Question

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.

Answer 1

a = 3.26 m/s², 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, m₂ is moving 6.52 m/s downward.
At time 2 s, m₂ stops for a moment. But m₁ 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/s²
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 m₁ also starts moving downward.
So, the string becomes tight again after \[\frac{2}{3} s\]