A sphere of mass m held at a height `2R` between a wedge of same m and a rigid wall, is released from, Assuming that all the surfaces are frictionless. Find the speed of the bodies when the sphere hits the ground.
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Explanation

If speed of sphere is v downwards then speed of wedge at this instant will be v cot `alpha` in horizontal direction.
Now,
Decrease in potention energy of sphere=increase in kinetic energy of both
`:. mgR=1/2 mv^(2) + 1/2 (v cos alpha)^(2)`
`=1/2mv^(2) alpha`
`:. v=sqrt(2gR) sin alpha=`speed of sphere
and speed of wedge `=v cot alpha`
`=sqrt(2gR) cos alpha`.