In the adjoining figure, block A is of mass (m) and block B is of mass 2m. The spring has force constant k. All the surfaces are smooth and the system is released form rest with spring unstreched.
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Answer

A, B, C, D

Explanation

(a) At maximum extension in the spring
`v_(A) = v_(B) =0` Therefore, appluing conservation of mechanical energy:
(momentarilt)
Therefore, appluing conservation of mechanical energy:
decreasing in gravitational energy of block B = incrasing in elastic potential
enerrgy of spring .
or `m_(B)gx_(M) =1/2Kx_(m)^(2)`
or `2mgx_(m) =1/2Kx_(m)^(2)`
:. `x_(m) =(4mg)/K`
(b) At `x=x_(m)/2 =(2mg)/K`
Let `v_(A) =v_(B) =v(say)`
The decrase in gravitational potential energy of block (B) =increase in elastic potential energy of spring + increase in kinetic inergy of both the blocks.
`:. m_(B) gx =1/2Kx^(2) +1/2(m_(A) + m_(B)) v^(2)`
or `(2m) (g) ((2mg)/k) = 1/2K ((2mg)/K)^(2) + 1/2(m+2m)v_(2)`
`:. v = 2gsqrtm/(2K)`
(c) At `x=x_(m/4)=(mg)/K`
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or `Kx=mg`
`a=(Net pulling force)/(Total mass) =(2mg-mg)/3m`
` =g/3`(downwards).