Three identical blocks A, B and C are placed on horizontal frictionless surface. The blocks B and C are at rest but A is approaching towards B with a speed 10 ms-1.The coefficient of restitution for all collisions is 0.5. The speed of the block C just after collision is approximately

For collision between A and B,              m×10=mvA+mvB⇒vB+vA=10                            …(i)                         e=12=vB-vA10 or vB - vA=5                     …(ii)Solving Eqs. (i) and (ii), we get                    vB = 7.5 ms-1Hence, A has given 75% of its speed to B and B will also transfer its 75% speed to C.∴                     vC=75100×7.5=5.625 ms-1≃5.6 ms-1