three identical blocks A , B and C are placed on horizontal frictionless surface. THE blocks B and C are at rest block A is approaching towards B with speed of 10m/s .If the coefficient of restitution for all collisions is 0.5 then the speed of the block C just after collision is approximately(1) 5.6m/s (b) 6.4 m/s © 3.2 m/s (4) 4.6m/s

Collision is not perfectly elastic so there is some loss in speedmA= mB=mC = mAcc. To linear momentumuA = vA + vB10 = vA + vB ........ (1)Nowe = vB - vA / uA0.5uA = vB- vA ........(2)From 1 & 22vB = 15vB = 15/2Now collision between B & CuB = vB+ vC 15/2 = vB + vC .........(3)Ore = vC - vB /uB uB*0.5= vC - vB........(4)From 3 & 42vC = 15/2 + 0.5*15/2On solving the equvC = 225/40vC = 5.6 m/sAns.