Question

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

A. `sqrt((60mg^(2))/(k))`
B. `sqrt((60mg^(2))/(k))`
C. `sqrt((10mg^(2))/(k))`
D. None of these

Answer 1

Correct Answer - B
`(B)` The initial extension in 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) (` sec. fig. `)` and comes to rest.

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