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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.
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Solution
a = 3.26 m/s2, T = 3.9 N
After 2 s, velocity of mass m1,
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 \[\frac{2}{3} s\]
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