| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2009 |
| Session | January |
| 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 composite lamina calculations using given formulae. Part (a) involves straightforward application of the centre of mass formula with areas and distances, part (b) uses the provided semicircle formula, and part (c) applies equilibrium geometry. All steps are routine textbook exercises 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 |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Mass ratio: MR = \(108\), \(18\pi\), \(108 + 18\pi\) | B1 | |
| \(x_i(\rightarrow)\) from \(AD\): \(4\), \(6\), \(\bar{x}\) | B1 | |
| \(y_i(\downarrow)\) from \(BD\): \(6\), \(-\dfrac{8}{\pi}\), \(\bar{y}\) | ||
| \(AD(\rightarrow): 108(4) + 18\pi(6) = (108 + 18\pi)\bar{x}\) | M1 | |
| \(\bar{x} = \dfrac{432 + 108\pi}{108 + 18\pi} = 4.68731\ldots = \underline{4.69}\) (cm) (3 sf) | A1 | AG |
| Total: (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y_i(\downarrow)\) from \(BD\): \(6\), \(-\dfrac{8}{\pi}\), \(\bar{y}\) | B1 oe | |
| \(BD(\downarrow): 108(6) + 18\pi\!\left(-\dfrac{8}{\pi}\right) = (108 + 18\pi)\bar{y}\) | M1, A1ft | |
| \(\bar{y} = \dfrac{504}{108 + 18\pi} = 3.06292\ldots = \underline{3.06}\) (cm) (3 sf) | A1 | |
| Total: (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Diagram showing \(12 - \bar{x}\) horizontally and \(\bar{y}\) vertically from \(B\) | M1 | |
| \(\tan\theta = \dfrac{\bar{y}}{12 - 4.68731\ldots} = \dfrac{3.06392\ldots}{12 - 4.68731\ldots}\) | dM1, A1 | |
| \(\theta = 22.72641\ldots = \underline{23}\) (nearest degree) | A1 | |
| Total: [12] |
## Question 5:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Mass ratio: MR = $108$, $18\pi$, $108 + 18\pi$ | B1 | |
| $x_i(\rightarrow)$ from $AD$: $4$, $6$, $\bar{x}$ | B1 | |
| $y_i(\downarrow)$ from $BD$: $6$, $-\dfrac{8}{\pi}$, $\bar{y}$ | | |
| $AD(\rightarrow): 108(4) + 18\pi(6) = (108 + 18\pi)\bar{x}$ | M1 | |
| $\bar{x} = \dfrac{432 + 108\pi}{108 + 18\pi} = 4.68731\ldots = \underline{4.69}$ (cm) (3 sf) | A1 | **AG** |
| **Total: (4)** | | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y_i(\downarrow)$ from $BD$: $6$, $-\dfrac{8}{\pi}$, $\bar{y}$ | B1 oe | |
| $BD(\downarrow): 108(6) + 18\pi\!\left(-\dfrac{8}{\pi}\right) = (108 + 18\pi)\bar{y}$ | M1, A1ft | |
| $\bar{y} = \dfrac{504}{108 + 18\pi} = 3.06292\ldots = \underline{3.06}$ (cm) (3 sf) | A1 | |
| **Total: (4)** | | |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Diagram showing $12 - \bar{x}$ horizontally and $\bar{y}$ vertically from $B$ | M1 | |
| $\tan\theta = \dfrac{\bar{y}}{12 - 4.68731\ldots} = \dfrac{3.06392\ldots}{12 - 4.68731\ldots}$ | dM1, A1 | |
| $\theta = 22.72641\ldots = \underline{23}$ (nearest degree) | A1 | |
| **Total: [12]** | | |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{4c8ebad3-0ebb-4dfe-8036-54b651deb9cf-08_781_541_223_687}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A uniform lamina $A B C D$ is made by joining a uniform triangular lamina $A B D$ to a uniform semi-circular lamina $D B C$, of the same material, along the edge $B D$, as shown in Figure 2. Triangle $A B D$ is right-angled at $D$ and $A D = 18 \mathrm {~cm}$. The semi-circle has diameter $B D$ and $B D = 12 \mathrm {~cm}$.
\begin{enumerate}[label=(\alph*)]
\item Show that, to 3 significant figures, the distance of the centre of mass of the lamina $A B C D$ from $A D$ is 4.69 cm .
Given that the centre of mass of a uniform semicircular lamina, radius $r$, is at a distance $\frac { 4 r } { 3 \pi }$ from the centre of the bounding diameter,
\item find, in cm to 3 significant figures, the distance of the centre of mass of the lamina $A B C D$ from $B D$.
The lamina is freely suspended from $B$ and hangs in equilibrium.
\item Find, to the nearest degree, the angle which $B D$ makes with the vertical.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2009 Q5 [12]}}