In the figure shown, the heavy ball of mass 2m rests on the horizontal surface and the lighter ball of mass m is dropped from a height h > 2l. At the instant the string gets taut, the upward the velocity of the heavy ball will be

Velocity of light ball, when it is descended by height 2l

$\mathrm{v}=\sqrt{2\mathrm{g}2\mathrm{l}}=2\sqrt{\mathrm{gl}}$

When string will taut then impulsive tension will be generated in the string.

Let velocity all both masses become v' after string taut.

$\mathrm{J}=2\mathrm{mv}\text{'} ---- \left(1\right)$

$\mathrm{J}=\mathrm{m}\times 2\sqrt{\mathrm{gl}}-\mathrm{mv}\text{'} ---- \left(2\right)$

(1) = (2)

From conservation of linear momentum.

$\mathrm{m}2\sqrt{\mathrm{gl}}=2{\mathrm{mv}}^{\mathrm{\prime }}+{\mathrm{mv}}^{\mathrm{\prime }}$

$2\mathrm{m}\sqrt{\mathrm{gl}}=3{\mathrm{mv}}^{\mathrm{\prime }}$

$\frac{2}{3}\sqrt{\mathrm{gl}}={\mathrm{v}}^{\mathrm{\prime }}$