Two balls having masses m and 2m are affixed to two light strings...

Let the velocities of the light ball of mass $m$ and the heavy ball of mass $2m$ just before the collision be $u_1$ and $u_2$ respectively. Since both are released from a horizontal position, they fall through a height $l$.

$u_1 = \sqrt{2gl}$

$u_2 = -\sqrt{2gl}$

For a perfectly elastic collision, the velocities after the collision are given by:

$v_1 = \dfrac{m_1 - m_2}{m_1 + m_2}u_1 + \dfrac{2m_2}{m_1 + m_2}u_2$

$v_2 = \dfrac{2m_1}{m_1 + m_2}u_1 + \dfrac{m_2 - m_1}{m_1 + m_2}u_2$

Substituting $m_1 = m$ and $m_2 = 2m$:

$v_1 = \dfrac{m - 2m}{3m}(\sqrt{2gl}) + \dfrac{4m}{3m}(-\sqrt{2gl}) = -\dfrac{5}{3}\sqrt{2gl}$

$v_2 = \dfrac{2m}{3m}(\sqrt{2gl}) + \dfrac{2m - m}{3m}(-\sqrt{2gl}) = \dfrac{1}{3}\sqrt{2gl}$

The maximum height reached by the heavy ball is:

$h_2 = \dfrac{v_2^2}{2g} = \dfrac{\dfrac{1}{9}(2gl)}{2g} = \dfrac{l}{9}$

For the light ball, the theoretical maximum height as a free projectile would be:

$h_1 = \dfrac{v_1^2}{2g} = \dfrac{\dfrac{25}{9}(2gl)}{2g} = \dfrac{25l}{9}$

Since the light ball is attached to a string of length $l$, it can only reach a maximum height of $2l$ if it completes the vertical circle. The condition to complete the vertical circle is that the velocity at the lowest point must be greater than or equal to $\sqrt{5gl}$.

$|v_1| = \sqrt{\dfrac{50}{9}gl}$

Since $\dfrac{50}{9} > 5$, the light ball completes the vertical circle. Thus, its maximum height is restricted by the string to the top of the circle, which is $2l$.

Answer: Light ball $2l$, heavy ball $l/9$