| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Centre of mass with variable parameter |
| Difficulty | Challenging +1.2 This is a standard centre of mass problem with composite solids requiring systematic application of formulas. Part (i) is routine calculation, part (ii) involves algebraic manipulation with a given answer to show, and part (iii) requires geometric insight about symmetry. The multi-step nature and the need to work backwards in part (ii) elevate it slightly above average, but the techniques are all standard M2 material with no novel problem-solving required. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass |
| Answer | Marks | Guidance |
|---|---|---|
| (i) | M1 | Table of moments idea |
| \(\pi 0.6^2 \times 0.6 \times 0.3 - 2\pi 0.6^{1/3} \times 3 \times 0.6/8\) | A1 | Correct elements |
| \(= (\pi 0.6^3 + 2\pi 0.6^3)d\) | A1 | Correct composite |
| \(d = 0.09 \text{ m}\) | A1 | [4] |
| (ii) | M1 | Table of moments idea (about \(O\)) |
| \(\frac{2}{3}\pi 0.6^3 \times \frac{1}{4} \times 0.6 - \pi \times 0.6^3 \times 0.3 + 0.484 \times 0.36 = 0\) | A1 | Correct elements |
| \(A = 3\pi/16 \text{ m}^2\) | A1 | [4] |
| OR | M1 | Table of moments idea (about \(O\)) |
| \([\frac{2}{3}\pi \times 0.6^3 + \pi \times 0.6^3] \times 0.09 = 0.484 \times 0.36\) | A1 | Correct elements |
| \(A = 3\pi/16\) | A1 | |
| (iii) Increase in length \([= 2 \times (0.6 - 0.48)] = 0.24\text{m}\) | B1 [1] | Remove cylinder with centre of mass at \(O\) |
**(i)** | M1 | Table of moments idea
$\pi 0.6^2 \times 0.6 \times 0.3 - 2\pi 0.6^{1/3} \times 3 \times 0.6/8$ | A1 | Correct elements
$= (\pi 0.6^3 + 2\pi 0.6^3)d$ | A1 | Correct composite
$d = 0.09 \text{ m}$ | A1 | [4]
**(ii)** | M1 | Table of moments idea (about $O$)
$\frac{2}{3}\pi 0.6^3 \times \frac{1}{4} \times 0.6 - \pi \times 0.6^3 \times 0.3 + 0.484 \times 0.36 = 0$ | A1 | Correct elements
$A = 3\pi/16 \text{ m}^2$ | A1 | [4]
**OR** | M1 | Table of moments idea (about $O$)
$[\frac{2}{3}\pi \times 0.6^3 + \pi \times 0.6^3] \times 0.09 = 0.484 \times 0.36$ | A1 | Correct elements
$A = 3\pi/16$ | A1 |
**(iii)** Increase in length $[= 2 \times (0.6 - 0.48)] = 0.24\text{m}$ | B1 [1] | Remove cylinder with centre of mass at $O$6 A uniform solid consists of a hemisphere with centre $O$ and radius 0.6 m joined to a cylinder of radius 0.6 m and height 0.6 m . The plane face of the hemisphere coincides with one of the plane faces of the cylinder.\\
(i) Calculate the distance of the centre of mass of the solid from $O$.\\[0pt]
[The volume of a hemisphere of radius $r$ is $\frac { 2 } { 3 } \pi r ^ { 3 }$.]\\
(ii)\\
\includegraphics[max width=\textwidth, alt={}, center]{a093cbad-3ba0-45ce-a617-d4ecc8cb1ec9-4_547_631_593_797}
A cylindrical hole, of length 0.48 m , starting at the plane face of the solid, is made along the axis of symmetry (see diagram). The resulting solid has its centre of mass at $O$. Show that the area of the cross-section of the hole is $\frac { 3 } { 16 } \pi \mathrm {~m} ^ { 2 }$.\\
(iii) It is possible to increase the length of the cylindrical hole so that the solid still has its centre of mass at $O$. State the increase in the length of the hole.
\hfill \mbox{\textit{CAIE M2 2011 Q6 [9]}}