| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2016 |
| Session | March |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Lamina suspended in equilibrium |
| Difficulty | Standard +0.3 This is a standard centre of mass problem requiring calculation of composite lamina COM using moments, then applying equilibrium conditions for suspension. The geometry is straightforward, calculations are routine, and the suspension angle problems follow directly from standard principles (vertical line through COM). Slightly easier than average due to methodical step-by-step nature with no novel insight required. |
| Spec | 6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2 \times 0.56 \times 0.28 + 1.2^2(0.56 + 1.2/2) = h(2 \times 0.56 + 1.2^2)\) | M1 | Moments about BC |
| \(h = 0.775\) | A1 | |
| \(2 \times 0.56 \times 1 + 1.2^2(1.2/2) = v(2 \times 0.56 + 1.2^2)\) | M1 | Moments about BAG |
| \(v = 0.775\) | A1 | Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(45°\) | B1 | Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\tan\theta = (0.56 + 1.2 - 0.775)/(1.2 - 0.775)\) | M1 | |
| \(\theta = 66.7°\) | A1 | Total: 2 |
## Question 4:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2 \times 0.56 \times 0.28 + 1.2^2(0.56 + 1.2/2) = h(2 \times 0.56 + 1.2^2)$ | M1 | Moments about BC |
| $h = 0.775$ | A1 | |
| $2 \times 0.56 \times 1 + 1.2^2(1.2/2) = v(2 \times 0.56 + 1.2^2)$ | M1 | Moments about BAG |
| $v = 0.775$ | A1 | **Total: 4** |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $45°$ | B1 | **Total: 1** |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\tan\theta = (0.56 + 1.2 - 0.775)/(1.2 - 0.775)$ | M1 | |
| $\theta = 66.7°$ | A1 | **Total: 2** |
---4\\
\includegraphics[max width=\textwidth, alt={}, center]{334b4bdf-6d9c-4208-9032-572eb7c5f9ee-2_549_579_1505_781}
A uniform lamina is made by joining a rectangle $A B C D$, in which $A B = C D = 0.56 \mathrm {~m}$ and $B C = A D = 2 \mathrm {~m}$, and a square $E F G A$ of side 1.2 m . The vertex $E$ of the square lies on the edge $A D$ of the rectangle (see diagram). The centre of mass of the lamina is a distance $h \mathrm {~m}$ from $B C$ and a distance $v \mathrm {~m}$ from $B A G$.\\
(i) Find the value of $h$ and show that $v = h$.
The lamina is freely suspended at the point $B$ and hangs in equilibrium.\\
(ii) State the angle which the edge $B C$ makes with the horizontal.
Instead, the lamina is now freely suspended at the point $F$ and hangs in equilibrium.\\
(iii) Calculate the angle between $F G$ and the vertical.
\hfill \mbox{\textit{CAIE M2 2016 Q4 [7]}}