| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Centre of mass of composite shapes |
| Difficulty | Challenging +1.2 This is a multi-part moments question requiring center of mass calculation for a circular segment (using standard formula), then two equilibrium cases with different force directions. While it involves several steps and careful geometry, the techniques are standard M2 content with no novel problem-solving required. The center of mass result is given to verify, and the moment calculations follow routine procedures once the geometry is established. |
| Spec | 6.04b Find centre of mass: using symmetry6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(OG\text{ (sector)} = \frac{2\times0.35\sin\frac{2\pi}{3}}{3\times\frac{2\pi}{3}}\) | B1 | \(0.096482\ldots\) |
| \(\text{Area (sector)} = \frac{0.35^2\times\frac{4\pi}{3}}{2}\) | B1 | \(0.256563\ldots\), \(0.116666\ldots\) |
| \(OG\text{ (triangle)} = \frac{2\times0.35\cos\frac{\pi}{3}}{3}\) and \(\text{Area (triangle)} = \frac{0.35^2\sin\frac{2\pi}{3}}{2}\) | B1 | \(0.053044\ldots\) |
| \(0.096482\ldots\times0.256563\ldots - 0.116666\ldots\times0.053044\ldots = d(0.256563\ldots+0.053044\ldots)\) | M1, A1 | |
| \(d = 0.06(00)\text{ m}\) | A1 | [6] AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(F\times0.35\cos\frac{\pi}{3} = 0.06\times14\) | M1 | |
| \(F = 4.8\) | A1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(F\sin\frac{\pi}{3}\times\left(0.35+0.35\sin\frac{\pi}{3}\right) + F\cos\frac{\pi}{3}\times0.35\cos\frac{\pi}{3} = 14\times0.06\) | M1, A1 | Moments about point of contact |
| \(F = 1.29\text{ N}\) | A1 | [3] |
| *OR*: \(F\times0.35 + \text{Fr}\times0.35 = 14\times0.06\) | (M1) | Moments about \(O\) |
| \(0.35F + 0.35F\sin\frac{\pi}{3} = 14\times0.06\) | (A1) | As \(\text{Fr} = F\sin\frac{\pi}{3}\) |
| \(F = 1.29\text{ N}\) | (A1) |
## Question 7:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $OG\text{ (sector)} = \frac{2\times0.35\sin\frac{2\pi}{3}}{3\times\frac{2\pi}{3}}$ | B1 | $0.096482\ldots$ |
| $\text{Area (sector)} = \frac{0.35^2\times\frac{4\pi}{3}}{2}$ | B1 | $0.256563\ldots$, $0.116666\ldots$ |
| $OG\text{ (triangle)} = \frac{2\times0.35\cos\frac{\pi}{3}}{3}$ and $\text{Area (triangle)} = \frac{0.35^2\sin\frac{2\pi}{3}}{2}$ | B1 | $0.053044\ldots$ |
| $0.096482\ldots\times0.256563\ldots - 0.116666\ldots\times0.053044\ldots = d(0.256563\ldots+0.053044\ldots)$ | M1, A1 | |
| $d = 0.06(00)\text{ m}$ | A1 | [6] AG |
### Part (ii) a:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $F\times0.35\cos\frac{\pi}{3} = 0.06\times14$ | M1 | |
| $F = 4.8$ | A1 | [2] |
### Part (ii) b:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $F\sin\frac{\pi}{3}\times\left(0.35+0.35\sin\frac{\pi}{3}\right) + F\cos\frac{\pi}{3}\times0.35\cos\frac{\pi}{3} = 14\times0.06$ | M1, A1 | Moments about point of contact |
| $F = 1.29\text{ N}$ | A1 | [3] |
| *OR*: $F\times0.35 + \text{Fr}\times0.35 = 14\times0.06$ | (M1) | Moments about $O$ |
| $0.35F + 0.35F\sin\frac{\pi}{3} = 14\times0.06$ | (A1) | As $\text{Fr} = F\sin\frac{\pi}{3}$ |
| $F = 1.29\text{ N}$ | (A1) | |7\\
\includegraphics[max width=\textwidth, alt={}, center]{81be887c-ab01-4327-a5df-f25c68a6fdb6-3_586_527_1030_810}
A uniform lamina $A B C$ is in the form of a major segment of a circle with centre $O$ and radius 0.35 m . The straight edge of the lamina is $A B$, and angle $A O B = \frac { 2 } { 3 } \pi$ radians (see diagram).\\
(i) Show that the centre of mass of the lamina is 0.0600 m from $O$, correct to 3 significant figures.
The weight of the lamina is 14 N . It is placed on a rough horizontal surface with $A$ vertically above $B$ and the lowest point of the arc $B C$ in contact with the surface. The lamina is held in equilibrium in a vertical plane by a force of magnitude $F \mathrm {~N}$ acting at $A$.\\
(ii) Find $F$ in each of the following cases:
\begin{enumerate}[label=(\alph*)]
\item the force of magnitude $F \mathrm {~N}$ acts along $A B$;
\item the force of magnitude $F \mathrm {~N}$ acts along the tangent to the circular arc at $A$.
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2014 Q7 [11]}}