| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Particle attached to lamina - find mass/position |
| Difficulty | Standard +0.3 This is a standard M2 centre of mass question involving composite bodies with straightforward calculations. Parts (a)-(b) use the standard formula for combined centres of mass with given positions. Part (c) applies equilibrium conditions (vertical line through suspension point). Part (d) involves adding a particle to achieve horizontal equilibrium. All techniques are routine for M2 students with no novel problem-solving required, making it slightly easier than average. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Rectangular lamina centre of mass: \(30\) cm from \(AD\) | B1 | Centre of rectangle |
| Triangular lamina centre of mass: \(5\) cm from \(DC\), so \(25\) cm from \(AB\)... distance from \(AD = 5\) cm | ||
| \((5+4)\bar{x} = 5 \times 30 + 4 \times 5\) | M1 | Moments equation about \(AD\) |
| \(9\bar{x} = 150 + 20 = 170\) | A1 | Correct equation |
| \(\bar{x} = \frac{170}{9} \approx 18.9\) cm from \(AD\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Rectangular lamina centre of mass: \(15\) cm from \(AB\) | B1 | Centre of rectangle |
| Triangular lamina centre of mass: \(10\) cm from \(AD\), \(25\) cm from \(AB\) | ||
| \(9\bar{y} = 5 \times 15 + 4 \times 25\) | M1 | Moments equation about \(AB\) |
| \(9\bar{y} = 75 + 100 = 175\) | A1 | Correct equation |
| \(\bar{y} = \frac{175}{9} \approx 19.4\) cm from \(AB\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(P\) is midpoint of \(AB\), so \(P\) is \(30\) cm from \(AD\) and \(0\) cm from \(AB\) | B1 | Position of P identified |
| Horizontal distance from \(P\) to CoM \(= 30 - \frac{170}{9} = \frac{100}{9}\) cm | M1 | Correct horizontal distance |
| Vertical distance from \(P\) to CoM \(= \frac{175}{9}\) cm | A1 | Correct vertical distance |
| \(\tan\theta = \frac{100/9}{175/9} = \frac{100}{175} = \frac{4}{7}\) | M1 | Use of tan with correct distances |
| \(\theta = \arctan\left(\frac{4}{7}\right) \approx 29.7°\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| For \(AB\) horizontal, CoM must lie vertically below \(P\), i.e. directly below \(P\) at \(x = 30\) cm from \(AD\) | M1 | Moments about \(P\) with added mass |
| \((9+m) \times 30 = 9 \times \frac{170}{9} + m \times 60\) | M1 | Moments equation |
| \(270 + 30m = 170 + 60m\) | A1 | Correct equation |
| \(100 = 30m\) | ||
| \(m = \frac{10}{3} \approx 3.33\) kg | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| The centre of mass of the uniform rectangular lamina \(ABCD\) is at its geometrical centre (intersection of diagonals), i.e. at \(30\) cm from \(AD\) and \(15\) cm from \(AB\) | B1 | Must reference geometrical centre or midpoint |
## Question 4:
### Part (a): Distance of centre of mass from AD
| Working/Answer | Mark | Guidance |
|---|---|---|
| Rectangular lamina centre of mass: $30$ cm from $AD$ | B1 | Centre of rectangle |
| Triangular lamina centre of mass: $5$ cm from $DC$, so $25$ cm from $AB$... distance from $AD = 5$ cm | | |
| $(5+4)\bar{x} = 5 \times 30 + 4 \times 5$ | M1 | Moments equation about $AD$ |
| $9\bar{x} = 150 + 20 = 170$ | A1 | Correct equation |
| $\bar{x} = \frac{170}{9} \approx 18.9$ cm from $AD$ | A1 | cao |
### Part (b): Distance of centre of mass from AB
| Working/Answer | Mark | Guidance |
|---|---|---|
| Rectangular lamina centre of mass: $15$ cm from $AB$ | B1 | Centre of rectangle |
| Triangular lamina centre of mass: $10$ cm from $AD$, $25$ cm from $AB$ | | |
| $9\bar{y} = 5 \times 15 + 4 \times 25$ | M1 | Moments equation about $AB$ |
| $9\bar{y} = 75 + 100 = 175$ | A1 | Correct equation |
| $\bar{y} = \frac{175}{9} \approx 19.4$ cm from $AB$ | A1 | cao |
### Part (c): Angle between AB and horizontal when suspended from P
| Working/Answer | Mark | Guidance |
|---|---|---|
| $P$ is midpoint of $AB$, so $P$ is $30$ cm from $AD$ and $0$ cm from $AB$ | B1 | Position of P identified |
| Horizontal distance from $P$ to CoM $= 30 - \frac{170}{9} = \frac{100}{9}$ cm | M1 | Correct horizontal distance |
| Vertical distance from $P$ to CoM $= \frac{175}{9}$ cm | A1 | Correct vertical distance |
| $\tan\theta = \frac{100/9}{175/9} = \frac{100}{175} = \frac{4}{7}$ | M1 | Use of tan with correct distances |
| $\theta = \arctan\left(\frac{4}{7}\right) \approx 29.7°$ | A1 | cao |
### Part (d): Mass m attached at B for AB horizontal
| Working/Answer | Mark | Guidance |
|---|---|---|
| For $AB$ horizontal, CoM must lie vertically below $P$, i.e. directly below $P$ at $x = 30$ cm from $AD$ | M1 | Moments about $P$ with added mass |
| $(9+m) \times 30 = 9 \times \frac{170}{9} + m \times 60$ | M1 | Moments equation |
| $270 + 30m = 170 + 60m$ | A1 | Correct equation |
| $100 = 30m$ | | |
| $m = \frac{10}{3} \approx 3.33$ kg | A1 | cao |
### Part (e): Use of uniform lamina
| Working/Answer | Mark | Guidance |
|---|---|---|
| The centre of mass of the uniform rectangular lamina $ABCD$ is at its geometrical centre (intersection of diagonals), i.e. at $30$ cm from $AD$ and $15$ cm from $AB$ | B1 | Must reference geometrical centre or midpoint |4 A uniform rectangular lamina $A B C D$ has a mass of 5 kg . The side $A B$ has length 60 cm and the side $B C$ has length 30 cm . The points $P , Q , R$ and $S$ are the mid-points of the sides, as shown in the diagram below.
A uniform triangular lamina $S R D$, of mass 4 kg , is fixed to the rectangular lamina to form a shop sign. The centre of mass of the triangular lamina $S R D$ is 10 cm from the side $A D$ and 5 cm from the side $D C$.\\
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\begin{enumerate}[label=(\alph*)]
\item Find the distance of the centre of mass of the shop sign from $A D$.
\item Find the distance of the centre of mass of the shop sign from $A B$.
\item The shop sign is freely suspended from $P$.
Find the angle between $A B$ and the horizontal when the shop sign is in equilibrium.
\item To ensure that the side $A B$ is horizontal when the shop sign is freely suspended from point $P$, a particle of mass $m \mathrm {~kg}$ is attached to the shop sign at point $B$.
Calculate $m$.
\item Explain how you have used the fact that the rectangular lamina $A B C D$ is uniform in your solution to this question.\\
(1 mark)
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\end{center}
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{9d039ec3-fd0a-40ae-9afe-7627439081df-11_2486_1714_221_153}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2011 Q4 [14]}}