A small bob $A$ of mass $m$ is attached to a massless rigid rod of length 1 m pivoted at point $P$ a

Question

A small bob $A$ of mass $m$ is attached to a massless rigid rod of length 1 m pivoted at point $P$ and kept at an angle of $60^{\circ}$ with vertical as shown in figure. At distance of 1 m below point $P$, an identical $\operatorname{bob} B$ is kept at rest on a smooth horizontal surface that extends to a circular track of radius $R$ as shown in figure. If $\operatorname{bob} B$ just manages to complete the circular path of radius $R$ upto a point $Q$ after being hit elastically by bob $A$, then radius $R$ is $\_\_\_\_$ m.

Professor PC Thomas Classes · Answer

Bob A falls from an angle of $60^\circ$. Its height drop is $h = L(1 - \cos 60^\circ) = 1(1 - 0.5) = 0.5 	ext{ m}$. The velocity of A at the bottom is $v_A = \sqrt{2gh} = \sqrt{2 	imes 10 	imes 0.5} = \sqrt{10} 	ext{ m/s}$. Since the collision is elastic and masses are identical, A transfers its entire velocity to B, so $v_B = \sqrt{10} 	ext{ m/s}$. For B to just complete the vertical circular track, its velocity at the bottom must be $v = \sqrt{5gR}$. Equating these gives $\sqrt{5 	imes 10 	imes R} = \sqrt{10} \implies 50R = 10 \implies R = \frac{10}{50} = \frac{1}{5} 	ext{ m}$.