| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| 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 using the given formula for a semicircle. Part (a) requires straightforward application of composite body techniques with clear geometry, and part (b) is a routine equilibrium problem using tan θ = horizontal distance / vertical distance. The 'show that' format and multi-step nature add slight complexity, but this remains easier than average as it follows a well-practiced template with no novel insight required. |
| Spec | 6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Mass ratios: \(4a^2 : \frac{\pi a^2}{2} : a^2\left(4 + \frac{\pi}{2}\right)\) | B1 | Or equivalent |
| Distances relative to \(BD\): \(a,\ -\frac{4a}{3\pi},\ (d)\) | B1 | Or equivalent. Condone sign errors |
| Moments about \(BD\) (or a parallel axis) | M1 | Dimensionally correct. All terms required. Condone sign errors. Accept in a vector equation |
| \(4a^2 \times a + \frac{\pi a^2}{2} \times \frac{-4a}{3\pi} = \left(4 + \frac{\pi}{2}\right)a^2 \times d\) | A1 | Correct unsimplified equation |
| \(a\left(4 - \frac{4}{6}\right) = \left(4 + \frac{\pi}{2}\right)d\) | Distance from \(AE = \frac{(28+6\pi)a}{3(8+\pi)}\) | |
| \(d = a \times \frac{10}{3} \times \frac{2}{(8+\pi)} = \frac{20a}{3(8+\pi)}\) | A1 | Obtain given answer from correct working. Condone -ve becoming positive with no explanation at the end |
| (5 marks total) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use trig to find relevant angle | M1 | Using the given value of \(d\) |
| \(\tan\theta = \frac{a}{d} = \frac{3(8+\pi)}{20}\left(= \frac{1}{0.598}\right)\) | A1 | Correct expression for required angle |
| \(\theta\ (= 59.12...) = 59°\) | A1 | NB: The question asks for the nearest degree |
| (3 marks total) |
# Question 3:
## Part 3a:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Mass ratios: $4a^2 : \frac{\pi a^2}{2} : a^2\left(4 + \frac{\pi}{2}\right)$ | B1 | Or equivalent |
| Distances relative to $BD$: $a,\ -\frac{4a}{3\pi},\ (d)$ | B1 | Or equivalent. Condone sign errors |
| Moments about $BD$ (or a parallel axis) | M1 | Dimensionally correct. All terms required. Condone sign errors. Accept in a vector equation |
| $4a^2 \times a + \frac{\pi a^2}{2} \times \frac{-4a}{3\pi} = \left(4 + \frac{\pi}{2}\right)a^2 \times d$ | A1 | Correct unsimplified equation |
| $a\left(4 - \frac{4}{6}\right) = \left(4 + \frac{\pi}{2}\right)d$ | | Distance from $AE = \frac{(28+6\pi)a}{3(8+\pi)}$ |
| $d = a \times \frac{10}{3} \times \frac{2}{(8+\pi)} = \frac{20a}{3(8+\pi)}$ | A1 | Obtain **given answer** from correct working. Condone -ve becoming positive with no explanation at the end |
| **(5 marks total)** | | |
## Part 3b:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use trig to find relevant angle | M1 | Using the given value of $d$ |
| $\tan\theta = \frac{a}{d} = \frac{3(8+\pi)}{20}\left(= \frac{1}{0.598}\right)$ | A1 | Correct expression for required angle |
| $\theta\ (= 59.12...) = 59°$ | A1 | NB: The question asks for the nearest degree |
| **(3 marks total)** | | |
---3. [The centre of mass of a semicircular lamina of radius $r$ is $\frac { 4 r } { 3 \pi }$ from the centre.]
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{88731f1c-5177-4096-841b-cd9c3f87782b-08_581_460_374_740}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows the uniform lamina $A B C D E$, such that $A B D E$ is a square with sides of length $2 a$ and $B C D$ is a semicircle with diameter $B D$.
\begin{enumerate}[label=(\alph*)]
\item Show that the distance of the centre of mass of the lamina from $B D$ is $\frac { 20 a } { 3 ( 8 + \pi ) }$.
The lamina is freely suspended from $D$ and hangs in equilibrium.
\item Find, to the nearest degree, the angle that $D E$ makes with the downward vertical.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2018 Q3 [8]}}