| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2009 |
| Session | June |
| 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 composite shapes (rectangle + triangle), finding the centroid using moments, then applying equilibrium for a suspended lamina. The calculations are straightforward with clearly defined shapes and dimensions, making it slightly easier than average but still requiring proper method. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Ratio of areas triangle : sign : rectangle \(= 1 : 5 : 6\) (1800 : 9000 : 10800) | B1 | |
| Centre of mass of triangle is 20cm down from \(AD\) (seen or implied) | B1 | |
| \(6 \times 45 - 1 \times 20 = 5 \times \bar{y}\) | M1A1 | |
| \(\bar{y} = 50\text{cm}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Distance of centre of mass from \(AB\) is 60cm | B1 | |
| Required angle is \(\tan^{-1}\frac{60}{50}\) | M1A1ft | (their values) |
| \(= 50.2°\) (0.876 rads) | A1 |
## Question 5:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Ratio of areas triangle : sign : rectangle $= 1 : 5 : 6$ (1800 : 9000 : 10800) | B1 | |
| Centre of mass of triangle is 20cm down from $AD$ (seen or implied) | B1 | |
| $6 \times 45 - 1 \times 20 = 5 \times \bar{y}$ | M1A1 | |
| $\bar{y} = 50\text{cm}$ | A1 | |
**Subtotal: (5)**
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Distance of centre of mass from $AB$ is 60cm | B1 | |
| Required angle is $\tan^{-1}\frac{60}{50}$ | M1A1ft | (their values) |
| $= 50.2°$ (0.876 rads) | A1 | |
**Subtotal: (4) Total: [9]**
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8e220b8a-46f1-4b9b-88a4-f032c7fbda50-07_564_910_207_523}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A shop sign $A B C D E F G$ is modelled as a uniform lamina, as illustrated in Figure 2. $A B C D$ is a rectangle with $B C = 120 \mathrm {~cm}$ and $D C = 90 \mathrm {~cm}$. The shape $E F G$ is an isosceles triangle with $E G = 60 \mathrm {~cm}$ and height 60 cm . The mid-point of $A D$ and the mid-point of $E G$ coincide.
\begin{enumerate}[label=(\alph*)]
\item Find the distance of the centre of mass of the sign from the side $A D$.
The sign is freely suspended from $A$ and hangs at rest.
\item Find the size of the angle between $A B$ and the vertical.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2009 Q5 [9]}}