(a) The final kinetic energy of neutron expressed as the fraction of its original kinetic energy is [tex]\boxed{\frac{K}{9}}[/tex].
(b) The number of collisions required to reduce the speed of the neutron is [tex]\boxed{10}[/tex].
Further Explanation:
Part (a):
Since the collision of the neutron with deuterons is an elastic collision. Therefore, the momentum as well as the energy of the system during the collision remains conserved.
Initially, the deuteron is at rest to its velocity is zero. The momentum of the system remains conserved.
Apply the conservation of the momentum before and after collision.
[tex]\left( {{m_n} \times {v_n}}\right)+\left( {{m_d}\times {v_d}}\right)=\left( {{m_n} \times v_n^'} \right) + \left( {{m_d} \times v_d^'}\right)[/tex]
Here, [tex]{m_n}\,\& \,{m_d}[/tex] is the mass of neutron and deuteron respectively, [tex]{v_n}\,\& \,{v_d}[/tex] is the initial velocity of the neutron and deuteron and [tex]v_n^'\,\& \,v_d^'[/tex] is the final velocity of the neutron and deuteron respectively.
Since the mass of the deuteron is twice that of the mass of neutron, the values can be substituted as:
[tex]mv + 0=mv_n^'+2mv_d^'[/tex]
[tex]v = v_n^' + 2v_d^'[/tex] ......(1)
Since the collision between the two bodies is elastic. Therefore, the final resultant velocity will be equal to that of the initial velocity.
[tex]\begin{aligned}v&= v_d^' - v_n^' \hfill\\v_d^' &= v + v_n^' \hfill\\\end{aligned}[/tex]
Substitute [tex]v + v_n^{'\,}[/tex] for [tex]v_d^'[/tex] in equation (1).
[tex]\begin{aligned}v&=v_n^' + 2\left( {v + v_n^{'\,}}\right)\hfill\\v&= - 3v_n^' \hfill\\v_n^'&= - \frac{v}{3}\hfill\\\end{aligned}[/tex]
It means that the neutron will start to move in the direction opposite to its initial velocity.
The initial kinetic energy of the neutron is:
[tex]K =\dfrac{1}{2}m{v^2}[/tex]
The final kinetic energy of the neutron will be.
[tex]\begin{aligned}K'&= \frac{1}{2}m{\left({v_n^'}\right)^2}\\&=\frac{1}{2}m\frac{{{v^2}}}{9}\\&= \frac{K}{9}\\\end{aligned}[/tex]
Thus, the final kinetic energy of neutron expressed as the fraction of its original kinetic energy is [tex]\boxed{\dfrac{K}{9}}[/tex].
Part (b):
From the above calculations, the speed of the particle is reduced to [tex]\dfrac{1}{3}{\text{rd}}[/tex] of its initial velocity after every collision.
After [tex]N[/tex] collisions, the speed will be reduced to [tex]\dfrac{1}{{59000}}[/tex] of its value. It can be expressed as:
[tex]\begin{aligned}{\left({\frac{1}{3}} \right)^N}&=\left({\frac{1}{{59000}}} \right)\\{3^N}&= 59000\\N&= \frac{{\ln \,59000}}{{\ln \,3}}\\&\approx 10\\\end{aligned}[/tex]
Thus, the number of collisions required to reduce the speed of the neutron is [tex]\boxed{10}[/tex].
Learn More:
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Answer Details:
Grade: High School
Subject: Physics
Chapter: Conservation of momentum
Keywords: Canadian nuclear reactors, heavy water moderators, collision, neutrons, deuterons, conservation of momentum, elastic collision, original kinetic energy, successive collisions.
