A block of mass `180 g` is suspended by a massless spring. The spring extends by `1.8 cm` due to the weight of block. A particle of mass `20 g` is dropped from a height `80 cm` on the block. The collision is completely inelastic . Find the maximum elongation of the spring.

Explanation

Here `m = 20 g , M = 180 g , h = 80 cm`
First , the spring extends by `x_(0)` due to the weight of the block
`Mg = kx_(0) rArr (180)/(1000) xx 10 = k xx (1.8)/(100) rArr k = 100 N//m`
Velocity of particle after falling `h`,
`v_(0) = sqrt(2 gh) = sqrt(2 xx 10 xx 0.8) = 4 m//s`
Applying the momentum conservation between `(C )` and `(D)`
`mv_(0) = (m + M)v rArr 20 xx 4 = (20 + 180)v rArr v = 0.4 m//s`
Applying the energy conservation between `(D)` and `(E)`
`(1)/(2) (m + M)v^(2) + (1)/(2) kx_(0)^(2) + (m + M) gx = (1)/(2) k(x + x_(0))`
`(1)/(2) xx 0.2 xx (0.4)^(2) + (1)/(2) xx 100 xx ((1.8)/(100))^(2) + 0.2 xx 10 x`
`= (1)/(2) xx 100 (x + (1.8)/(100))^(2)`
`0.016 + (1)/(2) xx 100 xx ((1.8)/(100))^(2) + (1)/(2) xx 100 xx 2x xx (1.8)/(100)`
`0.016 + 2x = 50 x^(2) + 1.8 x`
`50 x^(2) - 0.2 x - 0.016 = 0`
`x = (0.2 +- sqrt((0.2)^(2) + 4 xx 50 xx 0.016))/(100) = (0.2 +- 1.8)/(100)`
`x = (1)/(100) m = 2 cm`
Maximum extension of spring `= x_(0) + x = 3.8 cm`