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Moderators. Canadian nuclear reactors use heavy water moderators in which elastic collisions occur between the neutrons and deuterons of mass 2.0 u (see Example 8.11 in Section 8.4 ) (a) What is the speed of a neutron, expressed as a fraction of its original speed, after a head-on, elastic collision with a deuteron that is initially at rest? (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 neutron to $1 / 59,000$ of its original value?

(a) $v_{n 2}=\frac{1}{3} v_{n 1}$

(b) $K_{n 2}=\frac{1}{9} K_{n 1}$

(c) $n=10$ collisions

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in this question and neutron collides head on with a do trim. The mass of the doctrine is two times bigger than the mass off the neutral which is only one unit off atomic mass. In this collision, what happens is that initially the velocity off the new term is zero and the velocity off the neutron is V. After the collision, the velocity off the neutral will be the prime. Then in the first item, we have to deter mine. What is the value off the prime over V For that, we can use this equation which is suitable for that kind of situation. This equation a is our neutral and B is the Dutra which begins at rest the reform. What we have here adapted tow. Our problem is the following V prime, which is the new velocity is equal to the mass off the nutrient minus the mask off the neutral divided by the mass off the neutron, plus the mass off the do trim on. These multiplies the velocity off the neutral before the collision. Therefore, V prime, divided by V, is equal to the mass off the neutral, minus the mass of the neutral, divided by the mass off the neutron, plus the mass off the new trim. And this is one minus two over one plus two. Then we have minus 1/3. So this is the fraction minus one third. As the speed is just the absolute value off the velocity, we can say that the new speed divided by the old speed is one third. This is the answer to the first item. In the second item, we have to evaluate what is the new kinetic energy divided by the old kinetic energy. For that, we have to use the expression for dragnet IQ energy. So we have the following K Prime is the new kinetic energy. So it's the mass of the neutron times the new velocity of the neutrons squared, divided by true. Then we divided by the old kinetic energy which is the mass off the neutral times that old velocity squared, divided by two. As you can see, Ah, lot of simplifications will happen just like that. Then K prime, divided by K is just the prime squared divided by V squared and it can write it as v prime divided by V squared from the previous item. We know that the prime divided by V is minus one third. Therefore, we have minus one third squared and the result is 19 Therefore, K prime divided by K is equals to 1/9 on. This is the answer to the second item off this question. Now, for the last item, we have to think about the following. After each collision, the velocity of the nutrient is reduced by one third. Then after a lot of collisions, say big N, the velocity of the neutron will be reduced by one divided by 59,000. What is M? So we have to note the following after one collision, there is a reduction off one third after two collisions, there will be a reduction off one third on top of that one third, which amounts to 19 Then you can already see something going on After pre collisions will have one third off one third off one third, which is one divided by 27. The reform after and collisions the velocity will be reduced by a factor off one third to the power off em. Then all that you have to do is the following we want that the reduction factor to be equals to one by 59,000. Therefore, what we know is that one third to the end is equals to one divided by 59,000. Then one to the end, which is equals to one divided by three to the end is equals to one divided by 59,000. Then three to the end is equals to 59,000. And now to solve this, we can take the luxury item by taking the luxury tum off this equation. We got the following and times the longer it, um off three. Is he, of course, to the lab. A rhythm off 59,000 then and is equal to the logarithms off 59,000 divided by the longer Etem for free. And these results in an being approximately 10. And this is the answer to this question.

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