Recent content by cosmos17

p = \gamma m v E = \gamma m c^2 \frac{p}{E} = \frac{ \gamma m v}{ \gamma m c^2} \frac{p}{E} = \frac{v}{c^2} Amazing stuff.

Momentum. I think I have solved it. Solving for initial mass and momentum, we can make use of E_2=p^2 c^2+m^2 c^4 because this formula has a special property- it relates momentum, energy, and rest mass, which is what I'm trying to solve for. We know energy and momentum are conserved so we can...

Could you elaborate? I don't really understand.

Why does E = \gamma mc^2 presume that kinetic energy is conserved? The total energy is conserved because the kinetic energy is converted to mass during the collision, by my understanding.

It was my understanding that by inelastic collision, all we know is that the bodies stick together and there is a maximum kinetic energy loss. I didn't think that implied all kinetic energy was lost, otherwise the compound object would have zero momentum after the collision, which we know cannot...

Please correct me if I'm wrong, but I thought total energy is always conserved, but kinetic energy is lost during the collision. Hence why I use total relativistic energy, E = \gamma mc^2 .

Homework Statement A meteorite is approaching Earth at very high speed. In order to avoid an impact on earth, the world space agency has launched two space missions: one mission sends the astronaut Albert to the meteorite approaching earth. The other mission, lead by the space commander Trebla...

My name is Sasha and I'm a 19 year old engineering physics student at Queen's University. Because of this I feel like I'll be hanging around here often, so you might as well get to know me. I'd write some more but I really need to get back to studying for these finals. I'm a cool guy though, I...