Let's go through this problem step-by-step.

Problem Summary

1. Given: A block of mass $m$ moving with speed $v$ collides and sticks to a rod of mass $2m$ and length $L$ attached to a pivot.
2. Goal for Part (a): Derive an expression for the angular speed $\omega$ of the block-rod system immediately after the collision.
3. Goal for Part (b): Show that the height $h$ reached by the end of the rod after rotation can be derived as $h = \frac{3v^2}{20g}$.

Solution Outline for Part (a)

1. Step 1: Determine the moment of inertia of the combined system about the pivot.

   • The rod’s moment of inertia about the pivot (since it’s pivoted at one end) is $I_{\text{rod}} = \frac{2mL^2}{3}$.
   • When the block sticks to the end of the rod, the block’s moment of inertia about the pivot is $I_{\text{block}} = mL^2$.
   • Total moment of inertia $I_{\text{total}} = I_{\text{rod}} + I_{\text{block}} = \frac{2mL^2}{3} + mL^2 = \frac{5mL^2}{3}$.

2. Step 2: Use conservation of angular momentum to find $\omega$.

   • The initial angular momentum of the system (before collision) is due only to the moving block: $L_{\text{initial}} = m \cdot v \cdot L$
   • After the collision, the angular momentum is shared by the block and the rod: $L_{\text{final}} = I_{\text{total}} \cdot \omega$
   • By conservation of angular momentum, $L_{\text{initial}} = L_{\text{final}}$, so: $m \cdot v \cdot L = \frac{5mL^2}{3} \cdot \omega$
   • Solving for $\omega$: $\omega = \frac{3v}{5L}$

Solution Outline for Part (b)

1. Step 1: Use conservation of energy to determine the height $h$ reached by the block-rod system.

   • After the collision, the system has rotational kinetic energy and no potential energy at the pivot.
   • The initial rotational kinetic energy is: $K = \frac{1}{2} I_{\text{total}} \omega^2 = \frac{1}{2} \cdot \frac{5mL^2}{3} \cdot \left( \frac{3v}{5L} \right)^2$
   • Substitute $\omega = \frac{3v}{5L}$: $K = \frac{1}{2} \cdot \frac{5mL^2}{3} \cdot \frac{9v^2}{25L^2} = \frac{3mv^2}{10}$

2. Step 2: Equate the initial kinetic energy to the potential energy at maximum height.

   • At the maximum height, the kinetic energy is fully converted to gravitational potential energy.
   • The center of mass of the rod-block system rises to height $h$: $m_{\text{eff}} \cdot g \cdot h = K$
   • Setting ( K = mgh