| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Folded lamina |
| Difficulty | Standard +0.3 This is a standard M2 centre of mass question involving a folded lamina. Part (a) requires dividing the shape into two components (the main square minus triangle, and the folded triangle), finding their individual centres of mass, and using the composite body formula—all routine techniques. Part (b) applies the equilibrium condition that the centre of mass hangs directly below the suspension point, requiring basic trigonometry. While it involves multiple steps (6 marks for part a), each step follows a well-established procedure taught in M2 with no novel insight required. |
| Spec | 6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Mass ratios (must add up to \(16a^2\)): e.g. \(3(a^2)\), \(4(a^2)\), \(4.5(a^2)\), \(4.5(a^2)\) | B1 | Must be split into triangles and rectangles; cannot be awarded if they haven't got \(16\bar{x}\) in their moments equation |
| One correct distance e.g. \(0.5a\) or \(2a\) or \(2a\) or \(2a\) | B1 | Any single correct appropriate distance from \(AB\) |
| Moments about \(AB\): \(16\bar{x} = (3 \times 0.5a) + (4 \times 2a) + (4.5 \times 2a) + (4.5 \times 2a)\) | M1A2 | M1 for moments about \(AB\) or any other parallel line; A2 for correct equation (-1 e.e.o.o); A0 if not split into triangles and rectangles |
| \(16\bar{x} = \frac{55}{2}a\) | ||
| \(\bar{x} = \frac{55}{32}a\) ANSWER GIVEN | A1 | A1 for given answer; N.B. if masses don't add to \(16a^2\) then max B0B1M1A0A0A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\bar{y} = \frac{55}{32}a\) seen or implied | B1 | By symmetry |
| \(\tan\theta = \pm\left(\dfrac{\frac{55a}{32} - a}{4a - \bar{y}}\right)\) oe | M1A1 | M1A1 for correct expression; or reciprocal \(\left(= \frac{23a/32}{73a/32}\right)\) |
| \(\theta = 17°\), \(17.5°\) or better | A1 | In radians: 0.31 or better |
# Question 4:
## Part 4a:
| Answer/Working | Marks | Guidance |
|---|---|---|
| Mass ratios (must add up to $16a^2$): e.g. $3(a^2)$, $4(a^2)$, $4.5(a^2)$, $4.5(a^2)$ | B1 | Must be split into triangles and rectangles; cannot be awarded if they haven't got $16\bar{x}$ in their moments equation |
| One correct distance e.g. $0.5a$ or $2a$ or $2a$ or $2a$ | B1 | Any single correct appropriate distance from $AB$ |
| Moments about $AB$: $16\bar{x} = (3 \times 0.5a) + (4 \times 2a) + (4.5 \times 2a) + (4.5 \times 2a)$ | M1A2 | M1 for moments about $AB$ or any other parallel line; A2 for correct equation (-1 e.e.o.o); A0 if not split into triangles and rectangles |
| $16\bar{x} = \frac{55}{2}a$ | | |
| $\bar{x} = \frac{55}{32}a$ **ANSWER GIVEN** | A1 | A1 for given answer; N.B. if masses don't add to $16a^2$ then max B0B1M1A0A0A0 |
## Part 4b:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\bar{y} = \frac{55}{32}a$ seen or implied | B1 | By symmetry |
| $\tan\theta = \pm\left(\dfrac{\frac{55a}{32} - a}{4a - \bar{y}}\right)$ oe | M1A1 | M1A1 for correct expression; or reciprocal $\left(= \frac{23a/32}{73a/32}\right)$ |
| $\theta = 17°$, $17.5°$ or better | A1 | In radians: 0.31 or better |
> **N.B.** Instead of using $(4a - \bar{y})$ to find $\frac{73a}{32}$, they may start again to find distance of $G$ from $AE$. Then $\frac{73a}{32}$ scores the B1 and if their value is used correctly, can score M1.4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d5f79f83-ccfb-47a5-8100-88db81fd0862-07_728_748_214_639}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
The uniform square lamina $A B C D$ shown in Figure 2 has sides of length 4a. The points $E$ and $F$, on $D A$ and $D C$ respectively, are both at a distance $3 a$ from $D$.
The portion $D E F$ of the lamina is folded through $180 ^ { \circ }$ about $E F$ to form the folded lamina $A B C F E$ shown in Figure 3 below.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d5f79f83-ccfb-47a5-8100-88db81fd0862-07_709_730_1395_639}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
\begin{enumerate}[label=(\alph*)]
\item Show that the distance from $A B$ of the centre of mass of the folded lamina is $\frac { 55 } { 32 } a$.\\
(6)
The folded lamina is freely suspended from $E$ and hangs in equilibrium.
\item Find the size of the angle between $E D$ and the downward vertical.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2014 Q4 [10]}}