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[SOLVED] Two charged particles fired at each other.
A proton and an alpha particle (q = +2.00e, m = 4.00u ) are fired directly toward each other from far away, each with an initial speed of 0.172c. What is their distance of closest approach, as measured between their centers? (Hint: There are two conserved quantities. Make use of both.)
E_{i}=\frac{1}{2}(m_{p}+m_{\alpha})v^{2} + k_{e}(\frac{q1q2}{\infty})E_{f}=\frac{1}{2}(m_{p}+m_{\alpha})0^{2} + k_{e}(\frac{q1q2}{r})
As I've shown with the equations above, from my understanding, initially the particles only have kinetic energies since they are an infinite distance away. As they get closer their energies will be converted to potential energy and their velocities will momentarily reach 0 when all the energy is potential, which would give us r.
Subbing in values I get 1.11E-11 J for Ei
So I set Ei = Ef do some algebra and get
r = k_{e}(\frac{q1q2}{E_{i}})
When I do this I get 4.16E-17 m, but this is wrong (according to CAPA). What did I miss?
Homework Statement
A proton and an alpha particle (q = +2.00e, m = 4.00u ) are fired directly toward each other from far away, each with an initial speed of 0.172c. What is their distance of closest approach, as measured between their centers? (Hint: There are two conserved quantities. Make use of both.)
Homework Equations
E_{i}=\frac{1}{2}(m_{p}+m_{\alpha})v^{2} + k_{e}(\frac{q1q2}{\infty})E_{f}=\frac{1}{2}(m_{p}+m_{\alpha})0^{2} + k_{e}(\frac{q1q2}{r})
The Attempt at a Solution
As I've shown with the equations above, from my understanding, initially the particles only have kinetic energies since they are an infinite distance away. As they get closer their energies will be converted to potential energy and their velocities will momentarily reach 0 when all the energy is potential, which would give us r.
Subbing in values I get 1.11E-11 J for Ei
So I set Ei = Ef do some algebra and get
r = k_{e}(\frac{q1q2}{E_{i}})
When I do this I get 4.16E-17 m, but this is wrong (according to CAPA). What did I miss?
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