Two identical blocks `A` and `B` , each of mass `m` resting on smooth floor are connected by a light spring of natural length `L` and spring constant `k`, with the spring at its natural length. A third identical block `C` (mass `m`) moving with a speed `v` along the line joining `A` and `B` collides with `A`. The maximum compression in the spring is

Options

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

Answer

A

Explanation

Initial momentum of the system (block C) = mv
After striking with A, the block C comes to rest and now both block A and B moves with velocity V, when compression in spring is maximum. By the law of conservation of linear momentum
`mv=(m+m)V rArr V = (v)/(2)`
By the law of conservation of energy
K.E. of block C = K.E. of system + P.E. of system
`(1)/(2)mv^(2)=(1)/(2)(2m)V^(2)+(1)/(2)kx^(2)`
`rArr (1)/(2)mv^(2)=(1)/(2)(2m)((v)/(2))^(2)+(1)/(2)kx^(2)`
`rArr kx^(2)=(1)/(2)mv^(2)`
`rArr x = v sqrt((m)/(2k))`