Three blocks arranged with pullwy and spring as shown in fig. If the string connecting blocks `m_(2)` and `m_(3)` is cut at point A, find the acceleration of masses `m_(1), m_(2)` and `m_(3)`. Just after the string is cut at point A.

Explanation

Let us analyzes the system at equilibrium. The forces acting on the blocks are shown in fig.

Just after cutting the string at A, the tensions T and T' will be zero. But the spring force will remain unchanged just after cutting. the figure shows the forces acting on the blocks just after cutting the string.
The forces acting on `m_(1)` just before and just after cutting the string are same and just before cutting string `m_(1)` were in equilibrium. Hence, the acceleration of `m_(1)` will be zero just after cutting the string.
At equilibrium, `F_(S)=(m_1)g`
As `F_(S)` does not change just before and just after cutting the string, acceleration of `m_(2)`
`a_(2)=(m_(2)g+F_(S))/(m_2) = (m_(2)g+m_(1)g)/(m_2) = (1+(m_(1))/(m_2))g`
From the free-body diagram of block `m_(3)` just after cutting the string, only `m_(3)g` will act and tensions in the string will disappear. Hence, acceleration of `m_(3)`.
`a_(3)=(m_(3)g)/(m_3) = g`
Finally, acceleration of `m_(1)=0`
Acceleration of `m_(2)=(1+(m_1)/(m_2))g` and acceleration of `m_(3)=g`.