Answer:
6.12 m/s
Step-by-step explanation:
Using the law of conservation of momentum
momentum before collision = momentum after collision
m₁v₁ + m₂v₂ = (m₁ + m₂)V (since the train cars become attached to each other) where m₁ = mass of car 1 = 1,825 kg, m₂ = mass of car 2 = 2,645 kg, v₁ = initial velocity of car 1 = + 15.0 m/s (positive since it is moving in the positive x direction), v₂ = initial velocity of car 2 = 0 m/s (since it is initially stationary) and V = velocity of both cars after collision,
So, m₁v₁ + m₂v₂ = (m₁ + m₂)V
m₁v₁ + m₂(0 m/s) = (m₁ + m₂)V
m₁v₁ + 0 = (m₁ + m₂)V
V = m₁v₁/(m₁ + m₂)
substituting the values of the other variables into the equation, we have
V = 1,825 kg × 15.0 m/s/(1,825 kg + 2,645 kg)
V = 27375 kgm/s/ 4470kg
V = 6.124 m/s
V ≅ 6.12 m/s