Two blocks having masses of 400 g and 200 g are connected by a...

Using the principle of conservation of mechanical energy, the loss in potential energy of the system is equal to the gain in kinetic energy of the system.

$(m_1 - m_2)gh = \dfrac{1}{2}m_1v^2 + \dfrac{1}{2}m_2v^2 + \dfrac{1}{2}I\omega^2$

Since the string does not slip over the pulley, $\omega = \dfrac{v}{R}$. Substituting this into the equation gives:

$(m_1 - m_2)gh = \dfrac{1}{2} \left(m_1 + m_2 + \dfrac{I}{R^2}\right) v^2$

Given values are $m_1 = 0.4 \text{ kg}$, $m_2 = 0.2 \text{ kg}$, $h = 0.5 \text{ m}$, $R = 0.02 \text{ m}$, $I = 1.6 \times 10^{-4} \text{ kg-m}^2$, and taking $g = 9.8 \text{ m/s}^2$:

$(0.4 - 0.2) \times 9.8 \times 0.5 = \dfrac{1}{2} \left(0.4 + 0.2 + \dfrac{1.6 \times 10^{-4}}{(0.02)^2}\right) v^2$

$0.2 \times 4.9 = \dfrac{1}{2} \left(0.6 + \dfrac{1.6 \times 10^{-4}}{4 \times 10^{-4}}\right) v^2$

$0.98 = \dfrac{1}{2} (0.6 + 0.4) v^2$

$0.98 = \dfrac{1}{2} (1.0) v^2$

$v^2 = 1.96 \Rightarrow v = 1.4 \text{ m/s}$