Two balls possessing masses m and 2m are tied to two light strings...

Let the speed of each ball just before the collision be $v_0$. Since both balls are released from the horizontal position and fall through a height $l$, their speeds are given by:

$v_0 = \sqrt{2gl}$

Let the right direction be positive. The initial velocity of the light ball (mass $m$) is $u_1 = v_0$ and the initial velocity of the heavy ball (mass $2m$) is $u_2 = -v_0$.

Let $v_1$ and $v_2$ be the velocities of the light and heavy balls respectively just after the collision.

By the principle of conservation of linear momentum:

$m u_1 + 2m u_2 = m v_1 + 2m v_2$

$m(v_0) + 2m(-v_0) = m v_1 + 2m v_2$

$v_1 + 2v_2 = -v_0$ ... (i)

Since the collision is elastic, the coefficient of restitution is $e = 1$. The relative velocity of separation equals the relative velocity of approach:

$v_2 - v_1 = u_1 - u_2$

$v_2 - v_1 = v_0 - (-v_0) = 2v_0$ ... (ii)

Adding equations (i) and (ii):

$(v_1 + 2v_2) + (v_2 - v_1) = -v_0 + 2v_0$

$3v_2 = v_0 \Rightarrow v_2 = \dfrac{v_0}{3}$

Substituting $v_2$ into equation (ii):

$\dfrac{v_0}{3} - v_1 = 2v_0 \Rightarrow v_1 = \dfrac{v_0}{3} - 2v_0 = -\dfrac{5v_0}{3}$

Substituting $v_0 = \sqrt{2gl}$:

$v_1 = -\dfrac{5}{3}\sqrt{2gl} = -\sqrt{\dfrac{50gl}{9}}$

$v_2 = \dfrac{1}{3}\sqrt{2gl} = \sqrt{\dfrac{2gl}{9}}$

The negative sign for $v_1$ indicates that the light ball moves towards the left, and the positive sign for $v_2$ indicates that the heavy ball moves towards the right.