| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Smooth ring on rotating string |
| Difficulty | Standard +0.3 This is a standard conical pendulum problem with two configurations. Part (i) requires basic circular motion equations (T = mrω²) with straightforward geometry. Part (ii) involves resolving forces in two directions with given angles, which is routine for M2 level. The geometry is provided explicitly, and the problem follows standard textbook patterns without requiring novel insight or complex multi-step reasoning. |
| Spec | 3.03e Resolve forces: two dimensions3.03n Equilibrium in 2D: particle under forces6.05a Angular velocity: definitions6.05b 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 | |
| \(KE ( = 0.3 \times (8.19 \times 0.2)^2/2) = 0.402\) J | A1 | [5] Accept \(0.403\) J |
| (ii) \((0.9 - AB)/AB = 1/2\) | M1 | \(\alpha = \tan^{-1}0.5 = 26.565°\) or \(BC/(0.9-BC) = 1/2\) |
| \(AB = 0.6\) m | A1 | \(BC = 0.3\) 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 |
$KE ( = 0.3 \times (8.19 \times 0.2)^2/2) = 0.402$ J | A1 | [5] Accept $0.403$ J
**(ii)** $(0.9 - AB)/AB = 1/2$ | M1 | $\alpha = \tan^{-1}0.5 = 26.565°$ or $BC/(0.9-BC) = 1/2$
$AB = 0.6$ m | A1 | $BC = 0.3$ 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]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 7 N, and the bead rotates with angular speed $ω$ rad s$^{-1}$ in a horizontal circle about the vertical axis through $A$ and $C$.
\begin{enumerate}[label=(\roman*)]
\item Given that $B$ moves in a circle with centre $C$ and radius 0.2 m, calculate $ω$, and hence find the kinetic energy of $B$. [5]
\item Given instead that angle $ABC = 90°$, and that $AB$ makes an angle $\tan^{-1}(\frac{4}{3})$ with the vertical, calculate $T$ and $ω$. [6]
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2011 Q6 [11]}}