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. the kinetic energy of AB system at maximum compression of the spring is zero.
B. the kinetic energy of the AB system at maximum compression of the spring is `(mv^(2))/(4)`
C. The maximum compression of the spring is `v sqrt(m//k)`
D. the maximum compression of the spring is `v sqrt(m//2k)`

Answer

A, B, D

Explanation

Due to equal mass, C transfers its total momentum of A. Therefore A has the velocity v.At maximum compression the velocity of both the bodies are equal say, v. .
Conservation of linear momentum yields

`mv = (m + m)v.`
`rArr v. = (v//2)`
`:.` The K.E. of the system `= (1)/(2) (m + m)v^(2)`
`=(1)/(2) (2m) (v^(2))/(4) = (mv^(2))/(4)`
`(1)/(2) kx^(2) = Delta KE = (mv^(2))/(2) - (mv^(2))/(4)`
`rArr x= (sqrt((m)/(2k)))v`