Uranium fission within a nuclear reactor produces high-speed neutrons. These must be slowed down for them to trigger a chain reaction within the uranium. In a heavy-water reactor the neutrons of mass m make multiple elastic collisions with deuterium nuclei of 2m, losing speed at each collision.
(a) Suppose a fast-moving neutron of speed v makes a head-on elastic collision with a stationary deuterium nucleus. Show that its speed is reduced to v/3.
p=mv - We probably have to use this
Ek=0.5mv2 - Not sure if we have to use this, but the question IS in a section about kinetic energy. It just seems logical to use p=mv because using kinetic energy makes no sense.
The Attempt at a Solution
Ok it took me absolutely AGES to do this, but I don't think it's right:
PD1 = mv = 2m x 0 = 0 N s
PN1 = mv = 1m x v = mv
PD1 + PN1 = PD2 + PN2 if no external forces are applied
0 + mv = PD2 + PN2
mv = PD2 + PN2
PD2 = mv - PN2
PD2 = PN2 - not sure
Substitute above two equations together to show PN2:
PN2 = mv - PN2
2PN2 = mv
PN2 = mv/2
vN2 = PN2/2m
vN2 = (mv/2)/2m
vN2 = mv/4m - my rearranging may be wrong
vN2 = v/3m
But the answer is just v/3 not v/3m. Is this still right? Or is it just coincidence I got an answer so similar?