| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Framework or multiple rod structures |
| Difficulty | Standard +0.3 This is a standard two-rod framework problem requiring centre of mass calculation and moment equilibrium in two configurations. While it involves multiple steps and careful geometry with the 45° angle, the techniques are routine for M2 (Further Maths Mechanics): finding combined centre of mass using component method, then taking moments about the suspension point. The perpendicular force in part (b) adds mild complexity but follows the same approach. Slightly easier than average A-level due to the symmetric setup and straightforward geometry. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(0.212\) | B1 [1] | From \((0.6/2)\cos45\) |
| (ii) (a) \(0.3\cos45 \times (2 \times 7) = (2 \times 0.6\sin45) \times F\) | M1 | Moments about \(A\) |
| \(F = 3.5\) | A1 | [2] |
| (ii) (b) \(0.3\cos45 \times (2 \times 7) = 0.6F\) | M1 | Or Ans (i)\(\cos45\) |
| \(F = 4.95\) | A1 | [2] |
**(i)** $0.212$ | B1 [1] | From $(0.6/2)\cos45$
**(ii) (a)** $0.3\cos45 \times (2 \times 7) = (2 \times 0.6\sin45) \times F$ | M1 | Moments about $A$
$F = 3.5$ | A1 | [2]
**(ii) (b)** $0.3\cos45 \times (2 \times 7) = 0.6F$ | M1 | Or Ans (i)$\cos45$
$F = 4.95$ | A1 | [2]2 An object is made from two identical uniform rods $A B$ and $B C$ each of length 0.6 m and weight 7 N . The rods are rigidly joined to each other at $B$ and angle $A B C = 90 ^ { \circ }$.
\begin{enumerate}[label=(\roman*)]
\item Calculate the distance of the centre of mass of the object from $B$.
The object is freely suspended at $A$ and a force of magnitude $F \mathrm {~N}$ is applied to the rod $B C$ at $C$. The object is in equilibrium with $A B$ inclined at $45 ^ { \circ }$ to the horizontal.
\item (a)
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a093cbad-3ba0-45ce-a617-d4ecc8cb1ec9-2_401_314_799_995}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}
Calculate $F$ given that the force acts horizontally as shown in Fig. 1.\\
(b)
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a093cbad-3ba0-45ce-a617-d4ecc8cb1ec9-2_503_273_1446_1014}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
Calculate $F$ given instead that the force acts perpendicular to the rod as shown in Fig. 2.
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2011 Q2 [5]}}