Momentum, Ball being thrown on Ice

In summary, the problem involves a person standing on a sheet of ice catching a ball thrown at them horizontally. After setting the initial momentum equal to the final momentum, the final velocity is calculated to be 15.571 cm/s. The conversion from m/s to cm/s is requested in the problem. However, there may have been errors in the calculations as the expected answer is 0.0642 cm/s.
  • #1
Nickym707
2
0

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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  • #2
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.
 
  • #3
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...
 
  • #4
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.
 
  • #5
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
 
  • #6
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.
 

FAQ: Momentum, Ball being thrown on Ice

1. What is momentum?

Momentum is a measure of an object's motion, specifically its mass and velocity. It is calculated by multiplying an object's mass by its velocity.

2. How does momentum affect a ball being thrown on ice?

When a ball is thrown on ice, it will continue moving in the same direction and at the same speed due to its momentum. This is because there is very little friction on ice, so there is no force slowing down the ball's movement.

3. Can momentum be transferred between the ball and the ice?

Yes, momentum can be transferred between the ball and the ice. When the ball hits the ice, it exerts a force on the ice, causing it to move in the opposite direction. This transfer of momentum is what allows the ball to continue moving without slowing down.

4. Is momentum conserved in this scenario?

Yes, momentum is conserved in this scenario. This means that the total momentum of the ball and the ice before and after the ball is thrown remains the same, even though the direction and speed of each object may change.

5. How does the mass and velocity of the ball affect its momentum?

The greater the mass and velocity of the ball, the greater its momentum will be. This means that a heavier ball or a ball thrown with more force will have a greater impact on the ice and will be harder to stop or change direction.

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