A rigid body of mass m held at height H on two smooth wedges of mass M which are at horiz. frictionless floor. On releasing the body it moves down. The velcoity of recede of the wedges from each other when rigid body is at height h from the ground is   .

 can anyone show me the steps to find the veolcity.. RATES ASSUREd

sorry yarr no nickels

hc verma part 1:

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You missed an important part of the question bro. You forgot to include the angle of the wedge which have to be 45 degree to get your answer.

Initial energy is  mgH

Final Kinetic energy at an intermediate height h is : 1/2Mv² + 1/2Mv² + mgh + 1/2m(v₁)²

The velocity of two wedges are same by symmetry and i considered it to be v.

The velocity of the rod (v₁)  have to be found by geometry. As the angle of the wedge is 45 degree, v₁ will also be v. As the angle of the wedge is 45, the velocity of the rod is also v by common sense. But if the angle is not 45 , eg 30 i am not  able to find the velocity of the rod by geometry. Please tell me how to do it.  

Therefore as v₁ = v .

1/2Mv² + 1/2Mv² + mgh + 1/2m(v)² = Mv² + mgh + 1/2mv² = Final energy

By energy conservation (initial energy = final energy )

mgH = Mv² + mgh + 1/2mv²

mg(H-h) = v² (2M + m)/2

v = ( 2mg(H-h) / (2M + m) )^1/2

Velocity of recede of the two wedges is 2v as both are going in opposite direction.

Therefore, answer = 2v =  ( 8mg(H-h) / (2M + m) )^1/2

Hence the answer is proved. Wheres my rate bro ?