By Rhett Allain

One of the many awesome things about physics is you can study it anywhere. I’ll prove it with a fun problem that involves a cart and a ball. Not just any cart, but a cart that rolls along a table and shoots a ball into the air when you tug on a strung. Like this:

I think it’s cool that the ball lands in the cart. But how?

### Kinematics and Projectile Motion

Don’t worry. I won’t make you build a ball-launching cart. I can explain it right here. The ball leaving the launcher is an example of projectile motion because only one force acts on the ball—gravitational force. That means the ball accelerates at a rate of -9.8 m/s^{2} (or –*g*) vertically and at a constant rate horizontally. Wrapping your head around this requires understanding two important ideas. First, the horizontal motion (usually called the x-direction) and vertical motion (the y-direction) act independently of each other. The speed of the object in the x-direction does not change the motion in the y-direction. The second idea gets a little trickier, because it requires knowing the relationship between the position, velocity, and acceleration of the ball in the y-direction. Physicists call this a kinematic equation.

In this equation, *y*_{2} represents the final position in the y-direction and *y*_{1} represents the starting position. Of course *t* is the time and –*g* is the vertical acceleration. This ends your crash course kinematics. But why does the ball land back in the cart? Let me start with an expression for the position of the cart. At the moment the launcher fires the ball, the cart has an x-position of zero at a time of zero. Because the cart moves at constant speed (with an acceleration of zero), I can write the equation of motion as:

The subscript *c2* represents the final position of the cart. But what about the ball? Look closely and you’ll see that it moves both vertically and horizontally when launched. It isn’t fired straight up, but rather straight up with respect to the moving cart. This means the ball moves at the same horizontal velocity as the cart. With no forces acting on the ball in the horizontal direction, it should move at a constant speed in the y-direction. Let me show you with an equation, where the *b *subscript represents the position of the ball:

Look familiar? It should. It’s the same equation for the motion of the cart in the x-direction. That means that at any point in time, the ball is above the cart, so obviously it lands in the cart. That’s pretty awesome.

### Cart and Ball on a Ramp

Now for the challenge: What happens if the cart rolls down an incline? The launcher still fires the ball in a direction perpendicular to the cart, but then what? Let me show you a picture just to make sure you understand the situation.

Will the ball land in the cart? Will it land in front of the cart? How about behind it? Yes, I could just show you what happens. But think about it first, and apply what I told you earlier about kinematics. In the meantime, let this random picture serve as a barrier between your thoughts and the answer that follows.

If you’ve scrolled past the picture, I assume you have an answer supported by some kind of logical reasoning. Now for the answer. Here you go.

Yes, the ball lands right back in the cart. Why? Let me start with a diagram.

In this case, I put the x-axis in the same direction as the ramp. This means the cart moves only in the x-direction. However, it won’t move at a constant speed. Instead it accelerates because a component of the gravitational force acts in the same direction. Here’s the equation of motion for the cart, assuming it starts at *x* = 0 at time *t* = 0:

What about the ball? Once airborne, only gravitational force acts on it, and because the x-axis is not perpendicular to the gravitational force, the ball accelerates due to gravity in the x *and* y-directions. Further, since the ball launches perpendicular to the direction of the cart, it has the same x-velocity as the cart. That means the equation of motion in the x-direction looks like this:

The ball and the cart have the exact same equation of motion. The x-position of the ball and x-position of the cart never change, so the ball lands in the cart. Pretty awesome.

### Homework

Let me leave you with some homework in case you want to play around with this some more.

- What happens with a vertical ramp (θ is 90 degrees)?
- In the case of a horizontal cart, find an expression for the launch velocity (as a vector) for the ball.
- In the case of the inclined plane, use a horizontal a-axis and show that the ball lands back in the cart. This one is tricky, but not impossible.
- What happens if a small frictional force acts on the cart? How great can this force be and still allow the ball to come close to landing in the cart?
- Create a numerical model (I suggest using Python) to show that the ball lands back in the cart for the case of an inclined plane. If you can model it, you can understand it.

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