| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Equilibrium with applied force |
| Difficulty | Standard +0.3 This is a standard M2 centre of mass problem requiring decomposition into rectangles, calculation of composite centre of mass using standard formulas, then equilibrium analysis with limiting friction. All techniques are routine for M2 students with no novel insight required, making it slightly easier than average. |
| Spec | 6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(1.5\times0.4\times0.2 + 1\times1\times0.9 = (1\times1+1.5\times0.4)d\) or \(0.5\times0.4\times0.2+1\times1.4\times0.7=(0.5\times0.4+1\times1.4)d\) or \(1.5\times1.4\times0.7-1\times0.5\times0.9=(1.5\times1.4-1\times0.5)d\) | M1 | Table of moments idea; uses area or any weight/m² value |
| A1 | ||
| \(d = 0.6375\) | A1 [3] | Accept \(0.637\) or \(0.638\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(F\times1.5 = 120\times0.6375\) | M1 | Moments about \(O\) |
| \(F = 51\) | A1 | |
| \(F\times1.5 = 120\times(0.6375-0.4)\) | M1 | |
| \(F = 19\) | A1 | |
| \(51 > F > 19\) | M1, A1\(\checkmark\) [6] | Candidates consider both cases; \(\checkmark\) [cv(two values of \(F\))] accept \(>\) |
## Question 6:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $1.5\times0.4\times0.2 + 1\times1\times0.9 = (1\times1+1.5\times0.4)d$ or $0.5\times0.4\times0.2+1\times1.4\times0.7=(0.5\times0.4+1\times1.4)d$ or $1.5\times1.4\times0.7-1\times0.5\times0.9=(1.5\times1.4-1\times0.5)d$ | M1 | Table of moments idea; uses area or any weight/m² value |
| | A1 | |
| $d = 0.6375$ | A1 [3] | Accept $0.637$ or $0.638$ |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F\times1.5 = 120\times0.6375$ | M1 | Moments about $O$ |
| $F = 51$ | A1 | |
| $F\times1.5 = 120\times(0.6375-0.4)$ | M1 | |
| $F = 19$ | A1 | |
| $51 > F > 19$ | M1, A1$\checkmark$ [6] | Candidates consider both cases; $\checkmark$ [cv(two values of $F$)] accept $>$ |
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{6d3892e0-8c88-44ec-940f-c526d71a7fc6-3_720_723_1165_712}
The diagram shows the cross-section $O A B C D E$ through the centre of mass of a uniform prism. The interior angles of the cross-section at $O , A , C , D$ and $E$ are all right angles. $O A = 0.4 \mathrm {~m} , A B = 0.5 \mathrm {~m}$ and $B C = C D = 1 \mathrm {~m}$.\\
(i) Calculate the distance of the centre of mass of the prism from $O E$.
The weight of the prism is 120 N . A force of magnitude $F \mathrm {~N}$ acting along $D E$ holds the prism in equilibrium when $O A$ rests on a rough horizontal surface.\\
(ii) Find the set of possible values of $F$.
\hfill \mbox{\textit{CAIE M2 2012 Q6 [9]}}