Inelastic collision of two particles

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Homework Help Overview

The problem involves an inelastic collision between two particles of equal mass and initial speed, where they move together post-collision at a specified speed. The objective is to determine the angle between their initial velocity vectors.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need for momentum components in both x and y directions due to the angle involved in the collision.
  • Questions arise regarding whether the angles of the two particles are equal or if they should be treated as different angles.
  • Some participants suggest simplifying the problem by aligning one particle's motion with the x-axis to reduce the number of unknowns.
  • There is a discussion about the implications of assuming specific angles and how they affect the equations derived from momentum conservation.
  • Concerns are raised about the correctness of the equations being formed and the units involved in the calculations.

Discussion Status

The discussion is actively exploring various approaches to set up the equations needed to solve for the angle. Participants are providing guidance on how to structure the momentum equations and are questioning assumptions made about the angles involved. There is no explicit consensus yet, but several productive lines of reasoning are being developed.

Contextual Notes

Participants are navigating the complexities of the problem, including the conservation of momentum in an inelastic collision and the need for clarity on the angles involved. There are indications of confusion regarding the relationship between the angles and the resulting equations.

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Homework Statement



After a completely inelastic collision, two particles of the same mass m and same initial speed v are found to move away together with the speed (2/3)v. Find the angle between the initial velocity vectors of the objects.

The Attempt at a Solution



It is completely inelastic which means the two objects stick together and share the same final speed. Energy is not conserved but Momentum is so
mv + mv = (m+m)V

2mv = 2m(2v/3)
2mv=4mv/3
2=4/3

I definitely did something wrong here
 
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You need components I am pretty sure, because it is asking for angles. So makes your x components of momentum, and y components of momentum. So what I am saying is that your question asks for an angle, so that they collide at an angle, so you would need two equations one for everything in the y direction, and the other everything in the x direction.
 
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Should there be two different angles I am looking for here or are they equal? Meaning if I put the final velocity vector of the two particles along the x axis, is the angle in between the x-axis and the particle above the x-axis equal to the angle between the x-axis and the particle below the x axis?
 
So you are kind of on the right track, there is one initial angle θ and for simplicities sake, let's say one of the particles is moving perpendicular to the x axis, and the other is moving at an angle θ from the x axis, and let's say that the new angle formed when they collide is θ'. With this information you should be able to create two different equations, and the mass's and velocities should divide out, then you have two unknowns you could easily solve for.
 
attachment.php?attachmentid=69576&d=1399616444.png


Is this an accurate representation of what is going on? (my terrible drawing skills aside) the green and red vectors are the two particles before the collision and the blue one with angle θ' is the two of them moving while stuck together.
 

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    momentum vectors.png
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I am ending up with an equation for the x components:

vcosθ = (4/3)vcosθ'

and for the y components:

vsin + v = (4/3)vsinθ'
 
for simplicities sake I said, that you can make one of the particles move parallel to the x axis, then this eliminates one of the unknown angles.
 
Wouldnt that have to assume that the angle between the two starting vectors is 90 degrees?
 
No it wouldn't, imagine rotating the x and y-axis so that the one particle is moving parallel to the x axis. Instead of two angles, its one.
 
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  • #10
I think it would be simpler to take the final motion as being along an axis. Given that they have the same mass and initial speed it follows that they made the same angle to the axis on approach,
 
  • #11
ok so now I am getting

mvcosθ+mvcos(-θ) = 2/3vm

and

mvsinθ+mvsin(-θ) = 0

I still feel stuck here :[
 
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  • #12
Their components of velocity in opposite directions (which you made the 'sine component') will balance (or "annihilate") to give zero, but how will the components of velocity parallel to each other affect each other?
toothpaste666 said:
mvcosθ+mvcos(-θ) = 2/3v

Is that correct? On the left, you have units of momentum, on the right, units of velocity. What's missing?
 
  • #13
oops! thanks I fixed it. So the m's and v's cancel and it becomes cosθ+cos(-θ) = 2/3
 
  • #14
toothpaste666 said:
oops! thanks I fixed it. So the m's and v's cancel and it becomes cosθ+cos(-θ) = 2/3
So that means the mass after they collide (and stick together) will only be m? (That's what your math says, is that correct, though?)

Also, what is cos(-θ)? Is there another way to write that?
 
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  • #15
ok I am sorry i was being sloppy there

mvcosθ+mvcos(-θ) = 2/3v(m+m)

mvcosθ+mvcos(-θ) = 4/3vm

cosθ+cos(-θ) = 4/3

cosθ+cosθ = 4/3

2cosθ = 4/3

cosθ = 4/6

cosθ = 2/3

θ = cos^-1(2/3)

therefore the angle between them two starting vectors is double that?
 
  • #16
toothpaste666 said:
θ = cos^-1(2/3)

therefore the angle between them two starting vectors is double that?

Yes.
 
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  • #17
thanks guys
 

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