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Block A is hanging from vertical spring ...

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

Text Solution

Verified by Experts

The correct Answer is:
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))`
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