Inelastic collision of two particles

In summary, a problem discusses a completely inelastic collision between two particles with the same mass and initial speed. The two particles are found to move away together with a speed of (2/3)v, and the goal is to find the angle between their initial velocity vectors. After analyzing the conservation of momentum and using components of velocity, the final equation becomes cosθ + cos(-θ) = 4/3, which gives an angle θ = cos^-1(2/3) between the two starting vectors.
  • #1
toothpaste666
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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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  • #2
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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  • #3
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?
 
  • #4
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.
 
  • #5
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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  • #6
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'
 
  • #7
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.
 
  • #8
Wouldnt that have to assume that the angle between the two starting vectors is 90 degrees?
 
  • #9
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
 

1. What is an inelastic collision?

An inelastic collision is a type of collision between two particles or objects where there is a loss of kinetic energy. In this type of collision, the two particles stick together and move as one after the collision.

2. What is the difference between an inelastic collision and an elastic collision?

In an elastic collision, the two particles bounce off each other without any loss of kinetic energy. In an inelastic collision, there is a loss of kinetic energy and the particles stick together or deform upon impact.

3. What factors affect the amount of kinetic energy lost in an inelastic collision?

The amount of kinetic energy lost in an inelastic collision depends on the nature of the colliding particles, their masses, and their velocities. A larger mass or slower velocity will result in more kinetic energy being lost.

4. Can an inelastic collision still conserve momentum?

Yes, an inelastic collision still conserves momentum. The total momentum of the system before and after the collision remains the same, but the kinetic energy may be converted into other forms, such as heat or sound.

5. How is the coefficient of restitution related to inelastic collisions?

The coefficient of restitution is a measure of how much kinetic energy is conserved in a collision. In an inelastic collision, the coefficient of restitution is less than 1, indicating that there is a loss of kinetic energy. The lower the coefficient of restitution, the more inelastic the collision is.

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