A mass `m_(1)` is connected by a weghtless cable of water, whose mass is `m_(0)` at `t=0`. If the container release water in downward direction at constant rate b kg/s. with a velocity `v_(0)` relative to the container, determine the acceleration of `m_(1)` as a function of time.

Answer

`((m_(1)-m_(0)+kt)g+kv_(0))/((m_(1)+m_(0)-kt))`

Explanation

Writing equations:
`m_(1)g-T=m_(1)a` ...(i)
`T-mg=ma` ....(ii)
Adding equations (i) and (ii)
`g(m_(1)-m)=(m_(1)+m)a` ....(iii)
As `-(dm)/(dt)=k`
`-underset(m_(0))overset(m)intdm=underset(0)overset(t)intkdt`
`m_(0)-m=kt+C`
`m_(0)-kt=m`
By putting value of m in equation (iii) we get acceleration