(a)
   let u₁ and u₂i be the velocities of m₁ and m₂ just before the collision then from the
   law of conservation of energy we get            
   u₁ = - u₂
        = (2 g h)^{(1 / 2)}
        = …... m / s
(b)
   the velocites of the two blocks after collision be v₁, v₂
   then from the law of conservation of momentum we get
   (2.00 kg) v₁ + (3.60 kg) v₂ = (2.00 kg) u₁ + (3.60 kg) (- u₂)
   but for elastic head on collision we have
   v₁ + u₁ = v₂ + u₂
   on solving the above two equations we get
   v₁ = ........... m / s
   v₂ = ........... m / s
(c)
   the maximum heights to which m₁ reaches after collision is
   h₁ = v₁² / 2 g
       = .......... m
   h₂ = v₂² / 2 g
       = .......... m