Two unequal masses are connected on two sides of a light string passing over a light and smooth pulley as shown in figure. The system is released from rest. The larger mass is stopped 1.0 second after the system is set into motion  and then released immediately. The time elapsed before the string is tight again is: Take g = 10 $\mathrm{m}/{\mathrm{s}}^2$

Acceleration of the system:

$\mathrm{a} = \frac{{\mathrm{m}}_2-{\mathrm{m}}_1}{{\mathrm{m}}_2+{\mathrm{m}}_1}\mathrm{g} = \frac{10}{3}\mathrm{m}/{\mathrm{s}}^2$

Velocity of both the blocks at t = 1 s will be

${\mathrm{v}}_{\mathrm{o}} = \mathrm{at} = \frac{10}{3}\times 1 = \frac{10}{3}\mathrm{m}/\mathrm{s}$

Now at this moment, velocity of 2 kg block becomes zero, while that of I kg block is $\frac{10}{3}\mathrm{m}/\mathrm{s}$. upwards.

Hence, string becomes tight again when

Displacement of I kg block = displacement of 2 kg block

or ${\mathrm{v}}_0\mathrm{t}-\frac{1}{2}{\mathrm{gt}}^2 = \frac{1}{2}{\mathrm{gt}}^2$