A block of mass `4kg` is moving with a velocity of `7m//s` on a surface. It collides with another block of mass `3kg` elastically. The surface is smooth for `4kg` block but rough for `3kg` block`(mu=0.4)`. Find the time (in sec) after which next collision will occur.

Explanation

Let after collision velocities are `v_(1)`&`v_(2)`
applying conservation of momentum
`4xx7+0=4v_(1)+3v_(2)`
`4v_(1)+3v_(2)=28`…(1)
also `e=-((v_(2)-v_(1))/(u_(2)-u_(1)))`
`1=-((v_(2)-v_(1))/(0-7))`
`v_(2)-v_(1)=7`...(ii)
solving (1)&(2) `rArr" "v_(2)=8m//s`
`v_(1)=1m//s`
time after `2^(nd)` block stops `=(u)/(a)=(8)/(4)=2sec`
distance travelled by `2^(nd)` block till this moment `s=8t+(1)/(2)at^(2)`
`s=8xx2-(1)/(2)4xx2^(2)=8m`
so time elasped till `2^(nd)` collision `=(8)/(1)=8sec.`