Proton opposite directions available energy when collide

[accountkiller]

Homework Statement

Consider two beams of protons, moving in opposite directions, each with energy 110 GeV. What is the available energy E_a when these beams collide?

Homework Equations

E_a = [tex]\sqrt{2mc^2*E_m}[/tex]
E_a² = 2mc²(E_m + mc²)
E_a² = 2Mc²E_m + (Mc²)² + (mc²)²
p = m/v
E = p²/m

The Attempt at a Solution

I'm given this hint:
"In all particle collisions, momentum must be conserved. This is the reason that the available energy in collisions with a stationary target is less than the total energy of the collision: some of the energy must go into kinetic energy for the product to conserve momentum. If two beams of identical protons collide head-on, what is the total momentum of a pair of protons just before the collision? What is the minimum amount of energy that must go into kinetic energy to have the same total momentum after the collision?"

So the total momentum is just m/v, right? And I'm given energy, so I can find v from KE = 0.5mv² and plug it into momentum, giving me 3.64E-37 kg*m/s. Is it correct to do it like that? Then it asks for the energy, so I used E = p²/m = 2.27E-9 GeV but that did not give me the correct answer. I also tried using all three of the available energy equations above but none worked. How do I use the momentum hint?

Can anyone help? Thanks!

 

[anathema]

Hello, thank you for your question. You are on the right track in considering the conservation of momentum in this collision. However, there are a few things to keep in mind.

First, the total momentum of a pair of protons just before the collision would be the sum of the individual momenta of each proton. In this case, since both protons have the same mass and are moving in opposite directions, the total momentum would be zero.

Next, you are correct in using the equation KE = 0.5mv2 to find the velocity of each proton. However, since the protons are moving in opposite directions, the total kinetic energy of the system would be the sum of the kinetic energies of each proton. So the minimum amount of energy that must go into kinetic energy to conserve momentum would be twice the kinetic energy of one proton, or 221 GeV.

Finally, to find the available energy Ea when the beams collide, you can use the equation Ea = \sqrt{2mc^2*E_m}. Plugging in the values of m (proton mass), c (speed of light), and E_m (kinetic energy of one proton), you should get a value of approximately 221.2 GeV. This is the maximum amount of energy available for the collision.

I hope this helps clarify things! Let me know if you have any further questions.