| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2017 |
| Session | October |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Suspended lamina equilibrium angle |
| Difficulty | Standard +0.8 This is a multi-step centre of mass problem requiring: (1) finding the centre of mass of a composite lamina (equilateral triangle with circular hole removed) using the subtraction method, (2) applying equilibrium conditions for a suspended lamina where the centre of mass must lie vertically below the suspension point, and (3) using trigonometry to find the angle. While the techniques are standard M2 content, the problem requires careful coordinate geometry, multiple calculations, and spatial reasoning about the equilibrium configuration, making it moderately challenging but not exceptional for this module. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Mass ratio: \(\pi a^2\) : \(4a^2\sqrt{3}\) : \(a^2(4\sqrt{3}-\pi)\) | B1 | |
| Dist from \(D\): \(\frac{3a}{2}\) : \(\frac{2a}{3}\sqrt{3}\) : \(\bar{x}\) | B1 | |
| \(4\sqrt{3} \times \frac{2a}{3}\sqrt{3} - \pi\frac{3a}{2} = (4\sqrt{3}-\pi)\bar{x}\) | M1 | Moments about \(BC\). Condone sign errors. Accept in vector form. |
| A1ft | Correct unsimplified equation. Follow their clear statements of areas and mass ratios. | |
| \(\bar{x} = \dfrac{4\sqrt{3}\times\frac{2a}{3}\sqrt{3} - \pi\frac{3a}{2}}{(4\sqrt{3}-\pi)} = \dfrac{(16-3\pi)a}{2(4\sqrt{3}-\pi)}\) | A1 | \((0.86822...a =)\ 0.87a\) or better |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\tan\theta = \dfrac{2a}{\bar{x}}\) | M1 | Trig ratio of a relevant angle |
| \(\theta = \tan^{-1}\!\left(\dfrac{2a}{\bar{x}}\right)\) or \(\theta = 90°-\tan^{-1}\!\left(\dfrac{\bar{x}}{2a}\right)\) | A1ft | Follow their \(\bar{x}\) |
| \(\theta = 67°\) | A1 | Final answer. Q asks for acute angle and nearest degree. Do not ISW. SR: Final answer \(23°\) scores M1A0A0 |
## Question 5:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Mass ratio: $\pi a^2$ : $4a^2\sqrt{3}$ : $a^2(4\sqrt{3}-\pi)$ | B1 | |
| Dist from $D$: $\frac{3a}{2}$ : $\frac{2a}{3}\sqrt{3}$ : $\bar{x}$ | B1 | |
| $4\sqrt{3} \times \frac{2a}{3}\sqrt{3} - \pi\frac{3a}{2} = (4\sqrt{3}-\pi)\bar{x}$ | M1 | Moments about $BC$. Condone sign errors. Accept in vector form. |
| | A1ft | Correct unsimplified equation. Follow their clear statements of areas and mass ratios. |
| $\bar{x} = \dfrac{4\sqrt{3}\times\frac{2a}{3}\sqrt{3} - \pi\frac{3a}{2}}{(4\sqrt{3}-\pi)} = \dfrac{(16-3\pi)a}{2(4\sqrt{3}-\pi)}$ | A1 | $(0.86822...a =)\ 0.87a$ or better |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\tan\theta = \dfrac{2a}{\bar{x}}$ | M1 | Trig ratio of a relevant angle |
| $\theta = \tan^{-1}\!\left(\dfrac{2a}{\bar{x}}\right)$ or $\theta = 90°-\tan^{-1}\!\left(\dfrac{\bar{x}}{2a}\right)$ | A1ft | Follow their $\bar{x}$ |
| $\theta = 67°$ | A1 | Final answer. Q asks for acute angle and nearest degree. Do not ISW. SR: Final answer $23°$ scores M1A0A0 |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5ef8231d-5b95-4bbb-a8e2-788c708fa078-16_632_734_248_605}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
The uniform lamina $A B C$ is in the shape of an equilateral triangle with sides of length $4 a$. The midpoint of $B C$ is $D$. The point $E$ lies on $A D$ with $D E = \frac { 3 a } { 2 }$. A circular hole, with centre $E$ and radius $a$, is made in the lamina $A B C$ to form the lamina $L$, shown shaded in Figure 2.
\begin{enumerate}[label=(\alph*)]
\item Find the distance of the centre of mass of $L$ from $D$.
The lamina $L$ is freely suspended from the point $B$ and hangs in equilibrium.
\item Find, to the nearest degree, the size of the acute angle between $A D$ and the downward vertical.
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2017 Q5 [8]}}