Impulse of a ball hitting the wall

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Homework Help Overview

The problem involves calculating the change in momentum of a pool ball with a mass of 120g, which hits a wall at a speed of 9m/s and rebounds at 6m/s at a 45-degree angle. The original poster expresses uncertainty about their calculations and the resulting impulse value.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss breaking down the velocities into orthogonal components and the implications of directionality in momentum calculations. Questions arise about which velocity component to consider for the impulse calculation and how to handle the signs of the velocities.

Discussion Status

Some participants provide guidance on the approach to calculating change in momentum, suggesting methods for vector addition and clarification on the treatment of negative values. There is an ongoing exploration of the correct application of these concepts without a clear consensus on the original poster's calculations.

Contextual Notes

The original poster has provided a diagram to illustrate their calculations, but there is mention of potential missing details regarding the direction of the impulse and the need to multiply by mass, which may affect the final outcome.

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Homework Statement


Calculate the change in momentum of a 120g pool ball hitting at 9m/s and rebounding at 6m/s. It rebounds at an equal and opposite angle of 45 degrees.

Pi = 9m/s
Pf = 6m/s
angle = 45 degrees

Homework Equations





The Attempt at a Solution


I have done all my calcs in the picture below which shows a diagram

I have concluded that the impluse was 10.8166 N, somehow I don't think I have done this right
 

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miniradman said:
I have concluded that the impluse was 10.8166 N, somehow I don't think I have done this right
Well, you're right that you're wrong. Given that the impulse [itex]I = m\Delta v[/itex], consider breaking the velocities into orthagonal components. Which component will the normal force change? :wink:
 
welcome to pf!

hi miniradman! welcome to pf! :wink:
miniradman said:
I have done all my calcs in the picture below which shows a diagram

I have concluded that the impluse was 10.8166 N, somehow I don't think I have done this right

looks fine :smile:, except that you've not yet multiplied by the mass :wink:

(and do you need to specify the direction of the https://www.physicsforums.com/library.php?do=view_item&itemid=340" also?)
 
Last edited by a moderator:


tiny-tim said:
hi miniradman! welcome to pf! :wink:


looks fine :smile:, except that you've not yet multiplied by the mass :wink:

(and do you need to specify the direction of the https://www.physicsforums.com/library.php?do=view_item&itemid=340" also?)
I don't think so...

Also, I have trouble remembering which one to flip around and make a negative? Is there are certain rule of guideline I have to follow to do this?

Thanks ;)
 
Last edited by a moderator:
hi miniradman! :smile:
miniradman said:
I have trouble remembering which one to flip around and make a negative? Is there are certain rule of guideline I have to follow to do this?

i'm not sure i follow :confused:

if the momentum before is p = (a,b), and the momentum after is q = (c,d),

then the change in momentum is q - p = (c-a,d-b) :wink:
 
What I did, was I flipped the 6m/s around so it becomes -6m/s (beause I took them way like vectors). Because it is on an angle of 45 degrees, I added the vectors head to tail

is this right?
 
hi miniradman! :wink:

yes, you can either subtract the original vectors, tail-to-tail, or flip one so that it becomes adding, which is head-to-tail :smile:
 
awesome! Thanks mate ;)
 

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