1. In class we considered a situationin which a sub-atomic particle split into t
ID: 1761082 • Letter: 1
Question
1. In class we considered a situationin which a sub-atomic particle split into two smaller particles ofequal mass (mo). In the parentparticle’s rest frame, the two daughters were moving inopposite directions at 0.250 c. The whole experiment was observedin a frame in which the parent was moving at 0.650 c (we showed inclass that one daughter moved at 0.474 c, and the other at 0.774 cin this frame).
a. What is the rest mass of theparent particle (in terms of the rest mass,mo, of each of the twodaughters)?
b. Show that relativistic momentumand energy are conserved in this decay when viewed from the movingframe, but that Newtonian momentum is not. Note also that rest massis not conserved.
c. (BONUS 5 points) Show thatrelativistic momentum is also conserved in the lab frame, butNewtonian momentum is not conserved.
I'm not surehow to get started on this one. Any help would beuseful!
Explanation / Answer
In parent'srest frame,
speed of eachdaughter = .25 c
=1/(1-.25^2) = 1.0328
momentum ofeach daughter = p = m0v = .2582cm0
energy before= Mc^2 = energy after = 2(p^2c^2+m0^2c^4)
M =2(p^2/c^2+m0^2) =2m0(.2582^2+1) = 2.0656 m0
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Whenviewed from the moving frame,
parentspeed = V = .65c
= 1/(1-.65^2) = 1.3159
parentmomentum = MV = 1.7668 cm0
parent energy= m0 c^2 (1.7668^2+2.0656^2) = 2.72m0c^2
daughter1speed = v1 = .474c
=1/(1-.474^2) = 1.136
daughter1momentum = m0v1 = .538 cm0
daughter1energy = m0 c^2 (.538^2+1) = 1.136m0c^2
daughter2speed = v2 = .774c
=1/(1-.774^2) = 1.579
daughter2momentum = m0v2 = 1.222 cm0
daughter2energy = m0 c^2 (1.222^2+1) = 1.579m0c^2
daughter1energy + daughter2 energy = (1.136+1.579)m0c^2 = 2.71m0c^2 = parent energy.
daughter1momentum + daughter2 momentum = (.538+1.222)m0c = 1.76m0c= parent momentum
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