How Is Impulse Calculated in a Baseball Pitch?

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Impulse in a baseball pitch is calculated as the change in momentum of the ball. For a 0.150-kg baseball initially traveling at 40.0 m/s and then hit back at 50.0 m/s, the total impulse delivered by the bat is 90 kg*m/s. Two methods for calculating this impulse were discussed, both yielding the same result but differing in approach. The confusion arose regarding the signs of the velocities and whether to treat them as positive or negative based on direction. Ultimately, impulse is defined as the change in momentum, clarifying the relationship between the two concepts.
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Homework Statement


A 0.150-kg baseball, thrown with a speed of 40.0 m/s, is hit straight back at the pitcher with a speed of
50.0 m/s. (a) What is the magnitude of the impulse delivered by the bat to the baseball? (b) Find the magnitude of the average force exerted by the bat on the ball if the two are in contact for 2.00 * 10 ^-3 s.

Homework Equations


I = change in momemtum

The Attempt at a Solution


I don't know why doing it this way works out.

I = (0.150kg)(40m/s)
I = 6 kg*m/s

I = (0.150kg)(50m/s)
I = 7.5 kg*m/s

I (total) = 6 + 7.5 = 13.5 kg*m/s
 
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Well yes you get the right answer but your approach isn't right, you made two mistakes, but it got you to the right answer :-)
Firstly label your attempt with I1 and I2
Your I1 value is perfect if you assume your positive axis towards the batsman
That being said, do you think your I2 value is right?(consider the direction)
 
Suraj M said:
That being said, do you think your I2 value is right?(consider the direction)
I don't see a problem there, necessarily. It depends on exactly what master_333 was thinking.
Both velocity numbers in the calculation (40m/s, 50m/s) can be read as changes in velocity measured in the batter to pitcher direction, the first to halt the ball and the second to return it. The corresponding impulses therefore have the same sign and should be added.
 
Sorry I took I as momentum,
So I thought ##I_2## should be negative, that's the way we were taught to do it
Sorry I was wrong,:-/
 
Suraj M said:
Sorry I took I as momentum,
So I thought ##I_2## should be negative, that's the way we were taught to do it
Sorry I was wrong,:-/
Yes, I is momentum.
Here are two methods, both valid. In both cases, I will take the positive direction as batter to pitcher:
1.
Initial momentum of ball = m(-40m/s)=-40m m/s.
Final momentum of ball = m(50m/s) = 50m m/s.
Change = 50m-(-40m) = 90m.

2.
Momentum needed to stop ball = 40m
Momentum needed to return stationary ball = 50m
Total needed = 90m.

Since master_333 was surprised his method worked, it is unclear whether method 2 was applied or whether it was just luck.
 
haruspex said:
Yes, I is momentum.
Here are two methods, both valid. In both cases, I will take the positive direction as batter to pitcher:
1.
Initial momentum of ball = m(-40m/s)=-40m m/s.
Final momentum of ball = m(50m/s) = 50m m/s.
Change = 50m-(-40m) = 90m.

2.
Momentum needed to stop ball = 40m
Momentum needed to return stationary ball = 50m
Total needed = 90m.

Since master_333 was surprised his method worked, it is unclear whether method 2 was applied or whether it was just luck.

I don't understand why you did the change in momentum to get the correct answer.
 
master_333 said:
I don't understand why you did the change in momentum to get the correct answer.
The impulse is what changed the momentum. Impulse = change in momentum. What else would you do?
 
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