Calculate Loss of Energy from Bouncy Ball Lab

[reliquator]

Okay, so we did a lab where we dropped a bouncy ball (mass = 56.4 g) from 1meter above the ground, and measured the height that it bounced up to.

We need to calculate the:
a) loss of energy due to the first bounce for each of the balls
b) the speed with which the ball strikes the floor
c) how many bounces would it take before the ball has lost all of its energy

The speed with which the ball strikes the floor with. Just set a = -9.8 m/s^2, and have vi = 0, so vf^2 = vi^2 + 2ad, so you get vf to be sqrt of 19.6 m/s, so ~4.427 m/s, ok, so did I do that part right?

Now for energy lost, do I use the Ke1 + Pe1 = Ke2 + Pe2? I'm kind of confused, and as for how many bounces, what equation do I use? Thanks a lot in advance.

 

[daniel_i_l]

First calculate the original energy with mgh (h=1).
Then mesure the final hight after the first bounce (it should be less than the original height) and calculate the energy with mgh. The difference of the two energies is the energy loss.
With that energy loss you can easily calculate how many times it (the original energy) can lose that amount before going into "negative" energy.

 

[reliquator]

Okay, so thanks a lot Daniel. So if my initial energy was 1 joule, and the ball lost 0.3 joules of energy on the first bounce, it would take 4 total bounces until the ball stopped bouncing/exhausted all its energy? Thanks a lot, I just need 1 final answer.

Okay, if my initial energy before letting go was 1 joule, and the ball lost 0.3 joules on the first bounce (all the data I have), how do I figure out how many bounces it takes for the ball to lose all of its energy?

 

[reliquator]

Can anyone help? :( I really need to understand how to do this last problem...

If my initial energy before dropping a ball was 1 joule, and the ball lost 0.3 joules on the first bounce (all the data I have), how do I figure out how many bounces it takes for the ball to lose all of its energy?

 

[reliquator]

ball dropped from 1 meter, ball weights 0.0564 kg. so PE = (1 m)(9.8 m/s^2)(0.0564) = 0.55272 joules. KE is 0 (since v = 0), so energy is 0.55272 joules at the top.

The ball bounces up to 0.58 meters, so its energy there is 0.3205 joules, so it loses about 2321 joules of energy.


Okay, so if the initial energy of a ball is 0.55272 J, and it loses 0.2321 J on the first bounce, does it take 0.55272/0.2321409 = 2.38097 bounces for the ball to lose all of its energy? Is the energy lost per bounce constant? Thanks

 

[Astronuc]

If energy was conserved, when the ball bounced it would return to the same height.

However, there are a few mechanisms by which the ball looses energy. One, it looses energy due to its interaction with the air in which it travels, and this is a function of the balls velocity. However, such a loss may be neglible. The ball deforms and the surface which it hits deforms, and the deformation disperses some of the balls kinetic energy, and reduces the elastic recoil.

As the ball bounces to a new height, the difference in height can be used to show the change in gravitational potential energy, which is related to the energy lost as the ball falls, impacts the ground, recoils and travels vertical, until it reaches a new altitude where its vertical assent ceases, and it begins to fall again.

The energy loss may be less the next bounce.

If one takes a basket ball or tennis ball and drops its, the energy loss per bounce decreases, and ultimately the balls have very low amplitude bounces.

In some cases, it may be possible to determine an energy loss per length traveled, which means for each successive bounce, the energy loss decreases.

 

[reliquator]

Thanks for the insight Astronuc, but how can I solve the problem now? Is there an equation that takes into account the loss of energy with each successive bounce?

 

[reliquator]

Astronuc would you mind helping me again? This is due tomorrow and I really need some more help...

 

[reliquator]

can anyone help me? :(

 

[Astronuc]

Where are you now with the problem solution?

I think I have an idea. The ball starts at 1 m with a certain gravitational potential energy, mgh, which you correctly calculated to be 0.55272 J (and probably 0.553 J is better).

Now the ball recoils to 0.58 m, which would give it a gravitational potential energy of 0.322 J, which means it lost 0.231 J.

Another way of looking at this is that ball traveled 1 m down, and 0.58 m upward, and lost 0.231 J or an energy loss of 0.231 J/ 1.58 m = 0.146 J/m.

Now how far does the ball travel with all the bounces.

Well if the ball recoiled 0.58 m starting at 1 m, then perhaps the ball may achieve 0.58 the height of the previous bounce, so on the second bounce, it will achieve 0.58².

Then the total distance traveled is 1 + 2*0.58 + 2*0.58² + . . . . The terms of 0.58 represent a geometric series (2*r + 2*r² + 2*r³ + . . . = 2r/(1-r), r < 1).

So the total distance traveled by the ball is the initial height + sum of the geometric series.

Take the total distance and the energy loss per unit distance, and see what you can do with that.

 

[reliquator]

So wait, I get the 0.146 J/M part, but what about the "(r + r2 + r3 + . . . = 2r/(1-r), r < 1)." part? Do you take 1 + 0.58 + 0.58^2 + ?

Okay it loses 0.146 J/M, and it has 0.553 Joules to start with, so 0.553 = 0.146x, so it travels 3.79 m total. Now what?

 

[Astronuc]

Please re-read my previous post. I left out a factor of 2 in the geometric series. The ball travels up and down the same distance between bounces.

If one uses a spreadsheet, one can see the distance traveled or energy expended. At some point there is very little energy left in the ball. If one selects 10^-6 m as the cutoff criterion, one can determine the number of bounces.

 

[reliquator]

Okay, so it would be 0.553 x (0.58)^X = 10^(-4) right? But you can't set it to 0 or the equation wouldn't work? So X equals like 11.6 bounces, so 12 bounces before the ball expends all of its energy?

 

[Astronuc]

That's about right.

I went a little further, to about 10^-6 on the height and to 15 or 16 bounces.

In reality, a ball and surface undergo some elastic deformation and it stops bouncing when the elastic force is less than the weight of the ball.

 

[andrevdh]

What you are dealing here with is described by the coefficient of restitution given per definition by

$$e=v/u$$

where $v$ is the rebound speed of the ball after colliding with the ground with an impact speed of $u$. This coefficient is determined only by the two types of materials colliding with each other (sphere with a plane or sphere with sphere). As you can see it should take on values between zero and one.
The potential energy from dropping it onto the ground from $h_1$ is converted to $\frac{1}{2}mu^2$.
Therefore the speed of impact is given by

$$u=\sqrt{2gh_1}$$

Likewise the kinetic energy after impact is converted to potential energy where it reaches a height $h_2$

$$v=\sqrt{2gh_2}$$. Therefore

$$e=\frac{\sqrt{2gh_2}}{\sqrt{2gh_1}}$$

which gives

$$e=\sqrt{\frac{h_2}{h_1}}$$

The speed of the ball after successive bounces will therefore be

$$ue,\ ue^2,\ ue^3 \ ...$$
http://mysite.mweb.co.za/residents/andriesvdh/bouncing%20ball.htm"