Three identical blocks ,4, B and C are placed on horizontal friction less surface as shown in fig. (+). Theblocks B and C are at rest. The block A is approaching towards B with a speed t0 m,/s. The coefficient of restitution for all collisions is 0 . 5. The speed of block C just after collision is

For collision between A and Be=vB−vAuA−uB=vB−vA10−0=vB−vA10 ∴ VB−VA=10e=10×0⋅5=5 Applying the ]aw of conservation of momentum mAuA+mBuB=mAvA+mBvB m×10+0=mvA+mvB ∴ vA+vB=10 From eqs. (1) and (2), we get VB=7⋅5m/s For collision between B and C, we have vC−vB=7⋅5e=7⋅5×0⋅5=3⋅75 VC+VB=7⋅5 Solving eqs. (3) and (4), we get vC=5⋅6m/s