How long does it take the ball to come to a hault?

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In summary, the conversation discusses the calculation of the ball's horizontal speed with each bounce, with one person providing their assumptions and equations. Another person points out that the question is underspecified and gives their own assumptions. The first person acknowledges their assumptions and explains their reasoning. The second person then points out mistakes in the equations and suggests trying different values. Overall, the conversation highlights the importance of clarity and specificity in physics problems.
  • #1
iScience
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the question is on here http://imgur.com/z4EFJiO

ignore the silly answer i found it on facebook and started to solve it just for fun and i ended up with...

http://imgur.com/pBkXZIs

sorry for the messy format, but i explained all my calculations (you're going to have to click to enlarge, and then click again to enlarge further.. again i apologize)

one embarrassing thing though, i don't know how to calculate the sum :( ... if anyone would be so kind.. would you demonstrate how?
 
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  • #2
How does the ball's horizontal speed change with each bounce?
 
  • #3
well, if the ball loses one joule each time, i assumed it was referring to the friction and the inelasticity of each bounce, together simplifying it for me saying that it loses 1J w/ each bounce so i assumed it was the 1J in the direction of the vectorial sum of v. ie... just the direction of v.
 
  • #4
iScience said:
well, if the ball loses one joule each time, i assumed it was referring to the friction and the inelasticity of each bounce, together simplifying it for me saying that it loses 1J w/ each bounce so i assumed it was the 1J in the direction of the vectorial sum of v. ie... just the direction of v.
To me, the question is underspecified. There is no reason why it should lose energy in the same proportion in the two directions. In reality, I would expect most energy to be lost from the vertical component. Typically a ball stops bouncing long before it stops rolling. Forced to make an assumption, I would take it that all the lost energy comes from the vertical component. But that could be a wrong guess.
 
  • #5
haruspex said:
To me, the question is underspecified. There is no reason why it should lose energy in the same proportion in the two directions. In reality, I would expect most energy to be lost from the vertical component. Typically a ball stops bouncing long before it stops rolling. Forced to make an assumption, I would take it that all the lost energy comes from the vertical component. But that could be a wrong guess.

assuming my assumption is correct, is my final equation right? i posted this thread because i make a lot of mistakes and i wanted to check my answer
 
  • #6
iScience said:
assuming my assumption is correct, is my final equation right? i posted this thread because i make a lot of mistakes and i wanted to check my answer
Inside the sum you have an x which I think should be i, right? The sqrt expression it is in then gives you the ball's speed immediately after i bounces. Then you subtract that from the vertical speed just before the first bounce? That can't be right.
Your assumption that the energy is lost in equal proportion from the vertical and horizontal components leads to a very nasty sum. I really don't think this is what is intended.
 
  • #7
haruspex said:
Inside the sum you have an x which I think should be i, right?

yes, x(i)

haruspex said:
The sqrt expression it is in then gives you the ball's speed immediately after i bounces. Then you subtract that from the vertical speed just before the first bounce?

not quite; the first sqrt, ie sqrt(196.2) is the ball's velocity(y) right after its first bounce. and then i subtract from this, the ball's subsequent losses in velocities(y).
haruspex said:
That can't be right.
Your assumption that the energy is lost in equal proportion from the vertical and horizontal components leads to a very nasty sum. I really don't think this is what is intended.

just solving for fun... i honestly did have to assume sOMEthInG at the time due to the unspecific nature of the given problem, there were different assumptions i could've made, more likely ones that would be seen in nature, but i didn't see those at the time, er.. perhaps i did but anyhows i just went with this one :)
 
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  • #8
iScience said:
not quite; the first sqrt, ie sqrt(196.2) is the ball's velocity(y) right after its first bounce. and then i subtract from this, the ball's subsequent losses in velocities(y).
That's not how I read it. It is certainly wrong. √196.2 is the vertical velocity you computed for just before the first bounce, and √596.2 is the total speed at that time. Try putting x = 0 or 1; you'll see that the contribution to total time goes negative.
 

Related to How long does it take the ball to come to a hault?

1. How is the speed of the ball related to the time it takes to come to a halt?

The speed of the ball is directly related to the time it takes to come to a halt. The faster the ball is moving, the longer it will take to come to a complete stop.

2. Does the surface on which the ball is rolling affect its stopping time?

Yes, the surface on which the ball is rolling can greatly affect its stopping time. For example, a smooth surface will allow the ball to roll for a longer period of time compared to a rough surface which will cause the ball to slow down and stop more quickly.

3. How does the weight of the ball impact its stopping time?

The weight of the ball does play a role in its stopping time. A heavier ball will have more momentum and will take longer to come to a halt compared to a lighter ball.

4. Is air resistance a factor in the stopping time of a ball?

Yes, air resistance can affect the stopping time of a ball. The more surface area the ball has, the more air resistance it will encounter, causing it to slow down and eventually stop.

5. Can the shape of the ball affect its stopping time?

Yes, the shape of the ball can impact its stopping time. A ball with a rounder shape will have a more streamlined design and will encounter less air resistance, allowing it to roll for a longer period of time compared to a ball with a more irregular shape.

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