| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2013 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Two strings/rods system |
| Difficulty | Challenging +1.2 This is a standard two-string circular motion problem requiring resolution of forces and application of F=mrω². Part (i) involves setting up equations with equal tensions and solving simultaneously - a routine technique. Part (ii) requires recognizing that minimum angular speed occurs when one string becomes slack (tension = 0), which is a common exam pattern. The geometry is straightforward (3-4-5 triangle), and the problem follows predictable M2 circular motion methods without requiring novel insight. |
| Spec | 3.03d Newton's second law: 2D vectors6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(Tx(4/5) - Tx(3/5) = 0.2g\) | M1, A1 | Resolves vertically, 3 forces |
| \(T = 10\) | A1 | Maybe implied |
| \(Tx(4/5) + Tx(3/5) = 0.2v^2/(0.4 \times 3/5)\) | M1 | Resolves horizontally, N2L |
| \(v = 4.1(0)\ \text{ms}^{-1}\) | A1 [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(Tx(4/5) = 0.2g\) | B1 | \(T = 2.5\) |
| \(Tx(3/5) = 0.2\omega^2 x(0.4 \times 3/5)\) | M1 | N2L horizontally, single force |
| \(\omega = 5.59\ \text{rads}^{-1}\) | A1 [3] |
## Question 5:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $Tx(4/5) - Tx(3/5) = 0.2g$ | M1, A1 | Resolves vertically, 3 forces |
| $T = 10$ | A1 | Maybe implied |
| $Tx(4/5) + Tx(3/5) = 0.2v^2/(0.4 \times 3/5)$ | M1 | Resolves horizontally, N2L |
| $v = 4.1(0)\ \text{ms}^{-1}$ | A1 [5] | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $Tx(4/5) = 0.2g$ | B1 | $T = 2.5$ |
| $Tx(3/5) = 0.2\omega^2 x(0.4 \times 3/5)$ | M1 | N2L horizontally, single force |
| $\omega = 5.59\ \text{rads}^{-1}$ | A1 [3] | |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{6503ebb1-5649-4ca5-9500-da4fb28009dd-3_540_537_255_804}
A particle $P$ of mass 0.2 kg is attached to a fixed point $A$ by a light inextensible string of length 0.4 m . A second light inextensible string of length 0.3 m connects $P$ to a fixed point $B$ which is vertically below $A$. The particle $P$ moves in a horizontal circle, which has its centre on the line $A B$, with the angle $A P B = 90 ^ { \circ }$ (see diagram).\\
(i) Given that the tensions in the two strings are equal, calculate the speed of $P$.\\
(ii) It is given instead that $P$ moves with its least possible angular speed for motion in this circle. Find this angular speed.
\hfill \mbox{\textit{CAIE M2 2013 Q5 [8]}}