| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2010 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Two strings, two fixed points |
| Difficulty | Standard +0.8 This is a non-trivial circular motion problem requiring resolution of forces in two directions, application of Newton's second law for circular motion, and understanding of limiting cases. Part (i) involves simultaneous equations with trigonometry and centripetal force. Part (ii) requires conceptual understanding that at minimum speed, one string becomes slack. More challenging than standard single-string conical pendulum problems. |
| Spec | 6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(3T\cos 30° - T\cos 30° = 0.4g\) | M1 | Resolves vertically, 3 terms |
| \(T = 2.31\) | A1 | |
| \(\frac{0.4 \times 6^2}{r} = 4T\sin 30°\) | M1 | Newton's 2nd Law horizontally |
| \(r = 3.12\) | A1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(T_{PB} = 0\) | B1 | Resolves vertically, 2 terms |
| \(T\cos 30° = 0.4g \; (T = 4.62)\) | M1 | |
| \(\frac{0.4v^2}{3.12} = T\sin 30°\) | M1 | Newton's 2nd Law horizontally |
| \(v = 4.24 \text{ ms}^{-1}\) | A1 [4] |
## Question 5:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $3T\cos 30° - T\cos 30° = 0.4g$ | M1 | Resolves vertically, 3 terms |
| $T = 2.31$ | A1 | |
| $\frac{0.4 \times 6^2}{r} = 4T\sin 30°$ | M1 | Newton's 2nd Law horizontally |
| $r = 3.12$ | A1 **[4]** | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $T_{PB} = 0$ | B1 | Resolves vertically, 2 terms |
| $T\cos 30° = 0.4g \; (T = 4.62)$ | M1 | |
| $\frac{0.4v^2}{3.12} = T\sin 30°$ | M1 | Newton's 2nd Law horizontally |
| $v = 4.24 \text{ ms}^{-1}$ | A1 **[4]** | |
---(i) It is given that when the ball moves with speed $6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ the tension in the string $Q B$ is three times the tension in the string $P B$. Calculate the radius of the circle.
The ball now moves along this circular path with the minimum possible speed.\\
(ii) State the tension in the string $P B$ in this case, and find the speed of the ball.
\hfill \mbox{\textit{CAIE M2 2010 Q5 [8]}}