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?

• A

  $16 \mathrm{~ms}^{-1}$
• B

  $8 \mathrm{~ms}^{-1}$
• C

  $4 \sqrt{2} \mathrm{~ms}^{-1}$
• Correct answer
  D

  $4 \mathrm{~ms}^{-1}$

Solution

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

Hence, the correct answer is option: 'D'.