In the arrangment shown in figure , the mass of the body A is 4 m and that of body B is m . The height `h = 0*2` m . If the body B is released , then till what maximum height the body B will go up ? (g = `10 m//s^(2))`

Explanation

Let the body A go down with acceleration `a_(1)` and the body B go up with acceleration `a_(2)` ( just after release ) , then `a_(2) = 2a_ 2 a_(1) `
`4mg - 2T = 4ma_(1) " " ` …(i)
` T - mg = ma_(2) " " ` ….(ii)
Multiplying equation (ii) by 2 and adding it to equation (i)
` 2mg = 4 ma_(1) + 2ma_(2)`
` a_(1) = (2mg)/(8m) = g/4`
So, `a_(2) = 2a_(1) = g/2 `
The time till the body A reaches the floor, both bodies will be moving with above mentioned accelerations. Just after the body A strikes the floor, all strings will get slack , ie tension in acceleration will be g downwards .
From the constraint reletions , the body B will be at height 0.4 m from floor, when A strikes the floor. As the body B starts from rest , so we can write

`v^(2) = u^(20 + 2as = 0 + 2a_(2) (0.4) `
` v^(2) = 2(g/2) 0.4 = 4 `
or `rArr v = 2 m//s`
After this body B will retard and at highest point velocity of body B will again become zero
so , `v^(2) = u^(2) + 2as`
` 0 = 4-2 ( 10 h) rArr h = 0.2 m `
So height from ground is ` 0.4+ 0.2+ 0.6 m`