### Video Transcript

In an elastic collision, a
400-kilogram bumper car collides directly from behind with a second identical bumper
car that is traveling in the same direction. The initial speed of the leading
bumper car is 5.60 meters per second. And that of the trailing car is
6.00 meters per second. Assume that the mass of the drivers
is negligible. What is the final speed of the
leading bumper car? What is the final speed of the
trailing bumper car?

Letβs start out by drawing a
diagram of the situation. In this example, we have two bumper
cars weβve called the trailing car number one and the leading car number two. Car one is moving at an initial
speed weβve called π£ sub one π of 6.00 meters per second. And car two moves at an initial
speed weβve called π£ sub two π of 5.60 meters per second.

After these two cars collide, they
move at new speeds that weβve called π£ sub one π and π£ sub two π,
respectively. Itβs those final speeds of the
trailing and leading car we want to solve for. Weβre told that when these cars
collide, they collide elastically. That tells us two things, first
that momentum is conserved across this interaction, and so is kinetic energy.

Recalling that momentum is equal to
an objectβs mass times its velocity and is a vector and that kinetic energy is equal
to half an objectβs mass times its speed squared, we can write out the initial
momentum and kinetic energy of our system as well as the final momentum and kinetic
energy. Considering the momentum of our
system, initially, itβs equal to the mass of each car times its initial speed. And finally, itβs equal to the mass
of each car times its final speed.

Weβre told that car one and car two
are identical, which means their masses are the same. So we can cancel those terms out
from this expression. This leaves us with a simplified
statement for the conservation of momentum through this collision.

Now letβs consider the kinetic
energy of our system both initially and finally. The initial kinetic energy of our
system is the sum of the kinetic energies of car one and car two before they
collide. The final kinetic energy of the
system is different only in that weβre now using the final speeds each car
attains.

We see the expression one-half
times π appears in every term of this equation. So we can cancel that term out and
simplify our overall conservation expression. What remains is our simplified
conservation of kinetic energy expression for this collision. Knowing that we want to solve for
both the final speed of car one and car two, it doesnβt make much difference which
one we start with.

Just to pick one, letβs solve first
for the final speed of car two, the leading car in this collision. Weβll start doing this by
rearranging our conservation of momentum expression to solve for π£ sub one π. π£ sub one π is equal to π£ sub
one π plus π£ sub two π minus π£ sub two π. Weβre given both π£ sub one π and
π£ sub two π in the problem statement and can plug those in now.

When we do, we find that π£ sub one
π is equal to 11.6 meters per second minus π£ sub two π. Weβll now take this expression for
π£ sub one π and insert it into our conservation of kinetic energy expression. When we do, we get an expression
which is in terms of known values, π£ sub one π and π£ sub two π, and one unknown,
π£ sub two π, that we want to solve for.

Considering the left-hand side of
this expression, if we plug in 6.00 meters per second for π£ sub one π and 5.60
meters per second for π£ sub two π, when we square those terms and add them
together, we get 67.36 meters squared per second squared. Then, on the right-hand side, if we
multiply out this squared expression, then overall the right-hand side of our
expression simplifies to 134.56 meters squared per second squared minus 23.2 meters
per second times π£ sub two π plus two times π£ sub two π squared.

Weβre now going to rearrange this
whole equation into an expression in the form of a quadratic equation. Once weβve achieved this form, we
recall that the roots of a quadratic equation are equal to negative π plus or minus
the square root of π squared minus four ππ all over two π, where in our case π
is equal to two, π is equal to negative 23.2 meters per second, and π is 67.2
meters squared per second squared.

Of the roots of this equation, 6.00
meters per second and 5.60 meters per second, we know the leading car will have a
higher speed than the speed it started with. The final speed of the trailing car
is 11.6 minus 6.00 meters per second, 5.60 meters per second.