| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod hinged to wall with string support |
| Difficulty | Standard +0.3 This is a standard M2 moments problem requiring resolution of forces and taking moments about the hinge. Part (a) involves straightforward geometry (Pythagoras to find string length/angle), then moments about A to find tension. Part (b) is a simple extension using the same method with inequality. The multi-step nature and geometric setup make it slightly above average, but it follows a completely standard template for M2 equilibrium problems with no novel insight required. |
| Spec | 6.04a Centre of mass: gravitational effect6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(M(A): 3a \times T\cos\theta = 2amg + 4amg\) | M1 A1 A1 | |
| \(\cos\theta = \frac{2}{\sqrt{9+4}} = \frac{2}{\sqrt{13}}\) | B1 | |
| \(\frac{6}{\sqrt{13}}T = 6mg\) | ||
| \(T = mg\sqrt{13}\) * | A1 | (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(3a \times T \times \cos\theta = 2amg + 4aMg\) | M1 | |
| \(T = \frac{(2mg+4Mg)}{6}\sqrt{13} \leq 2mg\sqrt{13}\) | A1 | |
| \(mg + 2Mg \leq 6mg\) | ||
| \(M \leq \frac{5}{2}m\) * | A1 cso | (3 marks) |
# Question 6:
## Part (a):
| Working | Marks | Notes |
|---------|-------|-------|
| $M(A): 3a \times T\cos\theta = 2amg + 4amg$ | M1 A1 A1 | |
| $\cos\theta = \frac{2}{\sqrt{9+4}} = \frac{2}{\sqrt{13}}$ | B1 | |
| $\frac{6}{\sqrt{13}}T = 6mg$ | | |
| $T = mg\sqrt{13}$ * | A1 | (5 marks) |
## Part (b):
| Working | Marks | Notes |
|---------|-------|-------|
| $3a \times T \times \cos\theta = 2amg + 4aMg$ | M1 | |
| $T = \frac{(2mg+4Mg)}{6}\sqrt{13} \leq 2mg\sqrt{13}$ | A1 | |
| $mg + 2Mg \leq 6mg$ | | |
| $M \leq \frac{5}{2}m$ * | A1 cso | (3 marks) |6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{0a4e4cdd-bec4-4059-b9f7-9ce00bc34b71-20_721_958_127_495}
\captionsetup{labelformat=empty}
\caption{Figure2}
\end{center}
\end{figure}
Figure 2 shows a uniform rod $A B$ of mass $m$ and length 4a. The end $A$ of the rod is freely hinged to a point on a vertical wall. A particle of mass $m$ is attached to the rod at $B$. One end of a light inextensible string is attached to the rod at C , where $\mathrm { AC } = 3 \mathrm { a }$. The other end of the string is attached to the wall at D , where $\mathrm { AD } = 2 \mathrm { a }$ and D is vertically above A . The rod rests horizontally in equilibrium in a vertical plane perpendicular to the wall and the tension in the string is T .
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { T } = \mathrm { mg } \sqrt { } 13$.\\
(5)
The particle of mass $m$ at $B$ is removed from the rod and replaced by a particle of mass $M$ which is attached to the rod at B . The string breaks if the tension exceeds $2 \mathrm { mg } \sqrt { } 13$. Given that the string does not break,
\item show that $M \leqslant \frac { 5 } { 2 } m$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q6 [8]}}