The friction coefficient between the horizontal surface and each of the block shown in the figure is 0.2. The collision between the blocks is perfectly elastic. Find the separation between them (in cm) when they come to rest.  $(Takeg=10m/s^2)$

Velocity  of first  block before  collision
$v_1^2=1^2-2\left(2\right)\times 0.16=1-0.64$;   $v_1=0.6$
By  conservation of momentum  
$2\times 0.6=2v_1^{\text{'}}+4v_2^{\text{'}}$
also $v_2^{\text{'}}-v_1^{\text{'}}=v_1$  or elastic collision
It gives $v_2^{\text{'}}=0.4m/s$ ;  $v_1^{\text{'}}=-0.2m/s$
Now  distance moved  after collision $s_2=\frac{{\left(0.4\right)}^2}{2\times 2}$ and  $s_1=\frac{{\left(0.2\right)}^2}{2\times 2}$
 $s=s_1+s_2=0.05 m=5cm$