A 1.20-kg wooden block rests on a table over a large hole as in the figure below. A 5.25-g bullet with an initial velocity v, is fired upward into the bottom of the block and rem the collision. The block and bullet rise to a maximum height of 18.0 cm. M (a) Describe how you would find the initial velocity of the bullet using ideas you have learned in this chapter. (b) Calculate the initial velocity of the bullet from the information provided. (Let up be the positive direction. Indicate the direction with the sign of your answer.) m/s

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### Physics Problem: Bullet-Block Collision

**Problem Statement:**

A 1.20-kg wooden block rests on a table over a large hole as shown in the figure below. A 5.25-g bullet with an initial velocity \( v_i \) is fired upward into the bottom of the block and remains in the block after the collision. The block and bullet rise to a maximum height of 18.0 cm.

**Figure Description:**

The figure illustrates a wooden block of mass \( M \) positioned on a table with a large hole beneath it. A bullet of mass \( m \) is shown moving upward with an initial velocity \( v_i \) towards the block. 

* (a) Describe how you would find the initial velocity of the bullet using ideas you have learned in this chapter.
* (b) Calculate the initial velocity of the bullet from the information provided. (Let up be the positive direction. Indicate the direction with the sign of your answer.)

**Solution Approach:**

(a) **Determining Initial Velocity:**
   - **Concepts to Use:** 
     1. Conservation of Momentum: Before and after the collision.
     2. Conservation of Energy: For the movement of the block and bullet to the maximum height.
   - **Steps:**
     1. Calculate the combined mass of the block and bullet system.
     2. Apply conservation of momentum to find the velocity of the block and bullet system right after the collision.
     3. Use the conservation of energy principle to determine the initial velocity needed to reach the given height.

(b) **Calculating Initial Velocity:**
   \[
   m = 0.00525 \text{ kg}
   \]
   \[
   M = 1.20 \text{ kg}
   \]
   Total mass after collision:
   \[
   m + M = 1.20525 \text{ kg}
   \]
   Height (h):
   \[
   h = 0.18 \text{ m}
   \]
   Use the formula for potential energy (PE):
   \[
   PE = mgh
   \]
   Final kinetic energy (KE):
   \[
   KE = \frac{1}{2}mv^2
   \]
   Equate to find \( v \):
   \[
   \frac{1}{2}(M+m)v_f^2 = (M+m
Transcribed Image Text:### Physics Problem: Bullet-Block Collision **Problem Statement:** A 1.20-kg wooden block rests on a table over a large hole as shown in the figure below. A 5.25-g bullet with an initial velocity \( v_i \) is fired upward into the bottom of the block and remains in the block after the collision. The block and bullet rise to a maximum height of 18.0 cm. **Figure Description:** The figure illustrates a wooden block of mass \( M \) positioned on a table with a large hole beneath it. A bullet of mass \( m \) is shown moving upward with an initial velocity \( v_i \) towards the block. * (a) Describe how you would find the initial velocity of the bullet using ideas you have learned in this chapter. * (b) Calculate the initial velocity of the bullet from the information provided. (Let up be the positive direction. Indicate the direction with the sign of your answer.) **Solution Approach:** (a) **Determining Initial Velocity:** - **Concepts to Use:** 1. Conservation of Momentum: Before and after the collision. 2. Conservation of Energy: For the movement of the block and bullet to the maximum height. - **Steps:** 1. Calculate the combined mass of the block and bullet system. 2. Apply conservation of momentum to find the velocity of the block and bullet system right after the collision. 3. Use the conservation of energy principle to determine the initial velocity needed to reach the given height. (b) **Calculating Initial Velocity:** \[ m = 0.00525 \text{ kg} \] \[ M = 1.20 \text{ kg} \] Total mass after collision: \[ m + M = 1.20525 \text{ kg} \] Height (h): \[ h = 0.18 \text{ m} \] Use the formula for potential energy (PE): \[ PE = mgh \] Final kinetic energy (KE): \[ KE = \frac{1}{2}mv^2 \] Equate to find \( v \): \[ \frac{1}{2}(M+m)v_f^2 = (M+m
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