How can I solve for impulse without time or velocity?

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The discussion revolves around solving for impulse without having time or velocity provided. Participants suggest using the equation for impulse, which is the product of force and time, and mention that the force can be calculated using friction equations. There is uncertainty about whether the impulse should be considered negative due to the direction of friction. The conversation also touches on the need to determine how friction affects momentum over time and the average velocity during that period. Overall, the focus is on finding a way to calculate impulse while navigating the limitations of missing variables.
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Homework Statement


IMG_0043.jpg


Homework Equations


ΣF=ma
FF=μFN
impulse=ΣFΔt
momentum=mv
ΣFΔt=mΔv

The Attempt at a Solution


IMG_0044.jpg

IMG_0045.jpg


This is where I got stuck... How can I solve for impulse without time or velocity? Is there a way to manipulate the equations to love for either time or velocity?
 
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x2017 said:
How can I solve for impulse without time or velocity?
Hi x:

As I read the problem statement, the time to use is 1.5 seconds, and the force is calculated using FF=μFN.

Also, WRT (a), I think your teacher would like momentum in the units kg m/s.

Hope this helps.

Regards,
Buzz
 
Buzz Bloom said:
Hi x:

As I read the problem statement, the time to use is 1.5 seconds, and the force is calculated using FF=μFN.

Also, WRT (a), I think your teacher would like momentum in the units kg m/s.

Hope this helps.

Regards,
Buzz

Okay, thank you! I wasn't sure if the question meant friction while the athlete was exerting a price on the rock or afterwards (which I don't have a time for).
& thanks for the tip about the units!

c) impulse=ΣFΔt
= (18.64N)(1.5s)
=27.96Ns Is this supposed to be negative because friction is acting in the negative direction?

d) Yikes, still stuck on this one! Does it have anything to do with this equation? v'={M[V(1+e)-ev]+mv}/M+m
 
x2017 said:
Is this supposed to be negative because friction is acting in the negative direction?
Hi x:

My guess is that your teacher is likely to prefer a minus sign, but that's just a guess.

x2017 said:
d) Yikes, still stuck on this one! Does it have anything to do with this equation? v'={M[V(1+e)-ev]+mv}/M+m

I confess I have trouble parsing the equation in this linear form.

You have the momentum from (a). You also have the friction force you used to calculate (c). How much time will it take for the friction force to reduce the momentum to zero? How far will the stone travel in that amount of time? Since the velocity is being reduced linearly, what is the average velocity of the stone during the time between F1 becoming zero 0 and velocity becoming 0?

Regards,
Buzz
 
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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