Question

A bob of mass $m$, suspended by a string of length $l_1$, is given a minimum velocity required to complete a full circle in the vertical plane. At the highest point, it collides elastically with another bob of mass $m$ suspended by a string of length $l_2$, which is initially at rest. Both the strings are massless and inextensible. If the second bob, after collision acquires the minimum speed required to complete a full circle in the vertical plane, the ratio $\frac{l_1}{l_2}$ is:

Answer 1

For complite circular motion

$\Rightarrow \frac{{v_3}^2}{l^2}=g$

$\Rightarrow l^2=\frac{{v_3}^2}{g}$

By energy conservation,

$\Rightarrow \frac{1}{2}m{v_2}^2=mg\left(2l_2\right)+\frac{1}{2}m{v_3}^2$

$\Rightarrow {v_2}^2=4g{l}_2+{v_3}^2$

$=4g{l}_2+g{l}_2$

$\Rightarrow {v_2}^2=5g{l}_2⟶\left(1\right)$

For complete circular motion of $l_1$ radius circle $v_2$ should be

$\Rightarrow v_2-$

$\frac{{v_2}^2}{l_1}=g$ By $\left(1\right)$

$\frac{5g{l}_2}{l_1}=g$

$\Rightarrow \frac{l_1}{l_2}=5.$

Hence, the answer is $5.$