The system of two weights with masses `m_(1)` and `m_(2)` are connected with weightless spring as shown. The system is resting on the support S. the support S is quickly removed. Find the accelerations of each of the weights right after the support S is removed.

Options

A. `a_(1) = ms^(-2), a_(2) = 2g ms^(-2)`
B. `a_(1) = g//2 ms^(-2), a_(2) = 1/3 g ms^(-2)`
C. `a_(1)=0, a_(2) = 3/2 gms^(2)`
D. `a_(1) = 2/3 g ms^(-2), a_(2)=0`

Answer

C

Explanation

`F_(sp) = m_(1)g`
now if N = 0 then it will have no effect on mx instantly so `a_(1)=0` but `m_2` will exp. net downward force so `(m_(2g) + F_(sp)) = m_(2)a_(2) (m_(1) + m_(2))g = m_(2)a_(2)`
`rArr a_(2) = 3/2 g`