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Physics Problem

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Physics Problem

Unread postAuthor: PVC Arsenal 17 » Tue Dec 08, 2009 8:17 pm

This seemed like a rather easy physics question, but our class was split down the middle with the answers we came up with. If anyone here could provide the correct answer with an explanation, it would be much appreciated. The question is:

-In a slowpitch softball game, a .2kg softball crosses the plate at 15m/s at 45 degrees below the horizontal. The batter hits the ball toward center field, giving it a velocity of 40m/s at 30 degrees above the horizontal.

What is the impulse delivered in the y (vertical) direction?


Impulse is the change in momentum. You may use mΔv.


Thanks!

Edit: Consider "up" and "right" to be the positive directions.

Edit 2: I know many of you may disagree with what the diagram says it taking place in the problem (I do too) but consider that the case anyway. That's what our teacher said was happening so we have to solve the problem accordingly.
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Hopefully this clarifies what's taking place in the problem
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Last edited by PVC Arsenal 17 on Tue Dec 08, 2009 11:31 pm, edited 2 times in total.

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Unread postAuthor: Technician1002 » Tue Dec 08, 2009 8:35 pm

To solve; isolate the horizontal component of the vector and calculate only the vertical component. Take the rate the ball is falling as it reaches the bat and then the rate the ball is rising after impact. The delta is the change in velocity. From there the mass and energy for acceleration can be found. Do not try to say the ball lost energy from downward motion. The change in motion is a vector. You are looking for the change in vertical motion from a negative vector to a positive vector (falling to rising).
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Vertical change in velocity and the impulse to change it is the solution.
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Unread postAuthor: PVC Arsenal 17 » Tue Dec 08, 2009 8:47 pm

The way we were told to look at the problem was as follows: The ball is rising at a constant rate when it crosses the plate (before being hit) as if it was shot upwards from a lower point than the plate. After the batter hits the ball, it rises at an even faster constant rate.

The way I solved the problem was (40sin30 - 15sin45)(.2) = 1.88 kg m/s. I was told to try again... I'm damn sure this is the correct answer though.
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Unread postAuthor: Technician1002 » Wed Dec 09, 2009 2:28 am

PVC Arsenal 17 wrote:The way we were told to look at the problem was as follows: The ball is rising at a constant rate when it crosses the plate (before being hit) as if it was shot upwards from a lower point than the plate. After the batter hits the ball, it rises at an even faster constant rate.

The way I solved the problem was (40sin30 - 15sin45)(.2) = 1.88 kg m/s. I was told to try again... I'm damn sure this is the correct answer though.


The original statement has the vectors (UP/DOWN) change before the strike and after (falling/rising).

I'm wondering if the instructor is looking at the energy in relation to stationary. In the first case bringing the falling ball to a stop (in the vertical vector only) it gives up energy, not takes in energy.

Try to solve for the difference in energy and see if that is what the instructor is looking for. Understanding the problem will go a long way in reaching the proper result. It is possible (I haven't done the math) that the energy is close to zero or even a negative number much like you would get bouncing a ball off the floor.
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