# Two charged particles fired at each other.

1. Feb 1, 2008

### mellotron

[SOLVED] Two charged particles fired at each other.

1. The problem statement, all variables and given/known data
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.)

2. Relevant equations

$$E_{i}=\frac{1}{2}(m_{p}+m_{\alpha})v^{2} + k_{e}(\frac{q1q2}{\infty})$$

$$E_{f}=\frac{1}{2}(m_{p}+m_{\alpha})0^{2} + k_{e}(\frac{q1q2}{r})$$

3. 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?

Last edited: Feb 1, 2008
2. Feb 1, 2008

### mellotron

Figured out that momentum is conserved so I was wrong about my assumption that final kinetic energy would be 0, which in hindsight makes lots of sense.