Answer to Question #93424 in Mechanics | Relativity for Pritish Dixit

Question

1. Kinetic energy is conserved in an elastic collision by definition. Show, using the Galilean
transformation equations, that if a collision is elastic in one inertial frame it is elastic in
all inertial frames.
2. An electron whose speed relative to an observer in a laboratory is 0.800c is also being
studied by an observer moving in the same direction as the electron at speed of 0.500c
relative to the laboratory. What is the kinetic energy (in MeV) of the electron to each
observer ?
3. An observer on a spacecraft moving at 0.700c relative to the earth finds that a car takes
40.0min to make a trip. How long does the trip take to the driver of the car?

Answer 1

1)$m_1v'^2_1/2+m_2v'^2_2/2-(m_1u'^2_2/2+m_2u'^2_2/2)=0 ?$

For elastic collision we have

$m_1v^2_1/2+m_2v^2_2/2-(m_1u^2_2/2+m_2u^2_2/2)=0 ?$

From Galilean transformation we have v=v'+w $\implies$ v'=v-w

$m_1(v'_1+w)^2/2+m_2(v'_2+w)^2/2-(m_1(u'_1+w)^2/2+m_2(u'_2+w)^2/2)=0$

$m_1v'^2_1/2+m_2v'^2_2/2-(m_1u'^2_1/2+m_2u'^2_2/2)+w(P'_f+P'_i)=0 \mapsto (P'_f+P'_i)=0 ⟹$

$m_1(v'_1+w)^2/2+m_2(v'_2+w)^2/2-(m_1(u'_1+w)^2/2+m_2(u'_2+w)^2/2)=0 ⟹$

$m_1v'^2_1/2+m_2v'^2_2/2-(m_1u'^2_1/2+m_2u'^2_2/2)=0$

2) For first observer

$E=m_0c^2(\frac{1}{\sqrt(1-v^2/c^2)}-1)=9.1*10^{-31}*299 792 458^2(\frac{1}{\sqrt(1-0.82)}-1)=5.45*10^{-14} J =0.33MeV$

For second observer v₂=?

$v2= \frac{c2(v1-v)}{v1*v-c^2}=\frac{c^2(0,5c-0,8c)}{0,5c*0,8c-c2}=0,5c$

$E=m_0c^2(\frac{1}{\sqrt(1-v^2_2/c^2)}-1)=9.1*10^{-31}*299 792 458^2(\frac{1}{\sqrt(1-0.25)}-1)=1.27*10^{-14}J =0.079MeV$

3)

$t_0=t\sqrt{(1-\frac{v^2}{c^2})}=40*\sqrt{(1-0.49)}=28.6min$