To solve this problem, we'll use the principle of conservation of momentum, which states that the total momentum of a closed system remains constant before and after a collision, provided no external forces are acting on the system.
### Given Data
Before the collision:
- Mass of the first train car, [tex]\( m_1 = 600 \, \text{kg} \)[/tex]
- Mass of the second train car, [tex]\( m_2 = 400 \, \text{kg} \)[/tex]
- Velocity of the first train car, [tex]\( v_1 = 4 \, \text{m/s} \)[/tex]
- Velocity of the second train car, [tex]\( v_2 = 0 \, \text{m/s} \)[/tex]
After the collision:
- The two train cars stick together and move as a single unit.
### Step-by-Step Solution
1. Calculate the total mass after the collision:
[tex]\[
\text{Total mass} = m_1 + m_2 = 600 \, \text{kg} + 400 \, \text{kg} = 1000 \, \text{kg}
\][/tex]
2. Calculate the total momentum before the collision:
- The momentum of the first train car:
[tex]\[
\text{Momentum of first train car} = m_1 \times v_1 = 600 \, \text{kg} \times 4 \, \text{m/s} = 2400 \, \text{kg} \cdot \text{m/s}
\][/tex]
- The momentum of the second train car (since it's at rest, [tex]\( v_2 = 0 \)[/tex]):
[tex]\[
\text{Momentum of second train car} = m_2 \times v_2 = 400 \, \text{kg} \times 0 \, \text{m/s} = 0 \, \text{kg} \cdot \text{m/s}
\][/tex]
- The total momentum before the collision:
[tex]\[
\text{Total initial momentum} = (m_1 \times v_1) + (m_2 \times v_2) = 2400 \, \text{kg} \cdot \text{m/s} + 0 \, \text{kg} \cdot \text{m/s} = 2400 \, \text{kg} \cdot \text{m/s}
\][/tex]
3. Apply the principle of conservation of momentum:
Since the total momentum of the system is conserved, the total momentum after the collision is equal to the total momentum before the collision:
[tex]\[
\text{Total final momentum} = \text{Total initial momentum} = 2400 \, \text{kg} \cdot \text{m/s}
\][/tex]
### Conclusion
The total momentum of the system after the collision is:
[tex]\[
\boxed{2400 \, \text{kg} \cdot \text{m/s}}
\][/tex]
