Particle Collisions Homework: Solve for E_1 E_1' u_1 u_1

[Deadstar]

Homework Statement

A particle of rest mass $$m_1$$ moving with velocity $$u_1$$ along the x-axis collides with a stationary particle of rest mass $$m_2$$ stationary along the x-axis. If subsequently the particle with rest mass $$m_1$$ moves in the direction making an angle of $$60^{o}$$ relative to the x-axis (in the x - y plane), show that

$$E_1 E_1'(u_1 u_1' - 2) = 2m_2 (E_1' - E_1) - 2m_1^2$$

where $$E_1$$ and $$E_1'$$ are the total energies of the particle $$m_1$$ before and after the collision respectively and $$u_1$$ is its speed after the collision

The Attempt at a Solution

I don't need anyone to actually post the solution I just have some basic questions.

Is there any information missing from this question..? The things I'm unsure about are

Are the masses of the particles after the collision the same or different than before. I have them as different but then I end up with these extra masses in any sort of expression I can come up with.

Which direction does the stationary particle move in after the collision?

And carrying on from the above question, what is the momentum of $$m_2$$ after the collision?

I have it as,

$$p_{2'} = m_2' \gamma(u_2') (1, u_2' \cos(\alpha), u_2' \sin(\alpha), 0)$$

where $$\alpha$$ is the angle of deflection from the x-axis after the collision.
Now when I do the standard conservation of momentum calculations it gets in a mess fast. What am I missing here?

 

[fzero]

Are the masses of the particles after the collision the same or different than before. I have them as different but then I end up with these extra masses in any sort of expression I can come up with.

In an elastic collision, the masses of the particles don't change in the collision. It sounds like the problem involves an elastic collision even though they don't say that explicitly.

Which direction does the stationary particle move in after the collision?

You can make a rough guess about the direction, but since you are given the angle of deflection of the moving particle, you can use momentum conservation to determine the direction. Parameterize the momentum the same way you did particle 2, using the given angle. You'll be able to solve for the angle $$\alpha$$ below.

And carrying on from the above question, what is the momentum of $$m_2$$ after the collision?

I have it as,

$$p_{2'} = m_2' \gamma(u_2') (1, u_2' \cos(\alpha), u_2' \sin(\alpha), 0)$$

where $$\alpha$$ is the angle of deflection from the x-axis after the collision.
Now when I do the standard conservation of momentum calculations it gets in a mess fast. What am I missing here?

That looks right to me. Things will get easier if you set $$m_2' = m_2$$, etc., but some of the algebra will always be messy as soon as you try to solve for velocities.

 

[Deadstar]

In an elastic collision, the masses of the particles don't change in the collision. It sounds like the problem involves an elastic collision even though they don't say that explicitly.



You can make a rough guess about the direction, but since you are given the angle of deflection of the moving particle, you can use momentum conservation to determine the direction. Parameterize the momentum the same way you did particle 2, using the given angle. You'll be able to solve for the angle $$\alpha$$ below.



That looks right to me. Things will get easier if you set $$m_2' = m_2$$, etc., but some of the algebra will always be messy as soon as you try to solve for velocities.

Thank you! That the mass is the same was all I really needed to know. Will simplify things a lot.

 

[Deadstar]

Thank you! That the mass is the same was all I really needed to know. Will simplify things a lot.

Or not. I'm still having trouble with this Q. Since I'm a mathematician there might be something I'm missing here. Is the angle alpha I'm looking for just 30 degrees? I.e. should the two angles of deflection always add up to 90 degrees?

Another problem I'm having is really just what is the first step here.

I've tried three different ways, none of which have really gotten me anywhere.

I've tried using the conservation of momentum formula;

$$\overline{p_1} + \overline{p_2} = \overline{p_{1'}} + \overline{p_{2'}}$$

Then do the usual square both sides method but then I end up with

$$\gamma(u_1) = \gamma(u_1') \gamma(u_2') (-1 + \frac{u_1' u_2'}{2} \cos(\alpha) + \frac{\sqrt{3}}{2} u_1' u_2' \sin(\alpha))$$

which is not helping me.Then I tried using the equation...

$$\overline{p} \cdot \overline{p} = (E, \underline{p}) \cdot (E, \underline{p}) = -E^2 + \underline{p} \cdot \underline{p}$$

Which seemed promising since after a few steps I got

$$2m_2 (E_1 - E_1') + E_1 m_2 u_2' = \underline{p_{1'}} \cdot \underline{p_{2'}}$$

which had the $$2m_2 (E_1 - E_1')$$ term as in the equation I'm trying to derive.

But even with setting the unknown deflection angle as 30 degrees i still don't end up with anything close to what I'm after.Finally I just started with $$E_1 E_1'(u_1 u_1' - 2)$$ and tried to turn that into the right hand side of the equation in post one but I got nowhere with that one either.So how do I start this thing?!?

I should also say I've been using the $$m_1 \gamma(u_1) u_1 + m_2 \gamma(u_2) u_2 = m_1 \gamma(u_1') u_1' + m_2 \gamma(u_2') u_2'$$ equation but again it's not helping me much...

 

[fzero]

Since you are using 4-vectors,

$$\overline{p_1} + \overline{p_2} = \overline{p_{1}'} + \overline{p_{2}'}$$

contains all of the information about energy and momentum conservation. By equating components, you will obtain 3 equations for the 3 unknowns, $$u_1', u_2',\alpha$$. It will probably help to write each equation down separately and express the $$\gamma$$ factors in terms of the energies $$E_1,E_1'$$, etc. You should also remember to set $$u_2=0$$ from the initial conditions.

It also looks like they've used the square of the 4-momentum

$$p \cdot p = m^2$$. You will probably figure out what trick they used when you're further along in the algebra.

to rewrite the terms that are quadratic in the velocities $$u_1,u_1'$$.