A small bob A of mass m is attached to a mass | vedclass

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(B) $1$. Let $l = 1 \ m$ be the length of the rod. The velocity $V_A$ of bob $A$ at the lowest point is given by the conservation of mechanical energy

Vedclass · Accepted Answer

$ \frac{1}{5} $

Vedclass · Answer

(B) $1$. Let $l = 1 \ m$ be the length of the rod. The velocity $V_A$ of bob $A$ at the lowest point is given by the conservation of mechanical energy: $mgl(1 - \cos 	heta) = \frac{1}{2} m V_A^2$ $V_A = \sqrt{2gl(1 - \cos 60^{\circ})} = \sqrt{2 	imes 10 	imes 1 	imes (1 - 0.5)} = \sqrt{10} \ m/s$. $2$. Since the collision between identical bobs $A$ and $B$ is elastic, the velocities are exchanged. Thus, after the collision, bob $B$ moves with velocity $V_B = V_A = \sqrt{10} \ m/s$. $3$. For bob $B$ to just complete a vertical circular path of radius $R$, the minimum velocity at the bottom of the circular track must be $V_{min} = \sqrt{5gR}$. $4$. Equating the velocities: $\sqrt{10} = \sqrt{5gR} \implies 10 = 5 	imes 10 	imes R \implies 10 = 50R \implies R = \frac{10}{50} = \frac{1}{5} \ m$.

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