Moderators. Canadian nuclear reactors use heavy water...

Moderators. Canadian nuclear reactors use heavy water moderators in which elastic collisions occur between the neutrons and deuterons of mass $2.0 \mathrm{u}$ (see Example 8.11 in Section 8.4 ). (a) What is the spced of a neutron, expressed as a fraction of its original specd. after a hcad-on, clastic collision with a deuteron that is initially at rest? (b) What is its kinctic energy, expressed as a fraction of its original kinetic energy? (c) How many such successive collisions will reduce the speed of a neutron to $1 / 59,000$ of its original valuc?

Moderators. Canadian nuclear reactors use heavy water moderators in which elastic collisions occur between the neutrons and deuterons of mass $2.0 \mathrm{u}$ (see Example 8.11 in Section 8.4 ). (a) What is the spced of a neutron, expressed as a fraction of its original specd. after a hcad-on, clastic collision with a deuteron that is initially at rest?
(b) What is its kinctic energy, expressed as a fraction of its original kinetic energy? (c) How many such successive collisions will reduce the speed of a neutron to $1 / 59,000$ of its original valuc?

Key Concepts

Cumulative Effects of Successive Collisions

When a particle undergoes multiple collisions, the effects compound. Each collision reduces the speed (and hence kinetic energy) by a fixed fraction, leading to an exponential decrease over several interactions. Understanding this iterative process and its logarithmic nature is important in fields such as reactor physics and materials science.

Speed and Kinetic Energy Relationship

The kinetic energy of a particle is proportional to the square of its speed. When a particle’s speed is reduced by a certain fraction in a collision, its kinetic energy decreases by the square of that fraction. This relation is key in quantifying the energy loss per collision.

Elastic Collisions

These are collisions in which both momentum and kinetic energy are conserved. The analysis of elastic collisions allows one to predict the post-collision velocities of particles, making it a fundamental concept in classical mechanics and nuclear physics.

Conservation of Momentum and Kinetic Energy

In collision problems, especially elastic ones, the conservation laws of momentum and kinetic energy are essential. They provide the equations needed to determine the velocity changes of interacting particles, illustrating how energy and momentum are redistributed during the collision.

Head-On Collision Dynamics

A head-on collision, where two particles approach each other along the same line, simplifies the analysis and highlights the effect of mass differences on the final speeds. The formulas derived from such scenarios are crucial for understanding how factors like mass ratios influence the reduction in speed and kinetic energy.

Transcript

00:01 In this problem we have given canadian nuclear reactor uses heavy water moderators in which elastic collision occurs between the neutron and deutron of mass 2 u.

00:11 First of all, mass of neutron is 1 u and mass of deutron is 2 u.

00:18 So here collision happens between the neutrons and the deutrons so we have to find what is the speed of a neutron expressed as a fraction of initial speed so initially suppose this is neutron and this is deutron so suppose initially initial speed of neutron is vn initial and duetron is at rest but after the collision suppose speed of neutron is vn final and speed of deutron is vd final because vd d initial is zero initial velocity of deutron is zero so here we we will use momentum of conservation conservation of momentum so initially mass of neutron into speed vn initial should be equal to mass of neutron into speed of final after collision plus mass of deutron into vd final so from here this mass of neutron is 1 u so this will be equal to 1 u into v n initial is equal to 1 u into v n final plus 2 u v d final so from here we can write a equation which is v n final plus v plus twice of v d final will be equal to v n initial okay suppose this is number 1.

02:10 Now we know that relation between speed of approach and speed of separation.

02:22 So we will write vn1 vn initial minus vd initial is equal to minus of vn final minus vd final.

02:38 This is the standard result.

02:41 So from here we know that this is equal to 0.

02:44 So we can write vd final is equal to vd final is equal to vn initial plus vn final right this is equation number two now we will solve equation number one and two from equation number one and two we can write v n final plus twice of this is vd final so instead of vd final.

03:18 Final we will write v n initial plus vn final now it is equal to vn initial so from here if we solve this we will get vn final is equal to minus of 1 by 3 vn initial vn final will be equal to minus 1 by 3 vn initial vn final will be equal to minus 1 upon 3 vn initial so this is the speed of neutron after collision.

03:54 Now this is the velocity.

03:56 That means speed of neutron is one by third of initial speed of neutron and that this negative sign shows that the direction is reverse.

04:10 Right...

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