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Help with a momentum exchange please!

  1. Apr 12, 2009 #1
    Hi there,

    I am dealing with an inelastic, collinear momentum exchange of the form:

    m1v1+m2v2=m1u1+m2u2

    where m1 & m2 are known.

    v1 & v2 are both 0

    u1 & u2 are both unknown, however

    MOD(u1)+MOD(u2) is known (ie the sum of modulus of each speed, I don't know how to do straight brackets here.........)

    which initial speed is regarded as + or - is irrelevant (to me).

    I know that the exact speeds for u1 & u2 can be calculated, but I can't quite get me head around how (I'm more used to knowing one or the other, not their sum).

    Although I can find the answer by gradually increasing one of the values on a spreadsheet, I'd like to see the actual solution. I'm guessing it can be solved either simultaneously or with a bit of calculus, but I'm not very good and working these things out.

    Any help would be much appreciated.



    p.s. I know this looks like homework, but its not. It really does though doesn't it. Real world though, honest.
     
  2. jcsd
  3. Apr 12, 2009 #2

    mathman

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    Science Advisor
    Gold Member

    Since v1 and v2 are both 0, you immediately have:
    u1=-[m2/m1]u2

    You the substitute into your expression for|u1| + |u2| to get |u1|

    u1 = the result with a sign ambiguity.
     
  4. Apr 12, 2009 #3
    [tex]m_{1}u_{1}+m_{2}u_{2}=0[/tex] can be rearragned to [tex]\frac{u_{1}}{u_{2}}=-\frac{m_2}{m_1} [/tex]and taken the modulus we get [tex]\frac{|u_{1}|}{|u_{2}|}=\frac{m_{2}}{m_{1}}[/tex]
    Now, Lets call [tex]|u_{1}|+|u_{2}|=x[/tex] where x is known, and if we divide by [tex]|u_{2}|[/tex] and rearrange we get [tex]\frac{|u_{1}|}{|u_{2}|}=\frac{x}{|u_{2}|}-1[/tex]
    Therefore [tex]\frac{x}{|u_{2}|}-1=\frac{m_{2}}{m_{1}}[/tex] and rearragning for [tex]u_{2}[/tex] we get [tex]|u_{2}|=\frac{m_{1}}{m_{1}+m_{2}}x[/tex] and similarly for [tex]|u_{1}|=\frac{m_{2}}{m_{1}+m_{2}}x[/tex]
     
  5. Apr 13, 2009 #4
    That's brilliant, thank you very much for your help.
     
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