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Two masses of 1kg & 2kg are connected by a string going over a clamped light smooth pulley.The system is released from rest.The larger mass is stopped for a moment 1.0sec. after the system set in motion.Find the time elasped before the string is tight again.
8 years ago

Ramesh V
70 Points

First take the system and find the acceleration of system (each blocks)

the accn a = (M1 - M2)g / (M1 +M2)

SO, ACCLN = g/3

now after system is released the bigger block is stopped for a moment after 1 sec, hence after 1 sec ..

the vel V1 of the smaller block = g/3 * 1= g/3 m/sec

now again the system is free .so,all the thread which was winding now gets tightened ..

the string will be tight again when displacement of both particles are equal in magnitude.

So gT/3 -(1/2)gT2 = (1/2)gT2

Solving this we get T = 1/3 sec

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Naga Ramesh

IIT Kgp - 2005 batch

8 years ago
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