Momentum question -- What quantities are conserved in an elastic collision....

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SUMMARY

The discussion focuses on the conservation of momentum and kinetic energy in elastic collisions, specifically addressing the equations governing the velocities of two masses, m1 and m2. The key equations referenced include (m2 - m1)v0 = m2v2 + m1v1 and (m2 + m1)v0^2 = m2v2^2 + m1v1^2, where v0 is derived from the equation v0 = √(2gh). Participants express confusion regarding the separation of kinetic energy for mass m2 and the calculation of velocities v1 and v2 after the collision.

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Clara Chung
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Homework Statement


13.png


Homework Equations

The Attempt at a Solution


How do I calculate part d?
I know that (m2-m1)v0 = m2v2+m1v1
where v0 = root (2gh), v1 and v2 are the new velocity of the masses
(m2+m1)v02 = m2v22 + m1v12
I also know that v2-v1=2v0
but how do I separate the KE of mass 2?
 

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Clara Chung said:

Homework Statement


View attachment 223719

Homework Equations

The Attempt at a Solution


How do I calculate part d?
I know that (m2-m1)v0 = m2v2+m1v1
where v0 = root (2gh), v1 and v2 are the new velocity of the masses
(m2+m1)v02 = m2v22 + m1v12
I also know that v2-v1=2v0
but how do I separate the KE of mass 2?

It's ##m_1## that is the upper ball. Can you solve the equations to find ##v_1##?
 
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PeroK said:
It's ##m_1## that is the upper ball. Can you solve the equations to find ##v_1##?
I got a very strange answer...
v2-v1=2v0 ...(1)
(m2-m1)v0 = m2v2+m1v1...(2)
After substitution I get -v0=v1...
 
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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