| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Lamina suspended in equilibrium |
| Difficulty | Standard +0.8 This is a multi-part Further Maths Mechanics question requiring: (i) algebraic manipulation to find centre of mass using composite shapes (with a specific answer to show), (ii) equilibrium analysis with trigonometry, and (iii) another equilibrium scenario. The combination of symbolic algebra, geometric reasoning, and equilibrium conditions makes this moderately challenging, though the techniques are standard for FM M2. |
| Spec | 6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| M1 | Table of moments idea | |
| \((ah/2+0.5a)x = (ah/2)(-h/3)+(0.5a)(0.5/2)\) | A1 | Correct sum of parts |
| \(0.5a(1 + h)x = 0.5a(0.25 - h^2/3)\) | M1 | Must include cancelling of \(a\) |
| \(x = (3 - 4h^2)/(12 + 12h)\) | AG, A1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| \(3 - 4h^2 = 0\) | M1 | Uses \(x = 0\) |
| \(h = 0.866\) | A1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tan\theta = x/(a/2) = (a/2)/h\) | M1 | Correct trigonometry |
| \((2/18)/(a/2) = (a/2)/0.5\) | DM1 | Ratios accurately substituted |
| \(a = 0.471\) | A1 [3] | [9] |
**(i)**
| | M1 | Table of moments idea |
| $(ah/2+0.5a)x = (ah/2)(-h/3)+(0.5a)(0.5/2)$ | A1 | Correct sum of parts |
| $0.5a(1 + h)x = 0.5a(0.25 - h^2/3)$ | M1 | Must include cancelling of $a$ |
| $x = (3 - 4h^2)/(12 + 12h)$ | AG, A1 [4] | |
**(ii)**
| $3 - 4h^2 = 0$ | M1 | Uses $x = 0$ |
| $h = 0.866$ | A1 [2] | |
**(iii)**
| $\tan\theta = x/(a/2) = (a/2)/h$ | M1 | Correct trigonometry |
| $(2/18)/(a/2) = (a/2)/0.5$ | DM1 | Ratios accurately substituted |
| $a = 0.471$ | A1 [3] | [9] |6\\
\includegraphics[max width=\textwidth, alt={}, center]{98bbefd8-b3dd-49f1-8591-e939282cb05c-3_341_791_886_678}
A uniform lamina $A B C D E$ consists of a rectangle $B C D E$ and an isosceles triangle $A B E$ joined along their common edge $B E$. For the triangle, $A B = A E , B E = a \mathrm {~m}$ and the perpendicular height is $h \mathrm {~m}$. For the rectangle, $B C = D E = 0.5 \mathrm {~m}$ and $C D = B E = a \mathrm {~m}$ (see diagram).\\
(i) Show that the distance in metres of the centre of mass of the lamina from $B E$ towards $C D$ is
$$\frac { 3 - 4 h ^ { 2 } } { 12 + 12 h }$$
The lamina is freely suspended at $E$ and hangs in equilibrium.\\
(ii) Given that $D E$ is horizontal, calculate $h$.\\
(iii) Given instead that $h = 0.5$ and $A E$ is horizontal, calculate $a$.
\hfill \mbox{\textit{CAIE M2 2012 Q6 [9]}}