| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2023 |
| Session | October |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | L-shaped or composite rectangular lamina |
| Difficulty | Standard +0.3 This is a standard M2 centre of mass question involving composite laminas with different densities. Part (a) requires systematic calculation of centres of mass and moments (routine but multi-step), part (b) applies equilibrium with the lamina hanging (standard technique), and part (c) involves taking moments about a pivot with a horizontal force (textbook application). While it requires careful bookkeeping across multiple parts, all techniques are standard M2 material with no novel insight required, making it slightly easier than average. |
| Spec | 3.04b Equilibrium: zero resultant moment and force6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Mass ratios: \(PQUY = 16a^2\), \(RSTU = 2\times 4a^2\), \(VWXY = 2\times 4a^2\) | B1 | Correct mass ratios (accept 2:1:1) |
| Distances from \(PX\): \(2a\), \(5a\), \(a\) | B1 | Correct vertical distances |
| Moments about \(PX\) or a parallel axis | M1 | Dimensionally correct equation. All terms required. Allow for an equation within a vector equation. |
| \(16a^2 \times 2a + 8a^2 \times 5a + 8a^2 \times a = 32a^2 d\) \(\left(= (16+8+8)a^2 \times d\right)\) or equivalent for a parallel axis | A1 | Correct unsimplified equation. Allow for an equation within a vector equation. Could have \(y\) for \(d\) here. |
| \(80a = 32d \Rightarrow d = \frac{5}{2}a\) * | A1* | Obtain given answer from correct working. At least one stage of simplifying required. e.g. \(32a^3 + 40a^3 + 8a^3\) seen. Must get to \(d = \ldots\) in final line |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Moments about \(PQ\) or a parallel axis | M1 | Dimensionally correct equation. All terms required |
| \(16a^2 \times 2a + 8a^2 \times 3a + 8a^2 \times 5a = (16+8+8)a^2 \times h\) or equivalent for a parallel axis | A1ft | Unsimplified equation with at most one error. Follow their mass ratios. |
| A1 | Correct unsimplified equation | |
| \(\Rightarrow h = 3a\) from \(PQ\) | A1 | \(a\) from \(YT\), \(3a\) from \(XW\) |
| The working for the first 4 marks must be seen or used in part (b) | ||
| Correct use of trig. to find the tangent of a relevant angle | M1 | With *their* \(3a\), e.g. \(\tan\theta = \dfrac{3}{4 - \frac{5}{2}}\) |
| \(\tan\theta = 2\) | A1 | Correct only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Complete method to obtain an equation in \(M\) and \(F\) | M1 | e.g. Moments about \(Q\). Dimensionally correct equation. |
| \(3a \times Mg = 4a \times F\) | A1 | Correct unsimplified equation. Condone if \(a\) missing throughout |
| \(F = \dfrac{3}{4}Mg\) | A1 | Correct only |
| Total: [3] | Overall: (14) |
# Question 2:
## Part 2a:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Mass ratios: $PQUY = 16a^2$, $RSTU = 2\times 4a^2$, $VWXY = 2\times 4a^2$ | B1 | Correct mass ratios (accept 2:1:1) |
| Distances from $PX$: $2a$, $5a$, $a$ | B1 | Correct vertical distances |
| Moments about $PX$ or a parallel axis | M1 | Dimensionally correct equation. All terms required. Allow for an equation within a vector equation. |
| $16a^2 \times 2a + 8a^2 \times 5a + 8a^2 \times a = 32a^2 d$ $\left(= (16+8+8)a^2 \times d\right)$ or equivalent for a parallel axis | A1 | Correct unsimplified equation. Allow for an equation within a vector equation. Could have $y$ for $d$ here. |
| $80a = 32d \Rightarrow d = \frac{5}{2}a$ * | A1* | Obtain given answer from correct working. At least one stage of simplifying required. e.g. $32a^3 + 40a^3 + 8a^3$ seen. Must get to $d = \ldots$ in final line |
**Total: [5]**
## Part 2b:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Moments about $PQ$ or a parallel axis | M1 | Dimensionally correct equation. All terms required |
| $16a^2 \times 2a + 8a^2 \times 3a + 8a^2 \times 5a = (16+8+8)a^2 \times h$ or equivalent for a parallel axis | A1ft | Unsimplified equation with at most one error. Follow their mass ratios. |
| | A1 | Correct unsimplified equation |
| $\Rightarrow h = 3a$ from $PQ$ | A1 | $a$ from $YT$, $3a$ from $XW$ |
| **The working for the first 4 marks must be seen or used in part (b)** | | |
| Correct use of trig. to find the tangent of a relevant angle | M1 | With *their* $3a$, e.g. $\tan\theta = \dfrac{3}{4 - \frac{5}{2}}$ |
| $\tan\theta = 2$ | A1 | Correct only |
**Total: [6]**
## Part 2c:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Complete method to obtain an equation in $M$ and $F$ | M1 | e.g. Moments about $Q$. Dimensionally correct equation. |
| $3a \times Mg = 4a \times F$ | A1 | Correct unsimplified equation. Condone if $a$ missing throughout |
| $F = \dfrac{3}{4}Mg$ | A1 | Correct only |
**Total: [3] | Overall: (14)**
---2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f180f5f0-43c5-4365-b0d8-7284220b481e-04_784_814_260_646}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a template where
\begin{itemize}
\item PQUY is a uniform square lamina with sides of length $4 a$
\item RSTU is a uniform square lamina with sides of length $2 a$
\item VWXY is a uniform square lamina with sides of length $2 a$
\item the three squares all lie in the same plane
\item the mass per unit area of $V W X Y$ is double the mass per unit area of $P Q U Y$
\item the mass per unit area of $R S T U$ is double the mass per unit area of $P Q U Y$
\item the distance of the centre of mass of the template from $P X$ is $d$
\begin{enumerate}[label=(\alph*)]
\item Show that $d = \frac { 5 } { 2 } a$
\end{itemize}
The template is freely pivoted about $Q$ and hangs in equilibrium with $P Q$ at an angle of $\theta$ to the downward vertical.
\item Find the value of $\tan \theta$
The mass of the template is $M$\\
The template is still freely pivoted about $Q$, but it is now held in equilibrium, with $P Q$ vertical, by a horizontal force of magnitude $F$ which acts on the template at $X$. The line of action of the force lies in the same plane as the template.
\item Find $F$ in terms of $M$ and $g$
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2023 Q2 [14]}}