# Change in momentum help

1. Oct 13, 2006

### hbomb

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 [Broken]

Last edited by a moderator: May 2, 2017
2. Oct 14, 2006

### Fermat

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 ??

3. Oct 14, 2006

### hbomb

Home plate

4. Oct 15, 2006

### Fermat

that should be 38.69

5. Oct 15, 2006

### hbomb

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?

6. Oct 15, 2006