first consider that neutron and deuteron are two balls with having mass of #m_a and m_b# respectively
They collide with each other when their velocities are #v_(a1)# and #v_(b1)# respectively.
As the neutron only moves before the collision and the deuteron is standing before the collision. The velocity of deuteron is 0 before the collision. Therefore #v_(b1)# is 0.
After
Just assume that the ball b is not of equal mass
When the two balls have collided the velocity of the neutron is #v_(a2)# and the velocity of the deuteron is #v_(b2)#
Applying conservation of linear momentum , we get
#m_av_(a1) + m_bv_(b1) = m_av_(a2)+ m_bv_(b2)#
From conservation of Kinetic energy
#1/2m_av_(a1)^2 + 1/2m_bv_(b1)^2 = 1/2m_av_(a2)^2+ 1/2m_bv_(b2)^2#
And we get that
#v_(a2) = [m_a-m_b]/[m_a + m_b] v_(a1)+ [(2m_b)/(m_a + m_b)]v_(b1)#
#v_(b2) = [m_b-m_a]/[m_a + m_b] v_(b1)+ [(2m_a)/(m_a + m_b)]v_(a1)#
So we already know that #v_(b1)# is 0 so
#v_(a2) = [m_a-m_b]/[m_a + m_b] v_(a1)+0#
#v_(b2) = 0+ [(2m_a)/(m_a + m_b)]v_(a1)#
On taking the approximate that deuteron's weight is twice the weight of neutron so
To reduce the writing I would like to
#v_(a1) = v, m_a = m, m_b = 2m#
#v_(a2) = [(m-2m)/(m + 2m)]v#
#v_(a2) = (-mv)/(3m)#
#v_(a2) = -v/3#
#v_(b2) = [(2m)/(m + 2m)]v #
#v_(b2) = [(2m)/(3m)]v#
#v_(b2) = (2mv)/3m#
#v_(b2) = (2v)/3#
#v_(b2) = 2/3v#
#"Fractional energy loose" ~~ (1/2mv^2-1/2m[-v/3]^2)/(1/2mv^2)#
#(4/9mv^2)/(1/2mv^2)#
#(4/9)/(1/2)#
# 8/9#
Notice the approximation we made earlier this question that the deuteron's weight is twice the weight of neutron the if we want the answer more accurately.
Weight of an neutron = #1.674927351(74)×10^-27kg#
Weight of a deuteron atom = #3.34449439655xx10^-27kg #
#(3.34449439655xx10^-27kg) /(1.674927351(74)×10^-27kg) = 1.99679967884#
#2~~ 1.99679967884#
We can see that approximation was not very large
Using these calculate the fractional k.e lost.
#v_(a1) = v, m_a = m, m_b = 1.99679967884m#
#v_(a2) = [(m-1.99679967884m)/(m + 1.99679967884m)]v#
#v_(a2) *2.9968 = −0.9968v#
#v_(a2) = −0.332621v#
#"Fraction of K.E lost" = (1/2mv^2-1/2m[−0.332621v^2])/(1/2mv^2)#
= #2.221273/2 ~~ 8/9#
