A block of mass `1 kg` is horizontal thrown with a velocity of `10 ms^(-1)` on a stationary long of mass `2 kg` whose surface has a `mu = 0.5`. Plane rest on friction surface find the time when `m_(1)` comes to rest w.r.t. plane

Explanation

Acceleration of block `a_(1) = (mu m_(1)g )/(m_(1)) = mu g = 0.5 xx 10 = 5ms^(-2)`
(toward left)
Acceleration of block `a_(2) = (mu m_(1)g )/(m_(2)) = ( 0.5 xx 1 xx 10)/(2) = (5)/(2) m//s^(-2)`
(toward right)
The block comes to rest with respect block when the velocity of both and plane becomes equal
`v_(1) = v_(2) rArr v - a_(1) t = a_(2) t or 10 - 5 xx t = (5)/(2) t `
Which gives ` t = (4)/(3) sec`