The two bodies of mass `m_1 and m_2 (m_1 gtm_2)` respectively are tied to the ends of a massless 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))^2g`
B. `((m_2-m_1)/(m_1+m_2))^2g`
C. `(m_1g)/(m_1+m_2)`
D. zero

Answer

A

Explanation

In the pulley arrangement, `|a_1|=|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`
`:.` Acceleration of centre of mass
`a_(CM)=(m_1a_1+m_2a_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))^2g`