Kinetic energy of a bouncing ball

  • #1

Homework Statement


I'm trying to find the coefficient of restitution (COR) and energy loss upon impact of a bouncing ball, and I'm trying to understand how these relate to each other.

Homework Equations


COR=sqrt(2gh/2gH)
suvat equations (possibly)
1/2mv^2=mgH (using m=0.0585kg, g=9.8)
energy transfer efficiency = (useful output ÷ total input ) × 100

The Attempt at a Solution


In my case of a ball released from 1.2m then bouncing to 0.64m, COR=sqrt(2gh/2gH) = 0.73 (fairly elastic). I know that COR is a ratio of the ball velocity before/after impact, therefore also of the kinetic energy before/after. Now doesn't this mean that the final velocity/kinetic energy is 73% of the original?

I'm struggling to verify this because I can't measure the kinetic energy after the bounce. The suvat equations assume constant acceleration or time so I don't think I can use those. Given that I know the final velocity upon bouncing using mgH=1/2mv^2, v = sqrt(23.52)m/s, and I know the bounce height (0.64m) is there another way?
 

Answers and Replies

  • #2
mjc123
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I know that COR is a ratio of the ball velocity before/after impact, therefore also of the kinetic energy before/after. Now doesn't this mean that the final velocity/kinetic energy is 73% of the original?
Since kinetic energy is proportional to velocity squared, COR cannot be both the ratio of velocities and the ratio of kinetic energies. Which is it? Hint: what is it also the ratio of? see your equation above.
 
  • #3
Since kinetic energy is proportional to velocity squared, COR cannot be both the ratio of velocities and the ratio of kinetic energies. Which is it? Hint: what is it also the ratio of? see your equation above.
Sorry putting it into words is pretty confusing for me. I know that it's sqrt(2gh/2gH) which is derived from mgh=1/2mv^2, so I guess COR is the ratio of the square root of the velocities, seeing as v=sqrt(2gh). Unless you mean it's the ratio of the square root of the heights?
 
  • #4
haruspex
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Further to mjc's reply:
The suvat equations assume constant acceleration
There is constant acceleration up to just before the bounce, and again from just after the bounce. You can use SUVAT for any interval that does not include the bounce itself.
 
  • #5
Further to mjc's reply:

There is constant acceleration up to just before the bounce, and again from just after the bounce. You can use SUVAT for any interval that does not include the bounce itself.
Thanks I was confused about that. so using v^2=u^2+2as for the bounce. u=3.541750979m/s, which is consistent with the COR=0.73
 
  • #6
mjc123
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I know that it's sqrt(2gh/2gH) which is derived from mgh=1/2mv^2, so I guess COR is the ratio of the square root of the velocities, seeing as v=sqrt(2gh). Unless you mean it's the ratio of the square root of the heights?
No, it's the ratio of the velocities just before/after bounce. It is also the square root of the ratio of the potential energies at maximum height, and therefore the square root of the ratio of kinetic energies just before/after bounce.
u=3.541750979m/s
Do you think this is a reasonable number of significant figures to give in your answer?
 
  • #7
Maybe I should start a new thread but I was also wondering about the conservation of energy upon collision between ball and floor. The ball will lose energy due to heat, sound and friction at least, but how does the type of surface influence these energy losses? I know hard surfaces tend to 'reflect' more energy than softer surfaces, but is this just because of the tight structure of solids, and how is it 'reflected'? Also is friction the biggest cause of energy loss in this type of collision?
 
  • #8
haruspex
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Maybe I should start a new thread but I was also wondering about the conservation of energy upon collision between ball and floor. The ball will lose energy due to heat, sound and friction at least, but how does the type of surface influence these energy losses? I know hard surfaces tend to 'reflect' more energy than softer surfaces, but is this just because of the tight structure of solids, and how is it 'reflected'? Also is friction the biggest cause of energy loss in this type of collision?
In general, any body deforms elastically at first. [As the force (pressure, rather) increases, it may reach its elastic limit and deform plastically, but that is another story.]
You have been taught that a spring, say, has a constant k such that F=kx, where x is the deformation. That is not quite true. The "constant" depends whether the deformation is increasing or decreasing. When decreasing, so allowing the spring to return towards its relaxed length, the constant is less. As a result you get less work back out than you put in. The loss is as heat within the body.
In a bounce, this happens within both bodies. In the case of a rubber ball on a concrete floor almost all the deformation, hence most of the loss, is within the ball; for a rubber ball on a carpet the situation is rather different.

You ask if it is friction. Not in the usual sense, perhaps, but at the molecular level some coordinated motion is being transformed into random motion. E.g. think of pulling a rope that has some protrusions each side and these knock against obstacles. Some of the linear motion of the rope gets turned into lateral oscillations.
 

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