| 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 | 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 (triangle minus square) using the removal method, (2) applying equilibrium conditions for a suspended lamina to find an unknown dimension. Part (b) requires setting up and solving a trigonometric equation relating the geometry to the equilibrium angle, which goes beyond routine centre of mass calculations and demands spatial reasoning. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Mass ratio: \(ABC = 48\), \(PQRS = 16\), Result \(= 32\); Dist of c of m from \(AB\): \(\frac{8}{3}\), \(2\), \(\bar{x}\) | — | — |
| \(48\times\frac{8}{3} - 16\times2 = 32\bar{x}\) | M1 | Take moments about \(AB\). Must be subtracting the square from the triangle for mass & moments. Correct unsimplified equation. -1 each error. Could have a common factor of \(g\). |
| A2 | — | |
| \(\bar{x} = (8-2)\div2 = 3\) (cm) | A1 (4) | — |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\tan25 = \dfrac{\text{their }3}{AN}\) or \(\dfrac{AN}{\text{their }3} = \tan65\) | M1 | Use trig to find the distance \(AN\). Condone tan the wrong way up |
| A1 | Correct for their 3 | |
| Dist of \(G\) from \(DC = \dfrac{3}{\tan25} - 6\) | M1 | — |
| \((-)\bar{y} = 2\!\left(\dfrac{\text{their }3}{\tan25} - 6\right)\) | M1 | Moments equation in \(\bar{y}\) |
| A1ft | Correct unsimplified expression for \(\bar{y}\) | |
| Distance \(= 0.867…\) cm | A1 (6) | 0.87 or better – must be positive |
# Question 4:
## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| Mass ratio: $ABC = 48$, $PQRS = 16$, Result $= 32$; Dist of c of m from $AB$: $\frac{8}{3}$, $2$, $\bar{x}$ | — | — |
| $48\times\frac{8}{3} - 16\times2 = 32\bar{x}$ | M1 | Take moments about $AB$. Must be subtracting the square from the triangle for mass & moments. Correct unsimplified equation. -1 each error. Could have a common factor of $g$. |
| | A2 | — |
| $\bar{x} = (8-2)\div2 = 3$ (cm) | A1 (4) | — |
## Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $\tan25 = \dfrac{\text{their }3}{AN}$ or $\dfrac{AN}{\text{their }3} = \tan65$ | M1 | Use trig to find the distance $AN$. Condone tan the wrong way up |
| | A1 | Correct for their 3 |
| Dist of $G$ from $DC = \dfrac{3}{\tan25} - 6$ | M1 | — |
| $(-)\bar{y} = 2\!\left(\dfrac{\text{their }3}{\tan25} - 6\right)$ | M1 | Moments equation in $\bar{y}$ |
| | A1ft | Correct unsimplified expression for $\bar{y}$ |
| Distance $= 0.867…$ cm | A1 (6) | 0.87 or better – must be positive |
**Total: [10]**
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{82cadc37-4cb0-455e-9531-e09ec0c19533-07_737_823_223_532}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a lamina $L$. It is formed by removing a square $P Q R S$ from a uniform triangle $A B C$. The triangle $A B C$ is isosceles with $A C = B C$ and $A B = 12 \mathrm {~cm}$. The midpoint of $A B$ is $D$ and $D C = 8 \mathrm {~cm}$. The vertices $P$ and $Q$ of the square lie on $A B$ and $P Q = 4 \mathrm {~cm}$. The centre of the square is $O$. The centre of mass of $L$ is at $G$.
\begin{enumerate}[label=(\alph*)]
\item Find the distance of $G$ from $A B$.
When $L$ is freely suspended from $A$ and hangs in equilibrium, the line $A B$ is inclined at $25 ^ { \circ }$ to the vertical.
\item Find the distance of $O$ from $D C$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2014 Q4 [10]}}