A body `A` of mass moving with velocity `v` while passing through its mean position collides in perfect inelastically with a body `B` of same mass which is connected to a vertical wall through a spring whose spring constant is `k`. After collision it sticks to `B` and executes `S.H.M.` Find the amplitude of reulting motion:

Options

A. `sqrt((m)/(v))v`
B. `sqrt((m)/(2k))v`
C. `(m)/(k)sqrt(v)`
D. `(m)/(2k)sqrt(v)`

Answer

B

Explanation

Assuming the collision lests for a small interval. We can apply conservation of momentum and get the common velocity = `(v)/(2) K.E. =(1)/(2) (2m) ((v)/(2))^(2) = (1)/(4)mv^(2)`
This is also the total energy of system as the spring is unstretched at this moment. If the amplitude is `A`, total energy `=(1)/(2) kA^(2) :. (1)/(2) kA^(2) = (1)/(4)mv^(2)`
`:.A = sqrt((m)/(2k)).v`