| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2009 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Suspended lamina equilibrium angle |
| Difficulty | Standard +0.3 This is a standard M2 centre of mass question requiring straightforward application of the formula for composite bodies (taking moments about an axis), followed by a routine suspended lamina equilibrium calculation using tan θ = horizontal distance/vertical distance. The multi-part structure and need to set up coordinates adds slight complexity beyond pure recall, but all techniques are standard textbook exercises with no novel insight required. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass |
| Answer | Marks | Guidance |
|---|---|---|
| Rectangular lamina CoM from \(AD\): \(x = 10\) cm | B1 | Centre of rectangle |
| Circular lamina CoM from \(AD\): \(x = 5\) cm | B1 | Given in question |
| \(\bar{x} = \frac{8(10) + 2(5)}{8+2} = \frac{80+10}{10} = \frac{90}{10} = 9\) cm | M1 A1 | Correct weighted mean |
| Answer | Marks | Guidance |
|---|---|---|
| Distance from \(AB = 5\) cm | B1 | By symmetry |
| Answer | Marks | Guidance |
|---|---|---|
| \(P\) is midpoint of \(AB\), so \(P\) is at \((10, 0)\) from \(AD\) along \(AB\) | M1 | Identify position of \(P\) |
| Horizontal distance of CoM from \(P\): \(10 - 9 = 1\) cm | A1 | |
| Vertical distance of CoM from \(P\) (along \(AB\)): \(5\) cm | A1 | |
| \(\tan\theta = \frac{1}{5}\) | M1 | Correct ratio |
| \(\theta = 11.3°\) | A1 | Correct angle |
| Answer | Marks |
|---|---|
| The centre of mass of each uniform lamina is at its geometric centre | B1 |
# Question 4:
## Part (a):
| Rectangular lamina CoM from $AD$: $x = 10$ cm | B1 | Centre of rectangle |
|---|---|---|
| Circular lamina CoM from $AD$: $x = 5$ cm | B1 | Given in question |
| $\bar{x} = \frac{8(10) + 2(5)}{8+2} = \frac{80+10}{10} = \frac{90}{10} = 9$ cm | M1 A1 | Correct weighted mean |
## Part (b):
| Distance from $AB = 5$ cm | B1 | By symmetry |
|---|---|---|
## Part (c):
| $P$ is midpoint of $AB$, so $P$ is at $(10, 0)$ from $AD$ along $AB$ | M1 | Identify position of $P$ |
|---|---|---|
| Horizontal distance of CoM from $P$: $10 - 9 = 1$ cm | A1 | |
| Vertical distance of CoM from $P$ (along $AB$): $5$ cm | A1 | |
| $\tan\theta = \frac{1}{5}$ | M1 | Correct ratio |
| $\theta = 11.3°$ | A1 | Correct angle |
## Part (d):
| The centre of mass of each uniform lamina is at its geometric centre | B1 | |
|---|---|---|
---4 A uniform rectangular lamina $A B C D$ has a mass of 8 kg . The side $A B$ has length 20 cm , the side $B C$ has length 10 cm , and $P$ is the mid-point of $A B$.
A uniform circular lamina, of mass 2 kg and radius 5 cm , is fixed to the rectangular lamina to form a sign. The centre of the circular lamina is 5 cm from each of $A B$ and $B C$, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{851cb2a3-5bc8-4af9-b1fc-a143d37beebe-3_661_1200_589_406}
\begin{enumerate}[label=(\alph*)]
\item Find the distance of the centre of mass of the sign from $A D$.
\item Write down the distance of the centre of mass of the sign from $A B$.
\item The sign is freely suspended from $P$.
Find the angle between $A D$ and the vertical when the sign is in equilibrium.
\item Explain how you have used the fact that each lamina is uniform in your solution to this question.
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2009 Q4 [9]}}