To determine the minimum value of v, we need to consider the energy conservation principle. When the bullet passes through the pendulum bob, there is a transfer of kinetic energy. Initially, the bullet has kinetic energy given by (1/2)mv², where m is the mass of the bullet and v is its speed. After passing through the bob, the bullet emerges with a speed of v/2, so its kinetic energy is now (1/2)m(v/2)². The energy transferred to the bob is the difference between these two kinetic energies.

This energy transferred to the bob is then converted into gravitational potential energy as the bob swings through a vertical circle. At the highest point of the swing, when the bob momentarily comes to rest, all of the initial kinetic energy has been converted into potential energy. Therefore, we can equate the transferred energy to the gravitational potential energy: (1/2)m(v/2)² = mgh, where h is the maximum height of the swing.

By substituting the given values, solving for v, and considering that h = 2L (twice the length of the rod), where L is the length of the rod, we can find the minimum value of v

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.

To determine the minimum value of v, we need to consider the energy conservation principle. When the bullet passes through the pendulum bob, there is a transfer of kinetic energy. Initially, the bullet has kinetic energy given by (1/2)mv², where m is the mass of the bullet and v is its speed. After passing through the bob, the bullet emerges with a speed of v/2, so its kinetic energy is now (1/2)m(v/2)². The energy transferred to the bob is the difference between these two kinetic energies. This energy transferred to the bob is then converted into gravitational potential energy as the bob swings through a vertical circle. At the highest point of the swing, when the bob momentarily comes to rest, all of the initial kinetic energy has been converted into potential energy. Therefore, we can equate the transferred energy to the gravitational potential energy: (1/2)m(v/2)² = mgh, where h is the maximum height of the swing. By substituting the given values, solving for v, and considering that h = 2L (twice the length of the rod), where L is the length of the rod, we can find the minimum value of v.

Learn how to calculate the minimum speed of a bullet passing through a pendulum bob, applying energy conservation principles. This problem covers kinetic energy, potential energy, and energy transfer in a bullet-pendulum system. Suitable for undergraduate physics students.

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