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

Mass of block M =200g=0.20kg
Block of the particle m=120gm
=0.12kg

In the equilibrium condition, the spring is stretched by a distance
`x=1.00cm`
`=0.01m`
`rarr 0.2xxg=Kxx x`
`rarr 2=Kxx0.01`
`rarr k=200N/m`
The velocity with which the particle m will strike M is given by
`u=sqrt(2xx10xx0.45)`
`=sqrt9=3m/sec`
So, after the collision, the velocity of the particle and the block is
`V=(0.12xx3)/0.32=9/8 m/sec`
Let the spring be stretched through can extra deflection of `delta`.
`0-(1/2)xx0.32xx(81/64)`
`=0.32xx10xxdelta-(1/2)xx200xx(delta+0.1)^2-(1/2)xx200xx(0.01)^2`
Solving the above equation we can get
`delta=0.045=4.5cm