| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2013 |
| Session | January |
| Marks | 5 |
| 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 M2 centre of mass question requiring straightforward application of the formula for composite bodies (part a) and simple trigonometry with equilibrium conditions (part b). The perpendicular rod configuration simplifies calculations, and both parts follow routine textbook methods with no novel problem-solving required. Slightly above average difficulty only due to the two-part structure and need for careful coordinate setup. |
| Spec | 6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Diagram showing L-shape with dimensions: 0.5m vertical, 2kg mass; 2m horizontal BC, 3kg mass | M1 | Moments equation with lengths \(\frac{1}{4}\), 1 and (ratio of) masses 2, 3. Allow moments about a parallel axis. Use of length for mass is M0. |
| \(5\bar{y} = 2 \times 0.25(+0)\) | A1 | For distance from \(BC\) |
| \(\bar{y} = \frac{2 \times 0.25}{5} = 0.1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Diagram showing lamina suspended from \(A\) with \(G\) marked and angle \(\theta\) shown | M1 | Must suspend from \(A\) |
| \(\tan\theta = \frac{0.6}{0.5 - 0.1}\) | A1ft | Use of \(\tan\) with \(0.6\) and \(0.5 - \bar{y}\). Could be wrong way up. Must be using \(0.6\) |
| \(\theta = \tan^{-1}\left(\frac{6}{4}\right) = 56.3° \approx 56°\) | A1 | Correct way up. ft their \(\bar{y}\). Accept awrt 56.3 |
## Question 1:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Diagram showing L-shape with dimensions: 0.5m vertical, 2kg mass; 2m horizontal BC, 3kg mass | M1 | Moments equation with lengths $\frac{1}{4}$, 1 and (ratio of) masses 2, 3. Allow moments about a parallel axis. Use of length for mass is M0. |
| $5\bar{y} = 2 \times 0.25(+0)$ | A1 | For distance from $BC$ |
| $\bar{y} = \frac{2 \times 0.25}{5} = 0.1$ | | |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Diagram showing lamina suspended from $A$ with $G$ marked and angle $\theta$ shown | M1 | Must suspend from $A$ |
| $\tan\theta = \frac{0.6}{0.5 - 0.1}$ | A1ft | Use of $\tan$ with $0.6$ and $0.5 - \bar{y}$. Could be wrong way up. Must be using $0.6$ |
| $\theta = \tan^{-1}\left(\frac{6}{4}\right) = 56.3° \approx 56°$ | A1 | Correct way up. ft their $\bar{y}$. Accept awrt 56.3 |\begin{enumerate}
\item Two uniform rods $A B$ and $B C$ are rigidly joined at $B$ so that $\angle A B C = 90 ^ { \circ }$. Rod $A B$ has length 0.5 m and mass 2 kg . Rod $B C$ has length 2 m and mass 3 kg . The centre of mass of the framework of the two rods is at $G$.\\
(a) Find the distance of $G$ from $B C$.
\end{enumerate}
The distance of $G$ from $A B$ is 0.6 m .\\
The framework is suspended from $A$ and hangs freely in equilibrium.\\
(b) Find the angle between $A B$ and the downward vertical at $A$.\\
\hfill \mbox{\textit{Edexcel M2 2013 Q1 [5]}}