- #1
Froskoy
- 27
- 0
Homework Statement
A particle of mass [itex]m[/itex] moving at speed [itex]\frac{3}{5}c[/itex] collides with an identical particle at rest, and forms a new particle of mass [itex]M[/itex] which moves off at speed [itex]v[/itex]. Find v.
Homework Equations
E-P invariant: [itex]E_1^2-p_1c^2=E_2^2-p_2^2c^2=\mathrm{const.}[/itex]
Momentum: [itex]p^2=m^2c^2\left({\gamma_v^2-1}\right)[/itex]
Energy: [itex]E=\sum_i\gamma_im_ic^2[/itex]
The Attempt at a Solution
There is a single particle after the collision, so [itex]E_2^2-p_2^2c^2=M^2c^4[/itex]
where [itex]E_2=\gamma_vMc^2[/itex]
and [itex]p_2^2=M^2c^2\left({\gamma_v^2-1}\right)[/itex]
so
[tex]
\gamma_v^2M^2c^4-M^2c^4\left({\gamma_v^2-1}\right)^2=M^2c^4\\
\gamma_v^2-\left({\gamma_v^4-2\gamma_v^2+1}\right)=1\\
\left({\gamma_v^2-1}\right)\left({\gamma_v^2-2}\right)=0
[/tex]
Reject the [itex]\gamma_v^2=1[/itex] solution, since this would mean [itex]v=0[/itex].
The [itex]\gamma_v^2=2[/itex] solution gives [itex]v=\frac{\sqrt{2}}{2}c[/itex], which is incorrect - the answer is [itex]v=\frac{1}{3}c[/itex].
I expect there's something fundamentally wrong with my method, but not sure what? Thank you!