| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2024 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Folded lamina |
| Difficulty | Challenging +1.2 This is a standard M2 centre of mass question involving a folded lamina with typical multi-part structure. Part (a) requires systematic calculation of composite centre of mass using standard formulas (show-that format provides target). Part (b) involves moments about a suspension point with forces—routine application of equilibrium principles. While requiring careful bookkeeping across multiple steps, it demands no novel insight beyond standard M2 techniques, making it moderately above average difficulty. |
| Spec | 6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Correct mass ratios and distances (e.g. rectangle \(20a^2\), \(-\)triangle \(-\frac{9}{2}a^2\), \(+\)triangle \(\frac{9}{2}a^2\); distances \(2a\), \(3a\), \(2a\)) | B1 | Correct mass ratios for a correct division of the folded template and correct total of \(20a^2\) |
| Correct distances from \(AD\) seen or implied | B1 | B0B1 possible if incorrect masses but full set of correct distances |
| Moments about \(AD\) or a parallel axis | M1 | Dimensionally consistent. All terms for a correct division of \(L\) and no extras. Accept as part of a vector equation |
| \(40a^3 - \frac{27}{2}a^3 + 9a^3 = 20a^2 d\) or \(\frac{3}{2}a^3 + 16a^3 + 18a^3 = 20a^2 d\) or \(\frac{2}{3}a^3 + \frac{48}{2}a^3 + 9a^3 = 20a^2 d\) | A1 | Correct unsimplified equation for their axis |
| \(d = \frac{71}{40}a\) | A1* | Obtain given answer from correct working. Need at least one line of working to collect like terms e.g. \(20d = \frac{71}{2}a\). Final answer must be as printed |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Moments about \(S\) | M1 | A complete method to get an equation in \(W\) and \(F\) only. Need all terms and no extras. Dimensionally consistent |
| \(4W\times\frac{31}{40}a + W\times 3a = F\times 5a\) or \(\left(4W+W\right)\left(2.22a - a\right) = 5aF\) | A1 | Unsimplified equation with at most one error |
| A1 | Correct unsimplified equation | |
| \(F = \frac{61}{50}W\) | A1 | Accept \(1.22W\) or \(1.2W\) |
| [4] |
## Question 4:
**Part 4a:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct mass ratios and distances (e.g. rectangle $20a^2$, $-$triangle $-\frac{9}{2}a^2$, $+$triangle $\frac{9}{2}a^2$; distances $2a$, $3a$, $2a$) | B1 | Correct mass ratios for a correct division of the folded template and correct total of $20a^2$ |
| Correct distances from $AD$ seen or implied | B1 | B0B1 possible if incorrect masses but full set of correct distances |
| Moments about $AD$ or a parallel axis | M1 | Dimensionally consistent. All terms for a correct division of $L$ and no extras. Accept as part of a vector equation |
| $40a^3 - \frac{27}{2}a^3 + 9a^3 = 20a^2 d$ or $\frac{3}{2}a^3 + 16a^3 + 18a^3 = 20a^2 d$ or $\frac{2}{3}a^3 + \frac{48}{2}a^3 + 9a^3 = 20a^2 d$ | A1 | Correct unsimplified equation for their axis |
| $d = \frac{71}{40}a$ | A1* | Obtain given answer from correct working. Need at least one line of working to collect like terms e.g. $20d = \frac{71}{2}a$. Final answer must be as printed |
| **[5]** | | |
**Part 4b:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Moments about $S$ | M1 | A complete method to get an equation in $W$ and $F$ only. Need all terms and no extras. Dimensionally consistent |
| $4W\times\frac{31}{40}a + W\times 3a = F\times 5a$ or $\left(4W+W\right)\left(2.22a - a\right) = 5aF$ | A1 | Unsimplified equation with at most one error |
| | A1 | Correct unsimplified equation |
| $F = \frac{61}{50}W$ | A1 | Accept $1.22W$ or $1.2W$ |
| **[4]** | | |
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d5f06fe7-4d9c-4009-8931-3ecbc31fa5e5-10_552_680_255_447}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d5f06fe7-4d9c-4009-8931-3ecbc31fa5e5-10_547_494_255_1165}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
The uniform rectangular lamina $A B C D$, shown in Figure 2, has $D C = 4 a$ and $A D = 5 a$\\
The points $S$ on $A B$ and $T$ on $B C$ are such that $S B = B T = 3 a$\\
The lamina is folded along $S T$ to form the folded lamina $L$, shown in Figure 3.\\
The distance of the centre of mass of $L$ from $A D$ is $d$.
\begin{enumerate}[label=(\alph*)]
\item Show that $d = \frac { 71 } { 40 } a$
The weight of $L$ is $4 W$. A particle of weight $W$ is attached to $L$ at $C$.\\
The folded lamina $L$ is freely suspended from $S$.\\
A force of magnitude $F$, acting parallel to $D C$, is applied to $L$ at $D$ so that $A D$ is vertical.
\item Find $F$ in terms of $W$
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2024 Q4 [9]}}