Answer :
To solve this problem, we need to analyze it using principles of physics such as conservation of momentum, conservation of energy, and mechanics. Here’s a step-by-step explanation:
1. Given Data:
- Mass of the bullet, [tex]\( m_b = 0.023 \, \text{kg} \)[/tex]
- Velocity of the bullet before impact, [tex]\( v_b = 80 \, \text{m/s} \)[/tex]
- Mass of the block, [tex]\( m_{\text{block}} = 0.45 \, \text{kg} \)[/tex]
- Acceleration due to gravity, [tex]\( g = 9.8 \, \text{m/s}^2 \)[/tex]
2. Initial Momentum:
The initial momentum of the bullet before striking the block:
[tex]\[ p_{\text{bullet}} = m_b \cdot v_b \][/tex]
Plugging in the given values:
[tex]\[ p_{\text{bullet}} = 0.023 \, \text{kg} \cdot 80 \, \text{m/s} = 1.84 \, \text{kg m/s} \][/tex]
3. Final Velocity after Collision:
After the collision, the bullet comes to rest with respect to the block, so they move together. Using conservation of momentum:
[tex]\[ p_{\text{initial}} = p_{\text{final}} \][/tex]
Where:
[tex]\[ p_{\text{final}} = (m_b + m_{\text{block}}) \cdot v_f \][/tex]
Solving for the final velocity [tex]\( v_f \)[/tex]:
[tex]\[ v_f = \frac{p_{\text{bullet}}}{m_b + m_{\text{block}}} = \frac{1.84 \, \text{kg m/s}}{0.023 \, \text{kg} + 0.45 \, \text{kg}} = \frac{1.84}{0.473} = 3.89 \, \text{m/s} \][/tex]
4. Initial Kinetic Energy:
The initial kinetic energy of the bullet before impact is:
[tex]\[ KE_{\text{bullet}} = \frac{1}{2} m_b v_b^2 \][/tex]
Plugging in the numbers:
[tex]\[ KE_{\text{bullet}} = \frac{1}{2} \cdot 0.023 \, \text{kg} \cdot (80 \, \text{m/s})^2 = \frac{1}{2} \cdot 0.023 \cdot 6400 = 73.6 \, \text{J} \][/tex]
5. Kinetic energy of the combined mass just after the collision:
[tex]\[ KE_{\text{combined}} = \frac{1}{2} (m_b + m_{\text{block}}) v_f^2 \][/tex]
Plugging the values:
[tex]\[ KE_{\text{combined}} = \frac{1}{2} \cdot 0.473 \, \text{kg} \cdot (3.89 \, \text{m/s})^2 = \frac{1}{2} \cdot 0.473 \cdot 15.13 = 3.579 \, \text{J} \][/tex]
6. Height to which the block rises:
Using conservation of energy, the kinetic energy just after the collision is converted into potential energy at the highest point:
[tex]\[ (m_b + m_{\text{block}}) g h = KE_{\text{combined}} \][/tex]
Solving for [tex]\( h \)[/tex]:
[tex]\[ h = \frac{KE_{\text{combined}}}{(m_b + m_{\text{block}}) g} = \frac{3.579 \, \text{J}}{0.473 \, \text{kg} \cdot 9.8 \, \text{m/s}^2} = \frac{3.579}{4.6354} = 0.772 \, \text{m} \][/tex]
7. Amount of Heat Produced:
The heat produced is the difference between the initial kinetic energy of the bullet and the kinetic energy of the combined mass:
[tex]\[ \text{Heat produced} = KE_{\text{bullet}} - KE_{\text{combined}} \][/tex]
Plugging in the numbers:
[tex]\[ \text{Heat produced} = 73.6 \, \text{J} - 3.579 \, \text{J} = 70.021 \, \text{J} \][/tex]
### Summary of Results:
- Height to which the block rises: [tex]\( 0.772 \, \text{m} \)[/tex]
- Amount of heat produced: [tex]\( 70.021 \, \text{J} \)[/tex]
These detailed steps provide a comprehensive understanding of how we reach the numerical results.
1. Given Data:
- Mass of the bullet, [tex]\( m_b = 0.023 \, \text{kg} \)[/tex]
- Velocity of the bullet before impact, [tex]\( v_b = 80 \, \text{m/s} \)[/tex]
- Mass of the block, [tex]\( m_{\text{block}} = 0.45 \, \text{kg} \)[/tex]
- Acceleration due to gravity, [tex]\( g = 9.8 \, \text{m/s}^2 \)[/tex]
2. Initial Momentum:
The initial momentum of the bullet before striking the block:
[tex]\[ p_{\text{bullet}} = m_b \cdot v_b \][/tex]
Plugging in the given values:
[tex]\[ p_{\text{bullet}} = 0.023 \, \text{kg} \cdot 80 \, \text{m/s} = 1.84 \, \text{kg m/s} \][/tex]
3. Final Velocity after Collision:
After the collision, the bullet comes to rest with respect to the block, so they move together. Using conservation of momentum:
[tex]\[ p_{\text{initial}} = p_{\text{final}} \][/tex]
Where:
[tex]\[ p_{\text{final}} = (m_b + m_{\text{block}}) \cdot v_f \][/tex]
Solving for the final velocity [tex]\( v_f \)[/tex]:
[tex]\[ v_f = \frac{p_{\text{bullet}}}{m_b + m_{\text{block}}} = \frac{1.84 \, \text{kg m/s}}{0.023 \, \text{kg} + 0.45 \, \text{kg}} = \frac{1.84}{0.473} = 3.89 \, \text{m/s} \][/tex]
4. Initial Kinetic Energy:
The initial kinetic energy of the bullet before impact is:
[tex]\[ KE_{\text{bullet}} = \frac{1}{2} m_b v_b^2 \][/tex]
Plugging in the numbers:
[tex]\[ KE_{\text{bullet}} = \frac{1}{2} \cdot 0.023 \, \text{kg} \cdot (80 \, \text{m/s})^2 = \frac{1}{2} \cdot 0.023 \cdot 6400 = 73.6 \, \text{J} \][/tex]
5. Kinetic energy of the combined mass just after the collision:
[tex]\[ KE_{\text{combined}} = \frac{1}{2} (m_b + m_{\text{block}}) v_f^2 \][/tex]
Plugging the values:
[tex]\[ KE_{\text{combined}} = \frac{1}{2} \cdot 0.473 \, \text{kg} \cdot (3.89 \, \text{m/s})^2 = \frac{1}{2} \cdot 0.473 \cdot 15.13 = 3.579 \, \text{J} \][/tex]
6. Height to which the block rises:
Using conservation of energy, the kinetic energy just after the collision is converted into potential energy at the highest point:
[tex]\[ (m_b + m_{\text{block}}) g h = KE_{\text{combined}} \][/tex]
Solving for [tex]\( h \)[/tex]:
[tex]\[ h = \frac{KE_{\text{combined}}}{(m_b + m_{\text{block}}) g} = \frac{3.579 \, \text{J}}{0.473 \, \text{kg} \cdot 9.8 \, \text{m/s}^2} = \frac{3.579}{4.6354} = 0.772 \, \text{m} \][/tex]
7. Amount of Heat Produced:
The heat produced is the difference between the initial kinetic energy of the bullet and the kinetic energy of the combined mass:
[tex]\[ \text{Heat produced} = KE_{\text{bullet}} - KE_{\text{combined}} \][/tex]
Plugging in the numbers:
[tex]\[ \text{Heat produced} = 73.6 \, \text{J} - 3.579 \, \text{J} = 70.021 \, \text{J} \][/tex]
### Summary of Results:
- Height to which the block rises: [tex]\( 0.772 \, \text{m} \)[/tex]
- Amount of heat produced: [tex]\( 70.021 \, \text{J} \)[/tex]
These detailed steps provide a comprehensive understanding of how we reach the numerical results.
