### 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.