# Relativistic Inelastic Collision of Protons

• AnniB
This comes from E2 = (mc2)2 + (pc)2. We want to find the magnitude of p. The energy is the total energy of the system, so it should be 2mc2 + K. So (2mc2 + K)2 - p2c2 = (mc2)2, so p2c2 = 4(mc2)2 + 4Kmc2. The sqrt of that gives us the magnitude of p.
AnniB

## Homework Statement

Let us consider a perfectly inelastic collision between two protons: an incident proton with mass m, kinetic energy K, and momentum magnitude p joins with an originally stationary target proton to form a single product particle of mass M. Due to conservation of momentum, there must be some kinetic energy after the collision. Show that the energy available to create a product particle is

Mc2 = 2mc2(sqrt(1 + K/2mc2))

## Homework Equations

E0 = Ef
This is the only equation I can think of. Applying conservation of momentum in just seems like it would add a miscellaneous variable. The others are just figured out from the problem and a couple hints the teacher provided.
E1 = K + mc2
E2 = mc2
Ef = Mc2 + K

## The Attempt at a Solution

E1 + E2 = Ef
(E1 + E2)2 = Ef2
(Mc2 + K)2 = (K + mc2)2 + 2(K + mc2)(mc2) + (mc2)2
Solving this gave me:
(Mc2)2 + 2KMc2 = 4Kmc2 + 4(mc2)2

I can't get this to look like the equation I need. I feel like I either have one term too many, or I'm missing one.

AnniB said:

## Homework Statement

Let us consider a perfectly inelastic collision between two protons: an incident proton with mass m, kinetic energy K, and momentum magnitude p joins with an originally stationary target proton to form a single product particle of mass M. Due to conservation of momentum, there must be some kinetic energy after the collision. Show that the energy available to create a product particle is

Mc2 = 2mc2(sqrt(1 + K/2mc2))

## Homework Equations

E0 = Ef
This is the only equation I can think of. Applying conservation of momentum in just seems like it would add a miscellaneous variable. The others are just figured out from the problem and a couple hints the teacher provided.
E1 = K + mc2
E2 = mc2
Ef = Mc2 + K

It's very unlikely that the product particle will have the same kinetic energy as the incident proton, so your Ef is incorrect.

As a hint, note the factor of 2 in the formula for Mc2 is highly indicative that there's a frame in which this energy can be attributed in equal parts to the two reactant photons. What frame might this be?

Would it be one that is moving at half the speed of the approaching proton, so in the reference frame they are at rest after the collision?

What would that do to the energy, though? I tried calculating it how I thought it should be and still ended up with one term too many.

Okay, so I've *almost* got the right equation. There's just a extra term of K2/m2c4 that is keeping me from just writing one more step and being done.

I don't know if it's because the energy I used isn't right, but I've used 3 different values for it so far and this is the closest I've gotten.

It would help if you showed some work. What energy did you use? We can easily compute the available energy in the frame where the total momentum of the initial protons is zero.

After changing into the frame where both protons are moving towards each other so that there was only rest energy after the collision and so the total momentum should be zero, I have E0 = 2(K/4 + mc2), which I based off of 1/2mv2, but it probably is not good to assume that even applies in this case. I've used E0 = 2(K/2 + mc2) also, just to check. The first E0 I used was 2(K + mc2), but I know that can't be right because that's the kinetic energy from the frame of reference set up in the problem itself.

The hints I got from the professor were to set E1 = K + mc2 and to square the expression E1 + E2 = Ef, using the initial frames of the problem. When I tried to do that, though, I came up with answer that was even farther off than what I'm getting now.

First of all, we're dealing with relativistic equations, so if you wanted to relate kinetic energy to velocity, you need to use

$$K = \frac{p^2}{2m}, ~ p = \gamma m v.$$

However we don't really need that. In the lab frame, the total energy is 2mc2 + K. The momentum of the moving proton can be found by solving (K+mc2)2 +(pc)2 = (mc2)2. To find the total energy in the c.o.m. frame, we need to use energy-momentum conservation.

Ah, thank you! I don't know why I didn't think to use that formula earlier.

why do we solve (K+mc2)2 +(pc)2 = (mc2)2. for the momentum of the moving proton? I don't understand why this equation is true, since we have in general E^2 =p^2c^2+(mc^2)^2?

That was a typo. It should be (K+mc2)2-(pc)2=(mc2)2.

## 1. What is a relativistic inelastic collision of protons?

A relativistic inelastic collision of protons refers to a high-energy collision between two protons where they not only exchange energy but also undergo a change in their internal structure, resulting in the creation of new particles.

## 2. What is the significance of studying relativistic inelastic collision of protons?

Studying relativistic inelastic collision of protons can provide insights into the fundamental properties of matter and the forces that govern interactions between particles. It also has practical applications in fields such as particle accelerators and nuclear physics.

## 3. How is momentum conserved in a relativistic inelastic collision of protons?

In a relativistic inelastic collision of protons, momentum is conserved through the creation of new particles. The total momentum of the initial protons will be equal to the total momentum of the resulting particles, taking into account their mass and velocity.

## 4. What is the role of relativity in a relativistic inelastic collision of protons?

Relativity plays a crucial role in understanding and predicting the outcomes of a relativistic inelastic collision of protons. It takes into account the effects of high velocities and the resulting changes in mass and energy.

## 5. How is energy exchanged in a relativistic inelastic collision of protons?

In a relativistic inelastic collision of protons, energy is exchanged through the creation of new particles with different masses and energies. The total energy before and after the collision will be conserved, but it will be distributed differently among the resulting particles.

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