Physics I, Newton's Laws: Interesting pulley situation.

The problem: Find an expression for the acceleration of m₁ (the table is frictionless).

This is pretty straightforward, but it's easy to overlook something.

The acceleration in m₁ and m₂ are not equal. The block m₂ is accelerating half as fast as m₁. To understand this, draw a free-body diagram.

m₁ is pulled to the right by a tension force T. Since there is no friction force, the normal force and weight of m1 isn't important.

m₂ is pulled down by its weight, and pulled up by two tension forces, both equal to T.

a₁ = 2a₂
a₂ = a₁/2

F_{net (m1)} = T = m₁a₁
F_{net (m2)} = 2T - m₂g = m₂a₂

Since T = m₁a₁, plug that into the second expression. Also, since we're solving for a₁, eliminate a₂ by plugging in a₁/2

2m₁a₁ - m₂g = (m₂a₁)/2
4m₁a₁ - 2m₂g = m₂a₁
4m₁a₁ - m₂a₁ = 2m₂g
a₁(4m₁ - m₂) = 2m₂g

Solution: a₁ = (2m₂g)/(4m₁ - m₂)