| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2009 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Composite solid with hemisphere and cylinder/cone |
| Difficulty | Standard +0.3 This is a standard centre of mass problem for composite solids requiring knowledge of standard results (cone CM at h/4, cylinder at h/2) and straightforward moment calculations. Part (ii) involves basic equilibrium geometry with the vertical through CM passing through the pivot point. While it requires careful setup, it's a routine M2 question with no novel insights needed, making it slightly easier than average. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (i) \(8 \times 8 + 2 \times (16+3) = (8+2)\dot{y}\) | M1 | For taking moments about the base |
| A1 | ||
| Distance of centre of mass is \(10.2\) cm | A1 | AG [3] |
| (ii) \([\tan\theta_{\max} = r/\dot{y}]\) | M1 | \(\theta\) takes its max value when c.m. is vertically above point of contact |
| \(\tan\theta_{\max} = 4/10.2\) | A1\(\sqrt{}\) | |
| Greatest possible value is \(21.4\) | A1 | [3] |
## Question 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| **(i)** $8 \times 8 + 2 \times (16+3) = (8+2)\dot{y}$ | M1 | For taking moments about the base |
| | A1 | |
| Distance of centre of mass is $10.2$ cm | A1 | AG **[3]** |
| **(ii)** $[\tan\theta_{\max} = r/\dot{y}]$ | M1 | $\theta$ takes its max value when c.m. is vertically above point of contact |
| $\tan\theta_{\max} = 4/10.2$ | A1$\sqrt{}$ | |
| Greatest possible value is $21.4$ | A1 | **[3]** |
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\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{68acf474-5da2-4949-b3b2-fc42cd73bd4a-2_408_291_1027_927}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}
A uniform solid cylinder has mass 8 kg and height 16 cm . A uniform solid cone, whose base radius is the same as the radius of the cylinder, has mass 2 kg and height 12 cm . A composite solid is formed by joining the cylinder and cone so that the base of the cone coincides with one end of the cylinder (see Fig. 1).\\
(i) Show that the centre of mass of the composite solid is 10.2 cm from its base.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{68acf474-5da2-4949-b3b2-fc42cd73bd4a-2_401_444_1877_849}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
The composite solid is held with a point on the circumference of its base in contact with a horizontal table. The base makes an angle $\theta ^ { \circ }$ with the table (see Fig. 2, which shows a cross-section). When the cone is released it moves towards the equilibrium position in which its base is in contact with the table.\\
(ii) Given that the radius of the base is 4 cm , find the greatest possible value of $\theta$, correct to 1 decimal place.
\hfill \mbox{\textit{CAIE M2 2009 Q3 [6]}}