5 h 16 h 4 h 6.67 h 9 h Solution As the balls fall from a vertical distance of h

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As the balls fall from a vertical distance of h meter :- =>their final velocity at the ground before the strike :- v(i) = sqrt[2gh] -----------(i) =>KE(initial) before the collision = 1/2Mv^2 + 1/2mv^2 = 1/2 x 2gh x (M+m) = (M+m)gh ----(ii) As the steel ball shoots to a height of H meter it's velocity after collision:- =>v(f) = sqrt[2gH] KE(final) after the collision = 1/2 x m x 2gH = mgH As all the collisions are elastic in nature =>KE(initial) = KE(final) =>(M+m) gh = mgH =>H = [(M+m)/m] x h ----------------(X) is the required relation.

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