In terms of masses m₁, m₂ and g, find the acceleration of both the blocks shown in figure. Neglect all friction and masses of the pulley.

As the mass m₁ moves towards right through distance x, the mass m₂ moves down through distance x/2 Clearly, if the acceleration of m₁ is a, then that of m₂ will be a/2

Applying Newton’s second law to the motions of m₁ and m₂ we have 

T₁ = m₁a                  

And    m₂g – T₂ = m₂.a/2

Also,  T₂ = 2T₁ = 2m₁a

∴    m₂g – 2m₁a = m₂. a/2 

Or, 2m₂g – 4m₁a = m₂a or 2m₂g = (4m₁ + m₂)a    

∴ Acceleration of m₁ = a = (2m₂ g)/(4m₁+m₂ )                     m₂=a/2=(m₂ g)/(4m₁+ m₂ )