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

Options

A. the `K.E.` of the `A-B` system , at the maximum compression of the spring is zero
B. the `K.E.` of the `A-B` system at maximum compression of the spring is `mupsilon^(2)//4`
C. the maximum compression of the spring is `upsilonsqrt(m//k)`
D. the maximum compression of the spring is `upsilonsqrt(m//2k)`

Answer

B, D

Explanation

If `upsilon^(')` is the velocity acquired by `A` and `B` , then using law of conservation of linear momentum, we have `m upsilon=m upsilon^(')+m upsilon^(')` or `upsilon^(')=upsilon//2.` .According to law of conservation of total energy, we have
`(1)/(2)m upsilon^(2)=(1)/(2)m upsilon'^(2)+(1)/(2)m upsilon'^(2)=(1)/(2)kx^(2)`
where `x` is the displacement , we have
`x=upsilonsqrt(m//2k)`
At maximum compression, K.E. of `A-B` system will be
`=(1)/(2)m upsilon'^(2)+(1)/(2)m upsilon'^(2)`
`=m upsilon'^(2)=(m upsilon^(2))/(4)`