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((6mg^(2))/K)`
B. `sqrt((3mg^(2))/K )`
C. `sqrt((12mg^(2))/K )`
D. `2sqrt((mg^(2))/K)`

Answer

A

Explanation

Since the block is hanging in equilibrium hence extension of the spring is mg/K .
We can neglect finite forces like gravity at the time of collision and we can apply conaservation of linear momentum just before and just after the collision . Let v is the velocity of the block plus particle just after collision then
`"mu" = (m+m) v rArr v = u/ 2 `
Now the block and particle move upward and comes to rest when spring attains natural length . We can apply conservation of mechanical energy .
Loss in K.E + Loss in spring potential energy = gain in gravitational potentail energy .
` 1/.2 (2m) (u/2)^(2) +1/2K ((mg)/K)^(2) = (2m) g ((mg)/K)`
`rArr " " ("mu"^(2))/2 + (m^(2)g^(2))/K = 4 (m^(2)g^(2))/K rArr (u^(2))/2 = 3 (mg^(2))/K `
`rArr " " u = sqrt((6mg^(2))/K)`