Conservation of momentum of a neutron

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Homework Help Overview

The discussion revolves around the conservation of momentum in a nuclear reaction involving a He4 nucleus that breaks up into a neutron and a He3 nucleus. The original poster is trying to determine the speed of the He3 nucleus given the conditions of the problem.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to apply the Pythagorean theorem to find the speed of the He3 nucleus based on the neutron's motion. Some participants question the correctness of the original poster's assumptions about the neutron's velocity components and the resulting motion of the He3 nucleus.

Discussion Status

Participants are actively discussing the conservation of momentum principles, with some clarifying misconceptions about the directions of motion and the implications for the velocities of the nuclei involved. There is a recognition of the need to consider horizontal momentum conservation, and some guidance has been provided regarding the relationships between the velocities of the neutron and the He3 nucleus.

Contextual Notes

There are indications of confusion regarding the direction of velocities and the application of conservation laws, with some participants expressing uncertainty about the calculations and assumptions made in the problem setup.

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A He4 nucleus, with a mass of 4 amu moving with speed v breaks up into a neutron 1amu and a He3 nucleus 3amu . If the neutron moves in a direction perpendicular to the direction of motion of the original He4 nucleus with speed 3v, what is the speed of the He3 nucleus?

if the neutron moves down. The nucleus will be pushed upward with 1/3v. the nuetron also has the original 1v in the horizontal direction. so should i just use pythagorean theorem?: so i get sqrt(10)v/3?
 
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the nuetron also has the original 1v in the horizontal direction
Incorrect:
the neutron moves in a direction perpendicular to the direction of motion of the original He4 nucleus

The nucleus will be pushed upward with 1/3v
Incorrect.
 
which of those quotes are incorrect? should the horizontal velocity of the nucleus increase as well? I am so confused.
 
"the nuetron also has the original 1v in the horizontal direction"
"The nucleus will be pushed upward with 1/3v"

The above are incorrect. Horizontal momentum must be conserved, and the only thing moving horizontally is the nucleus. So yes, its velocity in x direction must increase.
 
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so the nucleus will be pushed +v/3 in the x direction so the total x direction is 4v/3.
In the y direction the velocity will be sqrt(3v^2 - v^2) for the neutron. and 1/3 that for the nucleus?
 
so the nucleus will be pushed +v/3 in the x direction so the total x direction is 4v/3.
Yep.

In the y direction the velocity will be sqrt(3v^2 - v^2) for the neutron. and 1/3 that for the nucleus?
In y-direction it is given in the problem that the neutron has 3v. And yes, 1/3 (although I'm not sure how you got the ratio, did you use the conservation of momentum?) of the neutron's velocity in y direction is the nucleus' velocity in y direction.
 
ahhh, now i get it, the velocity of the neutron is only in the y direction. (yes i used conservation of momentum) so the total speed of the neucleus is sqrt(v^2 + (4v/3)^2) which is 5v/3! thanks a lot
 

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