Time for ball to stop bouncing

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In summary, the time it takes for the ball to stop bouncing is: t=\sqrt{2h/g}, where h is the height of the first bounce and g is the gravitational constant.
  • #1
DeldotB
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Homework Statement


Hello!
A ball is dropped and falls to the floor (no horizontal velocity). It hits the floor and bounces with inelastic collisions. The velocity after each bounce is [itex] \mu [/itex] times the velocity of the previous bounce (here [itex]\mu [/itex] is the constant of restitution). The velocity of the first bounce is just [itex] v_0[/itex]. Find the time it takes for the ball to stop bouncing.

Homework Equations


Newtons Laws

The Attempt at a Solution



Well:
I know this will turn into a convergent geometric series. I am just trying to find what that series will look like.

using the formula [itex]h=x_0+v_0t+1/2at^2 [/itex] its easy to see that the time it takes for the ball to reach the ground is:

[itex] h=1/2gt^2[/itex] so [itex] t=\sqrt{2h/g}[/itex].
Using energy I also have: [itex] mgh=1/2mv_0^2[/itex] so [itex]gh=1/2v_0^2 [/itex]

Time for the next bounce: well, the ball now has an upward velocity of [itex] \mu v_0[/itex] and the height of the first bounce is [itex] h'=\mu v_0t-1/2gt^2[/itex].

I realize this is a simple problem but for some reason I'm not seeing it. If I solve this equation for time, (using quadratic formula) the resulting series for the times [itex]t=t_1+t_2+... [/itex] isn't geometric and actually quite complicated. Is my approach right?
 
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  • #2
Hi,
You don't want the height of the next bounce, but the time for the ball to reach the ground again. Easier to solve, too!
 
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  • #3
I solved the quadratic for the time. Is this not the right approach?
 
  • #4
Should lead to the same answer - with a lot more work.
What did you use for ##h'## ?
 
  • #5
BvU said:
Should lead to the same answer - with a lot more work.

Ahh, I see what you mean. So after the first bounce, I have:

Time for ball to reach ground again: [itex]0= \mu v_0 -1/2gt^2 [/itex] solvig for t yields: [itex] 2 \mu v_0/g[/itex] so

[itex]t_1 =2 \mu v_0/g[/itex]

Time for ball the reach the ground the third time:

[itex]t_2= 2 (\mu)^2 v_0/g[/itex]
and so on. Is this the right direction?
 
  • #6
And there you have your geometric sequence !
Make sure you have the right summation: the first t is only 'half a bounce'
 

Related to Time for ball to stop bouncing

1. How does the weight of the ball affect the time it takes to stop bouncing?

The weight of the ball does not significantly affect the time it takes to stop bouncing. The main factors that determine the time for a ball to stop bouncing are the initial height from which it is dropped and the elasticity of the ball's material.

2. Does the surface the ball bounces on affect the time it takes to stop bouncing?

Yes, the surface the ball bounces on can affect the time it takes to stop bouncing. A softer surface, such as a foam mat, will absorb more of the ball's energy and cause it to stop bouncing faster. A harder surface, such as concrete, will cause the ball to bounce for a longer period of time.

3. How does air resistance impact the time for a ball to stop bouncing?

Air resistance can have a small effect on the time for a ball to stop bouncing. As the ball bounces, it will push air molecules out of the way, creating a slight resistance. This can cause the ball to lose some of its kinetic energy and take slightly longer to come to a complete stop.

4. Can the shape of the ball affect the time it takes to stop bouncing?

Yes, the shape of the ball can affect the time it takes to stop bouncing. A ball with a more spherical shape will have a more predictable and consistent bounce, while a ball with an irregular shape may bounce in an unpredictable manner and take longer to come to a stop.

5. How does the temperature of the ball and its surroundings impact the time for it to stop bouncing?

The temperature of the ball and its surroundings can affect the time for it to stop bouncing. In colder temperatures, the ball's material may become stiffer and less elastic, causing it to bounce for a shorter amount of time. In warmer temperatures, the ball's material may become more elastic, resulting in a longer bounce time.

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