
|---|---|---|
| $\tan \phi = \frac{a}{61} = \frac{27}{61} \cdot \frac{a}{\frac{27}{a}}$ | M1 A1 ft | |
| $\phi = 23.87\ldots = 24°$ (accept 23.9), 0.417 radians | A1 | |\includegraphics{figure_1}
Figure 1 shows a uniform lamina $ABCDE$ such that $ABDE$ is a rectangle, $BC = CD$, $AB = 4a$ and $AE = 2a$. The point $F$ is the midpoint of $BD$ and $FC = a$.
\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $a$, the distance of the centre of mass of the lamina from $AE$. [4]
\end{enumerate}
The lamina is freely suspended from $A$ and hangs in equilibrium.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the angle between $AB$ and the downward vertical. [3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2011 Q4 [7]}}