1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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:


    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


    User Avatar
    Science Advisor

    Since v1 and v2 are both 0, you immediately have:

    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Help with a momentum exchange please!