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° with vertical as shown in figure. At distance of 1 m below point P, an identical 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 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.

vertical descent height of bob A.

$$ h = L - L \cos(60^\circ) = 1 - 1\left(\frac{1}{2}\right) = 0.5 \text{ m} $$

Apply conservation of energy

$$ \frac{1}{2} m v_A^2 = mgh \implies v_A = \sqrt{2 \cdot g \cdot 0.5} = \sqrt{g} $$

elastic collision between identical masses.

$$ \text{Velocities are exchanged, so } v_B = v_A = \sqrt{g} $$

minimum velocity required at the bottom to just complete the vertical circular track.

$$ v_B = \sqrt{5gR} $$

$$ \sqrt{g} = \sqrt{5gR} \implies g = 5gR \implies R = \frac{1}{5} = 0.2 \text{ m} $$