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Conservation of Momentum, 2D, Unknown Final Velocities

  • Thread starter Ocata
  • Start date
196
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1. The problem statement, all variables and given/known data

Momentume%20and%20Conservation%201_zpsymoqense.jpg


Conservation of momentum of x components
[itex] 10(20Cos(-30)) + 5(5Cos60) = 10v_{f}Cos(α) + 5v_{f}Cos(β) [/itex]

[itex] 185.7 = 10v_{f}Cos(α) + 5v_{f}Cos(β) [/itex]

Conservation of momentum of y components

[itex] 10(20Sin(-30)) + 5(5Sin60) = 10v_{f}Sin(α) + 5v_{f}Sin(β) [/itex]

[itex] -78.3 = 10v_{f}Sin(α) + 5v_{f}Sin(β) [/itex]


2. Relevant equations

Conservation of momentum in x:

(mv)1ix + (mv)2ix = (mv)1fx + (mv)2fx

Conservation of momentum in y

(mv)1iy + (mv)2iy = (mv)1fy + (mv)2fy


3. The attempt at a solution

x components
[itex] 185.7 = 10v_{f}Cos(α) + 5v_{f}Cos(β) [/itex]

and

y components
[itex] -78.3 = 10v_{f}Sin(α) + 5v_{f}Sin(β) [/itex]


With two equations, each with two unknown velocities and two unknown angles. It just seems like there is not enough information to algebraically solve for the final velocities and angles.
 
32,585
8,465
That is right, and there is no way to fix it. Collisions in at least two dimensions have some freedom. You can fix one more if you make the collision elastic, or fix the amount of energy lost in the process, but then the result will still depend on details of the collision process. Just have a look at billiard, then you'll see that different collisions give different results (here: depending on where you hit the other ball).
 
196
4
That is right, and there is no way to fix it. Collisions in at least two dimensions have some freedom. You can fix one more if you make the collision elastic, or fix the amount of energy lost in the process, but then the result will still depend on details of the collision process. Just have a look at billiard, then you'll see that different collisions give different results (here: depending on where you hit the other ball).

Thank you mfb.

If I establish the collision as 100% elastic, then

Ei = Ef

[itex]1/2(mv^{2})_{1i} + 1/2(mv^{2})_{2i}= 1/2(mv^{2})_{1f} + 1/2(mv^{2})_{2f}[/itex]

And Pi = Pf

Conservation of momentum in x:

(mv)1ix + (mv)2ix = (mv)1fx + (mv)2fx

Conservation of momentum in y

(mv)1iy + (mv)2iy = (mv)1fy + (mv)2fy


Now I have three equations. But before I continue, may I clarify one point?


Question is, is the conservation of energy component dependent just as conservation of momentum is component dependent? In other words, would I need:


[itex]1/2(mv^{2})_{1i} + 1/2(mv^{2})_{2i}= 1/2(mv^{2})_{1f} + 1/2(mv^{2})_{2f}[/itex]

Or, would I need to break kinetic energy into components?:

Conservation of energy of components in the x
[itex].5m(vCos(-30))^{2}_{1} + .5m(vCos(60))^{2}_{2} = .5m(vCosα)^{2}_{1} + .5m(vCosβ)^{2}_{2}[/itex]


Conservation of energy of components in the y
[itex].5m(vSin(-30))^{2}_{1} + .5m(vSin(60))^{2}_{2} = .5m(vSinα)^{2}_{1}+ .5m(vSinβ)^{2}_{2}[/itex]
 
32,585
8,465
Total energy is a single number, it does not have (meaningful) components.

[itex]1/2(mv^{2})_{1i} + 1/2(mv^{2})_{2i}= 1/2(mv^{2})_{1f} + 1/2(mv^{2})_{2f}[/itex]
Right.
 
196
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mfb, thank you.

What I have so far are these 3 equations:

Equation 1)
Conservation of momentum in the x component:

[itex]185.7 = 10vCos(α)_{1f} + 5vCos{β}_{2f} [/itex]


Equation 2)
Conservation of momentum in the y component:

[itex]-78.3 = 10vSin(α)_{1f} + 5vCos{β}_{2f} [/itex]


Equation 3)
Conservation of Energy

[itex]2062.5 = 5v^{2}_{1f} + 2.5v^{2}_{2f}[/itex]


Not sure what to do next at this point. I don't know the final x or y velocity components of either object, I don't know the final velocity magnitudes of either object, and I don't know their angles. Is there any method of substitution where I can find any of these unknowns?
 
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32,585
8,465
As I said, this is impossible. The collision can have different results even if the initial velocities are known.
For a real collision, it will depend on details of the collision process, especially the impact point.
 
196
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What might be the basic details needed in order to solve this scenario? One important detail, as you mention, is impact point. Are there any other details that would be necessary for solving a very theoretical situation? Can a problem like this be reduced in simplicity for the sake of learning the general idea (sort of like considering a rope to be massless when learning about tension and forces between blocks)?

For instance, how could this problem be solved with two spheres in zero gravity and no kind of external friction?
 
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32,585
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Can a problem like this be reduced in simplicity for the sake of learning the general idea (sort of like considering a rope to be massless when learning about tension and forces between blocks)?
You can learn that some problems don't have a unique answer.
What might be the basic details needed in order to solve this scenario? One important detail, as you mention, is impact point. Are there any other details that would be necessary for solving a very theoretical situation?
Everything about the colliding objects can be important.

For frictionless spheres: momentum transfer happens orthogonal to the contact plane only. This gives a fourth equation if you fix the impact parameter.
 
196
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Thank you. I will take note of this parameter and look forward to building up to this level. Appreciate you introducing this concept.
 

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