Answer to Question #99640 in Mechanics | Relativity for Nyx

Question

A neutron in a reactor makes an elastic headon collision with the nucleus of an atom initially at rest.
Assume: The mass of the atomic nucleus is
about 11.1 the mass of the neutron.
What fraction of the neutron’s kinetic energy is transferred to the atomic nucleus?
Don't round answer.

Answer 1

When the collision is elastic the energy of the system and total momentum is conserved so

$$E(n0) = E(n) +E(nuc)$$ $$P(n0)=P(n)+P(nuc)$$ $$E(n0)=(m(n)v(n0)^2)/2$$ $$E(n)=(m(n)v(n)^2)/2$$ $$E(nuc)=(m(nuc)v(nuc)^2)/2=(11.1m(n)v(nuc)^2)/2$$ $$P(n0)=m(n)v(n0)$$ $$P(n)=m(n)v(n)$$ $$P(nuc)=m(nuc)v(nuc)=11.1m(n)v(nuc)$$ $$P(n)=P(n0)-P(nuc)=m(n)v(n0)-11.1m(n)v(nuc)=m(n)v(n)$$ $$v(n) =(v(n0)-11.1v(nuc))$$

from the first equation

$$(m(n)v(n0)^2)/2=(m(n)(v(n0)-11.1v(nuc))^2)/2+(11.1m(n)v(nuc)^2)/2$$ $$v(n0)^2=(v(n0)-11.1v(nuc))^2+11.1v(nuc)^2$$ $$v(n0)^2=v(n0)^2-22.2v(n0)v(nuc)+123.21v(nuc)^2+11.1v(nuc)^2$$ $$22.2v(n0)v(nuc)=134.31v(nuc)^2$$ $$v(nuc)=(22.2/134.31)v(n0)=(1/6.05)v(n0)$$

the fraction of the neutron energy transmitted to the nucleus can be calculated as

$$f=E(nuc)/E(n0)=(11.1m(n)(((1/6.05v)(n0))^2))/(m(n)v(n0)^2)=0.303$$