| Exam Board | Edexcel |
|---|---|
| Module | FM2 AS (Further Mechanics 2 AS) |
| Year | 2021 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Frame with straight rod/wire components only |
| Difficulty | Standard +0.8 This is a Further Maths mechanics question requiring geometric reasoning to locate the centre of mass of a composite framework. Part (a) demands spatial visualization to prove the centre of mass coincides with point Q using the given ratios, while part (b) requires coordinate setup and the standard centre of mass formula. The geometric insight needed and multi-step calculation place it moderately above average difficulty. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| By symmetry, centre of mass at centre of square | B1 | Any clear explanation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Mass ratios \(40 : 32 : 72\) | B1 | Correct mass ratios seen or implied |
| Distances \(5a : 9a : (\bar{x})\) | B1 | Correct distances seen or implied |
| Moments about \(AB\): \((40 \times 5a + 32 \times 9a = 72\bar{x})\) | M1 | Moments equation for the whole framework about an axis parallel to \(AB\) or to \(BC\) |
| \(\bar{x} = \frac{61a}{9}\) | ||
| Complete method to find distance \(= \sqrt{\bar{x}^2 + \bar{x}^2}\) | M1 | Use of symmetry of the framework and Pythagoras with their \(\bar{x}\) to find the required distance |
| Distance \(= \dfrac{61\sqrt{2}a}{9}\) | A1 | Or equivalent. \(9.6a\) \((9.585...a)\) or better |
## Question 1:
**Part (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| By symmetry, centre of mass at centre of square | B1 | Any clear explanation |
**Part (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Mass ratios $40 : 32 : 72$ | B1 | Correct mass ratios seen or implied |
| Distances $5a : 9a : (\bar{x})$ | B1 | Correct distances seen or implied |
| Moments about $AB$: $(40 \times 5a + 32 \times 9a = 72\bar{x})$ | M1 | Moments equation for the whole framework about an axis parallel to $AB$ or to $BC$ |
| $\bar{x} = \frac{61a}{9}$ | | |
| Complete method to find distance $= \sqrt{\bar{x}^2 + \bar{x}^2}$ | M1 | Use of symmetry of the framework and Pythagoras with their $\bar{x}$ to find the required distance |
| Distance $= \dfrac{61\sqrt{2}a}{9}$ | A1 | Or equivalent. $9.6a$ $(9.585...a)$ or better |
---1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a7901165-1679-4d30-9444-0c27020e32ea-02_744_805_246_632}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A uniform rod of length $72 a$ is cut into pieces. The pieces are used to make two rigid squares, $A B C D$ and $P Q R S$, with sides of length $10 a$ and $8 a$ respectively. The two squares are joined to form the rigid framework shown in Figure 1.
The squares both lie in the same plane with the rod $A B$ parallel to the rod $P Q$.\\
Given that
\begin{itemize}
\item $A D$ cuts $P Q$ in the ratio $3 : 5$
\item $D C$ cuts $Q R$ in the ratio 5:3
\begin{enumerate}[label=(\alph*)]
\item explain why the centre of mass of square $A B C D$ is at $Q$.
\item Find the distance of the centre of mass of the framework from $B$.
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{Edexcel FM2 AS 2021 Q1 [6]}}