Centre of Mass and Collision Part 2

Concept:Energy conservation and elastic collision rule, followed by condition for completing a vertical circle.Explanation:Bob A loses height: $\Delta h = l(1 - \cos 60^\circ) = 1 \times (1 - 0.5) = 0.5$ m.Using energy conservation: $mg \Delta h = \frac{1}{2} m v_A^2$.So $v_A = \sqrt{2g \Delta h} = \sqrt{2g \times 0.5} = \sqrt{g}$.Collision is elastic and bobs are identical; velocities are exchanged.Thus, after collision, bob B has velocity $v_B = \sqrt{g}$.Bob B must just complete a vertical circle of radius $R$.Minimum velocity at bottom for completing circle: $\sqrt{5gR}$.Equate: $\sqrt{g} = \sqrt{5gR}$.Squaring: $g = 5gR$ → $R = \frac{1}{5}$ m.Answer:$R = \frac{1}{5}$ m, so option A.