A block of mass `(M = 2m)` is placed on smooth surface. Another block of mass `m` is given velocity `sqrt(9 gh)` in horizontal direction , as shown. Find the speed of `m` when it breaks off from `M`. Also , calculate the maximum height attained by it from its initial level. Take `h = 2m`.

Explanation

Applying the momentum and energy conservation
`mu = (m + M)v = 3mv rArr v = (u)/(3) = (1)/(3) sqrt(9 gh) = sqrt(gh)`
`(1)/(2) mu^(2) = (1)/(2) mv^(2) + (1)/(2) mv_(1)^(2) + (1)/(2) Mv^(2) + mgh`
`(1)/(2) m xx 9gh = (1)/(2) mgh + (1)/(2) mv_(1)^(2) + (1)/(2) xx 2 mgh + mgh`
`(1)/(2) mv_(1)^(2) = 2 mgh rArr v_(1) = sqrt( 4 gh)`
Speed of `m` , when it breaks off
`v_(0) = sqrt(v_(1)^(2) + v^(2)) = sqrt( 5 gh) = sqrt( 5 xx 10 xx 2) = 10 m//s`
Now the block of mass `m` will move as a free particle on a parabolic path.
`0 = v_(1)^(2) - 2 gh_(1)`
`h_(1) = (v_(1)^(2))/(2g) = 2h = 4m`
The maximum height attained by the smaller block from its initial level is
`h + h_(1) = 2 + 4 = 6m`