Air Hockey Momentum: Puck A & B Collision

  • Thread starter Chrisleo13
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In summary: Just remember to stay organized and keep track of your subscripts. It will save you a lot of confusion and time in the long run.
  • #1
Chrisleo13
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Homework Statement


Two pucks collide on an air hockey table. Puck A has a mass of 21 g and is initially traveling in the +x direction at 6.2 m/s. Puck B has a mass of 84 g and is initially at rest. After the pucks collide, puck A moves away at an angle of 49 above the +x axis, while puck B travels at an angle of 42 below the +x axis.

a. Calculate puck A's final speed.
b. Calculate puck B's final speed.
c. What fraction of kinetic energy was lost in this collision?

I really have no clue how to start this question. Thanks in advance.
 
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  • #2
Well you know that momentum is going to be conserved. So the initial momentum is going to equal the final momentum of the 2 pucks. You will also have to break it down into the x and y directions.
So the initial momentum in the x direction will equal the final momentum in the x direction. Same for the y. You have the masses, and the initial speed, and I surmise that you will end up with 2 equations and 2 unknowns, which means its cake.
 
  • #3
Thanks for the help, but I still have the equations wrong I guess. I understand what to do, but I guess I am setting them up wrong. If you could help me with that I would appreciate it.


Thanks
 
  • #4
[tex]p = mv[/tex]

[tex] p_{i} = p_{f}[/tex] In both x and y.

For x:
[tex] m_{1}v_{1ix} = m_{1}v_{1fx} + m_{2}v_{2fx}[/tex]
[tex] m_{1}v_{1i}cos(\Theta) = m_{1}v_{1f}Cos(\Theta) + m_{2}v_{2f}Cos(\Theta)[/tex]

For y:

[tex] m_{1}v_{1iy} = m_{1}v_{1fy} + m_{2}v_{2fy}[/tex]
[tex] m_{1}v_{1i}sin(\Theta) = m_{1}v_{1f}Sin(\Theta) + m_{2}v_{2f}sin(\Theta)[/tex]
[tex] 0 = m_{1}v_{1f}Sin(\Theta) + m_{2}v_{2f}sin(\Theta)[/tex]

0 because there is no initial velocity in the y direction.

So you have two equations there, and two unknowns. Solve one equation in terms of something, and plug that into the other equation in order to only have 1 unknown.

Make sure to keep track of your subscripts in order to not get confused. You don't want to add masses or velocities if they are for 2 different objects.
 
  • #5
Hmm, I should solve for something in the the X direction because in the Y direction Vi = 0.

Are there two separate ones for each Puck? For the final speed of A and B?
 
  • #6
It doesn't matter that Vi is 0 in the y direction equation. You are solving for Vf. The only two things you don't know are V1f and V2f. You have two equations though. Solve one of them for one of the Vf's and plug it into the other equation to get rid of the two unknowns in that equation.
 
  • #7
Alright, I think I solved one equation right to plug into the other.


I solved the first one

m1vix = m1vfx1 + m2vf2

vix = m1vfx1 + m2vf2 / m1

Am I on the right track?
 
  • #8
Yes and no. You did indeed solve for one of the knowns, but you did so incorrectly, and you dropped a couple of subscripts you need to keep track of, and where did the cosines go? Stay organized. Plus why would you solve your equation for vix when that is already one of your knowns? Solve it for one of your unknowns.

[tex] m_{1}v_{1i}cos(\Theta) = m_{1}v_{1f}Cos(\Theta) + m_{2}v_{2f}Cos(\Theta)[/tex]

[tex] \frac{m_{1}v_{1i}cos(\Theta) - m_{2}v_{2f}Cos(\Theta)}{m_{1}Cos(\Theta)} = v_{1f}[/tex]
 
Last edited:
  • #9
I plugged that equation into the first equation and when I canceled everything, I got m2v2cos(theta)s to cancel because 1 is negative and 1 is positive.
 
  • #10
I have pretty much helped you the most possible extent so this will more than likely be my last post. It seems you are missing some fundamental concept and you may just need to review things before you try to solve this. I will explain this one more time:

You have two equations. One for the x direction and one for the y direction.
1)
[tex]
m_{1}v_{1i}cos(\Theta) = m_{1}v_{1f}Cos(\Theta) + m_{2}v_{2f}Cos(\Theta)
[/tex]

2)
[tex]
0 = m_{1}v_{1f}Sin(\Theta) + m_{2}v_{2f}sin(\Theta)
[/tex]

You also have two unknowns, V1f and V2f. You solve one equation down to one of your unknowns, and then plug what you just solved for into the unknown of the other equation. Then you can get an answer for that unknown.
 
  • #11
I got it, thanks. I was plugging in the wrong mass. Thanks a lot though.
 
  • #12
Good, your welcome.
 

Related to Air Hockey Momentum: Puck A & B Collision

1. How does the momentum of the air hockey puck affect its speed and direction after a collision?

The momentum of the puck plays a crucial role in determining its speed and direction after a collision. According to the law of conservation of momentum, the total momentum of the system remains constant in the absence of external forces. So, when two pucks collide, their individual momentum vectors combine and result in a new momentum vector, which determines the direction and speed of the puck after the collision.

2. Does the mass of the puck affect the outcome of a collision?

Yes, the mass of the puck does affect the outcome of a collision. The more massive the puck, the greater its momentum, and the more force it exerts on the other puck during a collision. This force can result in a change in the speed and direction of both pucks after the collision.

3. Can the friction between the puck and the table affect its momentum?

Yes, the friction between the puck and the table can affect its momentum. Friction is a force that acts in the opposite direction of motion, so it can slow down the puck and decrease its momentum. This can result in a weaker collision with another puck, leading to a slower speed and less change in direction.

4. How does the elasticity of the puck affect its momentum during a collision?

The elasticity of the puck, or its ability to deform and return to its original shape, can affect its momentum during a collision. In a perfectly elastic collision, the pucks would bounce off each other with no loss of kinetic energy. However, in real-world scenarios, some energy is lost due to the deformation of the pucks, resulting in a decrease in momentum after the collision.

5. Can the angle at which the pucks collide affect their momentum?

Yes, the angle at which the pucks collide can affect their momentum. If the pucks collide head-on, their momenta will combine in a different direction than if they collide at an angle. This can result in a change in direction and speed of the pucks after the collision. Additionally, the angle of the collision can also affect the amount of energy transferred between the pucks, which can impact their individual momenta as well.

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