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This is a physics problem involving a ballistic pendulum, where a bullet is fired into a pendulum and gets embedded in it, causing the pendulum to swing to a certain angle. We are given:

- Mass of the bullet, \( m = 11 \, \text{g} = 0.011 \, \text{kg} \)
- Mass of the pendulum, \( M = 68 \, \text{kg} \)
- The pendulum deviates by an angle of \( \theta = 27^\circ \) after impact.
- The vertical distance of the pendulum's pivot point to the impact point, \( L = 92 \, \text{cm} = 0.92 \, \text{m} \).

We need to find the speed \( v_0 \) with which the bullet hits the pendulum.

### Step-by-Step Solution:

1. **Convert the height reached by the pendulum from angle deviation**:
   - When the pendulum reaches an angle \( \theta \), it gains potential energy due to the height it reaches.
   - The height \( h \) can be found using trigonometry:
     \[
     h = L(1 - \cos \theta)
     \]
   - Plugging in the values:
     \[
     h = 0.92 \, \text{m} \times (1 - \cos 27^\circ)
     \]

2. **Calculate the potential energy at the highest point**:
   - When the pendulum reaches the height \( h \), the potential energy (PE) of the pendulum-bullet system is:
     \[
     \text{PE} = (M + m) \cdot g \cdot h
     \]
   - Use \( g = 9.81 \, \text{m/s}^2 \).

3. **Determine the initial speed of the pendulum-bullet system after the collision**:
   - At the point of collision, the system has only kinetic energy (KE), which later converts to potential energy at the highest point.
   - The initial kinetic energy (just after collision) is:
     \[
     \text{KE} = \frac{1}{2} (M + m) v^2
     \]
   - Since KE converts to PE at the maximum height:
     \[
     \frac{1}{2} (M + m) v^2 = (M + m) \cdot g \cdot h
     \]
   - Solving for \( v \):
     \[
     v = \sqrt{2 \cdot g \cdot h}
     \]

4. **Use conservation of momentum to find the bullet's initial speed \( v_0 \)**:
   - The momentum just before impact is equal to the momentum just after impact.
   - So, \( m \cdot v_0 = (M + m) \cdot v \).
   - Solving for \( v_0 \):
     \[
     v_0 = \frac{(M + m) \cdot v}{m}
     \]

Substitute the values at each step to find the numerical result for \( v_0 \). Let me know if you'd like detailed calculations at each stage or further clarifications! 

---

### Additional Questions for Exploration:

1. How would the result change if the angle deviation was increased?
2. What happens to the potential energy if the length of the wire \( L \) changes?
3. How does conservation of energy play a role in ballistic pendulum problems?
4. How would air resistance affect the bullet’s speed calculation?
5. What would be the effect on \( v_0 \) if the bullet mass was doubled?

### Tip:
Always check unit conversions, especially with mass (e.g., grams to kilograms) and distance (e.g., centimeters to meters).

An 11 g bullet is fired with speed v into a 68 kg ballistic pendulum. The wires holding the pendulum deviate from the vertical by 27°. The vertical distance from the point of impact to the suspension points for the wires is 92 cm. Find the speed with which the bullet hits the pendulum.

Calculate the initial speed of a bullet fired into a ballistic pendulum. Given a bullet mass of 11 g, a pendulum mass of 68 kg, a 27° angle deviation, and a 92 cm vertical distance, this problem applies conservation of energy and momentum principles. Suitable for advanced high school and college physics students.

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