The system of mass `A` and `B` shown in the figure is released from rest with `x=0`, determine the velocity of mass `B` when `x=3m`. Also find the maximum displacement of mass `B`.

Explanation

Constraint relation gives
`2sqrt(x^(2)+16)+y=c`
Differentiating w.r.t. time, we get
`(dy)/(dt)= -(x)/(sqrt(x^(2)+dt))(dx)/(dt)`
Let `(dx)/(dt)=v=` velocity of `B`
`:. (dy)/(dt)=` velocity of `A=-(x)/(sqrt(x^(2)+16))v`
Minus sign indicates that it in upward direction.
`:.` Using energy conservation
`mg.3-(mg)/(sqrt(2)).2=(1)/(2)mv^(2)+(1)/(2)(m)/(sqrt(2))(v.(3)/(5))^(2)`
`rArr (3-sqrt(2))g=(v^(2)(25sqrt(2)+9))/(50sqrt(2))`
`rArr v=sqrt(((3-sqrt(2))(50sqrt(2)))/(25sqrt(2)+9))g=5 m//s`
Let `x_(1)` be the maximum displacement of `B`, then
`mgx_(1)-(mg)/(sqrt(2))2[sqrt(x_(1)^(2)+16)-4]=0`
`rArr x_(1)=8sqrt(2)m`.