How Do You Solve an Elastic Collision of Particles Using Conservation Laws?

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The discussion focuses on solving an elastic collision problem using conservation laws, specifically momentum and kinetic energy. Participants emphasize the importance of correctly defining variables and simplifying equations to avoid algebraic errors. One user suggests using simpler notation to make calculations easier, while another expresses frustration over repeated incorrect answers due to algebra mistakes. The conversation encourages sharing work for verification and offers resources for better understanding the symbols used in physics. Overall, the thread highlights the challenges of applying conservation laws in collision problems and the need for careful algebraic manipulation.
jaeeeger
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Homework Statement


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Homework Equations



conservation of momentum, m1v1i+m2v2i=m1v1f+m2v2f

and

conservation of KE, ½m1v1i^2+½m2v2i^2=½m1v1f^2+½m2v2f^2

The Attempt at a Solution



first, i defined my variables. v1i=-v2i, v1f=-0.750v1i, v1f=0.750v2i

I tried isolating V2f using the conservation of momentum equation, getting V2f=[-m1v2i+m2v2i-m1(0.750v2i)]/m2.

Then I use the conservation of kinetic energy equation and plug in my new v2f value. expand it out, then try to simply everything to get an answer for m2. I'm really bad at long and complicated algebra, I often cancel out items I'm not allowed to, and don't cancel out when I should. I've tried around 7-8 times now, all with varying answers, none of them right.
 
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Welcome to PF, Jaeeeger!
I'm just an old high school teacher - used to keeping the notation simple. It might be worth a try that way. Rather than V2f, I used "W" for the final velocity of the nm.
For conservation of momentum I wrote mv - nmv = -.75mv + nmW (1)
For conservation of energy (cancelling all the 1/2's right off):
mv² + nmv² = m(.75v)² + nmW² (2)
It doesn't looks so bad that way, does it? Cancel all the m's to make it even better. I solved (1) for W and substituted into (2) to get an equation in one unknown.

If you type in your work here (or scan and upload), we will make sure you get it right.
For the ² symbol, just copy one of mine and paste in your post.
More symbols to copy here: https://www.physicsforums.com/blog.php?b=346
 
Last edited by a moderator:
jaeeeger said:
I've tried around 7-8 times now, all with varying answers, none of them right.
If you know the correct answer, it is a good idea to provide it so others working along can verify they are on the right track.
 
NascentOxygen said:
If you know the correct answer, it is a good idea to provide it so others working along can verify they are on the right track.

The homework is online, I just know if I get it wrong.

Delphi51 said:
Welcome to PF, Jaeeeger!
I'm just an old high school teacher - used to keeping the notation simple. It might be worth a try that way. Rather than V2f, I used "W" for the final velocity of the nm.
For conservation of momentum I wrote mv - nmv = -.75mv + nmW (1)
For conservation of energy (cancelling all the 1/2's right off):
mv² + nmv² = m(.75v)² + nmW² (2)
It doesn't looks so bad that way, does it? Cancel all the m's to make it even better. I solved (1) for W and substituted into (2) to get an equation in one unknown.

If you type in your work here (or scan and upload), we will make sure you get it right.
For the ² symbol, just copy one of mine and paste in your post.
More symbols to copy here: https://www.physicsforums.com/blog.php?b=346

Okay thanks for the help. I'm fairly sure I just screwed up my algebra somewhere, I get n=-1, which is wrong.

I solved (1) for W, then plugged that into the conservation of energy.
My algebra just isn't as hot as it used to be.
4J5D7l.jpg
 
Last edited by a moderator:
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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