Three identical blocks A, B and C are at rest. But A is approaching towards B with a speed 10m/s. The coefficient of restitution for all collisions is 0.5. The speed of the block C just after collision is approximately

When A collide with B,m×10=mvA+mvB .....(i)e=12=vB−vA10vB−vA=5      ........(ii)Solving equation (I) and (ii), we getvB=7.5 m/si.e., A has transferred 75% of its speed to B. Similarly, B will also transfer 75% of its speed to C.vc=75100×7.5=5.625 m/s