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`        Mass of a pendulam is 10-2 kg. It is lifted to 5×10-2 m and released. At the lowest point of its path an extra mass of 10-3  kg  is attached with the pendulam. Find the height travelled by the pendulam At the opposite side.`
7 years ago

28 Points
```										Dear Subhajit,
Ans:- The velocity at the bottommost point is=√(2*10*0.05) m/s=1m/s
From the conservation of momentum we get the final velocity at the bottommost point after the extra mass is hanged is
=10-²*1/(10-²+10-³)
=0.91 m/s
Let the max height is =H m
Then H=0.91²/(2*10)
=4.1×10-² m
Please feel free to post as many doubts on our discussion 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 Subhajit !!!

Regards,
SOUMYAJIT IIT_KHARAGPUR
```
7 years ago
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