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**Closest approach of particle problem - Please help!!**

## Homework Statement

A proton (q = 1e, m = 1u) and an alpha particle (q = +2e, m = 4u) are fired directly toward each other from far away, each with a speed of 0.01c. What is their distance of closest approach, as measured between their centers?

[tex]e = 1.6 * 10^{-19}[/tex]

[tex]c = 3 * 10^8[/tex]

[tex]u = 1.661 * 10^{-27}[/tex]

This should be a simple problem, but I wanted to know if anyone got the same answer as I did.

## Homework Equations

Conservation of energy

[tex]K_i + U_i = K_f + U_f[/tex]

Conservation of momentum

[tex]m_1v_{1i} + m_1v_{2i} = m_1v_{1f} + m_2v_{2f}[/tex]

## The Attempt at a Solution

After making my conclusion that the proton will eventually turn around and reach a 0 velocity because the bigger particle (alpha particle) will make this "collision" similar to an elastic one.

I first have to find my final velocity of the alpha particle, v1.

[tex]m_1v_{1i} + m_1v_{2i} = m_1v_{1f} + m_2v_{2f}[/tex]

[tex](4u)(3 * 10^6 \frac{m}{s}) - (1u)(3 * 10^6 \frac{m}{s}) = (4u)v_{1f}[/tex]

[tex]9.0 * 10^6u\frac{m}{s} = (4u)v_{1f}[/tex]

[tex]v_{1f} = 2.25 * 10^6\frac{m}{s}[/tex]

Then I plugged that velocity into the energy

After plugging in and solving for my R (which is at minimum when the velocity of the proton is at 0), I get my R to be [tex]2.24 * 10^{-14} m[/tex]

Did anyone get this same answer? Thanks!