### Video Transcript

Train cars are coupled together by being bumped into one another. Suppose two loaded
train cars are moving toward one another, the first having a mass of one point five
zero times ten to the fifth kilograms and a velocity of zero point three zero metres
per second 𝑖 and the second having a mass of one point one zero times ten to the fifth kilograms and
the velocity of negative zero point one two meters per second 𝑖. What is their final velocity?

In this problem, we’ll assume that the train cars lose no energy to friction so
that their velocities when they encounter one another are equal to the initial velocities
given in the problem statement.

We’re told that the first train car has a mass of one point five
zero times ten to the fifth kilograms and is moving at zero point three zero meters per second in the positive 𝑖
direction. The second car has a mass of one point one zero times ten to the fifth kilograms and is moving in the negative 𝑖 direction at zero point one two meters per
second.

We’re asked to solve for the final velocity of the two cars after they’re coupled
together. Let’s call that 𝑣 sub 𝑓. This problem is all about momentum and how it is conserved, so let’s begin our
approach by recalling the conservation of momentum.

In its most simple form, the conservation of momentum tells us that the initial momentum in a system, which we represented as 𝑝 sub
𝑖, is equal to the final momentum in that system, 𝑝 sub 𝑓.

Recall that momentum is expressed as a mass of an object times its velocity, so
we can rewrite this equation as the initial mass of our system multiplied by its initial velocity is equal to the
final mass of our system multiplied by the final velocity.

Let’s begin to apply this conservation principle to our scenario by recalling
the values we were given in the problem statement. We’re told that the mass of our first train car, which we’ll call 𝑚 sub 1, is
one point five zero times ten to the fifth kilograms and that that car has velocity, we’ll call 𝑣 sub 1, of zero point three zero meters per second
in the positive 𝑖 direction.

The second train car, which we’ll call 𝑚 sub 2, has a mass of one point one
zero times ten to the fifth kilograms and a velocity, 𝑣 sub 2, of negative zero point one two meters per second in the
𝑖 direction

Looking back over at our conservation of momentum equation, we realize the 𝑚 sub 1, 𝑚 sub 2,
𝑣 sub 1, and 𝑣 sub 2 all apply to the initial momentum of our system, before the railcars have
encountered one another. So let’s write out the initial and final momentum of our system as best we can.

The initial momentum of our train car system is equal to the mass of car 1 times
its speed minus the mass of car 2 times its speed. Here we assume, as given in the problem, that motion to the right in our diagram is motion in the positive 𝑖 direction.

That’s why we subtract 𝑚 sub 2 𝑣 sub 2 from 𝑚 sub 1 𝑣 sub 1 and not the other way around. Now what’s the final momentum of our system, 𝑝 sub 𝑓? That’s equal to the coupled mass of these two cars times their final velocity, which is what we’re solving for, 𝑣 sub 𝑓.

Since the masses combine to form the final mass of our system, we can write our
final momentum as 𝑚 sub 1 plus 𝑚 sub 2 quantity times 𝑣 sub 𝑓.

And again we work with the sum of the masses for our final mass of the system
because the train cars couple together and combine. Since we’re looking to solve for 𝑣 sub 𝑓, let’s divide both sides of equation
by the quantity 𝑚 sub 1 plus 𝑚 sub 2.

When we do that, that quantity cancels out on the right side our equation, leaving
us with 𝑣 sub 𝑓 by itself. When we rewrite this expression for 𝑣 sub 𝑓, we see that the final speed of our railcars is equal to 𝑚 sub 1 𝑣 sub 1 minus 𝑚 sub 2 𝑣 sub 2 all
divided by the sum of the masses, 𝑚 sub 1 plus 𝑚 sub 2.

Now we plug in for these masses and speeds from the information given in our
problem statement, and we find out the final velocity of our train car system is equal to the mass
of car 1 multiplied by its initial velocity minus the mass of car 2 times its initial speed all divided
by the sum of the two masses, which equals zero point one two 𝑚 meters per second in the
positive 𝑖 direction.

This is the final velocity of the two railcars after they’re joined together.