| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Centre of mass with given condition |
| Difficulty | Challenging +1.2 This is a two-part centre of mass problem requiring standard techniques: (i) using the composite body formula with known COM positions for semicircle and triangle to find h, and (ii) applying equilibrium conditions with moments about a point. While it involves multiple steps and careful geometric reasoning, the methods are standard M2 content with no novel insights required. The given condition that COM is at O provides a clear starting constraint, making this moderately above average difficulty. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| (i) C of M semi-circle = \(4 \times 0.2/(3\pi)\) → \(\frac{\pi 0.2^2}{2} \times 4 \times \frac{0.2}{3\pi} = \frac{0.4h}{2} = \frac{h}{3} = 0.283\) | B1, M1, A1, A1 [4] | (0.08488...) Moments about a relevant point. |
| (ii) \(\tan\theta = 0.283/0.2\); \(\cos\theta = XD/0.2\) (= 0.5774); \(XD = 0.115\) m OR \(\tan\alpha = 0.2/0.283\); \(\sin\alpha = XD/0.2\) (= 0.5774); \(XD = 0.115\) m | M1, M1, A1 [3] | \(\tan ADO = h/0.2\), \(ADO = 54.75°\) For candidates ADO OR \(\tan DAO = 0.2/h\), \(DAO = 35.25°\) For candidate's DAO |
(i) C of M semi-circle = $4 \times 0.2/(3\pi)$ → $\frac{\pi 0.2^2}{2} \times 4 \times \frac{0.2}{3\pi} = \frac{0.4h}{2} = \frac{h}{3} = 0.283$ | B1, M1, A1, A1 [4] | (0.08488...) Moments about a relevant point.
(ii) $\tan\theta = 0.283/0.2$; $\cos\theta = XD/0.2$ (= 0.5774); $XD = 0.115$ m OR $\tan\alpha = 0.2/0.283$; $\sin\alpha = XD/0.2$ (= 0.5774); $XD = 0.115$ m | M1, M1, A1 [3] | $\tan ADO = h/0.2$, $ADO = 54.75°$ For candidates ADO OR $\tan DAO = 0.2/h$, $DAO = 35.25°$ For candidate's DAO2 A uniform lamina $A B C D$ consists of a semicircle $B C D$ with centre $O$ and diameter 0.4 m , and an isosceles triangle $A B D$ with base $B D = 0.4 \mathrm {~m}$ and perpendicular height $h \mathrm {~m}$. The centre of mass of the lamina is at $O$.\\
(i) Find the value of $h$.\\
(ii)\\
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The lamina is suspended from a vertical string attached to a point $X$ on the side $A D$ of the triangle (see diagram). Given the lamina is in equilibrium with $A D$ horizontal, calculate $X D$.
\hfill \mbox{\textit{CAIE M2 2011 Q2 [7]}}