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### 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 \( h = \frac{3v^2}{20g} \).

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### 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

A block of mass m is moving with speed v along a horizontal surface when it collides with a uniform rod of mass 2m and length L attached at one end to a pivot. The block sticks to the rod, and the block-rod system rotates so that the end of the rod reaches a height h. Derive an expression for the angular speed ω of the block-rod system immediately after the collision and show that the change in height h of the bottom of the rod can be given by the equation h = 3v^2 / 20g.

Learn how to derive the angular speed of a block-rod system immediately after a collision and show the change in height for a physics problem involving rotational dynamics, conservation of angular momentum, and energy. Includes step-by-step calculations and formulas, suitable for undergraduate physics students.

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