Question

A neutron of mass 1.67 x 10-27 kg, moving with velocity 2.0 x 104 m/s, makes a head-on collision with a boron nucleus of mass 17.0 x 10-27 kg. The boron nucleus is initially at rest. If the collision is elastic, what fraction of the original kinetic energy of the neutron is transferred to the boron nucleus?

Answer 1

let, denote neutron by 1 and boron nucleus by 2
v2 = (2*m1*u1)/(m1+m2) - ((m1-m2)*u2)/(m1+m2)
= (2*1.67*10^-27*2.0*10^4)/(1.67*10^-27 + 17.0*10^-27) - 0
= (6.68*10^-23)/(18.67*10^-27)
= (0.358*10^4) m/s

kinetic energy of boron nucleus = 0.5*m2*v2^2
initial kinetic energy of boron = 0.5*m1*u1^2

fraction transfer = (kinetic energy of boron nucleus)/(initial kinetic energy of boron)
= (0.5*m2*v2^2)/(0.5*m1*u1^2)
= (m2*v2^2)/(m1*u1^2)
= {17.0*10^-27*(0.358*10^4)}/{1.67*10^-27*(2.0*10^4)}
= 2.18/6.68
= 0.341

Answer: 0.341