Calculate Momentum/KE Transfer in Neutron-Boron Collision

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Homework Help Overview

The discussion revolves around a collision between a neutron and a boron nucleus, focusing on the transfer of kinetic energy in a perfectly elastic collision scenario. The problem involves concepts from momentum and kinetic energy conservation.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of conservation of momentum and kinetic energy equations, noting the challenge of dealing with two unknowns in the equations. There is mention of substitution methods and the potential need for the quadratic formula.

Discussion Status

The discussion is active, with participants exploring different approaches to solve the equations simultaneously. Some guidance has been provided regarding the use of substitution, and there is recognition of the complexity introduced by the two unknowns.

Contextual Notes

Participants are working under the constraints of needing to show their work and the requirement for a complete understanding of the equations involved before receiving further assistance.

richylai
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A neutron with a mass of 1.67×10^−27 kg and moving with a speed of 45000 m/s makes a head-on collision with a boron nucleus with a mass of 1.66×10^−26 kg. The boron nucleus is originally at rest.

If the collision is perfectly elastic, what fraction of the neutron's original kinetic energy is transferred to the boron nucleus?
 
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What equations apply? You need to show your work before we can offer tutorial help.
 
I used the conservation of momentum equation, M neutron * Vi,n = M neutron * Vf,n + M boron * Vf,b

Then I would use the KE equation .5(M neutron)(Vi,n)^2 = .5(M neutron)(Vf,n)^2 + .5(M boron)(Vf,b)^2

However, I realized that I would be messing with two unknowns and I have no idea how to approach that, unless I'm totally off with the equations I presented.
 
You indeed have two unknowns, but you also have two equations, which you can solve simultaneously.
 
Ah... Ok, I guess I have to do some kind of substitution.
 
richylai said:
Ah... Ok, I guess I have to do some kind of substitution.
That would be the way forward... :smile:
 
Oh no... Looks like I have to use the quadratic formula for this... -.-'
 

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