1. The problem statement, all variables and given/known data 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. 2. Relevant equations 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. 3. 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?