| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2019 |
| Session | March |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Centre of mass of composite shapes |
| Difficulty | Standard +0.3 This is a standard composite body centre of mass problem using the 'negative mass' technique. Students must recognize that the hole can be treated as removing mass, set up the moment equation about a convenient point, and solve for the distance. While it requires careful setup and algebraic manipulation, it follows a well-established method taught explicitly in M2 courses with no novel insight required. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.3g = \frac{24e}{0.6}\) | M1 | Note greatest speed occurs at the equilibrium position. Use \(T = \frac{\lambda x}{L}\) |
| \(e = 0.075\text{ m}\) | A1 | Fall \(= 0.275\text{ m}\) |
| PE Change \(= 0.3g \times 0.275\) | B1 | |
| \(\frac{0.3v^2}{2} = 0.3g \times 0.275 - \frac{24 \times 0.075^2}{2 \times 0.6}\) | M1 | Set up a 3 term energy equation |
| \(v = 2.18 \text{ m s}^{-1}\) | A1 | |
| 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.3g(0.2 + E) = \frac{24E^2}{2 \times 0.6}\) | M1 | Set up an energy equation. Note \(v = 0\) at the greatest distance |
| \(20E^2 - 3E - 0.6 = 0\) | M1 | Attempt to solve a 3 term quadratic equation |
| \(E = 0.264\) and so greatest distance is \(0.864\text{ m}\) | A1 | |
| 3 |
## Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.3g = \frac{24e}{0.6}$ | M1 | Note greatest speed occurs at the equilibrium position. Use $T = \frac{\lambda x}{L}$ |
| $e = 0.075\text{ m}$ | A1 | Fall $= 0.275\text{ m}$ |
| PE Change $= 0.3g \times 0.275$ | B1 | |
| $\frac{0.3v^2}{2} = 0.3g \times 0.275 - \frac{24 \times 0.075^2}{2 \times 0.6}$ | M1 | Set up a 3 term energy equation |
| $v = 2.18 \text{ m s}^{-1}$ | A1 | |
| | **5** | |
## Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.3g(0.2 + E) = \frac{24E^2}{2 \times 0.6}$ | M1 | Set up an energy equation. Note $v = 0$ at the greatest distance |
| $20E^2 - 3E - 0.6 = 0$ | M1 | Attempt to solve a 3 term quadratic equation |
| $E = 0.264$ and so greatest distance is $0.864\text{ m}$ | A1 | |
| | **3** | |5 A particle $P$ of mass 0.3 kg is attached to one end of a light elastic string of natural length 0.6 m and modulus of elasticity 24 N . The other end of the string is attached to a fixed point $O$. The particle $P$ is released from rest at the point 0.4 m vertically below $O$.\\
(i) Find the greatest speed of $P$.\\
(ii) Calculate the greatest distance of $P$ below $O$.\\
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b8e52188-f9a6-46fc-90bf-97965c6dd324-10_608_611_258_767}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}
Fig. 1 shows the cross-section of a solid cylinder through which a cylindrical hole has been drilled to make a uniform prism. The radius of the cylinder is $5 r$ and the radius of the hole is $r$. The centre of the hole is a distance $2 r$ from the centre of the cylinder.\\
\hfill \mbox{\textit{CAIE M2 2019 Q5 [8]}}