# PHYSICS - still cant get it

posted by winterWX

A 0.30-kg softball has a velocity of 12 m/s at an angle of 30° below the horizontal just before making contact with the bat. What is the magnitude of the change in momentum of the ball while it is in contact with the bat if the ball leaves the bat with a velocity of
17 m/s, vertically downward, Δp = __ kg·m/s, and 17 m/s, horizontally back toward the pitcher? Δp = __ kg·m/s

drwls, Friday, February 23, 2007 at 5:40am

All you have to do is calculate the change in momentum in the horizontal and vertical directions, separately. Initially, you have
Vx = -12 cos 30 = -10.39 m/s
Vy = -12 sin 30 = 6.0 m/s
if Vx is defined as positive in the direction of the pitcher and Vy is postive upward.
After the hit, Vx = +17 m/s and Vy = -17 m/s. Just do the subtractions to get delta V, and multiply by the mass of the softball to get delta p.

winterWx, Saturday, February 24, 2007 at 12:27am

Hmmm...This is what I did:

Vx1-Vx2=-10.39-17=-27.39 m/s
(-27.39)(.3 kg)=-8.217 kg*m/s

Vy1-Vy2=-6-(-17)=11 m/s
(11)(.3)=3.3 kg*m/s

But both of those are incorrect. What am I doing wrong???

* PHYSICS!!! - Marco, Saturday, February 24, 2007 at 7:38pm

You are wrong in your math.Look at part second.

o ..which part are you talking about? - winterWX, Saturday, February 24, 2007 at 9:26pm

how am i wrong in my math? can you please show me which parts i did wrong?

* PHYSICS!!! - winterWX, Saturday, February 24, 2007 at 10:29pm

I still can't get it. Oh well, it's due in an hour.

There are two questions here - one fopr eqach final velocity of the ball. When hit vertically upwards,

Vx1-Vx2 = -10.39 - = -10.39 m/s
px1 - px2 = (-10.39)(.3 kg)= -3.117 kg*m/s

Vy1-Vy2= -6-(17)=23 m/s
py1 = py2 = (23)(.3)= 6.9 kg*m/s
Note that the problem asks for the MAGNITUDE of the momentum change. This is
sqrt [(delta px)^2 + (delta py)^2]

1. Anonymous

Draw a picture!
Before hitting the bat, the horizontal component of momentum of the ball is
px= (.3)(.15)(cos(35)) and the vertical component is
py= -(.3)(.15)(sin(35)).

Afterwards, the horizontal component of momentum is 0 (it is going straight down) and the vertical component is -20(0.3)

The change in momentum vector is the difference between those. Then use the Pythagorean theorem to find the magnitude.

In the second part (back toward the pitcher), the momentum after the hit has horizontal component -(.3)(20) and vertical component 0.

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