How can I solve for impulse without time or velocity?

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Homework Help Overview

The discussion revolves around the concept of impulse in physics, specifically how to calculate it without having direct values for time or velocity. The original poster expresses confusion about manipulating relevant equations to derive these values.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between impulse, force, and time, questioning how to approach the problem without specific values for time or velocity. There is also discussion about the implications of friction on the calculations and whether certain values should be treated as negative.

Discussion Status

Some participants have provided insights regarding the use of specific values and units, while others have raised questions about the assumptions being made, particularly concerning the context of friction and its effects on momentum. There is an ongoing exploration of how to relate the equations to the problem at hand.

Contextual Notes

There is mention of a specific time value (1.5 seconds) and a friction force equation, but the original poster is unsure about the context in which these values apply. The discussion also highlights uncertainty about the directionality of forces and their impact on the calculations.

x2017
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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
 

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