What happens when an object collides with a stationary wall at some incident angle? If this object is a ball, we often say that it “reflects” off the wall just like light does with the incident angle equal to the reflected angle. Two questions:

- Is this true? Does the incident angle equal the reflected angle for a ball hitting a wall?
- Why would this “rule” be true and when would it not work?

Let’s take a look.

### Does incident angle equal reflected angle?

Of course this question depends on the types of objects colliding, but let’s just do a simple test. I could toss different balls at the floor and look at the reflected angle–but I’m not going to do that. The problem is that the velocity of the ball would change both before and after the collision. Oh sure, you could still do it but it would be a little more complicated.

Instead I am going to take this floating puck and push it along the floor (the puck has a fan in it so that it hovers with low friction). Using a video as recorded from above, I can get the following plot for the trajectory of this puck as it collides (x vs. y).

The slope of the trajectory line for the incident disk is -1.60 and the slope of the reflected 1.133. These aren’t exactly the same–but maybe it would be easier to look at them as angles. The angle of incidence is 57.9? and the reflected angle is 48.6?.

What about a few more tests? Here is that same hover disk with the same wall but at different incident angles. This is a plot of the incident trajectory slope vs. the reflected trajectory slope.

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If the law of reflection worked perfectly for this disk, the slope of this line would be 1.0–but it’s not. But why doesn’t it exactly work? Here is a plot of both the x and y position as a function of time. From the slopes of these lines we can get the x and y velocities.

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First look at the horizontal position. If you fit a linear function to the data, you would see that the x-velocity before the collision is 0.7 m/s and afterwards it is 0.37 m/s. So it slows down in the horizontal direction. For the vertical velocity, it goes from -1.09 m/s to 0.452 m/s. Oh, the disk also spins after the collision–but let’s not worry about that right now.

If the horizontal velocity didn’t change and the vertical velocity just changed directions–then you would have a perfect “reflection” collision. Of course, the changes in velocity depend on the types of objects colliding. I suspect that I could find a different set of materials that produces a better reflection.

### How does reflection work?

Start with a ball moving towards a wall with some initial velocity. When the ball comes in contact with the wall, there is a force exerted on the ball. Here is a diagram of the perfect collision.

When dealing with forces and momentum, we should of course consider the Momentum Principle:

In this special collision, the force from the wall is only perpendicular to the wall (in the y-direction). This means that there is no change in the x-component of momentum and only a change in the y-momentum. If this is a perfectly elastic collision such that the total kinetic energy is constant, then this y-momentum must have the same magnitude as before the collision (but in the opposite direction). This would make the reflected angle the same as the incident angle.

But what happens in our real collision case? It’s not a perfect collision so that the diagram might look like this:

For the non-perfect collision, the wall exerts two forces on the ball (or you could combine these into just one force if it made you happy). There is still a force pushing perpendicular to the wall, but there is also a frictional force parallel to the wall. This friction force does two things. First, it changes the momentum in the x-direction and second it exerts a torque on the disk. In the end, the x-momentum of the disk (or ball) changes and the ball acquires a spin. This is exactly what we see in the animation above.

But how do you get a “perfect” collision? You need two things. First, you need an elastic collision so that there is no kinetic energy lost. If you lose kinetic energy, there’s no way the y-velocity will remain the same. Second, you need to have no frictional forces on the object. These frictional forces will just change the x-velocity of the ball.

### Modeling a Ball-Wall Collision

You know I can’t stop without first making a numerical model. OK, so how do you model a ball colliding with a wall? The easiest way is with a spring. Here’s how my calculation will work.

- The ball moves along just normally with a constant velocity.
- If the center of the ball is closer to the wall than the radius of the ball, then there is a force pushing on the ball perpendicular to the wall.
- The strength of this force will be proportional to the amount the ball overlaps into the wall.
- When the ball is no longer “in contact” with the wall, the force goes back to zero.

What about a collision with friction? If I want to add a frictional force, I will just do the exact same thing except that the force from the wall won’t be completely perpendicular to the wall. There will be a small component of this force parallel to the wall and in the opposite direction to the velocity of the ball. I didn’t include loss of kinetic energy in the perpendicular direction–that’s a bit more complicated to model.

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Just press the “play” button to run the code. You can see that there are two balls. They are initially on top of each other, but after the collision they take a different path. The model isn’t perfect–but it mostly works. Go ahead and change the calculation a little bit to see if you can make a better model.

Why do I even care about balls colliding with walls? Trust me, there is a reason–but I’ll get to that in a future post.