Question

Consider the Atwood machine of the previous problem. The larger mass is stopped for a moment $2.0s$ after the system is set into motion. Find the time elapsed before the string is tight again.

Answer 1

$a-3.26m/s^2$

$T=3.9N$

After $2sec$ mass $m$, the velocity

$V=u+at=0+3.26\times 2=6.52m/s$ upward

At this time $m_2$ is moving $6.52m/s$ downward.

At time $2sec,m_2$ stops for a moment. But $m_1$ is moving upward with velocity $6$.

It will continue to move till final velocity (at highest point) because zero.

Here, $v=0;u=6.52$

$A=-g=9.8m/s^2$ [moving up ward $m_1$]

$V=u+at\Rightarrow 0=6.52+(-9.8)t$

$\Rightarrow t=6.52/9.8=0.66=2/3sec$

During this period $2/3sec$, $m-2$ mass also starts moving downward.

So the string becomes tight again after a time of $2/3sec$