Question

Canadian nuclear reactors use heavy water moderators in which elastic collisions occur between the neutrons and deuterons of mass 2.0 u. What is its kinetic energy, expressed as a fraction of its original kinetic energy? b) what is its kinetic energy expressed as a fraction of its original kinetic energy? c) how many such successive collisions will reduce the speed of a neuron to 1/59000 of its original value?

Answer 1

(a) The final kinetic energy of neutron expressed as the fraction of its original kinetic energy is $\boxed{\frac{K}{9}}$.

(b) The number of collisions required to reduce the speed of the neutron is $\boxed{10}$.

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.

$\left( {{m_n} \times {v_n}}\right)+\left( {{m_d}\times {v_d}}\right)=\left( {{m_n} \times v_n^'} \right) + \left( {{m_d} \times v_d^'}\right)$

Here, ${m_n}\,\& \,{m_d}$ is the mass of neutron and deuteron respectively, ${v_n}\,\& \,{v_d}$ is the initial velocity of the neutron and deuteron and $v_n^'\,\& \,v_d^'$ 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:

$mv + 0=mv_n^'+2mv_d^'$

$v = v_n^' + 2v_d^'$                           ......(1)

Since the collision between the two bodies is elastic. Therefore, the final resultant velocity will be equal to that of the initial velocity.

$\begin{aligned}v&= v_d^' - v_n^' \hfill\\v_d^' &= v + v_n^' \hfill\\\end{aligned}$

Substitute $v + v_n^{'\,}$ for $v_d^'$ in equation (1).

$\begin{aligned}v&=v_n^' + 2\left( {v + v_n^{'\,}}\right)\hfill\\v&= - 3v_n^' \hfill\\v_n^'&= - \frac{v}{3}\hfill\\\end{aligned}$

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:

$K =\dfrac{1}{2}m{v^2}$

The final kinetic energy of the neutron will be.

$\begin{aligned}K'&= \frac{1}{2}m{\left({v_n^'}\right)^2}\\&=\frac{1}{2}m\frac{{{v^2}}}{9}\\&= \frac{K}{9}\\\end{aligned}$

Thus, the final kinetic energy of neutron expressed as the fraction of its original kinetic energy is  $\boxed{\dfrac{K}{9}}$.

Part (b):

From the above calculations, the speed of the particle is reduced to $\dfrac{1}{3}{\text{rd}}$ of its initial velocity after every collision.

After $N$ collisions, the speed will be reduced to $\dfrac{1}{{59000}}$ of its value. It can be expressed as:

$\begin{aligned}{\left({\frac{1}{3}} \right)^N}&=\left({\frac{1}{{59000}}} \right)\\{3^N}&= 59000\\N&= \frac{{\ln \,59000}}{{\ln \,3}}\\&\approx 10\\\end{aligned}$

Thus, the number of collisions required to reduce the speed of the neutron is $\boxed{10}$.

Learn More:

1. A rocket being thrust upward as the force of the fuel being burned pushes downward is an example of which of Newton’s laws brainly.com/question/11411375

2. A 700-kg car, driving at 29 m/s, hits a brick wall and rebounds with a speed of 4.5 m/s brainly.com/question/9484203

3. For inelastic collisions which of the following statements can be true brainly.com/question/2990238

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.

Answer 2

M*U + 0 = m*v'1 + 2m*v'2 
the zero means deuteron has no velocity
where v'1 and v'2 are the post-collision velocities. 
The equatio becomes
U = v'1 + 2v'2
U = v'2- v'1 
v'2 = U + v'1 

U = v'1 + 2(U + v'1) = 2U + 3v'1 
U = -3V 
V = -U / 3 
The speed ratio is 1/3 

B) Since KE is proportional to the square of the speed, if the speed is 1/3, then KE is 1/9 

C) (1/3)ⁿ = 1/729 
3ⁿ = 729 
n = 6