| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2006 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Centre of mass of composite lamina |
| Difficulty | Standard +0.3 This is a standard composite lamina problem requiring knowledge of standard results (centroid of semicircle at 4r/3π from diameter) and systematic application of the composite body formula. The 'show that' format reduces difficulty as students know the target answer. The calculations are straightforward with Pythagoras (AC=5m), area calculations, and weighted averages—all routine techniques for Further Maths M2 students. |
| Spec | 6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Distance of centre of mass of semicircle from centre \(= \dfrac{10}{3\pi}\) | B1 | |
| \(2.5 \times 0.8 + \left(\dfrac{10}{3\pi}\right) \times 0.6\) | M1 | \(r\cos\theta + d\sin\theta\) |
| Distance is \(\left(2 + \dfrac{2}{\pi}\right)\)m | B1 (3) | From correct working (AG) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For using \(A\bar{x} = A_1\bar{x}_1 + A_2\bar{x}_2\) (3 terms required) | |
| \(\left(\frac{1}{2} \times 4 \times 3 + \frac{1}{2}\pi \times 2.5^2\right)\bar{x} = \frac{1}{2} \times 4 \times 3 \times (4/3) +\) | A1 | |
| \(\left(\frac{1}{2}\pi \times 2.5^2\right)\left(2 + \dfrac{2}{\pi}\right)\) | B1, A1 | |
| Distance is \(2.14\)m | A1 (5) | From correct working (AG) |
## Question 6:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Distance of centre of mass of semicircle from centre $= \dfrac{10}{3\pi}$ | B1 | |
| $2.5 \times 0.8 + \left(\dfrac{10}{3\pi}\right) \times 0.6$ | M1 | $r\cos\theta + d\sin\theta$ |
| Distance is $\left(2 + \dfrac{2}{\pi}\right)$m | B1 (3) | From correct working **(AG)** |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For using $A\bar{x} = A_1\bar{x}_1 + A_2\bar{x}_2$ (3 terms required) |
| $\left(\frac{1}{2} \times 4 \times 3 + \frac{1}{2}\pi \times 2.5^2\right)\bar{x} = \frac{1}{2} \times 4 \times 3 \times (4/3) +$ | A1 | |
| $\left(\frac{1}{2}\pi \times 2.5^2\right)\left(2 + \dfrac{2}{\pi}\right)$ | B1, A1 | |
| Distance is $2.14$m | A1 (5) | From correct working **(AG)** |
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\includegraphics[max width=\textwidth, alt={}, center]{0cb05368-9ddf-4564-8428-725c77193a1e-3_597_690_1416_731}
A large uniform lamina is in the shape of a right-angled triangle $A B C$, with hypotenuse $A C$, joined to a semicircle $A D C$ with diameter $A C$. The sides $A B$ and $B C$ have lengths 3 m and 4 m respectively, as shown in the diagram.\\
(i) Show that the distance from $A B$ of the centre of mass of the semicircular part $A D C$ of the lamina is $\left( 2 + \frac { 2 } { \pi } \right) \mathrm { m }$.\\
(ii) Show that the distance from $A B$ of the centre of mass of the complete lamina is 2.14 m , correct to 3 significant figures.
\hfill \mbox{\textit{CAIE M2 2006 Q6 [8]}}