The system of two weights with masses `m_(1)` and `m_(2)` is connected withh a weightless spring as shown. The system is resting on support `S`. The support `S` is quickly removed. The acceleration of each of the weights right after the support `S` is removed is (`a_(1)` is acceleration of `m_(1)` and `a_(2)` that of `m_(2)`)

Options

A. `a_(1)=0, a_(2)=((m_(1)+m_(2))g)/(m_(2))`
B. `a_(1)=0,a_(2)=((m_(1)+m_(2))g)/(m_(1))`
C. `a_(1)=((m_(1)+m_(2))g)/(m_(1)),a_(2)=0`
D. `a_(1)=0, ((m_(1)+m_(2))g)/(m_(1),a_(2))=0`

Answer

A

Explanation

Initially `m_(1)g=kx`

When the support is removed, spring force does not change.
New FBD,

For `m_(1)g-kg=m_(1)a_(1)impliesa_(1)=0`
For `m_(2):m_(2)g+kx=m_(2)a_(2)impliesa_(2)=((m_(1)+m_(2))g)/(m_(2))`