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Grade: 12

                        

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.

11 years ago

Answers : (1)

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
--
Please feel free to post as many doubts on our disucssion forum as you can. If you find any question difficult to understand - post it here and
we will get you the answer and detailed solution very quickly.We are all IITians and here to help you in your IIT JEE preparation. All the best.

Regards,
Naga Ramesh
IIT Kgp - 2005 batch

11 years ago
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