AP Physics Impulse Problem: Need Help

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Homework Help Overview

The discussion revolves around an impulse problem in AP Physics, focusing on the relationship between impulse and momentum. Participants are analyzing the components of momentum and the effects of a wall on a ball's motion.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the definition of impulse as the change in momentum and explore the components of momentum in both x and y directions. There are questions about whether the wall exerts force in the y direction and how to properly account for the components of velocity.

Discussion Status

There is an active exchange of ideas regarding the components of impulse and momentum. Some participants are clarifying their understanding of the problem setup, while others are questioning assumptions about the direction of forces involved. Guidance has been offered regarding the definition of axes and the implications for calculating impulse.

Contextual Notes

Participants express frustration with the clarity of instruction from their teacher, which may affect their understanding of the problem. There is a recognition that different conventions for defining axes can lead to varying interpretations of the problem.

BITI
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This problem is driving me crazy! I'm currently taking AP physics and my teacher gets confused with her own problems, anyways I want to see if anyone could give me a hand on this impulse problem.

Here is a link to the problem:
http://img101.exs.cx/img101/9262/lastscan.jpg

Thank you guys
 
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Impulse is the change in momentum.

so...
p1=mv1
p2=mv2

I=p1-p2 (impulse is the change in momentum)

p1=(.06kg)(25 m/s)sin45=1.06Ns
p2=(.06kg)(-25 m/s)sin45=-1.06Ms

I=p1-p2
I=1.06-(-1.06)
I=2.12Ns
 
what was your approach?

this is what i see so far

[tex]\vec{I} = \Delta \vec{p}[/tex]
 
Urban aren't your forgetting something?

[tex]\vec{I} = I_{x} \vec{i} + I_{y} \vec{j}[/tex]
 
The impulse of the wall has nothing to do with the "y direction"

The wall only pushs in the x direction so [tex]I_{y} \vec{j}[/tex] is zero
 
I am saying this:

[tex]\vec{p}_{o} = p_{xo} \vec{i} + p_{yo} \vec{j}[/tex]

[tex]\vec{p} = p_{x} \vec{i} + p_{y} \vec{j}[/tex]

You neglected the x component.
 
UrbanXrisis said:
p1=(.06kg)(25 m/s)sin45=1.06Ns
p2=(.06kg)(-25 m/s)sin45=-1.06Ms

The x componet is the "(25 m/s)sin45"
The y componet is 0
 
Where do you get 0??

[tex]\vec{p}_{o} = m|\vec{v_{o}}| \cos \theta \vec{i} + m|\vec{v_{o}}| \sin \theta \vec{j}[/tex]

[tex]\vec{p} = m|\vec{v}| \cos \theta \vec{i} + m|\vec{v}| \sin \theta \vec{j}[/tex]

What do you mean wall neglected? i am talking about the momentum of the ball, it's directed at a 45 degree angle, so it has a nonzero x and y component which happen to be the same because [itex]\sin \theta = \cos \theta[/itex] when theta is 45 degrees.
 
UrbanXrisis said:
The x componet is the "(25 m/s)sin45"
The y componet is 0

Depends on how you whose the axes.There's another thread with "Incline" in which u see that the axis cosing may alter your result,meaning that the modulus will be the same,but comonpents on two axis of coordinates will be generally different.
In two favorable cases,one of the components of the impulse is zero.It's less relevant whether it's denoted by "x",or by "y".

Daniel.
 
  • #10
But the wall does not act in the y direction does it? When I throw a ball at the wall, the wall only supplies the push backwards, it doesn't push it up nor down. The impulse goven by the wall should only concern the x direction. That's why I only did sin45 of 25 m/s
 
  • #11
UrbanXrisis said:
But the wall does not act in the y direction does it? When I throw a ball at the wall, the wall only supplies the push backwards, it doesn't push it up nor down. The impulse goven by the wall should only concern the x direction. That's why I only did sin45 of 25 m/s

Dextercioby said you're right... as long as you defined your axis. Some people use the "up" direction for the x-axis instead... But you should give a direction for your answer too since impulse is a vector...
 
  • #12
you guys are definitelly on top of it! This is my 1st year taking physics, my high school teacher is not a physics major, but a chemistry major, what is she doing in physics? Anyways, she usually get confused while explaining the problems, how could we possibly understand. :( I think we need one of you guys over in the school teaching some physics :)

Thank you guys I now have a better picture, how couldn;t I with this great discussion. Thank you guys for all...
 

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