(adsbygoogle = window.adsbygoogle || []).push({}); 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]

[tex]\frac{u_1-u_2}{\sqrt{1-\frac{u_1u_2}{c^2}}}=\frac{u-u_2}{\sqrt{1-\frac{uu_2}{c^2}}}[/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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: SR question

Can you offer guidance or do you also need help?

Draft saved
Draft deleted

**Physics Forums | Science Articles, Homework Help, Discussion**