Answer to Question #104005 in Mechanics | Relativity for Opatola

Question

Particles of mass m1 and m2 (m2>m1) are connected by a large inextensible string passing over a smooth fixed pulley. Initially both masses hang vertically with mass m2 at a height X above the floor. If the system is released from rest, what speed will mass m2 hit the floor and the mass m1 will rise a further distance (m2-m1)x/m1+m2 after this occurs

Answer 1

$$(m_2+m_1)a=(m_2-m_1)g\to a=\frac{(m_2-m_1)}{(m_2+m_1)}g$$

$$v^2=2ax$$

The speed is

$$v=\sqrt{\frac{(m_2-m_1)}{(m_2+m_1)}2gx}$$

For mass m₂ we can apply the conservation of energy principle:

$$m_2gh=0.5m_2v^2$$

$$h=\frac{(m_2-m_1)}{(m_2+m_1)}x$$