| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2007 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Suspended lamina equilibrium angle |
| Difficulty | Standard +0.3 This is a standard M2 centre of mass question involving composite shapes (subtraction of squares) and suspended lamina equilibrium. Part (a) requires routine application of the composite body formula with simple coordinates, and part (b) uses the standard tan θ = x̄/ȳ relationship. The geometry is straightforward with convenient coordinates (multiples of a), making calculations cleaner than average. Slightly easier than a typical M2 question due to the symmetric setup and standard method application. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.05a Angular velocity: definitions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(M(AF): \quad 4a^2 \cdot a - a^2 \cdot \frac{3a}{2} = 3a^2 \cdot \bar{x}\) | M1 A2,1,0 | Taking moments about AF or parallel axis, mass proportional to area. A1 A1 all correct; A1 A0 at most one error. Condone consistent lack of \(a\)'s for first three marks. NB: treating as rods rather than lamina is M0 |
| \(\bar{x} = \frac{5a}{6}\) | A1 | Accept 0.83a or better |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Symmetry \(\Rightarrow \bar{y} = \frac{5a}{6}\), or work from top to get \(\frac{7a}{6}\) | B1\(\sqrt{}\) | \(\bar{x} = \bar{y}\) = their 5a/6, or \(\bar{y}\) = distance from AB = \(2a\) - their 5a/6. Can be implied by working or awarded for clear statement in (a) |
| \(\tan q = \frac{5a/6}{2a - 5a/6} \quad \left(\frac{\bar{x}}{2a - \bar{y}}\right)\) | M1 A1\(\sqrt{}\) | Correct triangle identified and use of tan. \(\frac{2a-5a/6}{5a/6}\) is OK for M1. A1ft: tan \(\alpha\) expression correct for their 5a/6 and their \(\bar{y}\). Several candidates getting 45° without identifying correct angle is M0 unless clearly follows correctly from previous error |
| \(q \approx 35.5°\) | A1 | 35.5 (Q asks for 1 d.p.). Must suspend from point A; any other point is not a misread |
## Question 3:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $M(AF): \quad 4a^2 \cdot a - a^2 \cdot \frac{3a}{2} = 3a^2 \cdot \bar{x}$ | M1 A2,1,0 | Taking moments about AF or parallel axis, mass proportional to area. A1 A1 all correct; A1 A0 at most one error. Condone consistent lack of $a$'s for first three marks. NB: treating as rods rather than lamina is M0 |
| $\bar{x} = \frac{5a}{6}$ | A1 | Accept 0.83a or better |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Symmetry $\Rightarrow \bar{y} = \frac{5a}{6}$, or work from top to get $\frac{7a}{6}$ | B1$\sqrt{}$ | $\bar{x} = \bar{y}$ = their 5a/6, or $\bar{y}$ = distance from AB = $2a$ - their 5a/6. Can be implied by working or awarded for clear statement in (a) |
| $\tan q = \frac{5a/6}{2a - 5a/6} \quad \left(\frac{\bar{x}}{2a - \bar{y}}\right)$ | M1 A1$\sqrt{}$ | Correct triangle identified and use of tan. $\frac{2a-5a/6}{5a/6}$ is OK for M1. A1ft: tan $\alpha$ expression correct for their 5a/6 and their $\bar{y}$. Several candidates getting 45° without identifying correct angle is M0 unless clearly follows correctly from previous error |
| $q \approx 35.5°$ | A1 | 35.5 (Q asks for 1 d.p.). Must suspend from point A; any other point is not a misread |
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{778a0276-6738-40e6-90b2-a536ce5abe6a-04_568_568_205_685}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A uniform lamina $A B C D E F$ is formed by taking a uniform sheet of card in the form of a square $A X E F$, of side $2 a$, and removing the square $B X D C$ of side $a$, where $B$ and $D$ are the mid-points of $A X$ and $X E$ respectively, as shown in Figure 1.
\begin{enumerate}[label=(\alph*)]
\item Find the distance of the centre of mass of the lamina from $A F$.
The lamina is freely suspended from $A$ and hangs in equilibrium.
\item Find, in degrees to one decimal place, the angle which $A F$ makes with the vertical.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2007 Q3 [8]}}