| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2008 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Lamina with removed circle/semicircle |
| Difficulty | Standard +0.3 This is a standard M2 centre of mass question using the composite body method (triangle minus circle) with straightforward coordinate setup and equilibrium. The calculations are routine: find centroids, apply the formula with areas, then use equilibrium geometry. While multi-step (9+3 marks), it requires only direct application of standard techniques without novel insight or complex problem-solving. |
| 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 |
|---|---|---|
| Mass ratio | B1 B1 ft | |
| Triangle | Circle | \(S\) |
| 126 | \(9\pi\) | \(126 - 9\pi\) |
| (28.3) | (97.7) | |
| \(\overline{x}\) | 7 | 5 |
| \(\overline{y}\) | 4 | 5 |
| B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\overline{x} \approx 7.58\) (\(\frac{882 - 45\pi}{126 - 9\pi}\)) awrt 7.6 | M1 A1 ft A1 | (3 marks) [seen] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\overline{y} \approx 3.71\) (\(\frac{504 - 45\pi}{126 - 9\pi}\)) awrt 3.7 | M1 A1 ft A1 | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\theta \approx 15°\) | M1 A1 ft A1 | (3 marks) [12 marks total] |
**Part (a)**
Mass ratio | B1 B1 ft |
| | Triangle | Circle | $S$ |
|---|---|---|---|
| | 126 | $9\pi$ | $126 - 9\pi$ |
| | (28.3) | (97.7) | |
| $\overline{x}$ | 7 | 5 | $\overline{x}$ |
| $\overline{y}$ | 4 | 5 | $\overline{y}$ |
| | B1 |
$126 \times 7 = 9\pi \times 5 + (126 - 9\pi) \times \overline{x}$ ft their table values
$\overline{x} \approx 7.58$ ($\frac{882 - 45\pi}{126 - 9\pi}$) awrt 7.6 | M1 A1 ft A1 | (3 marks) [seen]
$126 \times 4 = 9\pi \times 5 + (126 - 9\pi) \times \overline{y}$ ft their table values
$\overline{y} \approx 3.71$ ($\frac{504 - 45\pi}{126 - 9\pi}$) awrt 3.7 | M1 A1 ft A1 | (3 marks)
**Part (b)**
$\tan \theta = \frac{\overline{y}}{2\overline{1} - \overline{x}}$
$\theta \approx 15°$ | M1 A1 ft A1 | (3 marks) [12 marks total]
Fits their $\overline{x}, \overline{y}$
---\includegraphics{figure_1}
A set square $S$ is made by removing a circle of centre $O$ and radius 3 cm from a triangular piece of wood. The piece of wood is modelled as a uniform triangular lamina $ABC$, with $\angle ABC = 90°$, $AB = 12$ cm and $BC = 21$ cm. The point $O$ is 5 cm from $AB$ and 5 cm from $BC$, as shown in Figure 1.
\begin{enumerate}[label=(\alph*)]
\item Find the distance of the centre of mass of $S$ from
\begin{enumerate}[label=(\roman*)]
\item $AB$,
\item $BC$. [9]
\end{enumerate}
\end{enumerate}
The set square is freely suspended from $C$ and hangs in equilibrium.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find, to the nearest degree, the angle between $CB$ and the vertical. [3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2008 Q4 [12]}}