- #1
mike217
- 16
- 0
A particle of rest mass m is accelerated to a kinetic energy K in a nuclear reactor. This particle is incident on a stationary target particle, also of rest mass m.
a) Show that the speed of the centre of mass (that is the speed of the frame in which the total momentum is zero) is [tex]\gamma v/(\gamma +1)[/tex] where v is the speed of the incident particle and [tex]\gamma =(1-v^2/c^2)^-1/2[/tex]. Verify that this expression reduces to the usual one in the non relativistic case when v<<c.
b) If the collision is perfectly inelastic-show by using the conservation of energy and momentum that the mass M of the resulting composite is [tex]M=m\sqrt{2(\gamma +1)}[/tex] Verify that this reduces to the usual value in the non relativistic case v<<c
I have the solution to this problem, but there are a couple of things I don't understand. How do you get the following expressions
[tex] (Ptot-Vcm*Etot/c^2)/\sqrt(1-(Vcm)^2/c^2)=0 [/tex]
and
[tex]
\gamma*m*v=MVcm/\sqrt(1-(Vcm)^2/c^2)=\gamma*M*v/((\gamma+1)*\sqrt((1-\gamma^2*v^2/c^2)/(\gamma+1)^2)
[/tex]
a) Show that the speed of the centre of mass (that is the speed of the frame in which the total momentum is zero) is [tex]\gamma v/(\gamma +1)[/tex] where v is the speed of the incident particle and [tex]\gamma =(1-v^2/c^2)^-1/2[/tex]. Verify that this expression reduces to the usual one in the non relativistic case when v<<c.
b) If the collision is perfectly inelastic-show by using the conservation of energy and momentum that the mass M of the resulting composite is [tex]M=m\sqrt{2(\gamma +1)}[/tex] Verify that this reduces to the usual value in the non relativistic case v<<c
I have the solution to this problem, but there are a couple of things I don't understand. How do you get the following expressions
[tex] (Ptot-Vcm*Etot/c^2)/\sqrt(1-(Vcm)^2/c^2)=0 [/tex]
and
[tex]
\gamma*m*v=MVcm/\sqrt(1-(Vcm)^2/c^2)=\gamma*M*v/((\gamma+1)*\sqrt((1-\gamma^2*v^2/c^2)/(\gamma+1)^2)
[/tex]
Last edited: