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## Homework Statement

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.

## Homework Equations

## 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?