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# 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.

28 Points
11 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)

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