| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | L-shaped or composite rectangular lamina |
| Difficulty | Standard +0.3 This is a standard M2 centre of mass question involving decomposing an L-shaped lamina into two rectangles, finding individual centres of mass, and applying the composite body formula. Part (b) adds a straightforward application of equilibrium conditions (vertical line through suspension point and centre of mass). The geometry is clearly specified and the method is routine textbook material, making it slightly easier than average. |
| Spec | 6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Mass ratios (Rectangle \(27\times9\): 243, Right triangle: 40.5, Top triangle: 40.5, Rectangle \(9\times18\): 162) | B1 | |
| Centres of mass: \((13.5, 4.5),\ (30,3),\ (3,30),\ (4.5,18)\) | B1 | |
| Take moments about AB: \(6\times13.5+1\times30+4\times4.5+1\times3 = 132 = 12\bar{x}\) | M1 | |
| A(2,1,0) | ||
| \(\bar{x} = 11\text{ cm}\) (solve for \(x\) co-ord) | A1 | |
| \(\bar{y} = 11\text{ cm}\) (using symmetry) | B1ft | |
| (7) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\tan\theta = \dfrac{\bar{x}}{36-\bar{y}}\) | M1 | |
| \(\tan\theta = \dfrac{11}{25} = 0.44\) | A1ft | |
| \(\theta = 24°\) | A1 | |
| (3) [10] |
## Question 5:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Mass ratios (Rectangle $27\times9$: 243, Right triangle: 40.5, Top triangle: 40.5, Rectangle $9\times18$: 162) | B1 | |
| Centres of mass: $(13.5, 4.5),\ (30,3),\ (3,30),\ (4.5,18)$ | B1 | |
| Take moments about AB: $6\times13.5+1\times30+4\times4.5+1\times3 = 132 = 12\bar{x}$ | M1 | |
| | A(2,1,0) | |
| $\bar{x} = 11\text{ cm}$ (solve for $x$ co-ord) | A1 | |
| $\bar{y} = 11\text{ cm}$ (using symmetry) | B1ft | |
| | **(7)** | |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\tan\theta = \dfrac{\bar{x}}{36-\bar{y}}$ | M1 | |
| $\tan\theta = \dfrac{11}{25} = 0.44$ | A1ft | |
| $\theta = 24°$ | A1 | |
| | **(3) [10]** | |5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c5760fa5-3c7f-4e29-87a2-b3b4145b9361-10_823_908_269_513}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
The uniform L-shaped lamina $A B C D E F$, shown in Figure 2, has sides $A B$ and $F E$ parallel, and sides $B C$ and $E D$ parallel. The pairs of parallel sides are 9 cm apart. The points $A , F$, $D$ and $C$ lie on a straight line.\\
$A B = B C = 36 \mathrm {~cm} , F E = E D = 18 \mathrm {~cm} . \angle A B C = \angle F E D = 90 ^ { \circ }$, and $\angle B C D = \angle E D F = \angle E F D = \angle B A C = 45 ^ { \circ }$.
\begin{enumerate}[label=(\alph*)]
\item Find the distance of the centre of mass of the lamina from
\begin{enumerate}[label=(\roman*)]
\item side $A B$,
\item side $B C$.
The lamina is freely suspended from $A$ and hangs in equilibrium.
\end{enumerate}\item Find, to the nearest degree, the size of the angle between $A B$ and the vertical.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2011 Q5 [10]}}