| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2017 |
| Session | March |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Lamina with attached triangle |
| Difficulty | Standard +0.3 This is a standard centre of mass question requiring decomposition of a composite lamina into rectangles, calculation of centroids using standard formulas, and a straightforward equilibrium application. The geometry is simple (right angles), the method is routine (split into two rectangles, use moments), and part (ii) is a direct application of tan θ = horizontal distance / vertical distance. Slightly easier than average due to the structured approach and standard techniques. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| 4(i) | A = 0.6 x 0.75 + 0.3 x 0.75/2 (= 0.5625) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| (0.6 + 0.3/3) | M1 | Takes moments about AB |
| Answer | Marks |
|---|---|
| x = 0.38 m (from AB) AG | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| x 0.25 | M1 | Takes moments about BC |
| Answer | Marks |
|---|---|
| y = 0.35 m (from BC) | A1 |
| Total: | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| 4(ii) | tanθ = 0.35/0.38 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| θ = 42.6° | A1 | |
| Total: | 2 | |
| Question | Answer | Marks |
Question 4:
--- 4(i) ---
4(i) | A = 0.6 x 0.75 + 0.3 x 0.75/2 (= 0.5625) | B1 | A = total area of the lamina
(cid:76) 1
0.5625x = 0.75 x 0.6 x 0.3 + 0.3 x 0.75 x
2
(0.6 + 0.3/3) | M1 | Takes moments about AB
(cid:76)
x = 0.38 m (from AB) AG | A1
(cid:76) 1
0.5625y = 0.75 x 0.6 x 0.375 + 0.3 x 0.75
2
x 0.25 | M1 | Takes moments about BC
(cid:76)
y = 0.35 m (from BC) | A1
Total: | 5
--- 4(ii) ---
4(ii) | tanθ = 0.35/0.38 | M1 | (cid:76) (cid:76)
tanθ = y/x where θ is the required angle
θ = 42.6° | A1
Total: | 2
Question | Answer | Marks | Guidance\includegraphics{figure_4}
The diagram shows a uniform lamina $ABCD$ with $AB = 0.75 \text{ m}$, $AD = 0.6 \text{ m}$ and $BC = 0.9 \text{ m}$. Angle $BAD =$ angle $ABC = 90°$.
\begin{enumerate}[label=(\roman*)]
\item Show that the distance of the centre of mass of the lamina from $AB$ is $0.38 \text{ m}$, and find the distance of the centre of mass from $BC$. [5]
\end{enumerate}
The lamina is freely suspended at $B$ and hangs in equilibrium.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find the angle between $BC$ and the vertical. [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2017 Q4 [7]}}