Three blocks A, B, and C of masses `3M, 2M`, and M are suspended vertically with the help of spring PQ and TU, and a string RS as shown in fig. If the acceleration of blocks A, B and C is `a_(1), a_(2)` and `a_(3)` , respectively, then

The value of acceleration `a_(3)` at the moment spring PQ is cut is

Explanation

The system is in equlibrium :
`kx_(3)=mg`..(i)

form fig.
`2mg+kx_(3)=kx_(1)` …(ii)
form fig
`:. 3mg=kx_(1)`
(a) When the spring between ceiling and block is cut, then the elongation of spring between B and C remains same just after cutting

`:. a_(c)=0 (kx_(3)=mg)`
For `(A+B)`
`kx_(3)+2mg=3mg`
`:. 3mg=2ma`
`:. a=3/2 g = 15 ms^(-2)`
`:. a_(A)=a_(B)=15 ms^(-2)`
For tension (from FBD of B),

`mg+kx_(3)-T=ma_(B)`
`mg+mg-T=(3mg)/(2)`
`:. T=(mg)/(2)`
(b) When string between A and B is cut, the elongation in spring do not change just after cutting the string.

`mg-kx_(1)=ma_(A)`
`mg-3mg=ma_(A)` `{kx_(1)=3mg}`
`-2mg=ma_(A)`
`a_(A)=-2g`
`:. a_(A)=2g` (upwards)
For B

`mg+kx_(3)=ma_(B)`
`mg+mg=ma_(B)` {`:. kx_(3)=mg`}
`:. a_(A)=2g`(downward)
for C,
`mg-kx_(3)=ma_(c)`
or `mg-mg=ma_(c)` `{kx_(3)=mg}`
`a_(c) =0`
`T=0`
(c ) when spring between B and C is cut.
for C, `mg=ma_(c)`
`:. a_(c)=g` (downward)

The acceleration of A and B will be equal, taking `"A+B"` together as system
`2mg-kx_(1)=2ma_(B)` {`:. a_(A)=a_(B)`}
`2mg-3mg=2ma_(B)` {`:. kx_(1)=3mg`}
`:. a_(B)=-(g)/(2)`
`:. a_(A)=a_(B)=g/2` (upward)

For tension in string between A and B, considering the FBD of B only

`T-(mg)=ma_(B)`
`T=mg+(mg)/(2)=(3mg)/(2)`.