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Distance of closest approach of particles

  1. Feb 3, 2009 #1
    1. The problem statement, all variables and given/known data

    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.)


    2. Relevant 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]


    3. 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]
     
  2. jcsd
  3. Feb 3, 2009 #2
    This might sound stupid (I am tired...) but, did not the problem state that the initial speed was 0.01c? As far as I can see, you used c as initial speed.
     
  4. Feb 4, 2009 #3
    Use K.E. lost =P.E. gained.Consider the electrical P.E. only since the gravitational P.E.can be considerd as negligible.Also,the question is phrased in such a way that you can take the original separation to be infinite.
     
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