Let’s carefully walk through this energy conservation problem involving the bullet-pendulum system.

Problem Recap & Setup:

1. A bullet with mass $m$ and speed $v$ passes through a pendulum bob.
2. After passing through, the bullet's speed decreases to $\frac{v}{2}$.
3. The energy transferred to the bob causes it to rise through a vertical circle to a maximum height $h = 2L$, where $L$ is the length of the pendulum rod.
4. We need to find the minimum speed $v$ such that the bob reaches the top of the swing.

---

Step 1: Kinetic Energy of the Bullet (Before and After Impact)

• Initial kinetic energy of the bullet: $K_i = \frac{1}{2} m v^2$

• Kinetic energy of the bullet after passing through the bob: $K_f = \frac{1}{2} m \left(\frac{v}{2}\right)^2 = \frac{1}{2} m \cdot \frac{v^2}{4} = \frac{m v^2}{8}$

---

Step 2: Energy Transferred to the Bob

The energy transferred to the bob equals the difference in the bullet's initial and final kinetic energy: $E_{\text{transferred}} = K_i - K_f = \frac{1}{2} m v^2 - \frac{m v^2}{8} = \frac{3}{8} m v^2$

---

Step 3: Gravitational Potential Energy of the Bob at the Top of the Swing

Let the mass of the bob be $M$. At the top of the vertical swing (height $h = 2L$), all transferred kinetic energy becomes potential energy: $E_{\text{transferred}} = M g h$

Given $h = 2L$, we have: $M g (2L)$

---

Step 4: Equating Transferred Energy to Gravitational Potential Energy

$\frac{3}{8} m v^2 = M g (2L)$

---

Step 5: Solving for $v$

Rearranging the equation to solve for $v$: $v^2 = \frac{16 M g L}{3 m}$ $v = \sqrt{\frac{16 M g L}{3 m}}$

---

Final Answer

The minimum speed $v$ required for the bullet to ensure the bob reaches the top of the vertical swing is: $v = \sqrt{\frac{16 M g L}{3 m}}$

---

Conclusion

This equation shows that the speed $v$ depends on the mass of the bullet $m$, the mass of the bob $M$, the gravitational acceleration $g$, and the length of the rod $L$.

---

Do you have any questions or want further clarifications? Here are five follow-up questions to consider:

1. What would change if the bullet embedded into the bob instead of passing through it?
2. How does the ratio $\frac{M}{m}$ affect the speed required for the bob to reach the top?
3. How does air resistance or friction impact the conservation of energy in this scenario?
4. What would happen if the bob only needed to reach a height $L$ instead of $2L$?
5. How does the outcome change if the bullet speed is initially less than the calculated minimum $v$?

Tip: Always analyze whether energy is conserved throughout the problem and account for any losses when they exist.