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(B) For a bob to complete a vertical circle,the minimum velocity at the highest point is $v_{top} = \sqrt{g \ell}$.Let the first bob have mass $m$ and

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$5$

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(B) For a bob to complete a vertical circle,the minimum velocity at the highest point is $v_{top} = \sqrt{g \ell}$.Let the first bob have mass $m$ and string length $l_1$. Its velocity at the highest point before collision is $u_1 = \sqrt{g l_1}$.The second bob has mass $m$ and string length $l_2$,and is initially at rest $(u_2 = 0)$.Since the collision is elastic and the masses are equal,the two bobs exchange their velocities.Therefore,the velocity of the second bob after collision is $v_2 = u_1 = \sqrt{g l_1}$.For the second bob to complete a vertical circle,its minimum velocity at the lowest point must be $\sqrt{5 g l_2}$.However,the collision occurs at the highest point of the second bob's path (which is the lowest point of its potential circular motion). Thus,the velocity acquired by the second bob after collision must be equal to the minimum velocity required at the lowest point to complete the circle.So,$v_2 = \sqrt{5 g l_2}$.Equating the two expressions for $v_2$: $\sqrt{g l_1} = \sqrt{5 g l_2}$.Squaring both sides: $g l_1 = 5 g l_2$.Therefore,$\frac{l_1}{l_2} = 5$.

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