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Help with velocity in relativistic momentum?

  1. Mar 31, 2005 #1
    help with velocity in relativistic momentum???

    a proton has 1836 times the rest mass of an electron. At what speed (in c) will an electron have the same momentum as a proton moving at .0180c?

    -- here's what i have so far------------------------

    1. by using the mass of a proton i determined the momentum of the electon as p proton= p electron

    p proton= (1.673e-27 * .0108*3e8)/(1-.0108^2)^1/2
    p proton= 9.035e-21

    i am having trouble converting the equation p= m1*v1/(1-(v1/c)^2)^1/2 into an equation that i can use to find the velocity of the electron. All of the answers i get seem to put c well over one... any ideas? any help would be greatly appreciated.
     
  2. jcsd
  3. Mar 31, 2005 #2

    jtbell

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    Staff: Mentor

    You need to re-do your arithmetic in the step above. Be very careful to avoid roundoff errors. It's best to calculate it in one continuous sequence on your calculator, without writing down any intermediate results.

    This should just take some algebra, to solve for v1. If fixing the number above doesn't improve your final result, post your algebra for this step and someone will probably be able to spot your problem.
     
  4. Mar 31, 2005 #3
    A proton moving at 0.0180c has momentum :

    [tex] \frac{mv}{\sqrt{1-\frac{v^2}{c^2}}} = \frac{(1.67x10^{-27})(5.4x10^6)}{\sqrt{1-\frac{2.916x10^{13}}{9x10^{16}}}} = 9x10^{-21}[/tex]

    You want [tex] p_{electron} = 9x10^{-21} kgm/s [/tex] given [tex] m = 9.1 x 10^{-31} [/tex]

    [tex] \frac{mv}{\sqrt{1-\frac{v^2}{c^2}}} = 9.1x10^{-21}[/tex]

    [tex] \frac{v}{\sqrt{1-\frac{v^2}{c^2}}} = 10^{10} [/tex]

    [tex] v^2 = 10^{20}(1-\frac{v^2}{c^2}) = 10^{20} - \frac{v^210^{20}}{3x10^8} = 10^{20} - \frac{10^{12}}{3}v^2[/tex]

    [tex] v^2+\frac{10^{12}}{3}v^2=10^{20} [/tex]

    [tex] v^2(1+\frac{10^{12}}{3}) = 10^{20} [/tex]

    [tex] v^2 = \frac{10^{20}}{1+\frac{10^{12}}{3}}} = \frac{10^{20}}{\frac{10^{12}+3}{3}} = \frac{3(10^{20})}{10^{12}+3} = \frac{300000000000000000000}{1000000000003} = 299999999.9991000000000027 \frac{m}{s} [/tex]

    In units of c, [tex] \frac{299999999.9991000000000027}{300000000} = 0.999999999997000000000009c [/tex]

    Hope this clarifies things a bit.
     
  5. Mar 31, 2005 #4
    thank you very much
     
  6. Mar 31, 2005 #5

    Andrew Mason

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    It is a little tricky.

    The momentum of a proton at .018 c is:

    [tex]p_p = \gamma m_{0p}v = \frac{m_{0p}v}{\sqrt{1 - v^2/c^2}}[/tex]

    For the proton at .018c, [itex]\gamma[/itex] works out to 1.00016. so the proton's momentum is:
    [tex]p_p = 1836m_e*1.00016*.018c = 33.05m_ec [/tex]

    So you have to find v for the electron so that it has a momentum of 33.05 electron masses x c.

    [tex]p_e = \gamma m_ev = \frac{m_ev}{\sqrt{1 - v^2/c^2}} = 33.05m_ec[/tex]

    [tex]\gamma\beta = 33.05[/tex] where [itex]\beta = v/c[/itex]

    [tex]\frac{\beta^2}{1 - \beta^2} = 33.05^2[/tex]

    [tex](33.05^2 + 1)\beta^2 = 33.05^2[/tex]

    v = .9995425c

    AM
     
    Last edited: Mar 31, 2005
  7. Mar 31, 2005 #6
    I was wrong? All that hard work? :'(

    I should stick to classical kinetics.
     
  8. Mar 31, 2005 #7

    jtbell

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    OK, I think I see what happened. You [smij44] wrote 0.0108 in your calculation, so that's the number I used. But just above it you wrote 0.0180, and that's apparently what you actually used. If the correct number is 0.0180, then your first step is OK.
     
  9. Mar 31, 2005 #8
    Can someone show me why Andrew Mason's answer and mine differ?
     
  10. Mar 31, 2005 #9

    jtbell

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    Staff: Mentor

    If someone else wants to try finding the error, s/he's welcome to try. For me, checking a long string of arithmetic is a royal pain in the butt. When one of my students does a problem that way, I give it back and say, "first do it algebraically, then plug in the numbers at the end."

    It's a lot easier to check algebra than to check arithmetic, you don't get roundoff errors when you do algebra, and if you're lucky, a lot of stuff cancels out so you don't have to do as much arithmetic if you wait till the end to do it.

    I suggest that you start over with

    [tex]p = \frac {mv} {\sqrt{1-v^2/c^2}}[/tex]

    and get an equation v = (some combination of m, p and c), or better yet, v/c = (some combination of m, p and c), then calculate the numeric answer in one shot on your calculator.

    I felt ambitious today, so I took this a step further. I started with

    [tex]\frac {m_e v} {\sqrt {1 - (v_e/c)^2}} = \frac {m_p v} {\sqrt {1 - (v_p/c)^2}}[/tex]

    and worked out a formula for the answer that contains only the quantities that were originally given: [itex]v_p / c = 0.018[/itex] for the proton, and the ratio of the proton and electron masses, [itex]m_p / m_e = 1836[/itex]:

    [tex]\frac{v_e}{c} = \frac {v_p/c} {\sqrt {\left(\frac{m_e}{m_p}\right)^2 + \left[1 - \left(\frac{m_e}{m_p}\right)^2 \right] \left(\frac{v_p}{c}\right)^2}}[/tex]

    Plugging in the numbers:

    [tex]\frac{v_e}{c} = \frac {0.018} {\sqrt{(1/1836)^2 + [1 - (1/1836)^2](0.018^2)}} = 0.999543[/tex]
     
  11. Mar 31, 2005 #10

    Andrew Mason

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    Don't be too hard on yourself. This problem illustrates the importance of avoiding plugging in numbers too early. If you work out the physics you will often find that variables cancel out and the problem becomes much simpler.

    Now, as far as where you went wrong, it appears that you forgot to use the value for the square of c in the denominator in this equality:

    [tex] v^2 = 10^{20}(1-\frac{v^2}{c^2}) = 10^{20} - \frac{v^210^{20}}{3x10^8}= 10^{20} - \frac{10^{12}}{3}v^2[/tex]

    it should be:

    [tex] v^2 = 10^{20}(1-\frac{v^2}{c^2}) = 10^{20} - \frac{v^210^{20}}{(3x10^8)^2}= 10^{20} - \frac{10^{4}}{9}v^2[/tex]

    At the end you also have to take the square root.

    aM
     
  12. Mar 31, 2005 #11
    Thanks alot.
     
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