How Does a Neutron's Speed Change After Colliding with a Deuterium Nucleus?

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Homework Help Overview

The discussion revolves around the behavior of a neutron colliding with a deuterium nucleus in the context of nuclear reactions, specifically focusing on the changes in speed of the neutron after an elastic collision. The problem is situated within the subject area of nuclear physics and momentum conservation.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of momentum and kinetic energy equations to analyze the collision. There are attempts to derive the final speeds of the neutron and deuterium, with some participants expressing uncertainty about their calculations and the correctness of their results.

Discussion Status

Multiple interpretations of the collision outcomes are being explored, with participants questioning the validity of their calculations and the units involved. Some guidance is offered regarding the cancellation of masses in the equations, but no consensus has been reached on the final results.

Contextual Notes

Participants mention an upcoming exam, indicating time constraints and the pressure to understand the problem thoroughly. There is also a reference to homework rules that may limit the extent of assistance provided.

Magma828
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Homework Statement



Uranium fission within a nuclear reactor produces high-speed neutrons. These must be slowed down for them to trigger a chain reaction within the uranium. In a heavy-water reactor the neutrons of mass m make multiple elastic collisions with deuterium nuclei of 2m, losing speed at each collision.

(a) Suppose a fast-moving neutron of speed v makes a head-on elastic collision with a stationary deuterium nucleus. Show that its speed is reduced to v/3.

Homework Equations



p=mv - We probably have to use this
Ek=0.5mv2 - Not sure if we have to use this, but the question IS in a section about kinetic energy. It just seems logical to use p=mv because using kinetic energy makes no sense.

The Attempt at a Solution



Ok it took me absolutely AGES to do this, but I don't think it's right:

PD1 = mv = 2m x 0 = 0 N s
PN1 = mv = 1m x v = mv

PD1 + PN1 = PD2 + PN2 if no external forces are applied

0 + mv = PD2 + PN2
mv = PD2 + PN2
PD2 = mv - PN2

PD2 = PN2 - not sure

Substitute above two equations together to show PN2:
PN2 = mv - PN2
2PN2 = mv
PN2 = mv/2
vN2 = PN2/2m
vN2 = (mv/2)/2m
vN2 = mv/4m - my rearranging may be wrong
vN2 = v/3m

But the answer is just v/3 not v/3m. Is this still right? Or is it just coincidence I got an answer so similar?
 
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v/3m cannot be right. it has units of m.kg/s
 
The answer i got was -v/3. So the neutron bounces backwards with speed v/3 and the deuterium goes forward with 2v/3. I didn't read through your answer very thoroughly, but anywhere with mass on the bottom will be wrong. All the masses cancel out in my calculations
 
dacruick said:
The answer i got was -v/3. So the neutron bounces backwards with speed v/3 and the deuterium goes forward with 2v/3. I didn't read through your answer very thoroughly, but anywhere with mass on the bottom will be wrong. All the masses cancel out in my calculations

Could you post your calculation please? You don't even have to explain it, but I've been trying to do this all night and I don't have a clue. Got an exam on thursday l0L!
 
bah fine. I'm a lazy individual.
v is initial velocity
v1 is final velocity of the neutron
v2 is the final velocity of the deuterium

mv = m . v1 + 2m . v2 ------------- (momentum)
mv² = m . v1² + 2m . v2² ------------- (energy)

All the masses cancel:

v = v1 + 2 . v2 ------------- (momentum)
v² = v1² + 2 . v2² ------------- (energy)

isolate the momentum equation for v2
v2 = (v - v1)/2
v2² = (v² - 2*v1*v + v1²)/4

sub that into your energy equation and isolate for v1.
 

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