Velocity of pions after moving far away from each other?

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Homework Help Overview

The problem involves a collision between a proton and an antiproton, resulting in the creation of two pions. The question focuses on determining the velocities of the pions after they have moved far apart, given their initial conditions and rest masses.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the conservation of energy and the significance of rest mass in the calculations. There are attempts to refine calculations by using more accurate values for constants and questioning the appropriateness of Newtonian versus relativistic approaches.

Discussion Status

The discussion is ongoing, with participants providing insights and suggestions for refining calculations. Some participants express uncertainty about their results and seek validation from others, while others explore different mathematical approaches.

Contextual Notes

Participants mention constraints such as the need for accuracy in calculations and the implications of using relativistic mass versus rest mass in their equations. There is also a concern about the proximity of calculated speeds to the speed of light.

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Homework Statement


A proton and an antiproton, each with an initial speed of 5.90
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107 m/s when they are far apart. When these two particles collide, they react to form two new particles: a positive pion (π+, charge +e) and a negative pion (π−, charge −e). Each pion has a rest mass of 2.5
multiply.gif
10-28 kg. These pions have enough energy that they move away from each other. When these two pions have moved very far away from each other, how fast is each pion going, v?

Homework Equations


E of system = kinetic + rest + potential energies
kinetic = 1/2mv^2
rest= mc^2
electric potential = (9x10^9)(q1 * q2)/radius

The Attempt at a Solution


Esys1 = Esys3
(2) (1/2) (1.7x10^-27) (5.9x10^7)^2 + (2) (1.7x10^-27) (3x10^8)^2 = (2) (1/2) (2.5x10^-28)(v final)^2 + (2) (2.5x10^-28) (3x10^8)^2
v = 1.03328 x 10^9
This answer turned out to be wrong! Can anyone see where I went wrong?
 
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Look at the magnitudes of the terms in your equation. The rest masses dominate. So when you take the difference in the rest masses you get a relatively small difference between two large numbers. That can turn a small numerical error into a much more significant one.
I would try using a more accurate value for c.
(What is the supposed answer?)
 
I don't know what the answer is, but I know it is supposed to be close to the speed of light... I tried plugging in 2.99792x10^8, and now my answer is 1.03258x10^9. Do you think this is correct? I only have one submission left!
 
physicslove22 said:
I don't know what the answer is, but I know it is supposed to be close to the speed of light... I tried plugging in 2.99792x10^8, and now my answer is 1.03258x10^9. Do you think this is correct? I only have one submission left!
With such great speeds, you should be using the generic relativistic mass, not adding rest mass to Newtonian KE, maybe? But that would yield a slightly smaller number.
 
Oh I forgot about that! I now have v = 2.99792x10^8! Do you think it's right?
 
physicslove22 said:
Oh I forgot about that! I now have v = 2.99792x10^8! Do you think it's right?
Still seems too close to c. Please post your working.
 
https://www.physicsforums.com/attachments/79662
 
Rather a lot of arithmetic operations doing it that way. Do it all algebraically until the final step. You should get ##c^2-v_2^2 = (c^2-v_1^2)(\frac {m_2}{m_1})^2##, which gives me about 2.967E8.
 
Oh ok! That certainly makes it simpler. Thank you!
 
  • #10
If you don't mind me asking, how did you get to that point? I tried it algebraically and got v = c at the end.
 
  • #11
physicslove22 said:
If you don't mind me asking, how did you get to that point? I tried it algebraically and got v = c at the end.
Post your algebra and I'll check it.
 
  • #12
https://www.physicsforums.com/attachments/79663
 
  • #13
Two different gammas?
 
  • #14
Ohhhhh... :mad:
 

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