To solve for the combined velocity of the two railroad cars after they collide and couple together, we can use the principle of conservation of momentum. The formula you provided is correct and can be expressed as follows:
\[ m_1 v_1 + m_2 v_2 = m_3 v_3 \]
Where:
- \(m_1\) = mass of the first car (6000 kg)
- \(v_1\) = velocity of the first car (5 m/s)
- \(m_2\) = mass of the second car (4000 kg)
- \(v_2\) = velocity of the second car (0 m/s, since it is stationary)
- \(m_3\) = combined mass of both cars after the collision (which is \(m_1 + m_2\))
- \(v_3\) = combined velocity after the collision (what we want to find)
First, calculate \(m_3\):
\[ m_3 = m_1 + m_2 = 6000 , \text{kg} + 4000 , \text{kg} = 10000 , \text{kg} \]
Now, substituting the values into the momentum equation:
\[ m_1 v_1 + m_2 v_2 = m_3 v_3 \]
Plug in the known values:
\[ 6000 , \text{kg} \cdot 5 , \text{m/s} + 4000 , \text{kg} \cdot 0 = 10000 , \text{kg} \cdot v_3 \]
This simplifies to:
\[ 30000 , \text{kg m/s} + 0 = 10000 , \text{kg} \cdot v_3 \]
So we have:
\[ 30000 = 10000 v_3 \]
Now, divide both sides by 10000 to solve for \(v_3\):
\[ v_3 = \frac{30000}{10000} = 3 , \text{m/s} \]
Therefore, the combined velocity of the railroad cars immediately after the impact is 3 m/s.