In the simplest form of Atwood machine,two masses are 0.3 kg and 0.6 kg.We consider massless pulley,string and frictionless surfaces.we let the arrangement move at t=0.
You can calculate that the tension in the string is 3.9 N and the acceleration~3.26
The larger mass is stopped 2 seconds later the arrangement started moving.
We are to find the timee elapsed before the string is taut again.
The Attempt at a Solution
I thought that the lighter mass would move upwards (at the instant bigger mass was stopped) with speed: v=at=(3.26x2) m/s=6.52 m/s
Then using total time of flight I got T=(2v/g)=1.33 s
Whereas the answer is 0.67 s
can anyone show me if I am wrong?