| Exam Board | Edexcel |
|---|---|
| Module | FM2 AS (Further Mechanics 2 AS) |
| Year | 2022 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Particle attached to lamina - find mass/position |
| Difficulty | Standard +0.8 This is a multi-part Further Maths mechanics question requiring: (a) centre of mass calculation for a composite wire framework (rectangle + semicircular arc) using standard formulas, (b) equilibrium geometry with trigonometry, and (c) finding an unknown mass using the centre of mass principle. While it involves several steps and Further Maths content, the techniques are standard applications of centre of mass formulas without requiring novel geometric insight or proof. |
| Spec | 3.04b Equilibrium: zero resultant moment and force6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Lengths: \(ABCD = 6a\), \(BEC = \pi a\), framework \(= 6a + \pi a\) | B1 | Any equivalent ratios |
| Distances from \(BC\): \(ABCD = \frac{1}{2}a\), \(BEC = -\frac{2a}{\pi}\), framework \(= \bar{x}\) | B1 | Or correct distances from a parallel axis |
| Moments about \(BC\) | M1 | Must be using framework. If \(BC\) included twice mark as misread |
| \(6a \times \frac{1}{2}a - \pi a \times \frac{2a}{\pi} = (6a + \pi a)\bar{x}\) | A1 | Correct unsimplified equation for their axis. Allow within a vector equation |
| \(\bar{x} = \frac{a}{6+\pi}\) | A1* | Correct given answer correctly obtained |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Angle \(DAE = \tan^{-1}\left(\frac{2a}{a}\right)\) | M1 | Correct relevant angle (or side if using cosine rule). Do not need to evaluate: accept \(\tan\alpha = \ldots\) or \(\alpha = \tan^{-1}\ldots\) (e.g. \(63.4°\) or \(90°-63.4°\)) |
| Angle \(DAG = \tan^{-1}\left(\dfrac{a - \frac{a}{6+\pi}}{a}\right) = \tan^{-1}\left(\frac{5+\pi}{6+\pi}\right)\) | M1 | Another correct relevant angle. Do not need to evaluate: accept \(\tan\beta = \ldots\) or \(\beta = \tan^{-1}\ldots\) (e.g. \(41.68°\) or \(90°-41.68°\)) |
| Angle \(= DAE - DAG\) | M1 | Correct method for finding the required angle |
| \(21.74637\ldots°\) | A1 | \(22°\) or better |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Moments about \(OA\) | M1 | Complete method to give an equation in \(k\) only |
| \(kMa\sin 45° = M\bar{x}\sin 45°\) | A1 | Correct equation in \(k\) only |
| \(k = \frac{1}{6+\pi}\) \((= 0.10939\ldots)\) | A1 | \(0.11\) or better |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Moments about \(O\): \(kM\begin{pmatrix}0\\a\end{pmatrix} - M\begin{pmatrix}\frac{-a}{6+\pi}\\0\end{pmatrix} = (k+1)M\begin{pmatrix}-\lambda\\\lambda\end{pmatrix}\) | M1 A1 | |
| \(k = \frac{1}{6+\pi}\) \((= 0.10939\ldots)\) | A1 |
## Question 2:
**Part 2(a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Lengths: $ABCD = 6a$, $BEC = \pi a$, framework $= 6a + \pi a$ | B1 | Any equivalent ratios |
| Distances from $BC$: $ABCD = \frac{1}{2}a$, $BEC = -\frac{2a}{\pi}$, framework $= \bar{x}$ | B1 | Or correct distances from a parallel axis |
| Moments about $BC$ | M1 | Must be using framework. If $BC$ included twice mark as misread |
| $6a \times \frac{1}{2}a - \pi a \times \frac{2a}{\pi} = (6a + \pi a)\bar{x}$ | A1 | Correct unsimplified equation for their axis. Allow within a vector equation |
| $\bar{x} = \frac{a}{6+\pi}$ | A1* | Correct given answer correctly obtained |
**Part 2(b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Angle $DAE = \tan^{-1}\left(\frac{2a}{a}\right)$ | M1 | Correct relevant angle (or side if using cosine rule). Do not need to evaluate: accept $\tan\alpha = \ldots$ or $\alpha = \tan^{-1}\ldots$ (e.g. $63.4°$ or $90°-63.4°$) |
| Angle $DAG = \tan^{-1}\left(\dfrac{a - \frac{a}{6+\pi}}{a}\right) = \tan^{-1}\left(\frac{5+\pi}{6+\pi}\right)$ | M1 | Another correct relevant angle. Do not need to evaluate: accept $\tan\beta = \ldots$ or $\beta = \tan^{-1}\ldots$ (e.g. $41.68°$ or $90°-41.68°$) |
| Angle $= DAE - DAG$ | M1 | Correct method for finding the required angle |
| $21.74637\ldots°$ | A1 | $22°$ or better |
**Part 2(c):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Moments about $OA$ | M1 | Complete method to give an equation in $k$ only |
| $kMa\sin 45° = M\bar{x}\sin 45°$ | A1 | Correct equation in $k$ only |
| $k = \frac{1}{6+\pi}$ $(= 0.10939\ldots)$ | A1 | $0.11$ or better |
**Part 2(c) alt:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Moments about $O$: $kM\begin{pmatrix}0\\a\end{pmatrix} - M\begin{pmatrix}\frac{-a}{6+\pi}\\0\end{pmatrix} = (k+1)M\begin{pmatrix}-\lambda\\\lambda\end{pmatrix}$ | M1 A1 | |
| $k = \frac{1}{6+\pi}$ $(= 0.10939\ldots)$ | A1 | |
---2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{99e1d643-7408-4793-9ebc-b33c91bc5fab-06_554_547_246_758}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Uniform wire is used to form the framework shown in Figure 2.\\
In the framework
\begin{itemize}
\item $A B C D$ is a rectangle with $A D = 2 a$ and $D C = a$
\item $B E C$ is a semicircular arc of radius $a$ and centre $O$, where $O$ lies on $B C$
\end{itemize}
The diameter of the semicircle is $B C$ and the point $E$ is such that $O E$ is perpendicular to $B C$.
The points $A , B , C , D$ and $E$ all lie in the same plane.
\begin{enumerate}[label=(\alph*)]
\item Show that the distance of the centre of mass of the framework from $B C$ is
$$\frac { a } { 6 + \pi }$$
The framework is freely suspended from $A$ and hangs in equilibrium with $A E$ at an angle $\theta ^ { \circ }$ to the downward vertical.
\item Find the value of $\theta$.
The mass of the framework is $M$.\\
A particle of mass $k M$ is attached to the framework at $B$.\\
The centre of mass of the loaded framework lies on $O A$.
\item Find the value of $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FM2 AS 2022 Q2 [12]}}