| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2016 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Frame with circular arc or semicircular arc components |
| Difficulty | Standard +0.3 This is a straightforward centre of mass problem requiring standard formula recall (centre of mass of semicircular arc at 2r/π from centre) and basic moment equilibrium. Part (i) is essentially showing a given answer using the formula, and part (ii) involves taking moments about the hinge—both routine applications with no novel problem-solving required. Slightly easier than average due to the 'show that' scaffolding. |
| Spec | 6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks |
|---|---|
| (ii) | OG = 0.4sin(π/2)/(π/2) |
| Answer | Marks |
|---|---|
| W = 21.2 N | B1 |
| Answer | Marks |
|---|---|
| A1 | 3 |
| 2 | = 0.25464... |
Question 2:
--- 2 (i)
(ii) ---
2 (i)
(ii) | OG = 0.4sin(π/2)/(π/2)
d = OG cos70 + 0.4sin70
d = 0.463 AG
0.463W = 15 × 0.8cos35
W = 21.2 N | B1
M1
A1
M1
A1 | 3
2 | = 0.25464...\includegraphics{figure_2}
A uniform wire has the shape of a semicircular arc, with diameter $AB$ of length $0.8 \text{ m}$. The wire is attached to a vertical wall by a smooth hinge at $A$. The wire is held in equilibrium with $AB$ inclined at $70°$ to the upward vertical by a light string attached to $B$. The other end of the string is attached to the point $C$ on the wall $0.8 \text{ m}$ vertically above $A$. The tension in the string is $15 \text{ N}$ (see diagram).
\begin{enumerate}[label=(\roman*)]
\item Show that the horizontal distance of the centre of mass of the wire from the wall is $0.463 \text{ m}$, correct to 3 significant figures. [3]
\item Calculate the weight of the wire. [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2016 Q2 [5]}}