Calculating Impulse in a Baseball Pitch and Hit

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Homework Help Overview

The discussion revolves around calculating the impulse experienced by a baseball when it is hit back after being pitched. The problem involves analyzing the velocities of the baseball before and after impact, taking into account their respective angles and speeds.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the decomposition of velocity vectors into their x and y components, questioning the correctness of angle adjustments and vector addition or subtraction.

Discussion Status

There are multiple interpretations of the calculations involved, with participants pointing out potential errors in the original poster's approach. Some guidance has been offered regarding the addition and subtraction of vector components, but no consensus on the final answer has been reached.

Contextual Notes

Participants note confusion regarding the definitions of the angles and the direction of the velocities, as well as the implications of the baseball's movement relative to the pitcher's and home plate.

hbomb
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A baseball pitcher delivers a fastball that crosses the plate with an angle of 5.67 degrees relative to the horizontal and a speed of 80.1 miles/hour. The ball (mass 0.145 kg) is hit back over the head of the pitcher at an angle of 37.09 degrees with respect to the horizontal and a speed of 108.5 miles/hour. What is the magnitude of the impulse received by the bat?

Ok, impulse is the change in momentum.
p=m(vf-vi)
But since the mass is pitched at an angle and is hit at an angle you need to decompose the velocity vector into it's y and x components.

vyi=(35.81)sin(180-5.67)=3.54 m/s
vxi=(35.81)cos(180-5.67)=-35.63 m/s

vyf=(48.50)sin(37.09)=29.25 m/s
vxf=(48.50)cos(37.09)=6.78 m/s

Add the x and y components to each other
y's=25.71 m/s
x's=42.41 m/s

Then find the velocity of the summed vectors
vf=sqr(vx^2+vy^2)=49.6 m/s

p=mv=(0.145 kg)(49.6 m/s)=7.192 kg m/s

When I enter this into my homework site for my class, it tells me this is the wrong answer. I'm I missing something here? At first I thought maybe I'm not suppose to find the difference between 180 and my angles for the intial velocties. So I tried it just with the original angles and I still get a wrong answer message. Please help.

http://img138.imageshack.us/img138/2597/baseballyg4.jpg
 
Last edited by a moderator:
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Hmm, I know very little about baseball (I'm a brit) but when that fastball is crossing the plate, is that the pitcher's plate,or the home plate ??
 
Home plate
 
hbomb said:
vxf=(48.50)cos(37.09)=6.78 m/s
that should be 38.69

hbomb said:
Add the x and y components to each other
y's=25.71 m/s
...
you subtracted instead of adding here.
 
What is the correct anwer, because I fixed the mistakes and I'm still getting a wrong answer message. Right now I have 29.25+3.54 for the y's because the vectors are going in the same direction. For the x's I have 38.69-35.63 because the vectors are going in opposite directions. I get 4.77 kg m/s for the impulse. Is this right?
 
You've been subtracting the x's instead of adding them.
Before impact, the ball has a speed of 35.63 m/s to the right (going by your diagram)
After impact, the ball has a speed of 38.69 m/s to the left (going by your diagram)
So the overall change in velocity (which is a vector) is the sum of the speeds = 74.32 m/s.

Edit: I get 11.78 kg m/s as the change in momemtum.
 
Last edited:

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