| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2018 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | L-shaped or composite rectangular lamina |
| Difficulty | Standard +0.3 This is a standard M2 centre of mass question involving a composite L-shaped lamina. Part (a) requires routine application of the formula for centre of mass by dividing into two rectangles, taking moments, and dividing by total area. Part (b) involves a straightforward equilibrium condition using tan(α) = horizontal distance / vertical distance from the suspension point. While it requires multiple steps and careful coordinate work, it follows a well-practiced method with no novel insight required, making it slightly easier than average. |
| Spec | 6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(36a^2(6),\ 6a^2(1),\ 30a^2(5)\) with \(3a,\ \frac{14a}{3},\ \bar{x}\) and \(3a,\ 5a,\ \bar{y}\) | B1 | Mass ratios for a valid combination of shapes |
| Distances from \(OD\) | B1 | Distance from \(OD\) or equivalent for a valid combination of shapes |
| Distances from \(OA\) | B1 | Distance from \(OA\) or equivalent for a valid combination of shapes |
| \(6 \times 3a - 1 \times \frac{14a}{3} = 5\bar{x}\) or \(6 \times 3a - 1 \times 5a = 5\bar{y}\) | M1 | Moments equation for a horizontal or vertical axis or a vector equation combining both. For a valid dissection. |
| \(\Rightarrow \bar{x} = \frac{8a}{3}\) | A1 | One correct distance |
| \(\bar{y} = \frac{13a}{5}\) | A1 | Both distances correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\theta = \tan^{-1}\left(\frac{\bar{x} - 2a}{6a - \bar{y}}\right)\) | M1 | Trig ratio of a relevant angle |
| \(\alpha = \tan^{-1}\left(\frac{4}{3}\right) - \tan^{-1}\left(\frac{\bar{x}-2a}{6a-\bar{y}}\right)\) | DM1 | Expression for the required angle for their \(\bar{x}, \bar{y}\) |
| \(\left(\text{or } \tan^{-1}\left(\frac{6a-\bar{y}}{\bar{x}-2a}\right) - \tan^{-1}\left(\frac{3}{4}\right)\right)\) or \(\left(90 - \theta - \tan^{-1}\frac{3}{4}\right)\) | A1ft | Correct unsimplified substituted expression |
| \(\alpha = 42°\) | A1 | The question asks for the answer to the nearest degree |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Correct method for lengths of all three sides of triangle \(CBG\) | M1 | \(\left(\sqrt{\frac{2701}{225}}a,\ \sqrt{\frac{2536}{225}}a,\ 5a\right)\) |
| Correct use of cosine rule | M1 | |
| \(\cos\alpha = \frac{25 + \frac{2701}{225} - \frac{2536}{225}}{2 \times 5 \times \sqrt{\frac{2701}{225}}}\) | A1ft | Correct unsimplified for their values of the correct distances |
| \(\alpha = 42°\) | A1 | The question asks for the answer to the nearest degree |
## Question 3(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $36a^2(6),\ 6a^2(1),\ 30a^2(5)$ with $3a,\ \frac{14a}{3},\ \bar{x}$ and $3a,\ 5a,\ \bar{y}$ | B1 | Mass ratios for a valid combination of shapes |
| Distances from $OD$ | B1 | Distance from $OD$ or equivalent for a valid combination of shapes |
| Distances from $OA$ | B1 | Distance from $OA$ or equivalent for a valid combination of shapes |
| $6 \times 3a - 1 \times \frac{14a}{3} = 5\bar{x}$ or $6 \times 3a - 1 \times 5a = 5\bar{y}$ | M1 | Moments equation for a horizontal or vertical axis or a vector equation combining both. For a valid dissection. |
| $\Rightarrow \bar{x} = \frac{8a}{3}$ | A1 | One correct distance |
| $\bar{y} = \frac{13a}{5}$ | A1 | Both distances correct |
---
## Question 3(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\theta = \tan^{-1}\left(\frac{\bar{x} - 2a}{6a - \bar{y}}\right)$ | M1 | Trig ratio of a relevant angle |
| $\alpha = \tan^{-1}\left(\frac{4}{3}\right) - \tan^{-1}\left(\frac{\bar{x}-2a}{6a-\bar{y}}\right)$ | DM1 | Expression for the required angle for their $\bar{x}, \bar{y}$ |
| $\left(\text{or } \tan^{-1}\left(\frac{6a-\bar{y}}{\bar{x}-2a}\right) - \tan^{-1}\left(\frac{3}{4}\right)\right)$ or $\left(90 - \theta - \tan^{-1}\frac{3}{4}\right)$ | A1ft | Correct unsimplified substituted expression |
| $\alpha = 42°$ | A1 | The question asks for the answer to the nearest degree |
**Alternative 3(b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct method for lengths of all three sides of triangle $CBG$ | M1 | $\left(\sqrt{\frac{2701}{225}}a,\ \sqrt{\frac{2536}{225}}a,\ 5a\right)$ |
| Correct use of cosine rule | M1 | |
| $\cos\alpha = \frac{25 + \frac{2701}{225} - \frac{2536}{225}}{2 \times 5 \times \sqrt{\frac{2701}{225}}}$ | A1ft | Correct unsimplified for their values of the correct distances |
| $\alpha = 42°$ | A1 | The question asks for the answer to the nearest degree |
---3.
\begin{figure}[h]
\begin{center}
\begin{tikzpicture}[scale=0.8]
% Define coordinates
\coordinate (O) at (0,0);
\coordinate (A) at (6,0);
\coordinate (B) at (6,3);
\coordinate (C) at (2,6);
\coordinate (D) at (0,6);
% Draw the lamina
\draw[thick] (O) -- (A) -- (B) -- (C) -- (D) -- cycle;
% Right angle marks at O, A, and D
\draw (0,0.4) -- (0.4,0.4) -- (0.4,0);
\draw (5.6,0) -- (5.6,0.4) -- (6,0.4);
\draw (0.4,6) -- (0.4,5.6) -- (0,5.6);
% Labels for points
\node[below left] at (O) {$O$};
\node[below right] at (A) {$A$};
\node[right] at (B) {$B$};
\node[above] at (C) {$C$};
\node[above] at (D) {$D$};
% Dimension labels
\node[above] at ($(D)!0.5!(C)$) {$2a$};
\node[below] at ($(O)!0.5!(A)$) {$6a$};
\node[right] at ($(A)!0.5!(B)$) {$3a$};
\node[left] at ($(O)!0.5!(D)$) {$6a$};
\end{tikzpicture}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The uniform lamina $O A B C D$ is shown in Figure 1, with $O A = 6 a , A B = 3 a , C D = 2 a$ and $D O = 6 a$ and with right angles at $O , A$ and $D$.
\begin{enumerate}[label=(\alph*)]
\item Find the distance of the centre of mass of the lamina
\begin{enumerate}[label=(\roman*)]
\item from $O D$,
\item from $O A$.
The lamina is suspended from $C$ and hangs freely in equilibrium with $C B$ inclined at an angle $\alpha$ to the vertical.
\end{enumerate}\item Find, to the nearest degree, the size of the angle $\alpha$.
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2018 Q3 [10]}}