| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2018 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Lamina with attached triangle |
| Difficulty | Standard +0.3 This is a standard M2 centre of mass question requiring calculation of composite lamina centroid using area ratios and moments, followed by equilibrium angle calculation. The similar triangles and given dimensions make calculations straightforward with no novel problem-solving required—slightly easier than average A-level. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass |
| Answer | Marks |
|---|---|
| Area ratios: \(9a^2 : \frac{9}{4}a^2 : 5 \times \frac{9}{4}a^2\) \((4 : 1 : 5)\) | B1 |
| Distances from \(AC\): \(a, \frac{a}{2}(\bar{y})\) | B1 |
| Moments about \(AC\): \(4 \times a + 1 \times \frac{a}{2} = 5\bar{y}\) | M1 A1 ft |
| \(\bar{y} = \frac{9}{10}a\) | A1 cso |
| (5) |
| Answer | Marks |
|---|---|
| Distances from vertical through \(B\): \(0, \frac{3}{2}a(\bar{x})\) | B1 |
| OR: | |
| Moments about vertical axis through \(B\): \(1 \times \frac{3}{2}a = 5\bar{x}\) | M1 |
| \(\bar{x} = \frac{3}{10}a\) | A1 |
| N.B. You may see \(\bar{x} = 3.3a\) (if distances measured from \(A\)) or \(\bar{x} = 2.7a\) (if distances measured from \(C\)) | |
| Required angle \(ABG: 45° + \theta\) (for their \(\bar{x}\); e.g. if \(\bar{x} < 0\), then \(45° - \theta\)) | M1 |
| Method for \(\theta\): \(\tan\theta = \frac{\bar{x}}{9a - \frac{a}{10}} = \left(-\frac{1}{7}\right)\) | M1 A1 ft |
| \(45° + \tan^{-1}\left(\frac{1}{7}\right) = 53°\) nearest degree | A1 |
| (7) | |
| [12] |
## 4a.
Area ratios: $9a^2 : \frac{9}{4}a^2 : 5 \times \frac{9}{4}a^2$ $(4 : 1 : 5)$ | B1 |
Distances from $AC$: $a, \frac{a}{2}(\bar{y})$ | B1 |
Moments about $AC$: $4 \times a + 1 \times \frac{a}{2} = 5\bar{y}$ | M1 A1 ft |
$\bar{y} = \frac{9}{10}a$ | A1 cso |
(5) | |
## 4b.
Distances from vertical through $B$: $0, \frac{3}{2}a(\bar{x})$ | B1 |
OR: | |
Moments about vertical axis through $B$: $1 \times \frac{3}{2}a = 5\bar{x}$ | M1 |
$\bar{x} = \frac{3}{10}a$ | A1 |
**N.B.** You may see $\bar{x} = 3.3a$ (if distances measured from $A$) or $\bar{x} = 2.7a$ (if distances measured from $C$) | |
Required angle $ABG: 45° + \theta$ (for their $\bar{x}$; e.g. if $\bar{x} < 0$, then $45° - \theta$) | M1 |
Method for $\theta$: $\tan\theta = \frac{\bar{x}}{9a - \frac{a}{10}} = \left(-\frac{1}{7}\right)$ | M1 A1 ft |
$45° + \tan^{-1}\left(\frac{1}{7}\right) = 53°$ nearest degree | A1 |
(7) | |
[12] | |
**Notes for Q4:**
4(a)
- First B1 for correct (unsimplified) area ratios seen
- Second B1 for correct distances from $AC$ (or from any parallel line)
- First M1 for a 'moments' equation about $AC$ (or any parallel line)
- First A1 ft for a correct equation, ft on their 'distances' and 'masses'.
- Second A1 cso for a correct GIVEN ANSWER with no errors seen
**N.B.**
Allow equivalent vector equation for two moments equations.
4(b)
- B1 for correct distances from vertical through $B$ (or from any parallel line) (May be seen in (a)) Allow missing $a$'s if they recover
- First M1 for 'moments' equation about vertical axis through $B$ (or from any parallel line) (May be seen in (a))
- First A1 for a correct $\bar{x}$ (allow recovery of missing $a$'s)
- Second M1 for use of correct angle (e.g. $45° + \theta$ or $45° - \theta$ if using tan method) oe
- Third M1 for appropriate equation in $\theta$ only, using their $\bar{x}$ and $\frac{9a}{10}$
- Second A1 ft for a correct equation
- Third A1 for $53°$ (nearest degree)
**ALTERNATIVE:**
- B1 for correct distances from vertical through $B$ (or from any parallel line) (May be seen in (a)) Allow missing $a$'s if they recover
- First M1 for a dim correct moments equation about vertical axis through $B$ (or from any parallel line) (May be seen in (a))
- First A1 for a correct $\bar{x}$ (allow recovery of missing $a$'s)
- Second M1 for identifying correct angle ($ABG = \alpha$)
- Third M1 for use of cos rule on triangle $ABG$ to give equation in $\alpha$ only, using their $\bar{x}$ and $\frac{9a}{10}$
- Second A1 ft for a correct equation
- Third A1 for $53°$ (nearest degree)
**OR:**
- B1 for correct distances from vertical through $B$ (or from any parallel line) (May be seen in (a)) Allow missing $a$'s if they recover
- First M1 for a dim correct moments equation about vertical axis through $B$ (or from any parallel line) (May be seen in (a))
- First A1 for a correct $\bar{x}$ (allow recovery of missing $a$'s)
- Second M1 for use of angle $ABG = $ angle $ABC$ - angle $GBC$
- Third M1 for use of cos rule on triangle $GBC$ to find angle $GBC$, using their $\bar{x}$ and $\frac{9a}{10}$
- Second A1 ft for a correct equation
- Third A1 for $53°$ (nearest degree)
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a3f28425-4acf-4878-b0e3-15b5bc8a92d7-12_702_1182_226_379}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
The uniform lamina $A B C$ is an isosceles triangle with $A B = B C , A C = 6 a$ and the distance from $B$ to $A C$ is $3 a$.
The uniform lamina $M N C$ is an isosceles triangle with $M N = N C$ and $M C = 3 a$. Triangles $A B C$ and $M N C$ are similar and are made of the same material.
The lamina $L$ is formed by fixing triangle $M N C$ on top of triangle $A B C$, as shown in Figure 2.
\begin{enumerate}[label=(\alph*)]
\item Show that the distance of the centre of mass of $L$ from $A C$ is $\frac { 9 } { 10 } a$
The lamina $L$ is freely suspended from $B$ and hangs in equilibrium.
\item Find, to the nearest degree, the size of the angle between $A B$ and the downward vertical.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2018 Q4 [12]}}