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Dear Bharath Devasani,
Ans:-From the law of conservation of linear momentum we have
mv=Mu .................(1)
where u velocity of the COM of the rod
Taking the law of conservation of angular momentum about the center we get,
mvL/2=1/12 ML²w Where w is the angular velocity of the rod.;
Hence wL=6mv/M...........(2)
Again for elastic collision, the relative velocity before and after collision will be the same.
Hence v=u+wL/2 ..............(3)
Solving I , 2 ,3 we get,
m:M=1:4
Again we know that the point which will be at rest immediately after the collision will be that point for which
v=wx
where x is the distance of that point from the point A
Putting this value in eq 2 and using the ratio m:M we get
x=2L/3(ans)
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All the best Bharat Devasani !!!
Regards,
Askiitians ExpertsSoumyajit Das IIT Kharagpur
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