The friction coefficient between the horizontal surfce and ech of the blocks shown in figure is 0.20. The collision between the blocks is perfectly elastic. Find the separation between the two blocks when they come to rest. Take `g=10 m/s^2`.

Explanation

Let velocilty of 2 kg block ion reaching the 4 kg block before collision `n=u_1`
Given `V_2=0` (velocity of 4 kg block)
`:.(1/2)mxxu_1^2-(1/2)mx(1)^2=-mxxmgxxS`
`rarr u_1=sqrt((1)^2-2xx0.20xx10xx0.16)`
`rarr =0.6m/sec`
`rarr u_1=0.6m/s`
`since it is a perfectly elastic collision.
Let `v_1,v_2rarr `velocity of 2 kg and 4 kg block after collision.
`m_1xxu_1+m_2u_2=m_1v_1+m_2v_2`
`rarr 2xx0.6xx4.0=2v_1+4v_2`
`2v_1+4v_2=1.2`.......i
`Again `v_1-V_2=+(u_1-u_2)`
`=+(0.6-0)`
`=-0.6` ...ii
Substractig ii from i
`3v_2=1.2`
`rarr v_2=0.4m/s`
`:. v_1=-0.6+0.4`
`=-0.2m/s`
`:.` Putting work energy principle for 1st 2kg block when comes to rest.
`(1/2)xx2xx(0)^2+(1/2)xx2xx(0.2)^2``
`=-2xx0.2xx10xxS_1`
`rarr S_1=1cm`
Putting work energy principle for 4 kg block
`(1/2)xx4(0)^2-(1/2)xx4xx(0.4)^2`
`=-4xx0.2xx10xxS_1`
`rarr 2xx0.4xx0.4=4xx0.2xx10xxS_2`
`rarr S_2=4cm`
distnce between 2kg and 4 kg block
`=S_1+S_2=1+4=5cm`