Two blocks A and B, each of mass m, are connected by a masslesss spring of natural length L and spring constant K. The blocks are initially resting on a smooth horizontal floor with the spring at its natural length, as shown in fig. A third identical block C, also of mass m, moves on the floor with a speed v along the line joining A and B, and collides elastically with A. Then, maximum compression of the spring is

Options

A. `v sqrt ((m)/(2k))`
B. `m sqrt ((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 blocl `A and B` moves with velocity momentum `V`, when compression in spring is maximum .
By the law of the conservation of linear momentum
`mv = (m + m) V implies V = (v)/(2)`
By the law conservation of energy
`KE`, of block C = K.E, of system + `PE` of system
` (1)/(2) mv^(2) =(1)/(2) (2m) V^(2) =(1)/(2) kx^(2)`
`(1)/(2) mv^(2) = (1)/(2) (2m) ((v)/(2))^(2) + (1)/(2) kx^(2)`
`implies kx^(2) = (1)/(2) mv^(2)`
` implies x = v sqrt((m)/(2k))`.