How Does Impulse Affect the Horizontal Speed of a Softball?

[*intheclouds*]

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=mv_f-mv_i
"change in" P=mv_f-mv_i

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=mv_f-mv_i.
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.

 

[asleight]

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=mv_f-mv_i
"change in" P=mv_f-mv_i

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=mv_f-mv_i.
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.

 

[*intheclouds*]

A collision? What type of equation is that?

 

[asleight]

A collision? What type of equation is that?

$$m_1\texbf{v}_{i,1}+m_2\textbf{v}_{i,2}=m_1\textbf{v}_{f,2}+m_2\textbf{v}_{f,2}$$

 

[*intheclouds*]

So, what would the 2.7 N*s be considered?

 

[asleight]

So, what would the 2.7 N*s be considered?

What are the units of Ns?

 

[*intheclouds*]

Im not really sure. I think it is Neutons*second. I am pretty sure it is the impulse.

 

[asleight]

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.

 

[*intheclouds*]

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... =]

 

[asleight]

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 ).

 

[*intheclouds*]

Oh wow so I was really close with my first answer...It was just negative and I needed it to be positive! Thank you. =]

 

[asleight]

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.