Impulse Problem: Find Mass, Height & Impulse

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A ball with a mass of 0.110 kg is dropped from a height of 1.25 m and rebounds to a height of 0.800 m, prompting a discussion on calculating the impulse imparted by the floor. The correct formula for impulse is clarified as J = Δp, with the need for precise definitions involving calculus for exact values. An initial attempt at calculating impulse using the integral of force over time is presented, but the results indicate a misunderstanding in the application of the equations. The conversation emphasizes the importance of correctly applying the impulse formula and understanding the relationship between force, time, and momentum change. Accurate calculations and definitions are essential for solving the problem effectively.
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Homework Statement


A ball of mass .110 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of .800 m. What impulse was given to the ball by the floor?


Homework Equations




I=ΣFΔt
Δp=I

The Attempt at a Solution



Could someone please tell me if this is the correct formula?

Thank you very much
 
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Hi chocolatelover! Both of your equations look right but need some minor tweaking.

For starters (I have no clue why, but...) the symbol for impulse is a capitol J.

So J = Δp for instance.

Your first equation should be extended a bit more.

J = ΣFΔt would give you an approximation for impulse however if you wanted the exact value of J you would need to make Δt -> 0, and sum up an infinite number tiny bits force.

A more concrete definition would be J = \int_{t_i}^{t_f}{F dt}
 
Thank you very much

Does this look right?
integral from 1.25 to .800 (.110kg)(9.8m/s^2)dt

=.8624-1.35
=-.4876

Thank you very much
 
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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