Answered

Consider a system where one train car is moving toward another train car at rest. When the train cars collide, the two cars stick together.

What is the total momentum of the system after the collision?

A. [tex]$800 \, \text{kg} \cdot \text{m/s}$[/tex]
B. [tex]$1,600 \, \text{kg} \cdot \text{m/s}$[/tex]
C. [tex]$2,400 \, \text{kg} \cdot \text{m/s}$[/tex]
D. [tex]$4,000 \, \text{kg} \cdot \text{m/s}$[/tex]



Answer :

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.