Ratio of String Lengths for Vertical Circular Motion Collision

To find the ratio $\frac{l_1}{l_2}$, we analyze the motion of the two bobs step-by-step:

1. Motion of the First Bob: A bob of mass $m$ suspended by a string of length $l_1$ is given the minimum speed at the lowest point required to complete a full vertical circle: $v_{\text{lowest}} = \sqrt{5gl_1}$

   Using conservation of mechanical energy between the lowest point and the highest point: $\frac{1}{2}m v_{\text{lowest}}^2 = \frac{1}{2}m v_{\text{top}}^2 + mg(2l_1)$ $\frac{1}{2}m (5gl_1) = \frac{1}{2}m v_{\text{top}}^2 + 2mgl_1$ $v_{\text{top}} = \sqrt{gl_1}$

   Thus, the speed of the first bob at its highest point, just before the collision, is: $u_1 = \sqrt{gl_1}$

2. Collision Between the Bobs: The second bob of mass $m$ is initially at rest hanging vertically from a string of length $l_2$, meaning it is located at the lowest point of its own path: $u_2 = 0$

   Since the collision is head-on and perfectly elastic between two bodies of equal mass ($m_1 = m_2 = m$), they exchange their velocities: $v_1' = u_2 = 0$ $v_2' = u_1 = \sqrt{gl_1}$

   Therefore, immediately after the collision, the second bob acquires a horizontal speed of: $v_2' = \sqrt{gl_1}$

3. Motion of the Second Bob: For the second bob to complete a full circle in the vertical plane from its lowest point, the minimum initial speed required is: $v_{\text{min}} = \sqrt{5gl_2}$

   According to the problem, the speed acquired by the second bob after the collision equals this minimum speed: $v_2' = \sqrt{5gl_2}$

   Equating the two expressions for the speed: $\sqrt{gl_1} = \sqrt{5gl_2}$

   Squaring both sides: $gl_1 = 5gl_2$

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