| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2013 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Folded lamina |
| Difficulty | Standard +0.8 This is a multi-part M2 centre of mass question requiring composite body techniques, understanding of folding transformations, and equilibrium of suspended bodies. Part (a) involves standard trapezium centre of mass calculation with given mass (5 marks suggests moderate working). Part (b) requires careful consideration of how folding affects the centre of mass position. Part (c) combines geometry with equilibrium conditions. While systematic, it requires multiple techniques and careful spatial reasoning beyond routine exercises, placing it moderately above average difficulty. |
| Spec | 6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| ABC: \(\frac{M}{1}\), ADE: \(\frac{4M}{9}\), BCED: \(\frac{5M}{9}\) | B1 | Correct mass ratios |
| \(\frac{h}{3}\), ADE: \(\frac{h}{3}(\frac{h}{3} + \frac{2h}{3}) = \overline{y}\) | B1 | Correct distance ratios |
| M1 | Moments equation. Condone sign slip | |
| \(M \cdot \frac{h}{3} - \frac{4M}{9} \cdot \frac{5h}{9} = \frac{5M}{9}\overline{y}\) | A1 | |
| \(\overline{y} = \frac{7h}{45}\) *Answer Given* | A1 | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| M1 | Moments equation for the folded shape. Requires correct mass ratios, and terms of correct structure. | |
| \(\frac{5M}{9} \cdot \frac{7h}{45} + \frac{4M}{9}\left(\frac{h}{3} - \frac{1}{3} \cdot \frac{2h}{3}\right) = M \overline{x}\) | A1 A1 | -1 each error, \(\frac{h}{9}\) |
| \(\overline{x} = \frac{11h}{81}\) | A1 | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tan\alpha = \frac{h}{3} - \overline{x} = \frac{h - 2a}{2a}\) | M1 | Use of tan in correct triangle. Allow reciprocal. |
| A1 ft | Correct unsimplified for their \(\overline{x}\) | |
| \(= \frac{8h}{27a}\) | DM1 A1 | Substitute and simplify |
| (4) | ||
| [13] |
**Part (a):**
| ABC: $\frac{M}{1}$, ADE: $\frac{4M}{9}$, BCED: $\frac{5M}{9}$ | B1 | Correct mass ratios |
| $\frac{h}{3}$, ADE: $\frac{h}{3}(\frac{h}{3} + \frac{2h}{3}) = \overline{y}$ | B1 | Correct distance ratios |
| | M1 | Moments equation. Condone sign slip |
| $M \cdot \frac{h}{3} - \frac{4M}{9} \cdot \frac{5h}{9} = \frac{5M}{9}\overline{y}$ | A1 | |
| $\overline{y} = \frac{7h}{45}$ *Answer Given* | A1 | (5) |
**Part (b):**
| | M1 | Moments equation for the folded shape. Requires correct mass ratios, and terms of correct structure. |
| $\frac{5M}{9} \cdot \frac{7h}{45} + \frac{4M}{9}\left(\frac{h}{3} - \frac{1}{3} \cdot \frac{2h}{3}\right) = M \overline{x}$ | A1 A1 | -1 each error, $\frac{h}{9}$ |
| $\overline{x} = \frac{11h}{81}$ | A1 | (4) |
**Part (c):**
| $\tan\alpha = \frac{h}{3} - \overline{x} = \frac{h - 2a}{2a}$ | M1 | Use of tan in correct triangle. Allow reciprocal. |
| | A1 ft | Correct unsimplified for their $\overline{x}$ |
| $= \frac{8h}{27a}$ | DM1 A1 | Substitute and simplify |
| | (4) |
| | [13] | |
---\includegraphics{figure_2}
A uniform triangular lamina $ABC$ of mass $M$ is such that $AB = AC$, $BC = 2a$ and the distance of $A$ from $BC$ is $h$. A line, parallel to $BC$ and at a distance $\frac{2h}{3}$ from $A$, cuts $AB$ at $D$ and cuts $AC$ at $E$, as shown in Figure 2.
It is given that the mass of the trapezium $BCED$ is $\frac{5M}{9}$.
\begin{enumerate}[label=(\alph*)]
\item Show that the centre of mass of the trapezium $BCED$ is $\frac{7h}{45}$ from $BC$. [5]
\end{enumerate}
\includegraphics{figure_3}
The portion $ADE$ of the lamina is folded through 180° about $DE$ to form the folded lamina shown in Figure 3.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the distance of the centre of mass of the folded lamina from $BC$. [4]
\end{enumerate}
The folded lamina is freely suspended from $D$ and hangs in equilibrium. The angle between $DE$ and the downward vertical is $\alpha$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find tan $\alpha$ in terms of $a$ and $h$. [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2013 Q6 [13]}}