Momentum, Ball being thrown on Ice

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Homework Help Overview

The problem involves a scenario where a person standing on ice catches a ball thrown horizontally, and the discussion centers around calculating the final velocity after the catch using the principle of conservation of momentum.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of momentum conservation, with some expressing confusion over unit conversions and the expected final velocity. There are attempts to set up equations based on initial and final momentum, and questions arise regarding discrepancies in calculated values.

Discussion Status

Multiple participants are exploring different interpretations of the problem and the calculations involved. Some have identified potential errors in their computations, while others are questioning the necessity of unit conversions. The discussion is ongoing, with no clear consensus on the correct approach or answer yet.

Contextual Notes

There is mention of specific unit requirements in the problem statement, which may be influencing the calculations. Participants are also considering the implications of the collision being inelastic.

Nickym707
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Homework Statement



You are standing on a sheet of ice that covers the football stadium parking lot in Buffalo; there is negligible friction between your feet and the ice. A friend throws you a ball of mass 0.400kg that is traveling horizontally at 10.5 m/s . Your mass is 65.0 kg .
If you catch the ball, with what speed do you and the ball move afterwards?


I set the momentum before equal to the momentum after, then solved for final velocity.
v2= (MaVa1+ MbVb1)/(Ma+Mb) and since the body isn't moving at all than the MbVb1 just is zero. you also want the 10.5 m/s into cm which is 1050 cm/s


I keep getting 16.1 and the answer is 6.42...
 
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This seems to be a relatively simple problem, though I am getting 15.571, rather than 16.1. Given the fact that this will be an inelastic collision with the person catching the ball, I change

m1iv1i + m2iv2i = m1fv1f + m2fv2f

to

m1v1 + m2v2 = m3v3

Then,

[STRIKE](65.0 kg)(0 m/s)[/STRIKE] + (.400 kg)(10.5 m/s) = (65.4 kg)v3
v3 = 15.571

Also, I may have missed something, but I do not understand why you wished to change m/s to cm/s.
 
Im sorry, I was unclear, the problem asks us to change it to cm/s. I'm having problems because the webwork is saying that the answer is not 15 or 16...
 
That's strange, especially considering that our answers would give 1500 and 1600 cm/s. The only way that I can think of that much of a change would be an error in units somewhere, seeing that the problem doesn't seem to want you to take energy loss into account.
 
danielu13 said:
This seems to be a relatively simple problem, though I am getting 15.571, rather than 16.1. Given the fact that this will be an inelastic collision with the person catching the ball, I change

m1iv1i + m2iv2i = m1fv1f + m2fv2f

to

m1v1 + m2v2 = m3v3

Then,

[STRIKE](65.0 kg)(0 m/s)[/STRIKE] + (.400 kg)(10.5 m/s) = (65.4 kg)v3
v3 = 15.571

Also, I may have missed something, but I do not understand why you wished to change m/s to cm/s.

I'm not sure how the answer is 15.571 since I've calculated it to be 0.0642
 
azizlwl said:
I'm not sure how the answer is 15.571 since I've calculated it to be 0.0642
I just looked back over my work, and I divided by 4.2 instead of 65.4, and just ended up with the reciprocal. I'm not really sure how I overlooked that, but at least I figured it out. I'm guessing he did something similar, as he came out with a similar answer to what I had.
 

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