Problem 16.55 Challenge

a) จาก \(K_{1} = E_{1} - E_{01}\)

\(K_{1} = m_{1}c^{2} - m_{01}c^{2}; E = mc^{2}\)

\(K_{1} = m_{1}c^{2} -\frac{m_{1}c^{2}}{\gamma}; m = \gamma m_{0}\)

\(K_{1} = (1 -\sqrt{1 -\frac{v^{2}}{c^{2}}})m_{1}c^{2}; \gamma =\frac{1}{1 -\frac{v^{2}}{c^{2}}}\)

\(K_{1} = (1 -\sqrt{1 -\frac{(0.800c)^{2}}{c^{2}}})(2.00 MeV/c^{2})c^{2}\)

\(K_{1} = 0.800 MeV \;\blacksquare\)

b) จาก \(K_{2} = (1 - \sqrt{1 -\frac{v^{2}}{c^{2}}})m_{2}c^{2}; a)\)

\(K_{2} = (1 -\sqrt{1 -\frac{(0.600c)^{2}}{c^{2}}})(1.00 MeV/c^{2})c^{2}\)

\(K_{2} = 0.200 MeV \;\blacksquare\)

c) จาก \(m_{01} = \frac{m_{1}}{\gamma}\)

\(m_{01} = m_{1}\sqrt{1 -\frac{v^{2}}{c^{2}}}\)

\(m_{01} = (2.00 MeV/c^{2})\sqrt{1 -\frac{(0.800c)^{2}}{c^{2}}}= 1.20 MeV/c^{2}\)

จาก \(m_{02} = \frac{m_{2}}{\gamma}\)

\(m_{02} = m_{2}\sqrt{1 -\frac{v^{2}}{c^{2}}}\)

\(m_{02} = (1.00 MeV/c^{2})\sqrt{1 -\frac{(0.600c)^{2}}{c^{2}}}= 0.800 MeV/c^{2}\)

\(\therefore m_{0} = m_{01} + m_{02}\)

\(m_{0} = (1.20 + 0.800) MeV/c^{2} = 2.00 MeV/c^{2} \;\blacksquare\)

d) จาก \(|Σ\vec{p}_{f}| = |\Sigma \vec{p}_{i}|\)

\(|\Sigma \vec{p}_{f}|^{2} = |\Sigma \vec{p}_{i}|^{2}\)

\(\frac{E_{1}^{2} - E_{01}^{2}}{c^{2}}+\frac{E_{2}^{2} - E_{02}^{2}}{c^{2}}=\frac{E^{2} - E_{0}^{2}}{c^{2}}; E^{2} = p^{2}c^{2} + E_{0}^{2}\)

\(E_{1}^{2} - E_{01}^{2} + E_{2}^{2} - E_{02}^{2} = E^{2} - E_{0}^{2}\)

\(E^{2} = E_{1}^{2} + E_{2}^{2} + E_{0}^{2} - E_{01}^{2} - E_{02}^{2}\)

\(E^{2} = (m_{1}^{2} + m_{2}^{2} + m_{0}^{2} - m_{01}^{2} - m_{02}^{2})c^{4}\)

\(E^{2} = (2.00^{2} + 1.00^{2} + 2.00^{2} - 1.20^{2}\)

\(- 0.800^{2}) (MeV)^{2}\)

\(E^{2} = 6.92 (MeV)^{2}\)

\(E = 2.63 MeV \;\blacksquare\)

e) จาก \(E^{2} = p^{2}c^{2} + E_{0}^{2}\)

\(E^{2} = (\gamma m_{0}v)^{2}c^{2} + m_{0}^{2}c^{4}; p = γm_{0}v\) และ \(E_{0} = m_{0}c^{2}\)

\(E^{2} =\frac{m_{0}^{2}c^{4}v^{2}}{c^{2} - v^{2}}+ E_{0}^{2}; \gamma ^{2} =\frac{c^{2}}{c^{2} - v^{2}}\)

\(\frac{m_{0}^{2}c^{4}v^{2}}{c^{2} - v^{2}}= E^{2} - m_{0}^{2}c^{4}\)

\(m_{0}^{2}c^{4}v^{2} = E^{2}c^{2} - E^{2}v^{2} - m_{0}^{2}c^{6} + m_{0}^{2}c^{4}v^{2}\)

\(v = c\sqrt{1 - (\frac{m_{0}c^{2}}{E})^{2}}\)

\(v = c\sqrt{1 - (\frac{(2.00 MeV/c^{2})c^{2}}{(2.63 MeV)})^{2}}= 0.649c \;\blacksquare\)

f) จาก \(Σ\vec{p}_{f}= \Sigma \vec{p}_{i}\)

îĵîĵ

\(m_{1}v_{1} + m_{2}v_{2} = \gamma m_{0}vcos\theta + \gamma m_{0}vsin\theta\)

จะได้ \(γm_{0}vcosθ = m_{1}v_{1} \cdots \cdots (1)\)

และ \(γm_{0}vsinθ = m_{2}v_{2} \cdots \cdots (2)\)

\((2)/(1)\) จะได้ \(tanθ = \frac{m_{2}v_{2}}{m_{1}v_{1}}\)

\(\theta = \tan ^{-1}(\frac{m_{2}v_{2}}{m_{1}v_{1}})\)

\(\theta = \tan ^{-1}(\frac{(1.00 MeV/c^{2})(0.600c)}{(2.00 MeV/c^{2})(0.800c)})\)

\(\theta =\tan ^{-1}(\frac{3}{8})\)

\(\theta = 20.6° \;\blacksquare\)