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Concept Question - Momentum

  • Thread starter LHC
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  • #1
LHC
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Hello,

I have a question about momentum that showed up in my physics textbook. It's a True/False question, and I have the answer, it's just that I'm not sure how it works. Could someone please explain this to me? Many thanks.

|Glider 1|-----> <------|Glider 2|
-------------------------------------------
Air Track
-------------------------------------------

Statement: In an isolated system, two gliders (m1 = m2) on an air track move toward each other at equal speed, collide, and then move away from each other at equal speed.

Question (T/F): For this collision, if the change in momentum of glider 1 is -1.4 kg×m/s [W], then the change in momentum of glider 2 is 1.4 kg×m/s [W].

The answer is true.

I'm not sure how the numbers even got there. I recognize that they just happen to be [tex]\sqrt{2}[/tex], so I'm assuming there's some type of calculation to be made with the formula for kinetic energy [tex]E_{k}=\frac{1}{2}mv^2[/tex]. However, I just don't see how it works. I thought that the change in momentum was simply [tex]\Delta p=2mv[/tex]
 

Answers and Replies

  • #2
malawi_glenn
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dont ask questions here....
 
  • #3
LHC
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Oh, ha, I'm so sorry. How foolish of me.
 
  • #4
malawi_glenn
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Oh, ha, I'm so sorry. How foolish of me.
a mentor will move this thread, so just relax and don't make a new one.
 
  • #5
tiny-tim
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For this collision, if the change in momentum of glider 1 is -1.4 kg×m/s [W], then the change in momentum of glider 2 is 1.4 kg×m/s [W].

The answer is true.

I'm not sure how the numbers even got there.
Hi LHC! :smile:

They just made up the 1.4. :smile:

Hint: is momentum conserved in this collision?

what does that mean? :wink:
 
  • #6
LHC
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Ohhhh! I get it. Thanks!
 
  • #7
dynamicsolo
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The thing that's funny about this T/F question is that the stated conditions are completely irrelevant. From Newton's Third Law or conservation of linear momentum, the change in linear momentum (also called the impulse) of glider 1 would be equal in magnitude and opposite in sign (since this is a one-dimensional collision) to the change in linear momentum of glider 2. Thus, the total momentum change of the two-glider system is zero (or the total linear momentum remains constant). That will be true regardless of the relative masses and initial velocities of the two gliders.
 

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