# Car Crash Physics

1. Aug 16, 2010

### Bussani

I've been sitting here thinking about collisions, deceleration, and conservation of momentum, and I just want to see if I'm on the right track.

For a start, let's say we crash a car into a solid, "unmovable" wall at 50 mph, and measure the results. If we then take two cars identical to that first one and crash them into each other in a head-on collision, each traveling at 50 mph, the result should be the same--both decelerate in the same amount of time as if they'd hit an invisible wall.

From there I started to wonder how this would look with a little relativity thrown in. A person in one of the cars should be able to claim that they're not moving and that the other car is coming towards them at 100 mph instead. The next logical question seemed to be whether a car going at 100 mph and crashing into an identical, stationary car would produce a practically identical result. This seems to make sense to me--car A, going at 100mph, should push car B as it hits it, probably causing both cars to now be moving in the direction car A was moving at closer to 50 mph than 100 mph. If that's right then the acceleration/deceleration felt by the two drivers should be roughly the same as the 50/50 head-on collision. Each car's speed would have changed by a factor of 50 mph.

Am I thinking about this the right way? Would the damage to each car be similar in all three of these examples, and would the forces felt by the people inside these cars actually be similar as well?

Thanks for any thoughts.

2. Aug 16, 2010

### AJ Bentley

Sounds all-right to me.

My initial reaction was 'no way!' there's twice as much energy and momentum involved with a two car collision - but your reasoning is correct. Although I doubt that any wall would have the 'immovable' characteristic you need - possibly the side of an aircraft carrier might work.

3. Aug 16, 2010

### ErikD

Mythbusters actually tested this:

Last edited by a moderator: Sep 25, 2014
4. Aug 16, 2010

### Bob S

Bussani-
Think about this. the energy of each car crashing to each other at v = 50 mph is ½mv2, so for both cars the total energy is mv2. But a car at 100 mph crashing into a stationary car represents a total energy of ½m(2v)2 = 2 mv2. Why the difference?

Bob S

5. Aug 16, 2010

### cabraham

Because the reference frames involved are non-inertial. Galilean transformations, like relativistic transformations, are valid for inertial frames only. At least that is what I understand. Am I mistaken? When the cars collide, both undergo acceleration, making the reference frames non-inertial.

Claude

Last edited: Aug 16, 2010
6. Aug 16, 2010

### Bussani

Hmm... Kinetic energy is mv2, like you say--mass * velocity2 / 2. This means that if you double the velocity, you quadruple the kinetic energy. What does this mean for our cars? Would the impact be worse, or is that extra energy accounted for by the distance the two cars slide after the 100 mph one collides with the stationary one?

7. Aug 17, 2010

### cjl

Well, if you ignore friction, then the cars after the collision (car going 2v with stationary car) will be going at speed v after the collision (50mph in this case). So, initially, you have a car of mass m going speed 2v, so your KE is equal to m*(2v)2 = 4mv2. Final KE is equal to 2m*v2 since you have 2 cars, each of mass m, traveling at speed 1v. So, your energy dissipated in the collision is 4mv2-2mv2=2mv2.

In the 2 moving car case, your final KE is effectively zero, but your initial KE is 2*(1m*(1v)2) = 2mv2. So, the energy dissipated in the collision is 2mv2 - 0 = 2mv2. This is exactly the same as calculated above, so the damage would be identical.

8. Aug 17, 2010

### Bussani

That makes a lot of sense. Thanks for the reply!