A body of mass 1 kg is suspended by a weightless string which passes over a frictionless pulley of mass 2 kg as shown in the figure. The mass is released from a height of 1.6 m from the ground. With what velocity does it strike the ground?

Given, mass of the body, \(m_1=1 \mathrm{~kg}\)
Mass of pulley, \(m_2=2 \mathrm{~kg}\)

According to conservation of energy,
\(m_1 g h=\frac{1}{2} m_1 v^2+\frac{1}{2} I \omega^2\)
\(\begin{aligned}=\frac{1}{2} \omega_1 v^2+ & \frac{1}{2} \times \frac{m_2 R^2}{2} \times\left(\frac{v}{R}\right)^2 \\ & {\left[\because I=\frac{m_2 R^2}{2} \text { and } \omega=\frac{v}{R}\right] }\end{aligned}\)
\(\Rightarrow \quad m_1 g h=\frac{1}{2} m_1 v^2+\frac{1}{4} \times m_2 v^2\)
\(\Rightarrow 1 \times 10 \times 1.6=\frac{1}{2} \times 1 \times v^2+\frac{1}{4} \times 2 \times v^2\)
\(\Rightarrow \quad v^2=4 \mathrm{~m} / \mathrm{s}\)