Video: Determining Speeds of Objects after Elastic Collision

In an elastic collision, a 400 kg 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 m/s and that of the trailing car is 6.00 m/s. 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?

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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?

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

After these two cars collide, they move at new speeds that we’ve called 𝑣 sub one 𝑓 and 𝑣 sub two 𝑓, respectively. It’s those final speeds of the trailing and leading car we want to solve for. We’re 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 object’s mass times its velocity and is a vector and that kinetic energy is equal to half an object’s 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, it’s equal to the mass of each car times its initial speed. And finally, it’s equal to the mass of each car times its final speed.

We’re 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 let’s 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 we’re now using the final speeds each car attains.

We see the expression one-half times π‘š 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 doesn’t make much difference which one we start with.

Just to pick one, let’s solve first for the final speed of car two, the leading car in this collision. We’ll start doing this by rearranging our conservation of momentum expression to solve for 𝑣 sub one 𝑓. 𝑣 sub one 𝑓 is equal to 𝑣 sub one 𝑖 plus 𝑣 sub two 𝑖 minus 𝑣 sub two 𝑓. We’re given both 𝑣 sub one 𝑖 and 𝑣 sub two 𝑖 in the problem statement and can plug those in now.

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

Considering the left-hand side of this expression, if we plug in 6.00 meters per second for 𝑣 sub one 𝑖 and 5.60 meters per second for 𝑣 sub two 𝑖, 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 𝑣 sub two 𝑓 plus two times 𝑣 sub two 𝑓 squared.

We’re now going to rearrange this whole equation into an expression in the form of a quadratic equation. Once we’ve achieved this form, we recall that the roots of a quadratic equation are equal to negative 𝑏 plus or minus the square root of 𝑏 squared minus four π‘Žπ‘ all over two π‘Ž, where in our case π‘Ž is equal to two, 𝑏 is equal to negative 23.2 meters per second, and 𝑐 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.

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