Elastic Neutron Collision (Conservation of Momentum and Energy?)

In summary, after a neutron collides with a helium nucleus, the helium nucleus rebounds at an angle 2 = 41° from the neutron's initial direction. The neutron's initial speed is 5.6 105 m/s.
  • #1
frownifdown
72
0
1. A neutron collides elastically with a helium nucleus (at rest initially) whose mass is four times that of the neutron. The helium nucleus is observed to rebound at an angle '2 = 41° from the neutron's initial direction. The neutron's initial speed is 5.6 105 m/s.

Determine the angle at which the neutron rebounds, '1, measured from its initial direction.

What is the speed of the neutron after the collision?

What is the speed of the helium nucleus after the collision?
2. Conservation of momentum and conservation of energy
3. I really don't even know how to start with this question. I assume it would have something to do with setting up a conservation of momentum equation then one of energy and solving them both but I'm not sure.
 
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  • #2
Yes, that is the way to go.
 
  • #3
voko said:
Yes, that is the way to go.

How would that look exactly? I have

mv(neutron)1 = mv(neutron)2 + 4mv(helium)2

Is that close?
 
  • #4
That is correct, if the v's are vectors. You may want to decompose them into xy components, taking x as the direction of the neutron's initial velocity.
 
  • #5
voko said:
That is correct, if the v's are vectors. You may want to decompose them into xy components, taking x as the direction of the neutron's initial velocity.

So what would I put for the equation then? Would I have to put the velocity of the helium multiplied by the sin and cos of 41?
 
  • #6
Yes, that is the components of helium's velocity after the collision. Do the same thing for all the other components, you should have two equations, one for the x components, another for the y components.
 
  • #7
Alright here is the work that I've done so far. I have the two equations but I don't really know how to proceed because I have so many unknowns. Also I'm not sure where I should put sin41 and cos41. Obviously the sin would go in the Y equation and cos in the X but I'm not sure where exactly it goes.

http://i.imgur.com/9dWvHPi.jpg
 
  • #8
Can anyone help me with this problem? It's been kicking my *** for days and it's due tonight
 
  • #9
What you have done so far is correct.

You have four unknowns, which you can reduce to three if you use the information on the angle of helium's velocity: ##V_{Hx} = V_H \cos \alpha, \ V_{Hy} = V_H \sin \alpha##. So you will have two equations and three unknowns. You can add to that conservation of energy, and the system should be solvable.
 
  • #10
voko said:
What you have done so far is correct.

You have four unknowns, which you can reduce to three if you use the information on the angle of helium's velocity: ##V_{Hx} = V_H \cos \alpha, \ V_{Hy} = V_H \sin \alpha##. So you will have two equations and three unknowns. You can add to that conservation of energy, and the system should be solvable.

What would my conservation of energy equation look like?
 
  • #11
I have Mneutron(5.6*10^5)^2 = Mneutron(5.6*10^5)^2 + 2Mhelium(Vhelium)^2

Not sure how to proceed if that is even right
 
  • #12
The neutron is said to collide elastically. What does that mean?
 
  • #13
voko said:
The neutron is said to collide elastically. What does that mean?

That energy of the system is conserved. Is that not what I did?
 
  • #14
frownifdown said:
I have Mneutron(5.6*10^5)^2 = Mneutron(5.6*10^5)^2 + 2Mhelium(Vhelium)^2

Not sure how to proceed if that is even right

Of course this is not right. Why would the neutron have the same velocity after the collision? That is what you need to find.
 
  • #15
Really I'm unsure about what to do with the masses. Do I need to put in a value for them? Or just do 1 and 2 for them after multiplying by half for all of the figures in the energy
 
  • #16
voko said:
Of course this is not right. Why would the neutron have the same velocity after the collision? That is what you need to find.

Ah good point. So then it is just that equation with the masses being...? Not sure on that. Would I put Vhelium as sinθ*Vhy?
 
  • #17
I would keep the M symbol to denote the mass of neutron, and 2M for helium. You can ditch the M, but when it is there, it serves as a sanity check. After you get the final system of equations, then you can drop it.
 
  • #18
voko said:
I would keep the M symbol to denote the mass of neutron, and 2M for helium. You can ditch the M, but when it is there, it serves as a sanity check. After you get the final system of equations, then you can drop it.

Ok I was thinking that I'd do that. Then what would I substitute in for Vh?
 
  • #19
Sorry I just have 10 minutes to get this answer in so I'm trying to hurry. Will take me from an 85% to a 97% on the hw. I really appreciate all the help you've given me so far though
 
  • #20
##V_H## is an unknown. You have two more: ##V_{Nx}, \ V_{Ny}##.

You have two equations for them. Add another from conservation of energy and solve the system.
 
  • #21
I just don't know what to put for Vhy or Vhx when putting it into the conservation of energy equation to solve for the speed of the neutron. And how would I find the angle? which is the first part of the question
 
  • #22
voko said:
##V_H## is an unknown. You have two more: ##V_{Nx}, \ V_{Ny}##.

You have two equations for them. Add another from conservation of energy and solve the system.


Add another equation? I'm sorry I'm struggling with this so much. Very frustrating
 
  • #23
You need to write the kinetic energies of neutron before the collision; it should be equal to the sum of the kinetic energies of n and He after the collision. I do not see why that is so hard.
 
  • #24
voko said:
You need to write the kinetic energies of neutron before the collision; it should be equal to the sum of the kinetic energies of n and He after the collision. I do not see why that is so hard.

Alright well I guess it's just different for me. Jesus. Well anyways thanks for the help.
 

FAQ: Elastic Neutron Collision (Conservation of Momentum and Energy?)

What is an elastic neutron collision?

An elastic neutron collision is a type of collision between two particles, where the total kinetic energy of the system is conserved and both particles remain intact after the collision.

How does the conservation of momentum apply to elastic neutron collisions?

In an elastic neutron collision, the total momentum of the system before and after the collision must be equal. This means that the sum of the momentum of the two particles before the collision is equal to the sum of their momentum after the collision.

What is the role of energy conservation in elastic neutron collisions?

Energy conservation states that the total energy of a system remains constant. In elastic neutron collisions, the kinetic energy of the particles before and after the collision must be equal, as no energy is lost in the collision.

How are the velocities of the particles calculated after an elastic neutron collision?

The velocities of the particles after an elastic neutron collision can be calculated using the conservation of momentum and energy equations. By equating the total momentum and kinetic energy before and after the collision, the velocities of the particles can be solved for.

What are some real-world applications of understanding elastic neutron collisions?

Understanding elastic neutron collisions is important in nuclear physics, as it helps scientists study the behavior of particles in nuclear reactions. It is also utilized in fields such as medical imaging, where neutron scattering can be used to create images of biological tissues.

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