Question

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.)

Answer 1

Answer:0.478 c

Explanation:

Given

mass of lighter Particle$(m_1)=3\times 10^{-28} kg$

mass of heavier Particle$(m_2)=1.51\times 10^{-27} kg$

speed of lighter particle$(v_1)=0.834 c$

Let speed of heavier particle$=v_2$

and Momentum of the particle is given by

$P=\frac{mv}{\sqrt{1-(\frac{v}{c})^2}}$

$P_1=\frac{m_1v_1}{\sqrt{1-(\frac{v_1}{c})^2}}$

$P_1=\frac{3\times 10^{-28}\times 0.834 c}{\sqrt{1-(\frac{0.834 c}{c})^2}}$

$P_1=8.219\times 10^{-28} kg c$

$P_2=\frac{m_2v_2}{\sqrt{1-(\frac{v_2}{c})^2}}$

as momentum is conserved therefore $P_1=P_2$

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

$v_2=0.478 c$