- #1
physicslove22
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Homework Statement
A proton and an antiproton, each with an initial speed of 5.90
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?