Actally the bullet is going
w = (u + v)/(1 + uv/c2)
=0.996c relative to guy on the ground.
I hope he was wearing safety goggles and hearing protection with that hot a load!
Great hypothesis curiousbystander. I'm no relativity expert either but you may be on to something. If the object is relativistic I don't think you can simultaneously measure the mass of the object at all points along its length (which you need to do calculate a classical center of mass)...
Momentum has nothing to do with it. How does momentum "carry it". I hope you guys are joking.
p=mv normally
and in this case
p=ymv (where y is relativity gamma)
I gave you v. and that's all that counts as Turin showed. It's going to drop just as fast no matter what m is, assuming...
I guess I did miss the point. You didn't convince me. Sure it may only drop a small fraction of a pm, but scale the dimensions and it eventually becomes significant. But even that doesn't matter.
It doesn't matter how little it drops (we're talking about simple ideal objects here - no tires...
Ok I messed up in the details.
Forget about the wheels of the car and replace the car with a block on a frictionless table and replace the ditch with a hole in the table.
Here is one that's been bugging me a lot:
Observer A is zooming along in a very fast car at 0.995c. The "rest length" of the car is about 6 meters. In front of the car is a deep ditch with a "rest length" of 2 meters across. The car has big tires and could drive over the ditch under normal...