| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2003 |
| Session | January |
| Marks | 9 |
| 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 composite shape into rectangle and triangle, finding individual centres of mass, then using the composite formula. Part (b) adds a particle and uses equilibrium conditions with moments. While multi-step, it follows routine M2 procedures without requiring novel insight—slightly easier than average A-level. |
| Spec | 6.04a Centre of mass: gravitational effect6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Shape | MR | CM |
| Square | \(48a^2\) | \(4a\) |
| Triangle | \(12a^2\) | \((-)\frac{1}{3} \times 4a\) |
| Pentagon | \(60a^2\) | \(\frac{}{x}\) |
| Marks: B1, B1ft | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(48a^2 \times 4a - 12a^2 \times \frac{4}{3}a = 60x\) | M1 A1 | |
| Solving to \(x = \frac{44}{15}a\) (*) | A1 | (6 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\lambda M \times 4a = M \times \frac{44}{15}a\) | M1 A1 | |
| \(\lambda = \frac{11}{15}\) | A1 | (3 marks) |
## Part (a)
| Shape | MR | CM |
|-------|----|----|
| Square | $48a^2$ | $4a$ |
| Triangle | $12a^2$ | $(-)\frac{1}{3} \times 4a$ |
| Pentagon | $60a^2$ | $\frac{}{x}$ |
Marks: B1, B1ft | B1 |
**Moment equation:**
$48a^2 \times 4a - 12a^2 \times \frac{4}{3}a = 60x$ | M1 A1 |
**Solving to $x = \frac{44}{15}a$ (*)** | A1 | **(6 marks)**
## Part (b)
$\lambda M \times 4a = M \times \frac{44}{15}a$ | M1 A1 |
$\lambda = \frac{11}{15}$ | A1 | **(3 marks)**
**(9 marks total)**
---\includegraphics{figure_2}
Figure 2 shows a uniform lamina $ABCDE$ such that $ABDE$ is a rectangle, $BC = CD$, $AB = 8a$ and $AE = 6a$. The point $X$ is the mid-point of $BD$ and $XC = 4a$. The centre of mass of the lamina is at $G$.
\begin{enumerate}[label=(\alph*)]
\item Show that $GX = \frac{14}{15}a$. [6]
\end{enumerate}
The mass of the lamina is $M$. A particle of mass $\lambda M$ is attached to the lamina at $C$. The lamina is suspended from $B$ and hangs freely under gravity with $AB$ horizontal.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the value of $\lambda$. [3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2003 Q4 [9]}}