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 when they come to rest. (Take `g=10m//s^2`).

Answer

C

Explanation

Retardation, `a=(mumg)/(m)=mug=0.2xx10=2m//s^2`
Velocity of first block before collision,
`v_1^2=1^2-2(2)xx0.16=1-0.64`
`v_1=0.6m//s`
By conservation of momentum, `2xx0.6=2v_1^'+4v_2^'`
also `v_2^'-v_1^'=v_1` for elastic collision
It gives `v_2^'=0.4m//s`, `v_1^'=-0.2m//s`
Now distance moved after collision
`s_1=((0.4)^2)/(2xx2)`
and `s_2=((0.2)^2)/(2xx2)`
`s=s_1+s_2=0.05m=5cm`