Invariant mass problem, elastic collision

[Silicon-Based]

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Homework Statement

Question b:

Homework Equations

E²=c²p²+m²c⁴

The Attempt at a Solution

We have c²p_initial²=E₀²-m²c⁴, and E_f²=c²p²+m²c⁴ for each outgoing proton. Combining those equations we get c²p²=E_f²-E₀²+c²p_initial². I don't know where to go from here.

 

[PeroK]

Is this for part a) or part b)?

What principles are you going to apply in solving this problem?

 

[Silicon-Based]

Is this for part a) or part b)?

What principles are you going to apply in solving this problem?

b). I would refrain from using conservation of momentum since that is specifically mentioned in part c (which I didn't include in the picture), and here it asks to equate the invariant masses.

 

[PeroK]

b). I would refrain from using conservation of momentum since that is specifically mentioned in part c (which I didn't include in the picture), and here it asks to equate the invariant masses.

Personally, I would never not make use of conservation of momentum. If momentum is not conserved, then is there a unique solution? So, I would reason that you must make use of conservation of momentum.

In any case, what else is conserved?

 

[Silicon-Based]

Personally, I would never not make use of conservation of momentum. If momentum is not conserved, then is there a unique solution? So, I would reason that you must make use of conservation of momentum.

In any case, what else is conserved?

When I use conservation of energy (E_f=(E₀+mc²)/2) the algebra seems to lead nowhere.

Also, presumably, I should ideally not introduce any gammas.

 

[PeroK]

When I use conservation of energy (E_f=(E₀+mc²)/2) the algebra seems to lead nowhere.

It says the total energy of the proton initially is $E_0$. Not the Kinetic Energy.

 

[Silicon-Based]

It says the total energy of the proton initially is $E_0$. Not the Kinetic Energy.

E₀ is the total energy of the proton in motion, E₀+mc² is the total energy of the system, which is equally distributed between both protons after the collision.

 

[PeroK]

E₀ is the total energy of the proton in motion, E₀+mc² is the total energy of the system, which is equally distributed between both protons after the collision.

Yes, of course. Anyway, what happens when you include conservation of momentum?

 

[Silicon-Based]

Yes, of course. Anyway, what happens when you include conservation of momentum?

I get results like mc²(E₀-mc²)=2γm²cos²(α)c², where γ is of the protons after the collision. But again, I should be able to solve the question without introducing any gammas, but I don't see a way to apply conservation of momentum without doing so.

 

[PeroK]

I get results like mc²(E₀-mc²)=2γm²cos²(α)c², where γ is of the protons after the collision. But again, I should be able to solve the question without introducing any gammas, but I don't see a way to apply conservation of momentum without doing so.

Yes, don't introduce gamma. You can relate momentum to energy and mass without gamma.

What about using $p_0 = 2p \cos \alpha$? That's what conservation of momentum in the original direction should have given you?

 

[Silicon-Based]

Yes, don't introduce gamma. You can relate momentum to energy and mass without gamma.

What about using $p_0 = 2p \cos \alpha$? That's what conservation of momentum in the original direction should have given you?

I end up with $$c^2p^2=\frac{(3E_0+mc^2)(E_0-mc^2)}{12-16\sin^2(\alpha)}$$

 

[PeroK]

I end up with $$c^2p^2=\frac{(3E_0+mc^2)(E_0-mc^2)}{12-16\sin^2(\alpha)}$$

Obviously, that doesn't look quite right. I suspect it's just your algebra.

Did you square the energy equation? After that it should be a couple of substitutions to get rid of the terms you don't want in the final expression. Note that you need to get rid of $E_0^2$.

 

[Silicon-Based]

Obviously, that doesn't look quite right. I suspect it's just your algebra.

Did you square the energy equation? After that it should be a couple of substitutions to get rid of the terms you don't want in the final expression. Note that you need to get rid of $E_0^2$.

E₀² doesn't cancel due to the factor of two in the equation of energy conservation. Getting sine out of cosine messes things up as well.

 

[PeroK]

E₀² doesn't cancel due to the factor of two in the equation of energy conservation. Getting sine out of cosine messes things up as well.

You'll need to post your attempt to see where you are going wrong. I get the expression in the question, so there is no mistake there.