Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: SR question

  1. Oct 25, 2008 #1
    1. The problem statement, all variables and given/known data

    Two particles of rest mass m1 and m2 are moving at velocities u1 and u2 respectively. They then collide to form a new particle of rest mass m moving at velocity u. Show that

    [tex]m^2=m_1 ^2+m_2 ^2+2m_1m_2\gamma_1\gamma_2(1-\frac{u_1u_2}{c^2})[/tex]

    where [tex]\gamma_n=\frac1{\sqrt{1-\frac{u_n ^2}{c^2}}}[/tex]

    2. Relevant equations

    3. The attempt at a solution

    I tried using conservation of energy and momentum to relate m to m1 and m2, but I am unable to get rid of u in the expression.

    1: [tex]\frac{mc^2}{\sqrt{1-\frac{u^2}{c^2}}}=\gamma_1m_1c^2+\gamma_2m_2c^2[/tex]

    2: [tex]\frac{mu}{\sqrt{1-\frac{u^2}{c^2}}}=\gamma_1m_1u_1+\gamma_2m_2u_2[/tex]

    From 1,

    [tex]m^2=\gamma_1 ^2m_1^2+\gamma_2 ^2m_2^2+2m_1m_2\gamma_1\gamma_2(1-\frac{u^2}{c^2})[/tex]

    When I made u the subject of eqn 2 and substituted it into the one above, I got quite a messy expression that didn't seem anywhere near the one in the question.

    I also tried working in the frame of particle 2, and obtained

    3: [tex]\gamma_1 'm_1c^2=\gamma'mc^2[/tex]

    4: [tex]\gamma_1 'm_1u_1 '=\gamma'mu'[/tex]

    (where prime indicates the value in the frame of particle 2)

    From 3 and 4,

    [tex]u_1 '=u'[/tex]


    When I proceeded to solve for u, I still got quite a messy expression.

    Is there a way to solve for m^2 without having to involve u? A later part of the question asks for an expression of u.
  2. jcsd
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted