In the arrangement shown, find the acceleration of each block, tension in the string and reaction in the pulley. The string and pulley are light (massless) `(m_(1)gtm_(2))`. Neglect the friction in pulley.

Explanation

Since `m_(2)gtm_(1)`, the block of mass `m_(1)` will move downward and the block of mass `m_(2)`, upward.


`darr:m_(1)g-T=m_(1)a` (i)
`uarr:T-m_(2)g=m_(2)a` (ii)
Solving (i) and (ii), we get
`a=((m_(1)-m_(2))g)/((m_(1)+m_(2)))`
`T=(2m_(1)m_(2)g)/(m_(1)+m_(2))`

`R-2T=0` [since the pulley is massless]
where `R` is the reaction in the pulley.