Block `A` is hanging from a vertical spring and is at rest. Block `B` strikes the 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. `sqrt((60mg^(2))/k)`
B. `sqrt((6mg^(2))/k)`
C. `sqrt((10mg^(2))/k)`
D. none of these

Answer

C

Explanation

The initial extension spring is `x_(0)=(mg)/k`. Just after collision of `B` with `A` the speed of combined mass is `v//2`. For the spring to just attain natural length the combined mass must rise up by `x_(0)=(mg)/k`(see figure) and comes to rest.

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