Prooving Elastic Collisions equations

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The discussion focuses on deriving the equations for final velocities in elastic collisions. The user is struggling to understand how to manipulate the conservation of momentum and kinetic energy equations to arrive at the provided formulas for v1f and v2f. A suggested approach involves collecting terms related to m1 and m2 separately, then dividing the left and right sides to establish a relationship between initial and final velocities. The user expresses a desire to comprehend the derivation rather than memorize the equations for exams. Understanding this derivation is crucial for mastering elastic collision problems in physics.
Lord Dark
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Homework Statement


hi everyone ,, how are you all ? ,, i have these two equations and i don't know how to get them :
v1f=((m1-m2)/(m1+m2))*v1i +(2m2/(m1+m2))*v2i
V2f=((2m1/m1+m2))*v1i+((m2-m1)/(m1+m2))*v2i


Homework Equations


m1v1i+m2v2i=m1v1f+m2v2f (conservation of momentum)
0.5m1v1^2i+0.5m2v2i^2=0.5m1v1f^2+0.5m2v2f^2 (conservation of Kinetic Energy)


The Attempt at a Solution


the book reach to taking m1 & m2 as common factor then says divide the kinetic by the momentum then I'll get the results above, but when i divide i get :
v2f=v1i+v1f ,, so someone help me to get the results above because i don't like memorizing and i know that I'll forget in the exam -_-
 
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m1v1i+m2v2i=m1v1f+m2v2f (conservation of momentum)
0.5m1v1^2i+0.5m2v2i^2=0.5m1v1f^2+0.5m2v2f^2 (conservation of Kinetic Energy)

In both the equations collect the terms containing m1 on one side and m2 on other side. Then divide left hand side and right hand side and equate. You will get relation between v1i, v1f ,v2i and v2f. Using this you can find the required result.
 
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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