# Video: Elastic and Inelastic Collisions

In this video we learn how to define elastic and inelastic collisions, in which cases momentum and kinetic energy are conserved, and how to categorize a collision using the coefficient of restitution.

14:52

### Video Transcript

In this video, we’re going to learn about elastic and inelastic collisions. We’ll learn what these two collision types are, how they differ from one another. And we’ll get to work with them practically in a few examples.

To start out, imagine that you are at a billiards hall and you’re performing some tricks on a billiard table. One trick you’ve been working on lately is chipping the cue ball up into the air and then landing it in a bowl placed on the table. The goal is to use the momentum of the ball to slide the ball across a line marked down on the table.

To understand this interaction, it will be helpful to know something about elastic and inelastic collisions. One thing we can say at the outset is that, for any type of collision, elastic or inelastic, linear momentum is conserved through the interaction.

If we have two or even more objects that run into one another, whether they stick together upon colliding or bounce off each other, we can say that the total momentum of the system before the interaction is the same as the momentum of the system after it. So for any type of collision, the conservation of linear momentum is an overarching principle.

When we talk about collisions, we may wonder what makes one collision elastic and another inelastic. Imagine that you take a blob of soft putty and you throw it against a wall. When the putty hits the wall, it will spread out a bit and stick to the spot where it hit.

When two objects that meet through a collision are joined together after the collision, then we call that an inelastic collision. One thing to notice about inelastic collisions is that when they happen, energy is lost in the interaction. In particular, kinetic energy is lost when the objects that interact change shape somehow, like the putty does in this collision.

In inelastic collisions, energy can be lost in other ways as well. It could be lost through sound. It could be lost through heat generated due to friction. However it’s lost, when the interacting objects combine together after they collide, we call that collision inelastic.

On the other hand, if particles or objects are able to collide without any kinetic energy being lost on the part of any of the interacting objects, we call that type of collision elastic. For an elastic collision, we can say that the initial kinetic energy of the system is the same as the final kinetic energy.

Examples of collisions that are perfectly 100 percent elastic are fairly rare. For this condition to be met, interacting particles couldn’t make a sound when they hit. They couldn’t change shape at all. And they couldn’t exert friction on one another. To meet these conditions, we often need to go to the very small, to the scale of atoms interacting in our atmosphere, or the very large, to massive objects interacting in space, which don’t physically contact one another but effectively collide through gravity.

It’s helpful to us practically to classify collisions as either inelastic or elastic. But in reality, collisions can exist on a spectrum between these two opposite extremes. There’s a tool known as the coefficient of restitution that quantifies just how elastic or inelastic any given collision is. Here’s how it works.

Say you want to know the coefficient of restitution of a ball colliding with a wall. If the speed of the ball as it approaches the wall is 𝑣 sub 𝑖 and the ball’s speed after it bounces off the wall is 𝑣 sub 𝑓, then the coefficient of restitution is equal to the ratio 𝑣 sub 𝑓 over 𝑣 sub 𝑖.

Let’s consider the limiting cases in the situation. Say we threw the ball at the wall. And instead of bouncing off, it hits the wall and attaches to it, splat! In that case, 𝑣 sub 𝑓, the final speed of the ball, is zero. If we look at our coefficient of restitution, 𝑣 sub 𝑓 being zero means that 𝐶 sub 𝑅 overall is zero. So when the coefficient of restitution is zero, that means our collision is perfectly inelastic.

At the other end of the spectrum, say we throw a ball against the wall and it bounces back with the speed 𝑣 sub 𝑓 equal in magnitude to 𝑣 sub 𝑖. In that case, our coefficient of restitution is one. And this result represents an interaction which is perfectly elastic, one where no kinetic energy is lost.

In real-life examples, 𝑣 sub 𝑓, the ball’s rebound speed, is typically between these extremes of zero and one. It’s neither perfectly inelastic nor perfectly elastic. Still, when collisions are perfectly inelastic or elastic or can be approximated as such, we have helpful tools for solving examples involving those collisions.

In particular, when a collision is classified as inelastic, we know that linear momentum in that interaction is conserved. And if the collision is elastic, not only is linear momentum conserved, but so is kinetic energy. It’s helpful to find a way to remember that, for inelastic collisions, kinetic energy is conserved, while for inelastic collisions it’s not.

One way that may seem silly but can be helpful is to focus on the capital 𝐸 in elastic and the capital 𝐸 in KE for kinetic energy. Keeping that connection in mind can remind us for which type of collision kinetic energy is conserved. Let’s get some practice now with elastic and inelastic collisions through a couple of examples.

Two hockey players of unequal masses collide with each other head-on, each moving at a speed of 15 meters per second. After the collision, the hockey players move in the same direction as each other, each at the speed of 5.0 meters per second. How many times greater is the mass of the more massive hockey player than the mass of the other hockey player?

If we call the mass of the larger player capital 𝑀 and the mass of the smaller player lowercase 𝑚, we want to solve for the ratio capital 𝑀 over lowercase 𝑚. We’ll start on our solution by drawing a diagram. When we sketch out both the initial and final conditions, that is, before and after the collision, we see that, initially, the larger and the smaller player approach one another at the same speed we’ve called 𝑣 sub 𝑖 of 15 meters per second.

After a head-on collision, both players move in the same direction at the same speed. We’ve called it 𝑣 sub 𝑓 of 5.0 meters per second. The direction of 𝑣 sub 𝑓 is the same as the original direction of the larger player. Based on all this, we want to solve for how many times more massive the larger player is than the small player.

Since this example involves a collision, the first thing we can ask ourselves is whether the collision is elastic or inelastic. Since the two players effectively stick together after the collision has occurred, we know that this is an inelastic collision. In a collision of this type, we know that the initial momentum involved is equal to the final momentum of the system. We can’t say, however, that kinetic energy is conserved. That’s only true for elastic collisions.

Knowing that momentum is a vector with both magnitude and direction, let’s write out the initial and final momentum of our two-hockey-player system. Recalling that an object’s momentum is equal to its mass times its velocity, if we decide that motion to the right is motion in the positive direction, we can write that the initial momentum of our system is equal to the mass of the larger player times his speed, 𝑣 sub 𝑖, minus the mass of the smaller player times his speed, 𝑣 sub 𝑖. The reason for the minus sign is because our second player, the smaller one, is moving in the negative direction.

We can write the final momentum of our system as capital 𝑀 plus lowercase 𝑚, since the players move together, multiplied by the final speed, 𝑣 sub 𝑓, which is positive because it’s to the right. Applying the principle of conservation of momentum, we can write that the quantity capital 𝑀 minus lowercase 𝑚 times 𝑣 sub 𝑖 is equal to the quantity capital 𝑀 plus lowercase 𝑚 times 𝑣 sub 𝑓.

If we rewrite this equation so that instead of grouping it by speed we group by the mass of the two different players, we see that the mass of the larger player times the difference in the initial and final speeds equals the mass of the smaller player times the sum of these two speeds. If we cross-multiply to get the mass ratio, the larger mass over the smaller mass, we see this is equal to the sum of the initial and final speeds divided by their difference.

Since we know both 𝑣 sub 𝑖 and 𝑣 sub 𝑓, we can plug in and solve for the ratio of the masses of the players. Plugging these values in, we see that the units of meters per second cancel out and that our numerator is equal to 20 and our denominator equal to 10. When we calculate this ratio then, we find it’s 2.0. That’s how many times more massive the larger player is compared to the smaller.

Now let’s look at an example involving an elastic collision.

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 𝑚 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 𝑎𝑐 all over two 𝑎, where in our case 𝑎 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.

Let’s summarize what we’ve learned about elastic and inelastic collisions. We’ve seen that collisions can be inelastic, where momentum only is conserved, or elastic, where momentum and kinetic energy are conserved. And we’ve seen that the coefficient of restitution tells us just how elastic or inelastic a collision is.