Find Coefficient of Restitution with just height of the ball after one bounce

In summary: By the way, if you work it symbolically instead of plugging in numbers you will find the relationship between e and the height ratio.
  • #1
x2017
84
1

Homework Statement


A ball is dropped from a height of 9.02 m. After hitting the ground, the ball then rebounds to a height of 1.78 m. What is the coefficient of restitution associated with the ball and ground impact?

Homework Equations


e=(V'-v')/(V-v)

The Attempt at a Solution


I thought maybe this was a trick question and you just do 9.02-1.78=7.24 and that was the answer but I am wrong... I need a bit of guidance on how to proceed with this question.

EDIT:
I just realized I should be using e=v'/v instead since there is only one ball... And I think I have realized that v' is -9.81? Not completely sure though because I also think that that might only be acceleration... CONFUSED lol
I thought that maybe I could use acceleration to find velocity, but I don't have the time...
 
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  • #2
x2017 said:
And I think I have realized that v' is -9.81? Not completely sure though because I also think that that might only be acceleration...
Yes, that's the acceleration. What SUVAT equation do you know that relates acceleration, distance and speeds?
 
  • #3
haruspex said:
Yes, that's the acceleration. What SUVAT equation do you know that relates acceleration, distance and speeds?

vf2=vi2+2ad perhaps?
Thanks for the reply, I always seem to forget about the SUVAT equations...

vf2=vi2+2ad
vf2-vi2=2(9.81)(9.02)

I used 9.02m and not 1.78m because it traveled down 9.02m when gravity was acting on it and I don't know what it's acceleration upwards is. Hopefully this is correct!

vf2-vi2=2(9.81)(9.02)
vf2-vi2=176.9724
vf-vi=13.30
Δv=13.30m/s

Is the above something I can do? So now I have the change in velocity of the ball...
But only on the way down correct? or is this all together?
 
  • #4
x2017 said:
vf2=vi2+2ad perhaps?
Thanks for the reply, I always seem to forget about the SUVAT equations...

vf2=vi2+2ad
vf2-vi2=2(9.81)(9.02)

I used 9.02m and not 1.78m because it traveled down 9.02m when gravity was acting on it and I don't know what it's acceleration upwards is. Hopefully this is correct!

vf2-vi2=2(9.81)(9.02)
vf2-vi2=176.9724
vf-vi=13.30
Δv=13.30m/s

Is the above something I can do? So now I have the change in velocity of the ball...
But only on the way down correct? or is this all together?
You have found the landing speed, yes. A few doubtful steps along the way, though.
You should be careful with signs. If up is positive, the acceleration is -9.81m/s2, and the distance is -9.02m. That way you get a positive Δ(v2).
Secondly, you cannot go straight from vf2-vi2 to vf-vi by square rooting. You need to plug in the value for vi (0) first.

Now you need to find the rebound speed by the same method.
 
  • #5
haruspex said:
You have found the landing speed, yes. A few doubtful steps along the way, though.
You should be careful with signs. If up is positive, the acceleration is -9.81m/s2, and the distance is -9.02m. That way you get a positive Δ(v2).
Secondly, you cannot go straight from vf2-vi2 to vf-vi by square rooting. You need to plug in the value for vi (0) first.

Now you need to find the rebound speed by the same method.

How do I get the acceleration for the rebound speed though? Since it's traveling upwards it's not gravity right? And I don't have enough information to solve for acceleration with Δv/Δt.
 
  • #6
x2017 said:
How do I get the acceleration for the rebound speed though? Since it's traveling upwards it's not gravity right?
why not? Does gravity stop acting on a body just because it is moving upwards?
 
  • #7
haruspex said:
why not? Does gravity stop acting on a body just because it is moving upwards?

No, I guess gravity is causing it to slow down as it approaches its peak height! Okay I understand that part now, thanks!

I solved for the rebound speed and got 5.91m/s.

e=v'/v
e=5.91/13.30
e=0.44 which comes up as correct!

This turned out to be pretty simple, thanks for helping me understand how to go about it haruspex! :)
 
  • #8
x2017 said:
No, I guess gravity is causing it to slow down as it approaches its peak height! Okay I understand that part now, thanks!

I solved for the rebound speed and got 5.91m/s.

e=v'/v
e=5.91/13.30
e=0.44 which comes up as correct!

This turned out to be pretty simple, thanks for helping me understand how to go about it haruspex! :)
Well done.

By the way, if you work it symbolically instead of plugging in numbers you will find the relationship between e and the height ratio. You don't actually need to calculate the velocities.
 
  • #9
nmarryat said:
What is that relationship between e and the height ratio?
How about you try what I suggested, solving the problem but entirely symbolically, i.e. not plugging in any numerical values.
 
  • #10
nmarryat said:
I only need the equation for some extra credit for a class that's not even relevant to my major. But thanks for the sarcasm.
There was no sarcasm, none at all. It was a serious suggestion for what would be far more useful to you than my merely providing an equation.
Indeed, that is how these forums work - we provide hints, corrections, advice, but require the student to show effort. You had effectively asked me to violate forum rules.
 

1. What is the coefficient of restitution?

The coefficient of restitution is a measure of the elasticity of a collision between two objects. It is a dimensionless number that ranges from 0 to 1, with 1 being a perfectly elastic collision and 0 being a perfectly inelastic collision.

2. How is the coefficient of restitution calculated?

The coefficient of restitution is calculated by dividing the velocity of the object after a collision by the velocity of the object before the collision. In the case of finding the coefficient of restitution with just the height of the ball after one bounce, the calculation would be: (height after bounce / initial height).

3. What is the significance of finding the coefficient of restitution with just the height of the ball after one bounce?

Finding the coefficient of restitution with just the height of the ball after one bounce can help determine the energy lost during a collision and the elasticity of the objects involved. It can also be used to predict the motion of the object after multiple bounces.

4. Can the coefficient of restitution be greater than 1?

Yes, the coefficient of restitution can be greater than 1. This indicates a super-elastic collision, where the objects involved gain energy during the collision.

5. How does the height of the ball after one bounce relate to the coefficient of restitution?

The height of the ball after one bounce is directly related to the coefficient of restitution. The higher the ball bounces, the closer the coefficient of restitution is to 1, indicating a more elastic collision. A lower bounce height would result in a lower coefficient of restitution, indicating a less elastic collision.

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