Question

Two blocks of masses 6 kg and 4 kg are attached to the two ends of a massless string passing over a smooth fixed pulley. If the system is released, the acceleration of the centre of mass of the system will be :

• A. g, vertical downwards
• B. g/5, vertical downwards
• C. g/25, vertical downwards
• D. zero

Answer 1

The correct option is C g/25, vertical downwards

basic formula for atwood machine : acceleration of masses $=\frac{(m_1-m_2)g}{m_1+m_2}=\frac{(6-4)g}{6+4}=0.2g$
6kg masses is going down and 4kg will go up ,hence acceleration of 4kg can be taken as -ve.
so acceleration of COM$=\frac{m_1{a}_1+m_2{a}_2}{m_1+m_2}=\frac{(6-4)0.2g}{6+4}=0.04g$