| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Lamina with removed circle/semicircle |
| Difficulty | Challenging +1.2 This is a standard centre of mass problem involving a composite shape (square minus quadrant) with straightforward application of the formula for composite bodies. Part (i) requires routine calculation using known formulae for centres of mass of standard shapes. Part (ii) involves algebraic manipulation and verification by substitution, which is mechanical rather than insightful. The multi-step nature and need to handle the equilibrium condition elevates it slightly above average difficulty. |
| Spec | 6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(OG_{\text{quadrant}} = 2\sin(\pi / 4) / (3\pi / 4)\), \(a^2(a\sqrt{2}/2) = \pi/4[2\sin(\pi/4) / (3\pi/4)] + (a^2 - \pi/4)x\), \(x = 2\sqrt{2}(3a^2 - 2) / (12a^2 - 3\pi)\) | B1, M1, A2, A1 | \(8 / (3\sqrt{2}\pi)\); –1 each error, min zero; There must be 3 moment terms; Other forms acceptable [5] |
| (ii) \(xcos 45° > 1\), \((6a^3 - 4) / (12a^2 - 3\pi) > 1\), \(3a^2(2 - a) < 3\pi / 2 - 2\) | B1, M1, A1, B1 | RHS = 2.712... compared with LHS = 2.709... and 2.76.. respectively; AG [4] |
**(i)** $OG_{\text{quadrant}} = 2\sin(\pi / 4) / (3\pi / 4)$, $a^2(a\sqrt{2}/2) = \pi/4[2\sin(\pi/4) / (3\pi/4)] + (a^2 - \pi/4)x$, $x = 2\sqrt{2}(3a^2 - 2) / (12a^2 - 3\pi)$ | B1, M1, A2, A1 | $8 / (3\sqrt{2}\pi)$; –1 each error, min zero; There must be 3 moment terms; Other forms acceptable [5]
**(ii)** $xcos 45° > 1$, $(6a^3 - 4) / (12a^2 - 3\pi) > 1$, $3a^2(2 - a) < 3\pi / 2 - 2$ | B1, M1, A1, B1 | RHS = 2.712... compared with LHS = 2.709... and 2.76.. respectively; AG [4]
7\\
\includegraphics[max width=\textwidth, alt={}, center]{9d377c95-09b8-4893-b29f-8517a5016e8b-4_597_1011_251_566}\\
$A B C D E$ is the cross-section through the centre of mass of a uniform prism resting in equilibrium with $D E$ on a horizontal surface. The cross-section has the shape of a square $O B C D$ with sides of length $a \mathrm {~m}$, from which a quadrant $O A E$ of a circle of radius 1 m has been removed (see diagram).\\
(i) Find the distance of the centre of mass of the prism from $O$, giving the answer in terms of $a , \pi$ and $\sqrt { } 2$.\\
(ii) Hence show that
$$3 a ^ { 2 } ( 2 - a ) < \frac { 3 } { 2 } \pi - 2$$
and verify that this inequality is satisfied by $a = 1.68$ but not by $a = 1.67$.
\hfill \mbox{\textit{CAIE M2 2011 Q7 [9]}}