 # 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. Askiitians Expert Soumyajit IIT-Kharagpur
28 Points
12 years ago

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)

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 Bharat Devasani !!!

Regards,