How Does Impulse Affect the Horizontal Speed of a Softball?

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Homework Help Overview

The discussion revolves around the effects of impulse on the horizontal speed of a softball after being hit by a bat. The problem involves a .24 kg softball that receives an impulse of 2.7 N*s when tossed upward and struck horizontally.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the appropriate equations related to impulse and momentum, with some expressing uncertainty about the correct application of these concepts. Questions arise regarding the interpretation of impulse and its units, as well as the initial conditions of the problem.

Discussion Status

There is an ongoing exploration of the relationship between impulse and momentum, with participants attempting to clarify their understanding of the equations involved. Some guidance has been offered regarding the interpretation of impulse, but no consensus has been reached on the correct approach to the problem.

Contextual Notes

Participants note confusion regarding the initial conditions of the softball's motion and the implications of the impulse received. There is also mention of treating the scenario as a collision, which introduces additional complexity to the discussion.

*intheclouds*
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Homework Statement



When tossed upward and hit horizontally by a batter, a .24kg softball receives an impulse of 2.7 N*s. With what horizontal speed does the ball move away from the bat?

Homework Equations



Here are all the equations that were in our notes for this section.
p (momentum) There was triangle P and triange t in my notes, so I wrote "change in"

p=mv
f=ma
f*"change in"t="change in"p
f*"change in"t=mvf-mvi
"change in" P=mvf-mvi

The Attempt at a Solution



I really was not sure what equation I should use because I didnt really know what the impulse was. I chose f*"change in"t=mvf-mvi.
2.7 N*s= (.24kg)(0m/s)-(.24kg)(v)
I got v=-11.25m/s. I know this was wrong, but I am not sure which equation I was supposed to use or if I entered the information correctly into the equation. Any help would greatly be appreciated.
 
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*intheclouds* said:

Homework Statement



When tossed upward and hit horizontally by a batter, a .24kg softball receives an impulse of 2.7 N*s. With what horizontal speed does the ball move away from the bat?

Homework Equations



Here are all the equations that were in our notes for this section.
p (momentum) There was triangle P and triange t in my notes, so I wrote "change in"

p=mv
f=ma
f*"change in"t="change in"p
f*"change in"t=mvf-mvi
"change in" P=mvf-mvi

The Attempt at a Solution



I really was not sure what equation I should use because I didnt really know what the impulse was. I chose f*"change in"t=mvf-mvi.
2.7 N*s= (.24kg)(0m/s)-(.24kg)(v)
I got v=-11.25m/s. I know this was wrong, but I am not sure which equation I was supposed to use or if I entered the information correctly into the equation. Any help would greatly be appreciated.

You can treat this as a collision.
 
A collision? What type of equation is that?
 
*intheclouds* said:
A collision? What type of equation is that?

m_1\texbf{v}_{i,1}+m_2\textbf{v}_{i,2}=m_1\textbf{v}_{f,2}+m_2\textbf{v}_{f,2}
 
So, what would the 2.7 N*s be considered?
 
*intheclouds* said:
So, what would the 2.7 N*s be considered?

What are the units of Ns?
 
Im not really sure. I think it is Neutons*second. I am pretty sure it is the impulse.
 
*intheclouds* said:
Im not really sure. I think it is Neutons*second. I am pretty sure it is the impulse.

1 N = 1 kg*m/s/s.

So, 2.7 N*s = 2.7 kg*m/s.

So, the impulse has a "mass" and a "velocity."

The ball isn't moving in the x-direction so its initial velocity is 0. Solve, the impulse goes away.
 
ok so in that equation you gave me:

(.24kg)(0m/s)+(.24kg)(0m/s)=(.24kg)(v)+.24(v)??

Its all going to work out to be 0? I am really confused. I know you are trying so hard to help me, but I am lost... =]
 
  • #10
*intheclouds* said:
ok so in that equation you gave me:

(.24kg)(0m/s)+(.24kg)(0m/s)=(.24kg)(v)+.24(v)??

Its all going to work out to be 0? I am really confused. I know you are trying so hard to help me, but I am lost... =]

0.24kg ( 0 m/s ) + Impulse = 0.24kg ( V m/s ).
 
  • #11
Oh wow so I was really close with my first answer...It was just negative and I needed it to be positive! Thank you. =]
 
  • #12
*intheclouds* said:
Oh wow so I was really close with my first answer...It was just negative and I needed it to be positive! Thank you. =]

You're welcome.
 

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