Velocity of pions after moving far away from each other?

[physicslove22]

Homework Statement

A proton and an antiproton, each with an initial speed of 5.90

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

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?

 

[haruspex]

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?)

 

[physicslove22]

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!

 

[haruspex]

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.

 

[physicslove22]

Oh I forgot about that! I now have v = 2.99792x10^8! Do you think it's right?

 

[haruspex]

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.

 

[physicslove22]

https://www.physicsforums.com/attachments/79662

 

[haruspex]

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.

 

[physicslove22]

Oh ok! That certainly makes it simpler. Thank you!

 

[physicslove22]

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.

 

[haruspex]

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.

 

[physicslove22]

https://www.physicsforums.com/attachments/79663

 

[haruspex]

Two different gammas?

 

[physicslove22]

Ohhhhh...