Problem 9.25

9.25.
Relativistic momentum must be conserved. For the total momentum to be zero, we must have
p₂ = p₁
or $\gamma_{1} m_{01} c^{2} = \gamma_{2} m_{02} c^{2}$
Using the values given for particle 1, we have
$p_{1} = p_{2} = \frac{2.5 \times 10^{-28} kg}{\sqrt{1 - (0.893)^{2}}} (.893 c)$
We then have
$\frac{(1.67 \times 10^{-27} kg) v_{2}}{\sqrt{1 - \frac{v_{2}^{2}}{c^{2}}}} = (4.960 \times 10^{-28} kg) c$
Squaring both sides and solving for v₂, we find
v₂ = .285 c