I Solving Relativistic Inelastic Collisions: 150 GeV Energy

[Eirik]

Hi! Hope I'm posting this in the right place! I'm practicing for exams and came over this question:

A proton with mass $m_p$ is accelerated to a relativistic velocity, with kinetic energy $K$. It collides completely inelastic with another proton, which has the same kinetic energy, $K$, and velocity in the opposite direction. How big does K have to be, in order for the composite particle to have energy equal to 150 GeV?

Intuitively I would use conservation of energy to solve this, which gives me:
$E=2(K+mc^2)$
This gives me K=75GeV.

However, I found a formula for essentially the exact same problem in my physics book, which states that:
$E=2mc^2(1+\frac{K}{mc^2})$
Which gives me about K=74.1 GeV.

The last formula should be the right one, but I don't really see how, as both protons have kinetic energy, $K$.. Any input?

 

[Eirik]

I suddenly realized I forgot to account for the fact that kinetic energy isn't conserved in inelastic collisions.. That probably has something to do with it doesn't it

 

[PeroK]

I suddenly realized I forgot to account for the fact that kinetic energy isn't conserved in inelastic collisions.. That probably has something to do with it doesn't it

The answer is simpler than that! Are the two answers really so different?

 

[Eirik]

The answer is simpler than that! Are the two answers really so different?

I must've typed something wrong into my calculator the first time I tried to solve it haha! It's the same equation. Hope I don't make mistakes as dumb as this one during my exam tomorrow. Thanks!

 

[Eirik]

@PeroK Is it okay if I ask another question on the same topic?

If the other proton had been at rest, is it true that
$E=K+mc^2+mc^2=2mc^2(1+\frac{K}{2mc^2})$

My physics book seems to think it's
$E=2mc^2*\sqrt{1+\frac{K}{2mc^2}}$

 

[PeroK]

@PeroK Is it okay if I ask another question on the same topic?

If the other proton had been at rest, is it true that
$E=K+mc^2+mc^2=2mc^2(1+\frac{K}{2mc^2})$

My physics book seems to think it's
$E=2mc^2*\sqrt{1+\frac{K}{2mc^2}}$

If the second proton is at rest, then the system has non-zero momentum, which is conserved. If you are looking for the total energy of the resultant particle after the collision, then that's just conservation of energy. But, if you are looking for the rest mass of the resultant particle, then that will be less than the total energy - as the resultant particle has some KE.

 

[Eirik]

If the second proton is at rest, then the system has non-zero momentum, which is conserved. If you are looking for the total energy of the resultant particle after the collision, then that's just conservation of energy. But, if you are looking for the rest mass of the resultant particle, then that will be less than the total energy - as the resultant particle has some KE.

That's what I thought. Looks like it's just a mistake in the textbook then, as they were referencing the total energy of the composite particle. Thanks again:)

 

[PeroK]

That's what I thought. Looks like it's just a mistake in the textbook then, as they were referencing the total energy of the composite particle. Thanks again:)

Yes, that is the formula for $Mc^2$, where $M$ is the mass of the composite particle.

 

[PAllen]

That's what I thought. Looks like it's just a mistake in the textbook then, as they were referencing the total energy of the composite particle. Thanks again:)

Are you sure? If they meant that the total rest energy of the composite particle, then the textbook formulas would be correct for both cases. The key is that for the first case, when total momentum of the system is zero, the total energy is available for result particle rest energy. In the second case, very little of the total energy is available for result rest energy due to conservation of momentum. Their formula for the second case is exactly the maximum rest energy of the composite result particle.