- #1

- 137

- 0

if two balls [1,2] are pulled away:

**oo ooo**, two balls [4,5] on the opposite side:**ooo→ oo**swing out. What happens if ball # 3 or 4 or 5 has mass = 2 ?**1) oo Ooo, 2) oo oOo, 3) oo ooO**
Last edited:

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- #1

- 137

- 0

Last edited:

- #2

- 391

- 1

- #3

Homework Helper

- 8,810

- 598

- #4

- 137

- 0

I suppose gaps between balls do not affect the outcome, only simplify solution.... it also helped to consider the balls to have spring like qualities.

case 1 and 3 should have similar outcome, case 2 is more difficult to break into separate collisions, but I suppose balls should have a certain rebound. Is it so?

Last edited:

- #5

- 137

- 0

I couldn't find previous thread, I can't see in what wayIn previous threads..., it alsohelpedto consider the balls to have spring like qualities, where the balls experience forces related tocompressions, decelerations, and accelerationsbetween balls during collision cycles.

Gaps between balls not only avoid complications but they let us also comprehend outcomes that are counter-intuitive: consider the simple 3-ball system case:

ball 2 hits ball 3 and stops, is hit by ball 1 and swings out with ball 3

Last edited:

- #6

Homework Helper

- 8,810

- 598

The older threads eventually link to each other and then onto this web page:I couldn't find previous thread, I can't see in what way that helps.

http://www.lhup.edu/~dsimanek/scenario/cradle.htm

- #7

- 137

- 0

. If balls are touching we cannot consider it anidealcradle, whereinitial and final velocity are the same.

link [Simanek] confirms what I said: {2.,2.}: "the balls aren't really touching, so a series of independent two-ball collisions occur", so inThe older threads eventually link onto this web page:http://www.lhup.edu/~dsimanek/scenario/cradle.htm

Case 1: v

Case 2 [v

- #8

- 137

- 0

1)oo Ooo, 2) oo oOo, 3) oo ooO

you are referring to this,

what do

- #9

- 105

- 0

you are referring to this,

oo→ Ooo, now, if balls 1,2 where tied, result would be [[itex]\equiv[/itex] O→ Ooo ] ,ooO oo→, (because the 2 loose balls, when hit, behave as if they were tied). But they are not, the question is : what is their[2 falling-hitting balls] behaviour when they are tied/not-tied touching/not-tied-not-touching/?

what doyouthink?

I'll take a swing at it :P

Just a guess :P

- #10

- 391

- 1

you are referring to this,

oo→ Ooo, now, if balls 1,2 where tied, result would be [[itex]\equiv[/itex] O→ Ooo ] ,ooO oo→, (because the 2 loose balls, when hit, behave as if they were tied). But they are not, the question is : what is their[2 falling-hitting balls] behaviour when they are tied/not-tied touching/not-tied-not-touching/?

what doyouthink?

No matter how you arrange this system what I said remains true, momentum and kinetic energy will remain constant. That means that if the middle ball has double the mass of all the others there is no effect on your problem, additionally you will get the same result whether or not the two balls you release are tied or touching.

- #11

- 137

- 0

if you think so, please, show your solutions. Let's sayThat means that if the middle ball has double the mass of all the others there isno effecton your problem,.

Last edited:

- #12

Homework Helper

- 8,810

- 598

Note that at the end of {2.,2.} is this statement:Simanek confirms what I said: {2.,2.}: "the balls aren't really touching, so a series of independent two-ball collisions occur"

Also mentioned throughout, is the fact that even if the middle balls end up at rest, they do get displaced slightly during each half cycle, but ideally end up back at their orignal position after a full two collision cycle. In reality, once the center balls are displaced, gravity will tend to recenter them. Using very long support lines would reduced the gravity / pendulum related effects on the middle balls.

Last edited:

- #13

Gold Member

- 21,934

- 5,477

if you think so, please, show your solutions.

Simply replace the m=2 ball with two m=1 balls and you've got the same thing.

Just as oo oooo becomes oooo oo

so does oo Ooo become ooO oo.

- #14

- 137

- 0

.. gaps between balls donotaffect the outcome, only simplify solution...

Sure, rcglrdNote that at the end of {2.,2.} is this statement:the surprising outcome is essentially the.samewhether or not they are touching

... gravity will tend to recenter them. .

Of course, in the end, gravity will recenter

The purpose of this thread is to see if we can work out some practical, general rules, and see if there can be any exceptions to the

If we strictly apply that

first general rule says that momentum passes on undisturbed if ball in the middle is smaller and not bigger, is it so?

Last edited:

- #15

Gold Member

- 21,934

- 5,477

first general rule says that momentum passes on undisturbed if ball in the middle is smaller and not bigger, is it so?

A small ball combined with a regular ball is essentially the same as one larger ball.

i.e:

That being said, it is

eg:

- #16

- 137

- 0

... kinetic energy and momentum must be conserved...

...aren't we forgetting [...momentum and kinetic energy will remain constant.' ..

Last edited:

- #17

- 391

- 1

Not if the balls have different masses, no.

- #18

- 7

- 0

Not if the balls have different masses, no.

that is not true, can you find out the oucome of this simple case?:

m1=2,v1=9

m2=1 v2=0

v1±v2 = v'2-v'1

Last edited:

- #19

- 391

- 1

[tex]

v_1 = v_1' + v_2'

[/tex]

[tex]

m_1 v_1 = m_1 v_1' + m_2 v_2' \Rightarrow m_1 v_1' + m_1 v_2' = m_1 v_1' + m_2 v_2' \Rightarrow m_1 = m_2

[/tex]

Therefore the two cannot both be, in the general case, true.

- #20

- 7

- 0

if conservation of velocity were true in every case then

in every case in this thread, because all collisions are head-on.

you haven't offered a single solution, so far.

- #21

- 391

- 1

Not only that but above I proved that your case of the two balls does NOT conserve velocity.

- #22

- 7

- 0

if you work out properly my example you'll see that net velocity of separation equals net velocity of approach , (the same as conservation of momentum means net momentum): v'_{ s} = v _{a},do you have any solution to the "cradle"?

Last edited:

- #23

- 391

- 1

I'd like to point out that your original formula (which was used in my last post) was changed after you made that post, but we might just chalk that up to a typo. In any case the person I was responding to stated that the sums of the velocities before/after the collision remained constant, that is not true and I was right to point that out.

Last edited:

- #24

- 7

- 0

(+) + (+), (+) + (-), (-)+(+), (-)+(-).

It was not a typo, I'm sorry if you did not appreciate it, I was only trying to help you when I saw from your reply that you did not know anything about it. It was just a hint, giving you the actual procedure in that particular example. bye.

Share:

- Replies
- 3

- Views
- 452

- Replies
- 13

- Views
- 863

- Replies
- 4

- Views
- 565

- Replies
- 4

- Views
- 232

- Replies
- 5

- Views
- 555

- Replies
- 1

- Views
- 435

- Replies
- 2

- Views
- 549

- Replies
- 19

- Views
- 577

- Replies
- 19

- Views
- 1K

- Replies
- 52

- Views
- 1K