Two blocks of masses `m_(1)` and `m_(2)` are connected with a light spring of force constant k and the whole system is kept on a frictionless horizontal surface. The masses are applied forces `F_(1)` and `F_(2)` as shown in fig. At any time the blocks have same acceleration `a_(0)` but in opposite direction. Now answer the following :

The value of `a_(0)` is

Options

A. `(F_(1)-F_(2))/(m_(1)+m_(2))`
B. `(F_(1)-F_(2))/(m_(1)-m_(2))`
C. `(F_(1)+F_(2))/(m_(1)-m_(2))`
D. `(F_(1)+F_(2))/(m_(1)+m_(2))`

Answer

B

Explanation

Let x be the compression in spring at any time

`F_(1)-kx=m_(1)a_(0), F_(2)-kx=m_(2)a_(0)`
Solve to get: `a_(0)=(F_(1)-F_(2))/(m_(1)-m_(2))`
and `kx=(m_(1)F_(2)-F_(1)m_(2))/(m_(1)-m_(2))`
Just after `m_(2)` is removed:
`a_(2)=(kx)/(m_2)= (F_(2)-m_(2)a_(0))/(m_2) =(F_2)/(m_2)-a_(0)`.