In the arrangement shown in fig the bodies have masses `m_(0), m_(1) and m_(2)` the friction is absent the masses of pulleys and the threads are negligible find the acceleration of the body `m_(1)`

Explanation

Let the acceleration of `m_(0)` with respect to the table is `a_(0)` and that of `m_(1)` and `m_(2)` is a in magnitude with respect to the movable pulley
Suppose `m_(1)` moves upward and `m_(2)` moves downward then their accelerations with respect to the table will be:
`a_(1)=a-a_(0)` upward
and `a_(2)=a+a_(0)` downward
By Newton's second law we have
`T_(0)=m_(0)a_(0)` ....(i)
`T-m_(1)g=m_(1)(a-a_(0))` ....(ii)
and `m_(2)g-T=m_(2)(a+a_(0))` ....(iii)
For movable pulley
`T_(0)-2T=0xxa_(0)` ....(iv)
Solving above equations for a, we get
`a=(4m_(1)m_(2)+m_(0)(m_(1)-m_(2))g)/(4m_(1)m_(2)+m_(0)(m_(1)+m_(2)))`