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bharath devasani Grade: 12
        A rod AB of mass M and length L is lying on a horizontal frictional surface. A particle of mass m travelling along the surface hits the end 'A' of the rod with a velocity v in a direction perpendicular to AB and comes to rest. The collision is completely elastic.

a)Find the ratio m/M.
b)A point on the rod is at rest immediately after the collision.Find the distance AP.
7 years ago

Answers : (1)

Askiitians Expert Soumyajit IIT-Kharagpur
28 Points
										

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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Regards,


Askiitians Experts
Soumyajit Das IIT Kharagpur

7 years ago
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