Relativity and kinetic energy, mass-energy systems

[Am.obrien]

Homework Statement

An object with mass of 900 kg Is moving at a speed of 0.85c when it collides with an object at rest with mass of 1400 kgs. Find the speed and mass of the composite object.

Homework Equations

mc^2/gamma +mc^2/ gamma=Mc^2

The Attempt at a Solution

My issue is that I am not sure which equation I really want to work with. I think the one I listed is for an inelastic collision where the combined mass doesn't move after the collision, and so it would be inappropriate here. I think I need to use momentum and energy equations, but like I said, what equation describes the relativistic mass-energy system for an inelastic collision where the objects keep moving?

 

[Simon Bridge]

Welcome to PF;
Trying to figure the right equation to use is not a useful way to think about physics.
How would you normally do the problem - without relativity.

 

[Am.obrien]

Welcome to PF;
Trying to figure the right equation to use is not a useful way to think about physics.
How would you normally do the problem - without relativity.

I Would use conservation of momentum, which I mentioned, but since I have two variables, I would also need an energy equation. So the question is, do I understand said equation correctly, that the composition mass stops moving after collision, and if so, what do I do for a composition function that continues to move.

 

[Simon Bridge]

The equation you wrote says $2mc^2=\gamma Mc^2$ ... so it is unlikely to be useful.
You want to conserve total energy? The gamma for each initial particle will be different, and the final object will also have its own gamma. You could pick a different regerence frame of course.

Do I understand correctly that the initial particles stick together and there is no radiation?

 

[Am.obrien]

The equation you wrote says $2mc^2=\gamma Mc^2$ ... so it is unlikely to be useful.
You want to conserve total energy? The gamma for each initial particle will be different, and the final object will also have its own gamma. You could pick a different regerence frame of course.

Do I understand correctly that the initial particles stick together and there is no radiation?

it
Why would I pick a different reference frame? - Why would the equation not be useful?

So the equation that I entered: Einitial for one object plus Einitial for the second object = Mc^2 - no gamma on the second side. - if we tried to use this equation for this problem, it would not simplify as you simplified it because the objects have different masses and different velocities. When the objects collide, they stick together. I am suppose to find the change in velocity and the change in mass. No radiation. A big part of my question was about having gamma on the right side. That was why I asked about the purpose of the equation. From my chapter, I seem to understand it is discussing a perfectly inelastic collision where the particles stop moving afterwards.

What would we do for a composite mass that kept moving?

 

[Simon Bridge]

Why would I pick a different reference frame?

Since you are free to pick any reference frame you like, it is a good idea to pick one where the maths is easiest. Something to bear in mind for later. It is very common to do relativistic collision calculations in the center of momentum reference frame, for eg., so you can exploit the symmetries. I suspect you are not expected to do that this time though.
In general, with anthing to do with relativty, it is best practice to have the reference frame in mind when you do the calculations.

Why would the equation not be useful?

The equation does not describe the system you are given. See below.

Einitial for one object plus Einitial for the second object = Mc^2

Um - E(mass 1) + E(mass 2) = E(final) would be correct, yes. But that's not what you wrote down.

You wrote:

mc^2/gamma +mc^2/ gamma=Mc^2

... which I read as: $$\frac{mc^2}{\gamma} + \frac{mc^2}{\gamma} = Mc^2$$... off which I can see:

• total energy of mass m is not $mc^2/\gamma$ ... you understand that total energy increases with speed? Yet these relations decrease with speed, so they must be incorrect.
• the two (initial) objects different masses and different speeds, but you wrote an equation giving them the same mass and speed. (Same variable name indicates same value and gamma depends on speed.)

The relations for total energy are: $$E_{tot} = \gamma mc^2 \\ E_{tot}^2 = m^2c^4 + p^2c^2$$ ... you should have these written down someplace.
You follow the usual method for conservation of energy and momentum, except that you are using the relativistic equations.

It helps to think of the maths as a language you can use to describe the situation you have.

 

[Am.obrien]

How does that help me find the change in mass?

 

[Simon Bridge]

You have to go through the steps using the correct equations... you'll have two equations and two unknowns.
The point of the exercize is that you construct the equations yourself.
Give it another shot with the above information and see where you get stuck.

[edit]
i.e. $$\frac{m_1c^2}{\gamma} + \frac{m_2c^2}{\gamma} = Mc^2$$... is still incorrect, and so is: $$\frac{m_1c^2}{\gamma_1} + \frac{m_2c^2}{\gamma_2} = Mc^2$$ ... $$E_1+E_2=E_3$$ ... vis: total energy of the first object + total energy of the second object is the total energy of the final object: this would be correct. You now need to expand the total energies in terms of the question.

I think that's as close as I can get without actually telling you the answer.
I'll need to see your next attempt before commenting further.

 

[Simon Bridge]

How did you get on?

OK: probably enough time has passed so...
The total energy is $E=\gamma mc^2$
This, with the momentum-energy relation, gives you: $p^2=(\gamma^2-1)m^2c^2$

All you have to do is sub the above equations for p and E - remembering to use a different index for different values. c^2 will appear as a common factor - so divide it out. For the problem of a mass approaching a stationary mass at speed u, and sticking together, you get:

Conserving total energy: $$\gamma_1m_1 + m_2 = \gamma M$$ ($\gamma_2=1$ because mass 2 is stationary.)

Conserving momentum: $$m_1\sqrt{\gamma_1^2-1} = M\sqrt{\gamma^2-1}$$ ($p_2=0$ because...)

Your two unknowns are $\gamma$ and $M$ ... $$\frac{v}{c}=\sqrt{1-\frac{1}{\gamma^2}}$$... is the speed you want.

Sub gamma back to get $M$, which is the mass you want.

You only needed to memorize two equations ... the total energy and the energy-momentum relation.
You did need to know E_0=mc^2 for the rest-energy... and you will sometimes need to know that the difference between the total energy and the rest-energy is deemed "kinetic energy" and given symbol T. Knowing rest-energy and total energy equations gives you $T=(\gamma-1)mc^2$ ... but you should see that I just decreased the number of equations you need to know for this part of your work.

Lastly: it is also very very useful to use units where c=1 wherever you can.