In the arrangement shown in fig neglect the masses of pulleys and string and also friction Calculate acceleration of blocks A and B

Explanation

Since string is same throughout, so the tension in it will be same everywhere Let `a_(1)` and `a_(2)` are the accelerations of blocks A and B respectively and T is the tension in the string from FBD
For block A, `m_(1)g-T=m_(1)a_(1)` …(i)
For block B, `2T-m_(2)g=m_(2)a_(2)` ....(ii)
and for pulley inside dotted box

`2T-T=0`
`rArr T=0`
:. From Eqns (i) and (ii) we get
`a_(1)=g` and `a_(2)=g`