| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2005 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Rotating disc/platform system |
| Difficulty | Challenging +1.2 This is a standard circular motion problem requiring resolution of forces and application of F=ma=mrω². Part (i) involves straightforward vertical/horizontal force resolution with given angle. Part (ii) requires geometric insight to relate the speeds of P and A through their radii, then similar force resolution. While it has multiple parts and requires careful geometry, the techniques are standard M2 content with no novel problem-solving required. |
| Spec | 6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Radius of path \(= 4 + 5\times7/25\) (\(= 5.4\) m) | B1 | |
| \((T\times(24/25)) = 24\times10\) (\(T=250\)) | M1 | For resolving forces vertically |
| M1 | For applying Newton's second law horizontally and using \(a = \omega^2 r\) | |
| \(24\omega^2\times5.4 = 250\times(7/25)\) | A1 ft | |
| \(\omega = 0.735\) | A1 | [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Radius of path \(= 2\times4\) | B1 | Using \(v\) is proportional to \(r\) |
| \(\sin\theta = 0.8\) | B1 | |
| \(T = 400\) | B1 ft | ft wrong \(\theta\) [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\dfrac{24v^2}{8} = 400\times\dfrac{4}{5}\) | M1 | For applying Newton's second law horizontally and using \(a = v^2/r\) |
| Speed is \(10.3\,\text{ms}^{-1}\) | A1 | [2] |
## Question 6:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Radius of path $= 4 + 5\times7/25$ ($= 5.4$ m) | B1 | |
| $(T\times(24/25)) = 24\times10$ ($T=250$) | M1 | For resolving forces vertically |
| | M1 | For applying Newton's second law horizontally and using $a = \omega^2 r$ |
| $24\omega^2\times5.4 = 250\times(7/25)$ | A1 ft | |
| $\omega = 0.735$ | A1 | **[5]** |
### Part (ii)(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Radius of path $= 2\times4$ | B1 | Using $v$ is proportional to $r$ |
| $\sin\theta = 0.8$ | B1 | |
| $T = 400$ | B1 ft | ft wrong $\theta$ **[3]** |
### Part (ii)(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{24v^2}{8} = 400\times\dfrac{4}{5}$ | M1 | For applying Newton's second law horizontally and using $a = v^2/r$ |
| Speed is $10.3\,\text{ms}^{-1}$ | A1 | **[2]** |
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{a20a6641-d771-4c89-b40f-168a0c61f99d-4_673_773_269_685}
A horizontal circular disc of radius 4 m is free to rotate about a vertical axis through its centre $O$. One end of a light inextensible rope of length 5 m is attached to a point $A$ of the circumference of the disc, and an object $P$ of mass 24 kg is attached to the other end of the rope. When the disc rotates with constant angular speed $\omega$ rad s $^ { - 1 }$, the rope makes an angle of $\theta$ radians with the vertical and the tension in the rope is $T \mathrm {~N}$ (see diagram). You may assume that the rope is always in the same vertical plane as the radius $O A$ of the disc.\\
(i) Given that $\cos \theta = \frac { 24 } { 25 }$, find the value of $\omega$.\\
(ii) Given instead that the speed of $P$ is twice the speed of the point $A$, find
\begin{enumerate}[label=(\alph*)]
\item the value of $T$,
\item the speed of $P$.
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2005 Q6 [10]}}