Finding the Maximum Mass Ratio for Elastic Collisions: A Quick Homework Problem

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The discussion centers on a homework problem involving elastic collisions between two objects of mass M and a smaller mass m. The user successfully applies the laws of conservation of energy and momentum to derive the post-collision velocities of both masses. The calculated velocities are v1 for mass m after colliding with B and v2 for mass B after the collision. The final result determines the maximum mass ratio M/m, ensuring that the velocity of m remains less than or equal to that of B, yielding a ratio of 2 + √5. The solution appears to be correct based on the calculations presented.
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Homework Statement


IMG_20171225_053813~2.jpg

Both object A and B have mass M and are moving.
An object with mass of m and moving with velocity u collide with the object B elastically. (m < M)
Find the following after m collide with B
1) Velocity of m after bouncing back
2) Velocity of B
Then m collide with A and bounce back
3) The highest value of M/m that velocity of m is lesser than or equal to that of B
(Surface has no friction)

Homework Equations

The Attempt at a Solution


Using the Law of Energy Conservation and the Law of Momentum Conservation

##\frac{1}{2} m u^2 = \frac{1}{2} m v_1^2 + \frac{1}{2} M v_2^2##

##m u = -m v_1 + Mv_2##

After solving the equation, I get this

##v_1 = \frac{M - m}{M + m} u## (Answer for (1))

##v_2 = \frac{2 m}{M + m} u## (Answer for (2))

Then object m hit object A. The velocity of m then change into

##(\frac{M - m}{m + M})^2 u##

Velocity of m should be equal to that of B so m cannot reach B

##(\frac{M - m}{m + M})^2 u = \frac{2 m u}{m + M}##

##\frac{M}{m} = 2 + \sqrt{5}## (Answer for (3))

Is that correct?
 

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Looks fine.
 
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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