The two bodies of mass `m_(1) and m_(2) (m_(1) gt m_(2))` respectively are tied to the ends of a muscles string, which passes over a light and frictionless pulley. The masses are initially at rest and then released. Then acceleration of the centre of mass of the system is

Options

A. `((m_(1)-m_(2))/(m_(1)+m_(2)))^(2)g`
B. `((m_(2)-m_(1))/(m_(1)+m_(2)))^(2)g`
C. `(m_(1)g)/(m_(1)+m_(2))`
D. zero

Answer

A

Explanation

In the pulley arrangement, `abs(a_(1))=abs(a_(2))=a=[(m_(1)-m_(2))/(m_(1)+m_(2))]g`
but `a_(1)` is in downward direction and `a_(2)` in the upward direction, ie, `a_(2)=-a_(1)`.
`therefore` Acceleration of centre of mass
`a_(CM)=(m_(1)a_(1)+m_(2)a_(2))/(m_(1)+m_(2))=(m_(1))/(m_(1)+m_(2))[(m_(1)-m_(2))/(m_(1)+m_(2))]g-m_(2)/(m_(1)+m_(2))[(m_(1)-m_(2))/(m_(1)+m_(2))]g=((m_(1)-m_(2))/(m_(1)+m_(2)))^(2)g`