A neutron makes a head-on elastic collision with a stationary deuteron. The fraction energy loss of the neutron in the collision is

Explanation

Neutron velocity =v, mass =m
Deuteron contains 1 neutron and 1 proton, mass = 2m

In elastic collision both momentum and K.E. are conserved `p_i=p_f`
`mv=m_(1)v_(2)+m_(2)v_(2)`
`rArr mv=mv_(1)+2mv_(2)` …(1)
By conservation of kinetic energy `1/2mv^(2)=1/2mv_(1)^(2)+1/2(2m)v_(2)^(2) …(2)`
By solving (1) and (2) we get `v_(1)=(m_(1)-m_(2))/(m_(1)+m_(2))v+(2m_(2))/((m_(1)+m_(2)) rArr v_(1)=((m_(1)+2m)/(3m)) v=-v/3`
`K_(i)=1/2mv^(2), K_(f)=1/2mv_(1)^(2)`
`rArr (K_(i)-K_(f))/(K_(i))=1-(v_(1)^(2))/(v^2)=1-1/9=8/9 ("Fractional change in K.E.")`