Concept Question - Understanding Momentum Conservation in Collisions

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SUMMARY

The discussion centers on the principle of momentum conservation in collisions, specifically involving two gliders of equal mass (m1 = m2) on an air track. It is established that when these gliders collide and move apart, the change in momentum of glider 1 (-1.4 kg×m/s) is equal in magnitude and opposite in sign to that of glider 2 (1.4 kg×m/s), confirming the truth of the statement regarding momentum conservation. The calculations are derived from Newton's Third Law and the conservation of linear momentum, which states that the total momentum change of the system remains zero during the collision.

PREREQUISITES
  • Understanding of Newton's Third Law of Motion
  • Familiarity with the concept of linear momentum
  • Basic knowledge of one-dimensional collisions
  • Ability to perform calculations involving momentum (Δp = mv)
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  • Study the principles of conservation of momentum in various collision types
  • Explore the mathematical derivation of impulse and momentum equations
  • Learn about one-dimensional elastic and inelastic collisions
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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 \sqrt{2}, so I'm assuming there's some type of calculation to be made with the formula for kinetic energy E_{k}=\frac{1}{2}mv^2. However, I just don't see how it works. I thought that the change in momentum was simply \Delta p=2mv
 
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dont ask questions here...
 
Oh, ha, I'm so sorry. How foolish of me.
 
LHC said:
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.
 
LHC said:
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:
 
Ohhhh! I get it. Thanks!
 
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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