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,

$\begin{array}{l}m\times 10=m{v}_A+m{v}_B\mathrm{.....}\left(i\right)\\ e=\frac{1}{2}=\frac{v_B-v_A}{10}\\ v_B-v_A=5\mathrm{........}\left(ii\right)\end{array}$

Solving equation (I) and (ii), we get

$v_B=7.5m/s$

i.e., A has transferred 75% of its speed to B. Similarly, B will also transfer 75% of its speed to C.

$v_c=\frac{75}{100}\times 7.5=5.625m/s$