| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2015 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Conical pendulum (horizontal circle) |
| Difficulty | Standard +0.8 This is a multi-part conical pendulum problem requiring resolution of forces in both vertical and horizontal directions, understanding of the constraint that normal force must be non-negative for contact, and solving simultaneous equations involving circular motion. Part (i) requires setting up an inequality (N ≥ 0), while part (ii) involves solving for two unknowns using Pythagoras and circular motion equations. More demanding than standard circular motion questions due to the constraint reasoning and algebraic manipulation required. |
| Spec | 6.05c Horizontal circles: conical pendulum, banked tracks6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T\cos\theta = 0.7g\) | M1 | \(\cos\theta = 0.14/0.5\) |
| \(T = 25\) N | A1 | |
| \(25 \times 0.48/0.5 \geqslant 0.7\omega^2 \times 0.48\) | M1 | Uses accn. \(= \omega^2 r\) |
| \(\omega \leqslant 8.45\ \text{rad s}^{-1}\) | A1 | [4 marks] Accept \(<\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T\cos\theta = 0.7g - 1.4\) | B1 | \(T\cos\theta = 5.6\) |
| \(T\sin\theta = 0.7 \times 2.5^2/(0.5\sin\theta)\) | M1 | |
| \(T\sin^2\theta = 8.75\) | A1 | |
| \(T\sin^2\theta / T\cos\theta = 8.75/5.6\) | M1 | \(T\sin^2\theta + T\cos^2\theta = 8.75 + 5.6^2/T\) |
| \(1 - \cos^2\theta = 1.5625\cos\theta\) | A1 | \(T^2 = 8.75T + 5.6^2\) |
| \(\cos\theta = 0.487(746..)\), \(T = 11.5\) N | A1 | \(\theta = 60.8°\), \(T = 11.5\) N |
| \(h = 0.244\) | A1 | [7 marks] |
## Question 7:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T\cos\theta = 0.7g$ | M1 | $\cos\theta = 0.14/0.5$ |
| $T = 25$ N | A1 | |
| $25 \times 0.48/0.5 \geqslant 0.7\omega^2 \times 0.48$ | M1 | Uses accn. $= \omega^2 r$ |
| $\omega \leqslant 8.45\ \text{rad s}^{-1}$ | A1 | **[4 marks]** Accept $<$ |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T\cos\theta = 0.7g - 1.4$ | B1 | $T\cos\theta = 5.6$ |
| $T\sin\theta = 0.7 \times 2.5^2/(0.5\sin\theta)$ | M1 | |
| $T\sin^2\theta = 8.75$ | A1 | |
| $T\sin^2\theta / T\cos\theta = 8.75/5.6$ | M1 | $T\sin^2\theta + T\cos^2\theta = 8.75 + 5.6^2/T$ |
| $1 - \cos^2\theta = 1.5625\cos\theta$ | A1 | $T^2 = 8.75T + 5.6^2$ |
| $\cos\theta = 0.487(746..)$, $T = 11.5$ N | A1 | $\theta = 60.8°$, $T = 11.5$ N |
| $h = 0.244$ | A1 | **[7 marks]** |7 A particle $P$ of mass 0.7 kg is attached to one end of a light inextensible string of length 0.5 m . The other end of the string is attached to a fixed point $A$ which is $h \mathrm {~m}$ above a smooth horizontal surface. $P$ moves in contact with the surface with uniform circular motion about the point on the surface which is vertically below $A$.\\
(i) Given that $h = 0.14$, find an inequality for the angular speed of $P$.\\
(ii) Given instead that the magnitude of the force exerted by the surface on $P$ is 1.4 N and that the speed of $P$ is $2.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, calculate the tension in the string and the value of $h$.
\hfill \mbox{\textit{CAIE M2 2015 Q7 [11]}}