Neutron, slow down!!!

Assume that the iron atoms are not moving at all.

To make the results of the collision more apparent, first switch to the frame that moves with the center of mass (CM). If the neutron has initial speed $v_0$ , the CM of the system moves with speed $\displaystyle \frac{v_0}{1+k}$ . Hence in the CM frame, the neutron has speed $\displaystyle \frac{k v_0}{1+k}$ and the iron atom $\displaystyle \frac{v_0}{1+k}$ . In this frame, the sum of the momenta is always 0, so after the collision occurs, both particles will still move colinearly but at some angle $\theta$ with respect to the original trajectory. And since the collision is elastic and net momentum is zero, the particles maintain their original velocities afterwards.

If we take the original trajectory to be along the x-axis, then the final velocity vector for the neutron becomes $\displaystyle \frac{k v_0}{1+k} (\cos{\theta}, \sin{\theta})$ . Bringing everything back into the laboratory frame yields the neutron's true vector $\displaystyle \frac{k v_0}{1+k} \left(\cos{\theta} + \frac{1}{1+k}, \sin{\theta}\right)$ .

We therefore have $\displaystyle v_f^2 = v_0^2 \left( \frac{k^2 \sin^2{\theta} + (1 + k \cos^2{\theta})}{(1+k)^2} \right)$ . The factor in the parentheses corresponds to how much kinetic energy is left in the neutron after the collision. But since this quantity depends on the scattering angle and there is no preference for a particular angle, we must consider only the average value over the range for $-\pi \leq \theta < \pi$ to obtain expected energy loss. Doing so yields a factor of $q = \displaystyle \frac{k^2 + 1}{(1+k)^2}$ .

In order for the energy to go down by a factor of $5 \times 10^{-8}$ would require on average $\ln{(5 \times 10^{-8})} / \ln{q} \approx$ 479 collisions.

assuming the collision to be two dimensional (along x and y axis) and let initial velocity of neutrons be v making angle p with x axis

then after collision velocity will get changed only in horizontal direction due to part of momentum transferred to iron atoms

that is vsinp component of velocity will remain same and vcosp will get changed

from momentum conservation and energy conservation we have velocity after the collision as

(55/57)cosp along x axis
{using (m2-m1)v/(m2+m1) along direction of collision }

net speed will be v1=sqrt ( (((55/57)cosp)^2) +((sinp)^2) ) *v

taking average of this function for p ranging from 0 to (pi)/2

that is (integral ((v1) dp) from 0 to ((pi)/2) ) /( (pi)/2 ) =0.982534462=q

the kinetic energy of neutrons after nth collisions would be (q^(2n)) * (initial kinetic energy)

so using the given values n=479 closest integer

hence the result

Use an well-known result in classical mechanics as a starting point: the ratio between the neutron's energy in the lab frame after and before the collision with the Fe atom (which is at rest in the lab frame) is related with $k$ (mass ratio between the Fe atom and the neutron) and $\phi$ (the angle change in the neutron motion in the center of mass frame):

$x(\phi)=\frac{E'}{E}=\frac{1+k^2+2k\cos{\phi}}{(1+k)^2}$

As we have no preference in $\phi$ , one can find the average $\bar{x}$ after each collision:

$\bar{x}=\frac{x(\phi=0)+x(\phi=\pi)}{2}=\frac{1+k^2}{(1+k)^2}$

So the needed number of collision is about:

$\frac{E_i}{E_f}=\bar{x}^n \Rightarrow n=\frac{\ln{(\frac{E_i}{E_f})}}{\ln{(\frac{1+k^2}{(1+k)^2})}} \approx 479$