| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Rotating disc with friction |
| Difficulty | Standard +0.3 This is a standard circular motion problem with friction requiring application of F=mv²/r and friction laws. Part (i) is straightforward; parts (ii) and (iii) involve resolving forces with tension included, requiring systematic application of Newton's second law in horizontal and vertical directions. The multi-part structure and need to consider limiting friction cases elevate it slightly above average, but the techniques are all standard M2 material with no novel insight required. |
| Spec | 3.03t Coefficient of friction: F <= mu*R model6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{0.2\times1.5^2}{0.5} = \text{Fr}\) | M1 | Newton's 2nd law, \(\text{acc} = \frac{v^2}{r}\) |
| \(\mu\left(=\frac{\text{Fr}}{0.2g}\right) = 0.45\) | A1 | [2] \(\text{Fr} = 0.9\text{ N}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(R = 0.2g - \frac{0.2g}{2}\sin30\) | M1 | \(1.5\) |
| \(\text{Fr} = 0.45\times1.5\) | A1 | \(0.675\) |
| \(\frac{0.2g}{2}\cos30 \pm 0.675 = 0.2\omega^2\cdot0.5\) | M1 | Either |
| \(\omega = 3.93\) (greatest) | A1 | |
| \(\omega = 1.38\) (least) | A1 | [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{0.2g}{2}\cos30 = 0.2\omega^2\cdot0.5\) | M1 | |
| \(\omega = 2.94\) | A1 | [2] |
## Question 6:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{0.2\times1.5^2}{0.5} = \text{Fr}$ | M1 | Newton's 2nd law, $\text{acc} = \frac{v^2}{r}$ |
| $\mu\left(=\frac{\text{Fr}}{0.2g}\right) = 0.45$ | A1 | [2] $\text{Fr} = 0.9\text{ N}$ |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $R = 0.2g - \frac{0.2g}{2}\sin30$ | M1 | $1.5$ |
| $\text{Fr} = 0.45\times1.5$ | A1 | $0.675$ |
| $\frac{0.2g}{2}\cos30 \pm 0.675 = 0.2\omega^2\cdot0.5$ | M1 | Either |
| $\omega = 3.93$ (greatest) | A1 | |
| $\omega = 1.38$ (least) | A1 | [5] |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{0.2g}{2}\cos30 = 0.2\omega^2\cdot0.5$ | M1 | |
| $\omega = 2.94$ | A1 | [2] |
---6 A horizontal disc with a rough surface rotates about a fixed vertical axis which passes through the centre of the disc. A particle $P$ of mass 0.2 kg is in contact with the surface and rotates with the disc, without slipping, at a distance 0.5 m from the axis. The greatest speed of $P$ for which this motion is possible is $1.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(i) Calculate the coefficient of friction between the disc and $P$.\\
$P$ is now attached to one end of a light elastic string, which is connected at its other end to a point on the vertical axis above the disc. The tension in the string is equal to half the weight of $P$. The disc rotates with constant angular speed $\omega \mathrm { rad } \mathrm { s } ^ { - 1 }$ and $P$ rotates with the disc without slipping. $P$ moves in a circle of radius 0.5 m , and the taut string makes an angle of $30 ^ { \circ }$ with the horizontal.\\
(ii) Find the greatest and least values of $\omega$ for which this motion is possible.\\
(iii) Calculate the value of $\omega$ for which the disc exerts no frictional force on $P$.
\hfill \mbox{\textit{CAIE M2 2014 Q6 [9]}}