collision is 1i ∆ [using Eq. ( . )] m m which is a positive quantity as expected. Consider next an elastic collision .
Using the above nomenclature with θ = θ = , the momentum and kinetic energy conservation equations are m v i = m v f + m v f ( . ) ( . ) From Eqs. ( .
) and ( . ) it follows that, or, )( Hence, ∴ ( . ) Substituting this in Eq. ( .
), we obtain ( . ) and ( . ) Thus, the ‘unknowns’ { v f , v 2f } are obtained in terms of the ‘knowns’ { m , m , v i }. Special cases of our analysis are interesting.
Case I : If the two masses are equal v f = v f = v i The first mass comes to rest and pushes off the second mass with its initial speed on collision. Case II : If one mass dominates, e.g. m > > m v f ~ − v i v f ~ The heavier mass is undisturbed while the lighter mass reverses its velocity. Example .
Slowing down of neutrons : In a nuclear reactor a neutron of high speed (typically m s – ) must be slowed to m s – so that it can have a high probability of interacting with isotope U and causing it to fission. Show that a neutron can lose most of its kinetic energy in an elastic collision with a light nuclei like deuterium or carbon which has a mass of only a few times the neutron mass. The material making up the light nuclei, usually heavy water (D O) or graphite, is called a moderator. Answer The initial kinetic energy of the neutron is while its final kinetic energy from Eq.
( . ) The fractional kinetic energy lost is while the fractional kinetic