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

#### Solution

a = 3.26 m/s^{2}, T = 3.9 N

After 2 s, velocity of mass m_{1}_{,}

v = u + at = 0 + 3.26 × 2

= 6.52 m/s upward

At this time, m_{2} is moving 6.52 m/s downward.

At time 2 s, m_{2} stops for a moment. But m_{1} is moving upward with velocity 6.52 m/s. It will continue to move till final velocity (at highest point) becomes zero.

Here, v = 0, u = 6.52 m/s

a = −g = − 9.8 m/s^{2}

v = u + at = 6.52 + (−9.8)t

\[\Rightarrow t = \frac{6 . 52}{9 . 8} \approx \frac{2}{3} \sec\]

After this time, the mass m_{1} also starts moving downward.

So, the string becomes tight again after \[\frac{2}{3} s\]