| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | String through hole/bead on string |
| Difficulty | Challenging +1.2 This is a two-part circular motion problem requiring resolution of forces, use of Newton's second law for circular motion, and geometric reasoning. Part (i) is relatively straightforward with given geometry, but part (ii) requires careful angle work and simultaneous equations. While it involves multiple steps and careful setup, the techniques are standard for M2 circular motion with no novel insights required—moderately above average difficulty. |
| Spec | 6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\theta = \sin^{-1}(0.2/0.7) = 16.6°\) with the vertical | B1 | \(73.4°\) with the horizontal |
| \(T\cos\theta = 0.3g\) | M1 | \(T = 3.13\) Resolves vertically |
| \(T + T\sin\theta = 0.3\omega^2 \times 0.2\) | M1 | Uses Newton's Second Law radially |
| \(\omega = 8.19\) | A1 | |
| \(\text{KE} ( = 0.3 \times (8.19 \times 0.2)^2/2) = 0.402 \text{ J}\) | A1 | [5] Accept 0.403 J |
| (ii) \((0.9 - AB)/AB = 1/2\) | M1 | \(\alpha = \tan^{-1}0.5 = 26.565°\) or \(BC/(0.9-BC) = 1/2\) |
| \(AB = 0.6 \text{ m}\) | A1 | \(BC = 0.3 \text{ m}\) |
| \(T\cos\alpha - T\sin\alpha = 0.3g\) | M1 | Resolves vertically |
| \(T = 6.71\) | A1 | |
| \(T\cos\alpha + T\sin\alpha = 0.3\omega^2 \times 0.6\sin\alpha\) | M1 | \(0.3\omega^2 \times 0.3\cos\alpha\) Uses Newton's Second Law radially |
| \(\omega = 10.6\) | A1 | [6] |
**(i)** $\theta = \sin^{-1}(0.2/0.7) = 16.6°$ with the vertical | B1 | $73.4°$ with the horizontal
$T\cos\theta = 0.3g$ | M1 | $T = 3.13$ Resolves vertically
$T + T\sin\theta = 0.3\omega^2 \times 0.2$ | M1 | Uses Newton's Second Law radially
$\omega = 8.19$ | A1
$\text{KE} ( = 0.3 \times (8.19 \times 0.2)^2/2) = 0.402 \text{ J}$ | A1 | [5] Accept 0.403 J
**(ii)** $(0.9 - AB)/AB = 1/2$ | M1 | $\alpha = \tan^{-1}0.5 = 26.565°$ or $BC/(0.9-BC) = 1/2$
$AB = 0.6 \text{ m}$ | A1 | $BC = 0.3 \text{ m}$
$T\cos\alpha - T\sin\alpha = 0.3g$ | M1 | Resolves vertically
$T = 6.71$ | A1
$T\cos\alpha + T\sin\alpha = 0.3\omega^2 \times 0.6\sin\alpha$ | M1 | $0.3\omega^2 \times 0.3\cos\alpha$ Uses Newton's Second Law radially
$\omega = 10.6$ | A1 | [6]6 A smooth bead $B$ of mass 0.3 kg is threaded on a light inextensible string of length 0.9 m . One end of the string is attached to a fixed point $A$, and the other end of the string is attached to a fixed point $C$ which is vertically below $A$. The tension in the string is $T \mathrm {~N}$, and the bead rotates with angular speed $\omega \mathrm { rad } \mathrm { s } ^ { - 1 }$ in a horizontal circle about the vertical axis through $A$ and $C$.\\
(i) Given that $B$ moves in a circle with centre $C$ and radius 0.2 m , calculate $\omega$, and hence find the kinetic energy of $B$.\\
(ii) Given instead that angle $A B C = 90 ^ { \circ }$, and that $A B$ makes an angle $\tan ^ { - 1 } \left( \frac { 1 } { 2 } \right)$ with the vertical, calculate $T$ and $\omega$.
\hfill \mbox{\textit{CAIE M2 2011 Q6 [11]}}