| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2013 |
| Session | June |
| Marks | 12 |
| 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 composite shapes (two identical rectangles forming an L-shape) with straightforward coordinate geometry. Parts (a) and (b) require routine application of the centre of mass formula for composite bodies, while part (c) involves using equilibrium conditions with a given angle. The question is slightly easier than average because the shapes are identical rectangles (simplifying calculations), the setup is clear, and the method is well-practiced in M2. It requires more steps than a basic question but no novel insight or difficult problem-solving. |
| Spec | 6.04b Find centre of mass: using symmetry6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Moments about OE: \(\frac{a}{2} \times 1 + (a+2) \times 1 = 1 \times 2\bar{x}\) | M1 | Moments equation. Needs to include both rectangles and the whole shape. |
| A2 | -1 each error. or equivalent | |
| \(2\bar{x} = \frac{3a}{2} + 2, \quad \bar{x} = \frac{3a}{4} + 1\) | A1 | Or simplified equivalent. |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Moments about OA: \(2 \times 1 + \frac{a}{2} \times 1 = 2y\) o.e. | M1 | Moments equation. Needs to include both rectangles and the whole shape. |
| A2 | -1 each error. or equivalent | |
| \(\bar{y} = 1 + \frac{a}{4}\) o.e. | A1 | Or simplified equivalent. |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Hanging in Equilibrium: \(\tan \alpha = \frac{\bar{x}}{\bar{y}} = \frac{\frac{3a}{4}+1}{1+\frac{a}{4}}\) | M1 | Condone reciprocal |
| A1ft | For their \(\bar{x}, \bar{y}\) | |
| \(\frac{4}{3} = \frac{\frac{3a}{4}+1}{\frac{a}{1+\frac{a}{4}}} = \frac{3a+4}{4+a}, \quad 4(4+a) = 3(3a+4)\) | DM1 | Their \(\tan a = \frac{4}{3}\) and solve for a |
| \(16 + 4a = 9a + 12, \quad 4 = 5a\) | ||
| \(a = \frac{4}{5}\) | A1 | Or 0.8 |
| (4) | ||
| [12] |
**(a)**
| Moments about OE: $\frac{a}{2} \times 1 + (a+2) \times 1 = 1 \times 2\bar{x}$ | M1 | Moments equation. Needs to include both rectangles and the whole shape. |
| | A2 | -1 each error. or equivalent |
| $2\bar{x} = \frac{3a}{2} + 2, \quad \bar{x} = \frac{3a}{4} + 1$ | A1 | Or simplified equivalent. |
| | (4) | |
**(b)**
| Moments about OA: $2 \times 1 + \frac{a}{2} \times 1 = 2y$ o.e. | M1 | Moments equation. Needs to include both rectangles and the whole shape. |
| | A2 | -1 each error. or equivalent |
| $\bar{y} = 1 + \frac{a}{4}$ o.e. | A1 | Or simplified equivalent. |
| | (4) | |
**(c)**
| Hanging in Equilibrium: $\tan \alpha = \frac{\bar{x}}{\bar{y}} = \frac{\frac{3a}{4}+1}{1+\frac{a}{4}}$ | M1 | Condone reciprocal |
| | A1ft | For their $\bar{x}, \bar{y}$ |
| $\frac{4}{3} = \frac{\frac{3a}{4}+1}{\frac{a}{1+\frac{a}{4}}} = \frac{3a+4}{4+a}, \quad 4(4+a) = 3(3a+4)$ | DM1 | Their $\tan a = \frac{4}{3}$ and solve for a |
| $16 + 4a = 9a + 12, \quad 4 = 5a$ | | |
| $a = \frac{4}{5}$ | A1 | Or 0.8 |
| | (4) |
| | [12] | |\includegraphics{figure_2}
The uniform L-shaped lamina $OABCDE$, shown in Figure 2, is made from two identical rectangles. Each rectangle is 4 metres long and $a$ metres wide. Giving each answer in terms of $a$, find the distance of the centre of mass of the lamina from
\begin{enumerate}[label=(\alph*)]
\item $OE$, [4]
\item $OA$. [4]
\end{enumerate}
The lamina is freely suspended from $O$ and hangs in equilibrium with $OE$ at an angle $\theta$ to the downward vertical through $O$, where $\tan \theta = \frac{4}{3}$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find the value of $a$. [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2013 Q5 [12]}}