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A particle (P) of mass m is released from rest at point A on a curved smooth wedge of mass M,free to move on smooth horizontal surface .Velocity of wedge w.r.t ground is V1 and velocity of particle w.r.t wedge is V2 at,when particle is point B.Momentum conservation equation in horizontal direction can be written as 


Asked by napaliwal 19th March 2018, 6:51 PM
Answered by Expert
There are two questions. First one can be answered only after seeing the appropriate figure related to this question. It is not clear how point-A and point-B are located.
Second question based on conservation of angular momentum : r x p = constant, ie m×v×r = constant
m is mass, v - tangential velocity, r is radial distance
since m is constant, 10×10 = v×5, this gives final tangential speed as 20 m/s

Answered by Expert 21st March 2018, 7:20 AM
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