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Physics Question on laws of motion

Two masses ′m1′ and ′m2′(m1>m2)'m_1'\,and\,'m_2'(m_1>m_2) connected to the ends of a light inextensible string which passes over the surface of a smooth fixed pulley. If the system is released from rest, the acceleration of the center of mass of the system will be (g =acceleration due to gravity)

A

g(m1−m2)m1+m2\frac{g(m_1-m_2)}{m_1+m_2}

B

g(m1−m2)2(m1+m2)2\frac{g(m_1-m_2)^2}{(m_1+m_2)^2}

C

g(m1+m2)m1−m2\frac{g(m_1+m_2)}{m_1-m_2}

D

g(m1+m2)2(m1−m2)2\frac{g(m_1+m_2)^2}{(m_1-m_2)^2}

Answer

g(m1−m2)2(m1+m2)2\frac{g(m_1-m_2)^2}{(m_1+m_2)^2}

Explanation

Solution

(acm)y=FcmM=(m1+m2)g−2Tm1+m2(a_{cm})_y = \frac{F_{cm}}{M} = \frac{(m_1+m_2)g - 2T}{m_1+ m_2} →(1) \rightarrow (1) But T=2m1m2gm1+m2T = \frac{2m_1m_2g}{m_1+m_2} →(2)\rightarrow (2)