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Elastic Neutron Collision (Conservation of Momentum and Energy?)

  • #1
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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Answers and Replies

  • #2
6,054
390
Yes, that is the way to go.
 
  • #3
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
6,054
390
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
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
6,054
390
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
6,054
390
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
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
6,054
390
The neutron is said to collide elastically. What does that mean?
 
  • #13
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
6,054
390
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
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
6,054
390
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
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
6,054
390
##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 in to 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
##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
6,054
390
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
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.
 

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