An unstable particle at rest spontaneously breaks into two fragments of unequal mass. The mass of the first fragment is 3.00 10-28 kg, and that of the other is 1.51 10-27 kg. If the lighter fragment has a speed of 0.834c after the breakup, what is the speed of the heavier fragment? (Assume the speeds are measured in a frame at rest with respect to the original particle.)

Question

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Respuesta :

nuuk nuuk · Answer 1 · 2019-10-24T05:49:02+02:00

Answer:0.478 c

Explanation:

Given

mass of lighter Particle[tex](m_1)=3\times 10^{-28} kg[/tex]

mass of heavier Particle[tex](m_2)=1.51\times 10^{-27} kg[/tex]

speed of lighter particle[tex](v_1)=0.834 c[/tex]

Let speed of heavier particle[tex]=v_2[/tex]

and Momentum of the particle is given by

[tex]P=\frac{mv}{\sqrt{1-(\frac{v}{c})^2}}[/tex]

[tex]P_1=\frac{m_1v_1}{\sqrt{1-(\frac{v_1}{c})^2}}[/tex]

[tex]P_1=\frac{3\times 10^{-28}\times 0.834 c}{\sqrt{1-(\frac{0.834 c}{c})^2}}[/tex]

[tex]P_1=8.219\times 10^{-28} kg c[/tex]

[tex]P_2=\frac{m_2v_2}{\sqrt{1-(\frac{v_2}{c})^2}}[/tex]

as momentum is conserved therefore [tex]P_1=P_2[/tex]

[tex]8.219\times 10^{-28} kg c=\frac{1.51\times 10^{-27}\times v_2}{\sqrt{1-(\frac{v_2}{c})^2}}[/tex]

[tex]v_2=0.478 c[/tex]