Two blocks `A` and `B`. each of mass `m`, are connected by a massless spring of natural length `I`. and spring constant `K`. The blocks are initially resting in 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

Options

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

Answer

A

Explanation

After striking with A, the block C comes to rest and block A moves with velocity v, when compression in spring is maximum, both A and B will be moving with common velocity v,

From law of conservation of energy,
KE of block C = KE of system + PE of system
`1/2mv^(2) = 1/2(2m)V^(2) + 1/2kx^(2)`
`rArr 1/2mv^(2) = 1/2(2m)(v/2)^(2) + 1/2kx^(2)`
`rArr kx^(2) = 1/2mv^(2)`
`rArr x = vsqrt(m/(2k)`