# Elastic Collisions in 1D

I need a nice model to predict the final velocity when two balls collide elastically. Don’t worry why I need this—just trust me.

After working on this for a short bit and making an error, I realized what I need to do. I need to blog about it. A blog is the perfect place to work things out.

So, here is the situation. A ball of mass 10 kg (ball A) is moving with a speed of 0.1 m/s in the positive x-direction. This collides with a 1 kg ball (ball B) moving at 0.1 m/s in the negative x-direction. What is the final velocity of the two balls if the collision is perfectly elastic.

For a perfectly elastic collision, the following two things are true:

• Momentum is conserved. The total momentum before the collision is equal to the total momentum after the collision.
• Kinetic energy is conserved. The total kinetic energy is the same before and after the collision.

In one dimension, I can write this as the following two equations. I’m going to drop the “x” notation since you already know it’s in the x-direction. Also, I am going to use A1 for the velocity of A before the collision and A2 for after. Same for ball B.

$m_Av_{A1}+m_B v_{B1} = m_Av_{A2}+m_B v_{B2}$

$\frac{1}{2}m_A v_{A1}^2+\frac{1}{2}m_B v_{B1}^2 = \frac{1}{2}m_A v_{A2}^2+\frac{1}{2}m_B v_{B2}^2$

That’s two equations and two unknowns (the two final velocities). Before solving this, I want to find the answer with a numerical calculation.

Numerical Solution

Here’s the basic plan (I’m not going over all the deets).

• Model the two masses as points with springs on them (not really going to show the springs).
• When the two masses “overlap” there is a spring force pushing them apart. The strength of this force depends on the amount they overlap.
• Calculate the position and force on each ball (the force would be the zero-vector in cases where they aren’t “touching”).
• Update the momentum of the balls.
• Update the position of the balls.
• Repeat until you get bored.

Oh, make sure you set your fake spring constant high enough. If it’s too low, the two masses can just pass through each other (which would still be an elastic collision).

Here is what it looks like.

Here is the code (you should take a look). Oh, the final velocities are 0.0636 m/s for ball A and 0.2639 m/s for ball B. Also, here is a plot of the momentum so you can see momentum is conserved.

What about the kinetic energy? Here you go.

Actually, notice that KE is NOT conserved. During the collision there is a decrease in the total KE because of the elastic potential energy. I just thought that was cool.

Analytical Solution

Now let’s get to solving this sucker. I’m going to start with a trick—a trick that I’m pretty sure will work (but not positive). Instead of having the two balls moving towards each other at a speed of 0.1 m/s each, I am going to use the reference frame that has ball B with an initial speed of 0 m/s and ball A with a speed of 0.2 m/s.

Since I am switching reference frames, I am going to rename the velocities. I am going to call ball A velocity C1 and C2 and then ball B will be D1 and D2 (for final and initial). Technically, I should use prime notation – but I think it will just get messy.

So, here is how it looks in the new reference frame.

In general, the initial C velocity would be:

$v_{C1}=v_{A1}-v_{B1}$

Now I get the following for the momentum and kinetic energy conservation equations.

$m_A v_{C1}=m_A v_{C2}+m_B v_{D2}$

$\frac{1}{2}m_A v_{C1}^2=\frac{1}{2}m_A v_{C2}^2 +\frac{1}{2}m_B v_{D2}^2$

Now we have two equations two unknowns. I’m going to cheat. I worked this out on paper and I’m just going to take a picture of it.

Here is the final solution (in case you can’t read it).

$v_{C2} = v_{C1}-\frac{m_B}{m_A}v_{D2}$

$v_{D2} = \frac{2v_{C1}}{\frac{m_B}{m_A}+1}$

So, you can get a value for vD2 and then plug that into vC2. After that, you can convert them back to the stationary reference frame to get vA2 and vB2.

Boom. It works. Here is my calculation. Just to be clear, it looks like this:

The output looks like this:

Winning. That agrees with my numerical model.

# MacGyver Season 1 Episode 16 Science Notes: Hook

Quicksand

Yes, it’s true. You don’t really sink all the way down in quicksand—that’s because the density of the stuff is greater than the density of water. Essentially, you float.

Here is a nice video on quicksand.

Pool shot.

I. Love. This. So, MacGyver is there trying to score a nice shot in pool (the game, not the pool). He starts thinking about all the physics to get the perfect shot—one with some curve. Here is what goes through his mind.

That’s pretty awesome, right? Let’s go over some of the key equations.

First, why does the ball turn? If you want to turn, you have to have a sideways force. In this case, the sideways force is a frictional force on the ball as it spins and slides on the table.

Once you know the frictional force, you can use this to find the new vector velocity after some short time. This is basically the numerical version of the definition of acceleration. Here’s that equation.

$\vec{v}_3= \vec{v}_2+\frac{\vec{F}_f}{m}\Delta t$

Oh, vectors. Look at the vector notation. Winning.

Next there is the changing angular speed of the ball. Since the ball is spinning with a frictional force, the ball would slow down. We describe the change in angular motion by using the angular momentum principle. This states:

$\vec{\tau} = \frac{\Delta \vec{L}}{\Delta t}$

Where L is the angular momentum vector. For a rigid object, the angular momentum is the product of the moment of inertia and the angular velocity (for most cases).

$\vec{L}=I\vec{\omega}$

Where I is the moment of inertia (or as I like to call it, the rotational mass) for a sphere.

$I = \frac{2}{5}MR^2$

That’s pretty much all the equations you see.

Of course the amazing part is that humans can make these very complicated shots WITHOUT doing the calculations. I don’t know how that works.

Zip-tie nunchucks.

Two broken pieces of a pool stick and a towel. Boom. Nunchuck. But this is all I can think of.

Magnet phone tracker

How do you track a fleeing truck? You stick your smart phone on the bottom using a magnet. Oh, this works so well that Spider-Man used this same trick the following year in Spider-Man: Homecoming when he left his phone in the Vulture’s car.

There is a small problem with those car magnets. They are like refrigerator magnets in that they have weird magnetic domains. Actually, you should try this experiment.

• Grab a fridge magnet.
• Flip it around and put it on the fridge.
• Oh, it doesn’t work!

The magnetic domains in these flat magnets are such that they stick on one side but not the other. That makes it tough to use for a magnetic phone tracker.

A better method would be to run a wire (or zip tie) through the magnet and around the phone. Like this.

Stab pepper spray with a knife

Yes, if you poke a hole in a can of pepper spray it will get pepper spray all over the place.

Circuit board knife

MacGyver uses a broken circuit board to cut through zip ties. Zip ties aren’t that strong anyway—it seems very plausible that you could sharpen a circuit board to cut through one of these things.

Control a car remotely

Is it possible to control a car with a computer? Sadly, this is real.

Moonshine spray

MacGyver sprays 155 proof moonshine and threatens to light it. Yes, 155 proof can catch on fire.