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When dealing with collisions, a key principle to consider is the conservation of momentum. According to the law of conservation of momentum, the total momentum of a system remains constant if no external forces act upon it. Here's a step-by-step explanation of how this applies to the collision of the two train cars:
1. Initial Setup:
- Imagine two train cars are involved in this scenario. One train car is moving towards another train car that is initially at rest.
- Let’s denote the mass of the moving train car as \( m_1 \) and its velocity as \( v_1 \). The second train car, which is at rest, has a mass \( m_2 \) and an initial velocity of \( v_2 = 0 \).
2. Before the Collision:
- The initial momentum of the first train car is \( p_1 = m_1 \times v_1 \).
- The initial momentum of the second train car is \( p_2 = m_2 \times v_2 \), which is \( p_2 = m_2 \times 0 = 0 \).
3. Total Initial Momentum:
- The total initial momentum of the system is the sum of the momenta of both cars before the collision:
[tex]\[ p_\text{initial} = p_1 + p_2 = m_1 \times v_1 + 0 = m_1 \times v_1. \][/tex]
4. After the Collision:
- After the collision, the two train cars stick together and move as a single unit. Let’s denote the final velocity of the combined mass as \( v_f \).
- The combined mass after the collision is \( m_1 + m_2 \).
5. Total Final Momentum:
- The total momentum after the collision is given by the combined mass multiplied by their common velocity:
[tex]\[ p_\text{final} = (m_1 + m_2) \times v_f. \][/tex]
6. Conservation of Momentum:
- According to the conservation of momentum:
[tex]\[ p_\text{initial} = p_\text{final}. \][/tex]
- Thus:
[tex]\[ m_1 \times v_1 = (m_1 + m_2) \times v_f. \][/tex]
Given the options provided and knowing that the total momentum of the system after the collision is exactly one of these values, we compare with our calculated options:
- [tex]$800 \, \text{kg} \cdot \text{m} / \text{s}$[/tex]
- [tex]$1,600 \, \text{kg} \cdot \text{m} / \text{s}$[/tex]
- [tex]$2,400 \, \text{kg} \cdot \text{m} / \text{s}$[/tex]
- [tex]$4,000 \, \text{kg} \cdot \text{m} / \text{s}$[/tex]
The correct answer, based on the conservation of momentum, is:
[tex]\[ 1,600 \, \text{kg} \cdot \text{m} / \text{s}. \][/tex]
Therefore, the total momentum of the system after the collision is [tex]\( 1,600 \)[/tex] kg·m/s.
1. Initial Setup:
- Imagine two train cars are involved in this scenario. One train car is moving towards another train car that is initially at rest.
- Let’s denote the mass of the moving train car as \( m_1 \) and its velocity as \( v_1 \). The second train car, which is at rest, has a mass \( m_2 \) and an initial velocity of \( v_2 = 0 \).
2. Before the Collision:
- The initial momentum of the first train car is \( p_1 = m_1 \times v_1 \).
- The initial momentum of the second train car is \( p_2 = m_2 \times v_2 \), which is \( p_2 = m_2 \times 0 = 0 \).
3. Total Initial Momentum:
- The total initial momentum of the system is the sum of the momenta of both cars before the collision:
[tex]\[ p_\text{initial} = p_1 + p_2 = m_1 \times v_1 + 0 = m_1 \times v_1. \][/tex]
4. After the Collision:
- After the collision, the two train cars stick together and move as a single unit. Let’s denote the final velocity of the combined mass as \( v_f \).
- The combined mass after the collision is \( m_1 + m_2 \).
5. Total Final Momentum:
- The total momentum after the collision is given by the combined mass multiplied by their common velocity:
[tex]\[ p_\text{final} = (m_1 + m_2) \times v_f. \][/tex]
6. Conservation of Momentum:
- According to the conservation of momentum:
[tex]\[ p_\text{initial} = p_\text{final}. \][/tex]
- Thus:
[tex]\[ m_1 \times v_1 = (m_1 + m_2) \times v_f. \][/tex]
Given the options provided and knowing that the total momentum of the system after the collision is exactly one of these values, we compare with our calculated options:
- [tex]$800 \, \text{kg} \cdot \text{m} / \text{s}$[/tex]
- [tex]$1,600 \, \text{kg} \cdot \text{m} / \text{s}$[/tex]
- [tex]$2,400 \, \text{kg} \cdot \text{m} / \text{s}$[/tex]
- [tex]$4,000 \, \text{kg} \cdot \text{m} / \text{s}$[/tex]
The correct answer, based on the conservation of momentum, is:
[tex]\[ 1,600 \, \text{kg} \cdot \text{m} / \text{s}. \][/tex]
Therefore, the total momentum of the system after the collision is [tex]\( 1,600 \)[/tex] kg·m/s.
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