A 2,511 kg railroad car moving at 4.192 m/s to the east collides with a 8,199 kg railroad car moving at 3.781 m/s to the west. If the cars couple together, what is their velocity after the collision?

Since we are dealing with two objects colliding with each other, we can use the Conservation of Momentum equation to help us find the final velocity.

m₁v₁ + m₂v₂ = m₁v₁ + m₂v₂
where the left-hand side of the equation is before the collision and the right-hand side of the equation is after the collision

If the cars are coupled together after the collision, this means that we are dealing with an inelastic collision.

Let's make the positive direction to the East and the negative direction to the West.

The railroad car moving to the East will be labeled as Car 1 and the railroad car moving to the West will be labeled as Car 2.

Listing out the variables we know from the question itself, we have:

m₁ = 2,511 kg
m₂ = 8,199 kg
v₁(i) = 4.192 m/s to the East
v₂(i) = 3.781 m/s to the West

We are trying to solve for v(f), and since the cars couple together after the collision, they will have the same final velocity.

Let's substitute the known variables into the Conservation of Momentum equation.

(2511)(4.192) + (8199)(-3.781) = (2511)(v) + (8199)(v)

Since the final velocity is the same for both cars, we can factor out "v".

(2511)(4.192) + (8199)(-3.781) = (2511 + 8199)(v)

Simplify both sides of the equation.

10526.112 - 31000.419 = 10710v
-20474.307 = 10710v

Divide both sides of the equation by 10,710 to isolate v.

-1.9117 = v

The final velocity of both railroad cars after the collision is -1.9117 m/s, meaning that they both traveled to the West at a speed of -1.9117 m/s.