A block of mass `180 g` is placed on a spring (spring constant `k = 120 N//m`) fixed from below. A ball of mass `20 g` is dropped from height `20 m` and the collision is completely inelastic. Find the maximum compression of the spring. Neglect the initial compression of the spring due to the block.

Explanation

Here, `m = 20 g , M = 180 g , h = 2m , k = 120 N//m`
The velocity of ball after falling `h`,
`v_(0) = sqrt(2gh) = sqrt( 2 xx 10 xx 20) = 20 m//s`
Applying momentum conservation between `(B)` and `(C )` ltbrlt `mv_(0) = (m + M)v rArr 20 xx 20 = (20 + 180)v rArr v = 2 m//s`
Let the maximum compression of the spring be `x`.
Applying the energy conservation between `(C )` and `(D)`
`(1)/(2) (m + M)v^(2) + (m + M) gx = (1)/(2) kx^(2)`
`(1)/(2) xx 0.2 xx (2)^(2) + 0.2 xx 10 x = (1)/(2) xx 120 x^(2)`
`60 x^(2) - 2x - 0.4 = 0`
`x = (2 +- sqrt(4 + 4 xx 60 xx 0.4))/(120) = ( 2 +- 10)/(120)`
`x = (12)/(120) = 0.1 m = 10 cm`