Calculating Mass in Relativistic Systems

In summary, the conversation revolves around a question regarding the mass of a ball and the incorrect calculation of a square root of a negative number. The conversation also touches upon the use of variables in calculations and the importance of following a formal layout to avoid confusion or misunderstanding. The conversation ends with a discussion about the relationship between velocity and mass, specifically in regards to the astronaut's mass changing due to her velocity.
  • #1
ttttrigg3r
49
0
Hello. I've uploaded the question onto pictures and linked them. I hope this method of asking a question works.
I'm trying to find the mass of the ball, but I get the square root of a negative number. Can someone take a look to see if I'm doing this problem correctly?

IMAG0012.jpg

IMAG0013.jpg

 
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  • #2
What immediately occurs to me:
In ##p^0=E/c## is that E the rest-mass energy, total energy, or kinetic energy?
(Closer look - you seem to have calculated gamma ... apologies: it is difficult for me to see your reasoning.)

It can help see where you went wrong (or right!) if you leave the actual numbers out until the end.

eg. In space-station coordinates: ##P_i = (M+m)\gamma_i(1,0,0,u)^t;\; P_f=M(1,0,0,0)^t+m\gamma_f(1,0,0,v)^t## (...or something) you have to find ##v## from the conservation laws and knowing ##M##, ##m##, ##\gamma_i##, and ##u##.

The laws give you two equations.
If you do this as a list, formally, you are less likely to get the numbers mixed up at the end.
 
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  • #3
OH I see. The mass of the astronaut in the final momentum is not 100kg like I thought. Have to give it the Variable M. Ok This will be a piece of cake then. Thanks.
 
  • #4
No - I just used variables to keep track of what's what.
Remember that the questions says to assume the astronaught's mass does not change.
All the masses I used are rest-masses.

What I am trying to get you to do is go back over the calculation more formally.
 
  • #5
No actually the question says it is the mass of the ball that does not change.
Here is my answer. I'm about 90% sure of it. Can you please check it out and see if what I did is good?
IMAG0016.jpg
 
  • #6
ttttrigg3r said:
No actually the question says it is the mass of the ball that does not change.
Oh so it does - I'm corrected. I find it ambiguous anyway ... why would it change?
Here is my answer. I'm about 90% sure of it. Can you please check it out and see if what I did is good?
Why do you make people work so hard to read your working? It'll cost you marks.
 
  • #7
Is my handwriting/work too sloppy? I will try to make it much neater.
 
  • #8
It's the layout - your working is all over the place so I cannot easily follow your reasoning. Someone marking your work could easily think you've left something out of misunderstood something. At least there's boxes around your final answers but those are worth, at most, one mark each.

You can reality check your results though ... should the speed of the ball be very high to cancel the momentum? Why would the astronaught's mass change (and is that a reasonable mass for an astronaught?)
 
  • #9
Ok here is what I've concluded about the velocity of the ball and the astronaut's mass. The velocity of the ball must be extremely high (close to c=1) to balance out the fact the the astronaut has velocity 0 in the final momentum in the x direction. Also the astronaut's mass in the beginning is 100kg whereas her final mass is 50.5 kg. This is because in the beginning the astronaut was moving with v=3/5 which is a big enough velocity to make her much more massive than when she is standing still. 50.5kg is about 110lbs which makes sense for a human being.

ps. Is the following statement correct?
As velocity increases approaching c=1, the more massive you get.
 
  • #10
So the masses quoted are not rest-masses, but kinetic energy contributions seen in the space-station frame? (Nobody literally gains mass by virtue of going fast.)

This also suggests a very small astronaut doesn't it?

I always thought the m in the 4-momentum was rest-mass though.
If the initial mass is not rest mass them don't we put ##\gamma_i (M+m)=101kg##?
 

What is conservation of four-momentum?

The conservation of four-momentum is a fundamental principle in physics that states that the total four-momentum of a closed system remains constant over time, regardless of the interactions between its components.

Why is conservation of four-momentum important?

Conservation of four-momentum is important because it is a fundamental law of nature that governs the behavior of particles at the most fundamental level. It allows us to make predictions about the outcomes of interactions between particles and helps us understand the dynamics of the universe.

How is conservation of four-momentum related to energy and momentum conservation?

Conservation of four-momentum is closely related to energy and momentum conservation. In fact, conservation of four-momentum is a more general concept that includes both energy and momentum conservation as special cases. This is because four-momentum takes into account both the energy and momentum of a particle, as well as its mass.

Can conservation of four-momentum be violated?

No, conservation of four-momentum is a fundamental law of nature that has been extensively tested and has never been found to be violated. It is a universal principle that applies to all interactions between particles, from the microscopic level to the largest scales of the universe.

How is conservation of four-momentum applied in practical situations?

Conservation of four-momentum is applied in many practical situations, such as in particle accelerators, where it is used to predict the behavior of particles in collisions. It is also used in cosmology to understand the evolution of the universe, and in nuclear reactions to calculate the energy and momentum of particles produced in the reaction.

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