| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2004 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Toppling on inclined plane |
| Difficulty | Standard +0.3 This is a standard centre of mass problem requiring calculation of the centroid of a trapezium using integration or the standard formula, followed by a toppling condition (centre of mass above edge). The geometry is straightforward and the techniques are routine for M2 level, making it slightly easier than average. |
| Spec | 6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| (i) Centre of mass of triangle is at height 0.2m | B1 | |
| For taking moments about the base | M1 | |
| \((0.4 \times 0.3 + \frac{1}{2} \times 0.6 \times 0.3)\bar{y} = 0.4 \times 0.3 \times 0.15 + \frac{1}{2} \times 0.6 \times 0.3 \times 0.2\) | ||
| For correct total area and correct (unsimplified) moment of the rectangle | A1 | |
| For the correct (unsimplified) moment of triangle | A1 ft | |
| Distance of centre of mass is 0.171 or \(\frac{6}{35}\) m | A1 | 5 marks |
| (ii) Centre of mass of triangle is \(\frac{2y}{3}\) from interface | B1 | |
| For using 'moment about the interface = 0' or equivalent or 'moment about \(AD = 0.4A'\) AND with areas in terms of \(y\) | M1 | |
| \([0.4y \times 0.2 = \frac{1}{2} \times 2y \times y \times \frac{2y}{3}\) or \(0.4(0.4y + \frac{1}{2} \times 2y \times y) = 0.4y \times 0.2 + \frac{1}{2}2y \times y(0.4 + \frac{2y}{3})]\) | ||
| For LHS of the above | A1 | |
| For RHS of the above | A1 ft | |
| Value of \(y\) is or \(\frac{\sqrt{3}}{5} \approx 0.346\) | A1 | 5 marks |
**(i)** Centre of mass of triangle is at height 0.2m | B1 |
For taking moments about the base | M1 |
$(0.4 \times 0.3 + \frac{1}{2} \times 0.6 \times 0.3)\bar{y} = 0.4 \times 0.3 \times 0.15 + \frac{1}{2} \times 0.6 \times 0.3 \times 0.2$ | |
For correct total area and correct (unsimplified) moment of the rectangle | A1 |
For the correct (unsimplified) moment of triangle | A1 ft |
Distance of centre of mass is 0.171 or $\frac{6}{35}$ m | A1 | 5 marks
**(ii)** Centre of mass of triangle is $\frac{2y}{3}$ from interface | B1 |
For using 'moment about the interface = 0' or equivalent or 'moment about $AD = 0.4A'$ AND with areas in terms of $y$ | M1 |
$[0.4y \times 0.2 = \frac{1}{2} \times 2y \times y \times \frac{2y}{3}$ or $0.4(0.4y + \frac{1}{2} \times 2y \times y) = 0.4y \times 0.2 + \frac{1}{2}2y \times y(0.4 + \frac{2y}{3})]$ | |
For LHS of the above | A1 |
For RHS of the above | A1 ft |
Value of $y$ is or $\frac{\sqrt{3}}{5} \approx 0.346$ | A1 | 5 marks7\\
\includegraphics[max width=\textwidth, alt={}, center]{81411376-b926-4857-bc9b-ac85d7957f3d-3_327_1006_1037_573}
A light container has a vertical cross-section in the form of a trapezium. The container rests on a horizontal surface. Grain is poured into the container to a depth of $y \mathrm {~m}$. As shown in the diagram, the cross-section $A B C D$ of the grain is such that $A B = 0.4 \mathrm {~m}$ and $D C = ( 0.4 + 2 y ) \mathrm { m }$.\\
(i) When $y = 0.3$, find the vertical height of the centre of mass of the grain above the base of the container.\\
(ii) Find the value of $y$ for which the container is about to topple.
\hfill \mbox{\textit{CAIE M2 2004 Q7 [10]}}