| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Session | Specimen |
| Marks | 13 |
| 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 (square + triangle), basic coordinate geometry, and equilibrium of a suspended lamina. The calculations are straightforward with clearly defined shapes and standard formulas, making it slightly easier than average but still requiring proper method. |
| Spec | 6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| Height of \(\Delta = \sqrt{15^2 - 9^2} = 12\) cm | M1 A1 | |
| Areas: Rectangle \(= 324\), Triangle \(= 108\), Combined \(= 432\) | M1 A1 | |
| Distance of CM from \(AE\): Rectangle \(= 9\), Triangle \(= 18 + \frac{1}{3}\cdot 12 = 22\), Combined \(= \bar{x}\) | B1 B1 ft | |
| \(9 \cdot 324 + 22 \cdot 108 = 432\bar{x}\) | M1 A1 | |
| \(\bar{x} = 12.25\) cm | A1 | (9) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| Distance of \(G\) from \(BD = 9\) cm | B1 | |
| \(\tan\theta = \frac{18 - 12.25}{9}\) | M1 A1 | |
| \(\theta = 32.6°\) | A1 | (4) |
## Question 7:
### Part (a):
| Answer/Working | Marks | Notes |
|---|---|---|
| Height of $\Delta = \sqrt{15^2 - 9^2} = 12$ cm | M1 A1 | |
| Areas: Rectangle $= 324$, Triangle $= 108$, Combined $= 432$ | M1 A1 | |
| Distance of CM from $AE$: Rectangle $= 9$, Triangle $= 18 + \frac{1}{3}\cdot 12 = 22$, Combined $= \bar{x}$ | B1 B1 ft | |
| $9 \cdot 324 + 22 \cdot 108 = 432\bar{x}$ | M1 A1 | |
| $\bar{x} = 12.25$ cm | A1 | **(9)** |
### Part (b):
| Answer/Working | Marks | Notes |
|---|---|---|
| Distance of $G$ from $BD = 9$ cm | B1 | |
| $\tan\theta = \frac{18 - 12.25}{9}$ | M1 A1 | |
| $\theta = 32.6°$ | A1 | **(4)** |
---7.\\
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A uniform plane lamina $A B C D E$ is formed by joining a uniform square $A B D E$ with a uniform triangular lamina $B C D$, of the same material, along the side $B D$, as shown in Fig. 2. The lengths $A B , B C$ and $C D$ are $18 \mathrm {~cm} , 15 \mathrm {~cm}$ and 15 cm respectively.
\begin{enumerate}[label=(\alph*)]
\item Find the distance of the centre of mass of the lamina from $A E$.
The lamina is freely suspended from $B$ and hangs in equilibrium.
\item Find, in degrees to one decimal place, the angle which $B D$ makes with the vertical.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q7 [13]}}