To answer the questions, let's go through the conservation of momentum principle.
### Step-by-Step Solution
#### Given:
1. Mass of the first train car ([tex]\( m_1 \)[/tex]) = 600 kg
2. Mass of the second train car ([tex]\( m_2 \)[/tex]) = 400 kg
3. Velocity of the first train car before collision ([tex]\( v_1 \)[/tex]) = 4 m/s
4. Velocity of the second train car before collision ([tex]\( v_2 \)[/tex]) = -4 m/s (assuming it's traveling towards the first train, hence the negative sign for opposite direction).
#### Total Momentum Before Collision:
Momentum is calculated by the formula:
[tex]\[ \text{momentum} = \text{mass} \times \text{velocity} \][/tex]
For the first train car:
[tex]\[ p_1 = m_1 \times v_1 = 600 \, \text{kg} \times 4 \, \text{m/s} = 2400 \, \text{kg} \cdot \text{m/s} \][/tex]
For the second train car:
[tex]\[ p_2 = m_2 \times v_2 = 400 \, \text{kg} \times (-4) \, \text{m/s} = -1600 \, \text{kg} \cdot \text{m/s} \][/tex]
To find the total momentum of the system before the collision, we sum the individual momenta of the two train cars:
[tex]\[ \text{Total Momentum Before Collision} = p_1 + p_2 = 2400 \, \text{kg} \cdot \text{m/s} + (-1600) \, \text{kg} \cdot \text{m/s} = 800 \, \text{kg} \cdot \text{m/s} \][/tex]
So, the total momentum of the system before the train cars collide is:
[tex]\[ 800 \, \text{kg} \cdot \text{m/s} \][/tex]
#### Total Momentum After Collision:
According to the law of conservation of momentum, the total momentum of the system remains constant if no external forces act on it. Therefore, the total momentum of the system after the collision must be the same as it was before the collision.
So, the total momentum of the system after the train cars collide is:
[tex]\[ 800 \, \text{kg} \cdot \text{m/s} \][/tex]
### Summary:
1. Total momentum before the collision: [tex]\( 800 \, \text{kg} \cdot \text{m/s} \)[/tex]
2. Total momentum after the collision: [tex]\( 800 \, \text{kg} \cdot \text{m/s} \)[/tex]
This completes the step-by-step solution based on the given information.
