**Physics Problem: Conservation of Momentum in a Collision****Scenario:**A railroad cart weighing 18,000 kg collides with a stationary cart weighing 12,000 kg. After the collision, the carts lock together and move as one unit. Their speed immediately after the collision is 8.5 m/s.**Question:**What was the speed of the first cart before the collision?**Solution Approach:**This problem can be solved using the principle of conservation of momentum, which states that the total momentum of a closed system remains constant if no external forces are acting on it.The momentum before the collision equals the momentum after the collision:\[ m_1 \times v_1 + m_2 \times v_2 = (m_1 + m_2) \times v_f \]Where:- \( m_1 = 18,000 \, \text{kg} \) (mass of the first cart)- \( v_1 \) = initial speed of the first cart (unknown)- \( m_2 = 12,000 \, \text{kg} \) (mass of the stationary cart)- \( v_2 = 0 \, \text{m/s} \) (initial speed of the second cart)- \( v_f = 8.5 \, \text{m/s} \) (final speed of the combined carts)Substituting the known values, you can solve for \( v_1 \).

Name: **Physics Problem: Conservation of Momentum in a Collision****Scenari
Uploaded: 2024-03-20T06:57:37.000Z
Channel: Bartleby
Description: **Physics Problem: Conservation of Momentum in a Collision****Scenario:**A railroad cart weighing 18,000 kg collides with a stationary cart weighing 12,000 kg. After the collision, the carts lock together and move as one unit. Their speed immediatel