A wedge (free to move) of mass M has one face making an angle α with horizontal and is resting on a smooth rigid floor. A particle of mass m hits the inclined face of the wedge with a horizontal velocity v0. It is observed that the particle rebounds in vertical direction after impact. Neglect friction between particle and the wedge and take M = 2m, v0 = 10 ms-1, tan α = 2, g = 10 ms-2. Calculate the coefficient of restitution for the impact.

Along common tangent direction, the speed of particle remains unchanged. So, we have      v0cos⁡α=vsin⁡α⇒v=v0cot⁡α  ………(1)Also, we can conserve linear momentum of the' particle + wedge' system along horizontal direction. So,mv0+0=2mV⇒ V=v02   ……(2)Coefficient of restitutione=( Relative velocity of separation )n-line ( Relative velocity of approach )n-line ⇒ e=Vsin⁡α+vcos⁡αv0sin⁡α=Vv0+vv0cot⁡αUsing (1) and (2), we get  e=12+cot2⁡αGiven that tan α=2⇒ cot⁡α=12⇒ e=12+14=34=0.75