Block `A` is hanging from vertical spring of spring constant `K` and is rest. Block `B` strikes block `A` with velocity `v` and sticks to it. Then the value of `v` for which the spring just attains natural length is

Options

A. (a) `sqrt((60mg^2)/(k))`
B. (b) `sqrt((6mg^2)/(k))`
C. (c) `sqrt((10mg^2)/(k))`
D. (d) None of the above

Answer

B

Explanation

The initial extension in spring is `x_0=mg//k`
Just after collision of B with A, the speed of the combined mass is `v//2`. For the spring to just attain natural length, the combined mass must rise up by `x_0=mg//k` and come to rest.

Applying conservation of energy between initial and final states,
`1/22m(v/2)^2+1/2k((mg)/(k))^2=2mg((mg)/(k))`
Solving, we get `v=sqrt((6mg^2)/(k))`