Impulse Problem: Find Mass, Height & Impulse

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SUMMARY

The discussion centers on calculating the impulse experienced by a ball of mass 0.110 kg dropped from a height of 1.25 m that rebounds to a height of 0.800 m. The correct formula for impulse is identified as J = Δp, where J represents impulse. The user is advised to refine their approach by using the integral form J = ∫F dt for precise calculations, particularly as Δt approaches zero. The final calculation presented yields an impulse of -0.4876 Ns.

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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
 

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