- #1

KillerZ

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## 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, with an initial speed of 0.01c. What is their distance of closest approach, as measured between their centers? (Hint: There are two conserved quantities. Make use of both.)

## Homework Equations

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

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

## The Attempt at a Solution

How does this look?

[tex]e = 1.60 * 10^{-19} C[/tex]

[tex]c = 3.00 * 10^8 m/s[/tex]

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

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

[tex](4u)(3 * 10^8 m/s) - (u)(3 * 10^8 m/s) = (4u + u)v_{f}[/tex]

[tex]v_{f} = \frac{(4u)(3 * 10^8 m/s) - (u)(3 * 10^8 m/s)}{(4u + u)}[/tex]

[tex]v_{f} = 1.80 * 10^8 m/s[/tex]

[tex]\frac{1}{2}(m_{1}+m_{2})v_{f}^{2} + \frac{1}{4\pi\epsilon_0}(\frac{q1q2}{\infty}) = \frac{1}{2}(m_{1}+m_{2})v_{i}^{2} + \frac{1}{4\pi\epsilon_0}(\frac{q1q2}{r})[/tex]

[tex]\frac{1}{2}(4u + u)(1.80*10^8 m/s)^{2} = \frac{1}{2}(4u + u)(3.00*10^8 m/s)^{2} + \frac{1}{4\pi\epsilon_0}(\frac{2(1.60*10^{-19} C)(1.60*10^{-19} C)}{r})[/tex]

[tex]4\pi\epsilon_0(\frac{1}{2}(4u + u)(1.80*10^8 m/s)^{2} - \frac{1}{2}(4u + u)(3.00*10^8 m/s)^{2}) = (\frac{2(1.60*10^{-19} C)(1.60*10^{-19} C)}{r})[/tex]

[tex]r= \frac{2(1.60*10^{-19} C)(1.60*10^{-19} C)}{4\pi\epsilon_0(\frac{1}{2}(4u + u)(1.80*10^8 m/s)^{2} - \frac{1}{2}(4u + u)(3.00*10^8 m/s)^{2})}[/tex]

[tex]r = -1.92 * 10^{-18} m[/tex]