Kinetic energy of a bouncing ball

In summary: In the case of a metal ball bouncing on a rubber floor, most of the deformation is within the floor.This is true, however, the metal ball will start to heat up more quickly as a result.
  • #1
Amarillo99
4
0

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?
 
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  • #2
Amarillo99 said:
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
mjc123 said:
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
Further to mjc's reply:
Amarillo99 said:
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
haruspex said:
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
Amarillo99 said:
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.
Amarillo99 said:
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
Amarillo99 said:
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, anybody 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.
 

1. What is kinetic energy?

Kinetic energy is the energy that an object possesses due to its motion. It is defined as the energy an object has while in motion and is directly proportional to the mass and velocity of the object.

2. How is kinetic energy calculated?

The formula for calculating kinetic energy is: KE = 1/2 * m * v^2, where m is the mass of the object and v is its velocity.

3. What is the relationship between the height of a bouncing ball and its kinetic energy?

The height of a bouncing ball directly affects its kinetic energy. As the ball bounces, it loses some of its potential energy due to its decrease in height. This potential energy is then converted into kinetic energy, causing the ball to bounce higher and with more force.

4. Why does a ball bounce?

A ball bounces due to the conservation of energy. When the ball hits the ground, its kinetic energy is converted into potential energy as it compresses. When the ball is released, the potential energy is converted back into kinetic energy, causing the ball to bounce.

5. How does the surface affect the kinetic energy of a bouncing ball?

The surface that the ball is bouncing on can affect its kinetic energy. A softer surface, such as a carpet, will absorb some of the energy, causing the ball to bounce lower. A harder surface, such as concrete, will cause the ball to bounce higher as less energy is absorbed.

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