coefficient of restitution problem

Dear TSR!

I am trying to solve a problem involving the coefficient of restitution. I thought I understood it perfectly fine, but after finding this question I don’t. DOH!
The idea is..
Two balls are dropped. One large, one small. The small back is on top of the larger ball. There are two events to calculate. The impact of the larger ball with the ground. And the impact between the smaller ball and the larger ball.

You can see it better in the video link here 0 – 10 secs
[video="youtube;wF6pBViU6ws"]http://www.youtube.com/watch?v=wF6pBViU6ws&feature=rela ted[/video]

I already know the following (via experimental results)
1) E1 (CoR with the large ball and the ground) = 0.6
2) E2 (CoR with the two balls together) = 0.5
3) Initial height of the two balls = 1m
4) Mass of large ball = 0.5kg
5) Mass of small ball = 0.05kg
6) After the second impact the larger ball reaches a high of 0.05m above the ground.

Also postive indicates downwards travel, negative indicates upwards.

First event… (pretty sure this is right)

Ball 1 hits the ground at 4.42 m/s
It rebounds at -2.66 m/s, because 4.42*0.6 = 2.66

Second event… (think something has gone wrong here)

Ball 1 is traveling at -2.66 m/s
Ball 2 is traveling at 4.42 m/s
Therefore approach speed is 7.08 m/s because 4.42 + 2.66 = 7.08
APP speed x E2 = 7.08*0.5 = 3.54 m/s separation speed.
Ball 1 (larger ball) reaches 0.05m above the ground after this. So (2*g*0.05)^0.5 = new speed v1 = -1m/s
There for Ball 2 must be traveling at –v2 = app speed + v1 = - 4.54 m/s

This is where I become confused…

During the experiment the smaller ball went flying off. Much faster than the -4.54 m/s I found above. Where am I making an error? Am I missing some events? Something wrong with my algebra? Have I not considered something else?

Also how can I do this with Conservation of momentum? What would be the correct formula for the COR/COM for these two events?

Reply 1

ghostwalker

17

Original post by got-it-wrong

...

If that's you in the video, I'd query the experiemental data. The larger ball seemed to rise ~0.25m, which makes a significant difference to your calculation, which is correct in principle, as best I can tell.

If it's not you, I'd still query the data.

(edited 11 years ago)

Original post by ghostwalker

If that's you in the video, I'd query the experiemental data. The larger ball seemed to rise ~0.25m, which makes a significant difference to your calculation, which are correct in principle, as best I can tell.

If it's not you, I'd still query the data.

Looks like I need to find a new party trick. :mad:

I use a rubber duck and a football.

Reply 3

got-it-wrong

OP

Original post by ghostwalker

If that's you in the video, I'd query the experiemental data. The larger ball seemed to rise ~0.25m, which makes a significant difference to your calculation, which is correct in principle, as best I can tell.

If it's not you, I'd still query the data.

The video was just an example I found online....

Do you think the calculations I have done are correct, and my data may be wrong?

Reply 4

ghostwalker

17

Original post by got-it-wrong

The video was just an example I found online....

Do you think the calculations I have done are correct, and my data may be wrong?

Yes. It's possible that the final velocity is sensitive to one of the items of data that is difficult to measure, and that a small variation may produce a large change in the result.

The theoretical maximum height is 9 times the original, with a velocity of 3 times the velocity with which the larger ball hits the ground, if it's a perfectly elastic collision.

Reply 5

got-it-wrong

OP

Original post by ghostwalker

Yes. It's possible that the final velocity is sensitive to one of the items of data that is difficult to measure, and that a small variation may produce a large change in the result.

The theoretical maximum height is 9 times the original, with a velocity of 3 times the velocity with which the larger ball hits the ground, if it's a perfectly elastic collision.