A block at rest is suddenly burst into two pieces of mass `2 kg` and `4 kg` , respectively . The pieces fly apart with a relative of `30 m//s`, determine the speed of each. If the coefficient of friction for surface is `0.5` , find the separation between pieces when they come at rest.

Explanation

By the momentum conservation
`0 = 2v_(1) - 4v_(2)`
`v_(1) = 2v_(2)` (i)
Given relative velocity
`v_(1) + v_(2) = 30` (ii)
Solving , `v_(2) = 10 m//s , v_(1) = 20 m//s`
Retardation on the rough surface
`a = (mu mg)/(m) = mu g = 0.5 xx 10 = 5 m//s^(2)`
For `2 kg : v^(2) = u^(2) - 2as`
`0 = v_(1)^(2) - 2 mu gd_(1)`
`rArr d_(1) = (v_(1)^(2))/(2 mu g) = ((20)^(2))/(10) = 40 m`
For `4 kg : 0 = v_(2)^(2) - 2 mu g d_(2)`
`rArr d_(2) = (v_(2)^(2))/(2 mu g) = ((10)^(2))/(10) = 10 m`
Separation between pieces `= d_(1) + d_(2) = 50 m`