simple but ....

le lo naa kuchh bhee...

m2>m1

If m1 is stopped the string will be taut the very next second since m1 was being pulled by m2

if we consider m1>m2

Then m1 is pulling m2. Since m1 stops it stops pulling m2. m2 will be having a velocity as it was accelerated by m1 before it stopped. The velocity is upwards. Therefore the string will become taut again after v becomes v(with opposite direction.

for m1>m2

well just before stopping they will have their velocities equal=v

when m1 is stopped its velocity goes to zero

then the tension vanishes and regains again when distance between them = length of thread again

after stopping m1 faces g downwards with zero velocity

m2 also faces g downwards with v velocity upwards

so let the string become taut after time t

vt-(1/2)gt²=1/2gt²

In this way we can obtain the time

also if the lighter mass is stopped in the above case

we have (let t again be time of tension regain)

vt+1/2gt²+1/2gt²=0

which gives t=0 which means the string will instantly become taut

tx..me 2 had da same doubt