| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2010 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Lamina with attached triangle |
| Difficulty | Moderate -0.3 This is a straightforward centre of mass problem requiring standard techniques: decomposing the L-shaped lamina into two rectangles, finding individual centroids, using the composite formula, then applying equilibrium conditions for a suspended lamina. The calculations are routine with no conceptual surprises, making it slightly easier than average but still requiring proper method and multiple steps for 6 marks total. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (i) 2mx0.45 + mx0.3 = 3mv | M1 |
| v = 0.4m (from AB) | A1 | |
| 2mx0.45 + mx(0.9+0.3) = 3mh | M1 | Table of values idea |
| h = 0.7m (from AD) | A1 |
| Answer | Marks |
|---|---|
| (ii) tan α = 0.4/0.7 | M1 |
| α = 29.7° | A1ft |
| [2] | Accept 0.519 radians |
Question 1:
1 | (i) 2mx0.45 + mx0.3 = 3mv | M1 | Table of values idea
v = 0.4m (from AB) | A1
2mx0.45 + mx(0.9+0.3) = 3mh | M1 | Table of values idea
h = 0.7m (from AD) | A1
[4]
(ii) tan α = 0.4/0.7 | M1
α = 29.7° | A1ft
[2] | Accept 0.519 radians\includegraphics{figure_1}
$ABCD$ is a uniform lamina with $AB = 1.8$ m, $AD = DC = 0.9$ m, and $AD$ perpendicular to $AB$ and $DC$ (see diagram).
\begin{enumerate}[label=(\roman*)]
\item Find the distance of the centre of mass of the lamina from $AB$ and the distance from $AD$. [4]
\end{enumerate}
The lamina is freely suspended at $A$ and hangs in equilibrium.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Calculate the angle between $AB$ and the vertical. [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2010 Q1 [6]}}