This problem involves a collision and subsequent rotational motion of a block-rod system. Let's break down the solution steps based on the image you've provided.

Problem Breakdown

1. Given data and setup:

   • A block of mass $m$ moving with speed $v$ collides and sticks to a rod of mass $2m$ and length $L$, which is initially vertical and pivoted at one end.
   • The system rotates after the collision, reaching a maximum height $h$ for the end of the rod.
   • We're asked to derive expressions for the angular speed $\omega$ immediately after the collision and the change in height $h$ using energy and angular momentum principles.

2. Parts of the problem:

   • (a) Derive the angular speed $\omega$ of the block-rod system immediately after the collision.
   • (b) Derive the height $h$ to which the end of the rod rises.

Solution Outline and Verification

Part (a): Deriving the Angular Speed $\omega$

1. Angular Momentum Conservation: Since there's no external torque about the pivot, we can use conservation of angular momentum. Right before the collision, only the block has linear momentum.

   • The initial angular momentum of the block about the pivot is: $L_{\text{initial}} = m \cdot v \cdot L$

2. Moment of Inertia of the System: After the collision, the block sticks to the rod, so we need the total moment of inertia $I_{\text{total}}$ about the pivot.

   • For the rod of mass $2m$ and length $L$ pivoted at one end, the moment of inertia is: $I_{\text{rod}} = \frac{1}{3} (2m) L^2 = \frac{2}{3} m L^2$
   • The block contributes an additional moment of inertia, since it is at a distance $L$ from the pivot: $I_{\text{block}} = m \cdot L^2$
   • Thus, the total moment of inertia is: $I_{\text{total}} = \frac{2}{3} m L^2 + m L^2 = \frac{5}{3} m L^2$

3. Applying Conservation of Angular Momentum: After the collision, the angular momentum is given by the product of $I_{\text{total}}$ and $\omega$: $L_{\text{final}} = I_{\text{total}} \cdot \omega = \frac{5}{3} m L^2 \cdot \omega$ Setting $L_{\text{initial}} = L_{\text{final}}$: $m \cdot v \cdot L = \frac{5}{3} m L^2 \cdot \omega$ Solving for $\omega$: $\omega = \frac{3v}{5L}$

Part (b): Determining the Height $h$

1. Using Conservation of Energy: After the collision, the system has rotational kinetic energy. At its highest point, all this kinetic energy is converted into gravitational potential energy.

   • Initial kinetic energy (rotational) is: $K_{\text{rot}} = \frac{1}{2} I_{\text{total}} \omega^2 = \frac{1}{2} \cdot \frac{5}{3} m L^2 \cdot \left(\frac{3v}{5L}\right)^2$
   • Simplify this expression: $K_{\text{rot}} = \frac{1}{2} \cdot \frac{5}{3} m L^2 \cdot \frac{9v^2}{25L^2} = \frac{15 m v^2}{50} = \frac{3 m v^2}{10}$

2. Gravitational Potential Energy at Height $h$: The potential energy when the center of mass reaches height $h$ is given by: $U = m_{\text{total}} \cdot g \cdot h$ Since we’re calculating the height $h$ for the bottom of the rod, the center of mass of the system effectively rises by a height proportional to $h$.

3. Equating Energy: By setting the rotational kinetic energy equal to the potential energy gain: [ \frac{3 m v^2}{10} = (m_{\text{total}})