I am so pumped up that MythBusters is back on. Not only do I like the show, but it offers so many blogging opportunities. Their latest show featured car crashing myths. One of the myths from the episode was a redo of the myth where two trucks crash head on simultaneously crushing a smaller car in the middle.

The first test was very similar to the previous time they tested this, but faster. They towed two 18 wheelers to crash together around 50 mph and smash a stationary car. The results were impressive. However, they did not have the result of the car completely contained in the collision.

For the next test, they changed it so that a rocket sled crashed into a stationary car next to a stationary wall. Adam made a claim (or statement) something about if the rocket car was going 100 mph into a stationary car, it would be equivalent to two 50 mph cars colliding. Actually, this is not true. I can’t remember if Adam actually said this or if he really meant it, however it is still and interesting situation.

Let me start with a couple of somewhat similar situations.

### Two cars, in space going the same speed, smashing a stationary object

Remember that in space, no one can hear you scream. Also, this makes it a little simpler to keep track of all the interactions. I will assume this is far from any massive objects so that gravity can be neglected. Clearly, there is no air resistance. The lines behind the cars are zooming motion they make. Assume they already turned off their rockets. The smiley face in the middle is some target car or object. Finally, I am looking at this in the frame where the target is stationary. Two things must be true:

The total momentum vector before and after the collision must be the same. This will be a constant value as long as there are no external forces, like this situation. So the total momentum vector is the zero vector. This is because the target object is not moving, and the momentums of the two cars are opposite each other.

For the energy, this will be constant as long as there is no work done on the system. Here the system consists of the two cars and the target. No work done. If these cars are moving at normal car speeds (or even normal rocket speeds) that are much less than the speed of light, then the mass energy will not really change. Before the collision all the energy is essentially kinetic energy. After the collision this energy can be either in kinetic energy of the moving objects or in something I will call structural energy in the deformation of the vehicles.

To look at the different things that can happen, let me simplify by looking at the component of momentum in the direction of motion (since this is 1-d problem). This will give the initial momentum as:

m_{c} is the mass of the car(s), m_{t} is the mass of the target. The target is at rest, so its component of momentum is zero in that direction. The energy before the collision is:

I neglected the mass energy of the objects since it doesn’t really change. So, I can have any situation in which the total component of momentum is zero and the total energy is m_{c}v^{2}. One such case would be in the two cars simply “bounce” off the target.

For this case, it is obvious that the momentum and the energy are the same as before.

Another case is when the two cars smash and stop at the target. In this case, the final momentum would be:

Forgive me for not putting units with the velocity – but you see this would indeed give the same initial momentum. What about the energy? Clearly the kinetic energy is zero so:

I am calling structural energy the change in the system of the cars. Clearly it takes some energy to twist all that metal and stuff. This is the key point though. If the mythbusters are trying to smash up this target car, they want to look at how much structural energy (I made that word up) goes into the thing. In this case, it is m_{c}v^{2} energy goes into changing all the cars (who knows how that energy is divided up). Now for the next case.

### One car moving at 2v_{1} crashing into an identical stationary car and target car

Here I am starting one car with twice the speed of the previous case. So, the initial momentum is:

The initial momentum is NOT zero in this case. Duh, that was obvious. I guess I didn’t even need to say that. Oh well, what about the energy before the collision?

So, by doubling the speed of one of the cars and making the other stationary, the initial energy is twice as much. This happens because the kinetic energy depends on the square of the velocity. Also, the initial momentum is not zero so the final momentum will not be zero. What if all the stuff sticks together during the collision? How much energy would go into structural energy? If everything sticks together, what would be its final speed? According to momentum:

So, it wouldn’t be stationary. The energy equation for this case would be:

All the energy can not be transfered into structural energy because the stuff still has to be moving after the collision in order to conserve momentum. Ok, well, what about the mythbusters case? It was different because they didn’t do it in space. Exactly. So let me look at that case and see how it is different.

### One moving, the other fixed to the ground and not in space

If I again take the two cars plus the target as the system, there is a huge difference in this situation: external forces. Not gravity, that doesn’t really do anything since the motion is perpendicular to gravity and the normal force from the ground essentially is opposite of it. However, the force of the ground on the stationary car IS important. This means that for this system, the initial and final momentum are not the same. The momentum principle can be written as:

Where the net force on the system is the force the ground exerts on the stationary car when it gets hit and delta t is the time the collision lasts (short). This is why the final momentum is not the same as the initial. In the frame of the Earth, the final momentum is zero (vector). What about energy?

In this system, the external forces don’t do any work. The force from the ground doesn’t actually move, so there is no work. This means that all the energy of the initial moving car can go into structural energy so that:

Ok, one more case. Is there anyway to make one object stationary and the other moving and have everything work? What if I did this in space (for simplicity) and had one of the cars stationary? One thing would be to look at the collision in the reference frame of one of the cars. This would be a little different.

### Reference frame of one of the moving cars

If you look at the first case I did, and pretend like you are moving with the right car (which I labeled as “B”), then you would see the other car (A) moving towards you at a speed of 2v_{1}. But you would also see the target moving towards you at a speed v_{1}. This would give a total initial momentum of:

If everything sticks together, it will be moving at a speed of (in the frame of car B):

So, the final speed of the mangled stuff is the same initial speed in the frame of the target as the two moving cars. This makes sense. If in the frame of the target, everything is stopped at the end and in the frame of the car B, the target is moving towards you at speed v_{1} then after it should still be v_{1}. So, the two frames agree. What about energy?

From the energy (before and after) equation, I can solve for the structural energy:

This is the EXACT same structural energy from before. That is good. The two cases should agree on something measurable like that. So, in summary, what is the point of this post? First, making one car stationary and the other moving twice as fast is NOT the same thing. Second, the MythBusters are still awesome. Third, if you move to a different reference frame, stuff should still work.

### Update

There was an algebra mistake as pointed out by the commenters (thanks!). I fixed it. Strange this is that I derived that for the last case the E-structural was 2m(v1)^2 and I said this was IDENTICAL as viewed from the other frame (which it isn’t). Odd. When I fixed it, the two are the same. I guess that’s what happens when you blog on vacation.