In the arrangement shown in figure, m_A=2kg and m_B=1kg. String is light and inextensible. Find the acceleration of centre of mass of both the blocks. Neglect friction everywhere.

In the arrangement shown in figure, m_A=2kg and m_B=1kg. String is lig

1.	In the arrangement shown in figure, m_A=2kg and m_B=1kg. String is light and inextensible. Find the acceleration of centre of mass of both the blocks. Neglect friction everywhere.
Answer» Solution :NET pulling force on the system is `(m_(A) - m_(B))g` or `(2-1)g=g` TOTAL mass being pulled is `m_(A) + m_(B)` or 3 kg `therefore a = ("Net pulling force")/("Total mass") =g/3`= acceleration of each block. Now, `veca_(CM) =(m_(A)veca_(A) + m_(B)veca_(B))/(m_(A) + m_(B)) =((2)(a) -(1)(a))/(1+2) =a/3 = g/9` downwards.