| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2004 |
| 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 involving a composite lamina made of simple shapes (triangle and rectangle). Part (a) requires finding centres of mass of individual shapes and combining them using the standard formula—straightforward application of bookwork. Part (b) involves using the hanging equilibrium condition (centre of mass directly below suspension point) and basic trigonometry. While it requires multiple steps, all techniques are routine for M2 students with no novel problem-solving required. Slightly easier than average due to the symmetric setup and standard shapes. |
| Spec | 6.04c Composite bodies: centre of mass |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Mass Ratio: Rectangle 6, Triangle 12, Decoration 18; Ratio 1:2:3 | B1 | |
| CM from BG: Rectangle \(-1\frac{1}{2}\), Triangle \(2\), Decoration \(\bar{x}\) | B1 | |
| \(18 \times \bar{x} = -6 \times 1\frac{1}{2} + 12 \times 2\) | M1 A1 | |
| \(\bar{x} = \frac{5}{6}\) | A1 | accept exact equivalents |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Identification and use of correct triangle | M1 | |
| \(\tan\theta = \dfrac{1}{3 + \bar{x}}\) | M1 A1ft | ft their \(\bar{x}\) |
| \(\theta = 14.6°\) | A1 | cao |
## Question 3:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Mass Ratio: Rectangle 6, Triangle 12, Decoration 18; Ratio 1:2:3 | B1 | |
| CM from BG: Rectangle $-1\frac{1}{2}$, Triangle $2$, Decoration $\bar{x}$ | B1 | |
| $18 \times \bar{x} = -6 \times 1\frac{1}{2} + 12 \times 2$ | M1 A1 | |
| $\bar{x} = \frac{5}{6}$ | A1 | accept exact equivalents |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Identification and use of correct triangle | M1 | |
| $\tan\theta = \dfrac{1}{3 + \bar{x}}$ | M1 A1ft | ft their $\bar{x}$ |
| $\theta = 14.6°$ | A1 | cao |
---3.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\includegraphics[alt={},max width=\textwidth]{8e694174-b9a9-4018-8896-31a3b4f0d344-3_860_565_269_740}
\end{center}
\end{figure}
Figure 1 shows a decoration which is made by cutting the shape of a simple tree from a sheet of uniform card. The decoration consists of a triangle $A B C$ and a rectangle $P Q R S$. The points $P$ and $S$ lie on $B C$ and $M$ is the mid-point of both $B C$ and $P S$. The triangle $A B C$ is isosceles with $A B = A C , B C = 4 \mathrm {~cm} , A M = 6 \mathrm {~cm} , P S = 2 \mathrm {~cm}$ and $P Q = 3 \mathrm {~cm}$.
\begin{enumerate}[label=(\alph*)]
\item Find the distance of the centre of mass of the decoration from $B C$.
The decoration is suspended from $Q$ and hangs freely.
\item Find, in degrees to one decimal place, the angle between $P Q$ and the vertical.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2004 Q3 [9]}}