Answer:
greatest speed=0.99c
least speed=0.283c
Explanation:
To solve this problem, we have to go to frame of center of mass.
Total available energy fo π + and π - mesons will be difference in their rest energy:
[tex]E_{0,K_{0} }-2E_{0,\pi } =497Mev-2*139.5Mev\\[/tex]
=218 Mev
now we have to assume that both meson have same kinetic energy so each will have K=109 Mev from following equation for kinetic energy we have,
K=(γ-1)[tex]E_{0,\pi }[/tex]
[tex]K=E_{0,\pi}(\frac{1}{\sqrt{1-\beta ^{2} } } -1)\\\frac{1}\sqrt{1-\beta ^{2} }}=\frac{K}{E_{0,\pi}}+1\\ {1-\beta ^{2}=\frac{1}{(\frac{K}{E_{0,\pi}}+1)^2}}\\\beta ^{2}=1-\frac{1}{(\frac{K}{E_{0,\pi}}+1)^2}}\\\beta = +-\sqrt{\frac{1}{(\frac{K}{E_{0,\pi}}+1)^2}}\\\\\\beta =+-\sqrt{1-\frac{1}{(\frac{109Mev}{139.5Mev+1)^2}}[/tex]
[tex]u'=+-0.283c[/tex]
note +-=±
To find speed least and greatest speed of meson we would use relativistic velocity addition equations:
[tex]u=\frac{u'+v}{1+\frac{v}{c^{2} } } u'\\u_{max} =\frac{u'_{+} +v_{} }{1+\frac{v}{c^{2} } } u'_{+} \\u_{max} =\frac{0.828c +0.9c }{1+\frac{0.9c}{c^{2} } } 0.828\\ u_{max} =0.99c\\u_{min} =\frac{u'_{-} +v_{} }{1+\frac{v}{c^{2} } } u'_{-}\\u_{min} =\frac{-0.828c +0.9c }{1+\frac{0.9c}{c^{2} } } -0.828c\\u_{min} =0.283c[/tex]
