Question

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 1

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
Now, `a_(CM) = (m_(A)a_(A) + m_(B)a_(B))/(m_(A) + m_(B)`
`=(2(a)-1(a))/(1+2)=a/3 = g/3` (downwards)