At the ends of a weightless and inextensible thread, thrown over a fixed block, weights of 100 g and 150 g

At the ends of a weightless and inextensible thread, thrown over a fixed block, weights of 100 g and 150 g are suspended with what accelerations the loads move. Block friction neglected

Given: m1 = 100 g; m2 = 150 g.
Find the accelerations of both bodies.
Decision:
Since m2> m1, the acceleration of a body with mass m2 will be directed downward, and a body with mass m1 will be directed upward. These bodies are acted upon by the forces of gravity F and the tension force of the thread T. Let us use the law of dynamics and write it for each body. m1a1 = F1 + T1 m2a2 = F2 + T2. These equations will change after superimposing them on the y-axis: m1a1 = m1g – T1; -m2a2 = m2g – T2. The speeds and accelerations will be equal: a1 = a2 = a. We do not take into account the influence of the thread. This means that the tension force of the thread during the transition through the block changes direction without changing the module. T’1 = T’2; T’1 = T1; T’2 = T2. From this we obtain T1 = T2 = T. Hence, we get m1a = m1g + T; m2a = T – m2g. Using a system of these equations, we get an equation with a highlighted a: a = (m2-m1) / (m1 + m2). We count by substituting known values. a = (150-100) / (100 + 150) = 50/250 = 5/25 = 1/5 = 0.2
Answer: 0.2