Find the acceleration of each block in the figure shown below, in terms of their masses `m_(1), m_(2)` and g. Neglect any friction.

Explanation

Let T be the tension in the string that is assumed to be massless.
For mass `m_(1)`, the FBD shows that `N_(1) = m_(1) g`
Where `N_(1)` is the force applied upward by plane on the mass `m_(1)` .If acceleration of `m_(1)`along horizontal is `a_(1)`, then `T= m_(1) a_(1)` ......(i)

For mass `m_(2)`, the FBD shows that
`m_(2) g - 2T = m_(2) a_(2) ` ...(ii)
Where `a_(2)` is vertical acceleration of mass `m_(2)`.
Note that upward tension on `m_(2)` is 2T applied.

by both sides of the string from constrain relation
`a_(2) = ( a_(1))/( 2)`
Thus, the acceleration of `m_(1)` its twice that `m_(2)`. with this input, solving (i) and (ii) we find
`a_(1) = ( 2 m_(2) g)/( 4m_(1) + m_(2))`
`a_(2) = ( m_(2) g)/( 4m_(1) + m_(2))`