| Exam Board | Edexcel |
|---|---|
| Module | FM2 AS (Further Mechanics 2 AS) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Frame with straight rod/wire components only |
| Difficulty | Standard +0.3 This is a standard centre of mass problem for a composite framework with uniform rods. It requires systematic application of the centre of mass formula with clear geometry, but involves only routine calculations with no conceptual surprises. The 'show that' in part (a) provides a target to verify, making it slightly easier than average for Further Maths. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass |
| Answer | Marks | Guidance |
|---|---|---|
| \(M(AF)\) | M1 | 2.1 |
| \(2Add = 4a \times 4a + 4a \times 6a + 2 \times 3a \times 6a (= 76a^2)\) | A1 | 1.1b |
| \(d = \frac{19a}{6}\) * | A1* | 2.2a |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\bar{y} = 2a\) | B1 | 1.1b |
| \(D^2 = \text{their } \bar{x}^2 + \text{their } \bar{y}^2 \left( = \frac{361}{36}a^2 + 4a^2 \right)\) | M1 | 1.1b |
| \(D = \sqrt{\frac{505}{36}}a = \frac{\sqrt{505}}{6}a\) | A1 | 1.1b |
| (3) | ||
| (7 marks) |
**1a**
| $M(AF)$ | M1 | 2.1 |
| $2Add = 4a \times 4a + 4a \times 6a + 2 \times 3a \times 6a (= 76a^2)$ | A1 | 1.1b |
| $d = \frac{19a}{6}$ * | A1* | 2.2a |
| | (4) | |
**Notes for 1a:**
- M1: Moments about $AF$ or a parallel axis. All terms required. Dimensionally correct but allow consistent cancelling of a factor of $a$ or $2a$.
- A1: Unsimplified equation with at most one error
- A1: Correct unsimplified equation. E.g. $12d = 2 \times 4a + 2 \times 6a + 2 \times 3 \times 3a (= 38a)$
- A1*: Obtain given answer including "$d =$" from correct exact working
**1b**
| $\bar{y} = 2a$ | B1 | 1.1b |
| $D^2 = \text{their } \bar{x}^2 + \text{their } \bar{y}^2 \left( = \frac{361}{36}a^2 + 4a^2 \right)$ | M1 | 1.1b |
| $D = \sqrt{\frac{505}{36}}a = \frac{\sqrt{505}}{6}a$ | A1 | 1.1b |
| | (3) | |
| | (7 marks) | |
**Notes for 1b:**
- B1: Seen or implied
- M1: Correct use of Pythagoras to find $D$ or $D^2$
- A1: Any equivalent form. Accept $3.7a$ or better. ($3.7453675\ldots$)
---1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{fd8bc7b5-adee-4d67-b15d-571255b00b83-02_586_824_244_623}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A uniform rod of length $24 a$ is cut into seven pieces which are used to form the framework $A B C D E F$ shown in Figure 1.
It is given that
\begin{itemize}
\item $A F = B E = C D = A B = F E = 4 a$
\item $B C = E D = 2 a$
\item the rods $A F , B E$ and $C D$ are parallel
\item the rods $A B , B C , F E$ and $E D$ are parallel
\item $A F$ is perpendicular to $A B$
\item the rods all lie in the same plane
\end{itemize}
The distance of the centre of mass of the framework from $A F$ is $d$.
\begin{enumerate}[label=(\alph*)]
\item Show that $d = \frac { 19 } { 6 } a$
\item Find the distance of the centre of mass of the framework from $A$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FM2 AS 2024 Q1 [7]}}