I think the secret is here

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Kirk
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I think the secret is here

Post by Kirk »

Condition 3 The barge is approaching shore at 20 feet per second. The thrower sees the ball leave at 40 feet per second and the observer on the barge sees it approaching at 60 feet per second then bounces and departs at 60 feet per second. The thrower sees it approaching at 80 feet per second.

velocity and ke are not a straight line
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re: I think the secret is here

Post by sleepy »

Very interesting.This assumes that there is no loss of momentum at any stage of the transaction,and that there is a complete transfer of momentum between the barge and the ball.But the real hang-up is that both the thrower and the barge had to put energy into the system.My guess is that the amount of energy needed to move the barge far outweighs the extra energy picked up by the ball.
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Mark
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Re: I think the secret is here

Post by Mark »

Kirk wrote:The barge is approaching shore at 20 feet per second. The thrower sees the ball leave at 40 feet per second and the observer on the barge sees it approaching at 60 feet per second then bounces and departs at 60 feet per second. The thrower sees it approaching at 80 feet per second.
I think not, assuming the obvious(?) conditions; no gravity, no air resistance, best ball ever made, and both thrower and observer are standing "still" or seated. (and, it's the ball that bounces :-)
For it to be true, wouldn't the thrower need to be moving toward the barge at 20 fps after the bounce?

[Unless the unknown conditions 1 and 2 were applicable.]
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re: I think the secret is here

Post by ME »

Ideally the best part where the thrower avoids the ball which will hit the sliding panel behind him. The ball bounces back towards barge, hitting it again and comes back at 120 ft/s.
If we're lucky then after a couple of round trips a lot of the barge's momentum will be transferred to the ball and shoots off when the sliding panel cleared its path.
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Re: I think the secret is here

Post by Kirk »

Mark wrote:
Kirk wrote:The barge is approaching shore at 20 feet per second. The thrower sees the ball leave at 40 feet per second and the observer on the barge sees it approaching at 60 feet per second then bounces and departs at 60 feet per second. The thrower sees it approaching at 80 feet per second.
I think not, assuming the obvious(?) conditions; no gravity, no air resistance, best ball ever made, and both thrower and observer are standing "still" or seated. (and, it's the ball that bounces :-)
For it to be true, wouldn't the thrower need to be moving toward the barge at 20 fps after the bounce?

[Unless the unknown conditions 1 and 2 were applicable.]
All second order effects are ignored in the example. The 80 fps is to a stationary observer on the bank.
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Re: re: I think the secret is here

Post by Kirk »

ME wrote:Ideally the best part where the thrower avoids the ball which will hit the sliding panel behind him. The ball bounces back towards barge, hitting it again and comes back at 120 ft/s.
If we're lucky then after a couple of round trips a lot of the barge's momentum will be transferred to the ball and shoots off when the sliding panel cleared its path.
Give the man a cigar. You see it.
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Tarsier79
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re: I think the secret is here

Post by Tarsier79 »

That is what you have to prove, Kirk. An imbalance between KE and momentum transfer. How do you do it?
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Mark
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Re: I think the secret is here

Post by Mark »

Kirk wrote:The 80 fps is to a stationary observer on the bank.
Are you saying that the barge velocity gets added twice? If so, why?
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re: I think the secret is here

Post by Furcurequs »

That is the upper limit for a perfectly elastic collision where the ball is much less massive than the barge.

It would be about the same with a baseball and a bat. The upper limit of the speed of the hit ball will be about the speed of the pitch plus two times the speed of the bat. That's the theoretical best you could do with such a collision.

As the original scenario was stated, if you are standing on the barge, you will see the ball coming at you at 60 fps (that's the 40 fps speed of the ball coming toward you relative to the ground plus your 20 fps speed going toward the ball relative to the ground, you just add those together) and ideally you will then see it bounce away from you at 60 fps, also. Relative to you, when it bounces it leaves at approximately the same speed it arrived.

That means the person on the ground who threw the ball will see it coming back at the 20 fps speed of the barge (relative to him and the ground) plus the 60 fps speed it leaves the barge - and thus 80 fps. The ball, then, approximately doubles its speed and ends up with about 4 times the kinetic energy. ...but it does get it from slowing down the much more massive barge by a very small amount.

Now, if you think of a golf club colliding with a stationary golf ball. Relative to the head of the club, the ball is coming toward it at the speed of the swing. It will then bounce away from the head of the club at around the same speed, but of course the head of the more massive club barely slows down, either, so the ball will be moving nearly double the speed of the head of the club (relative to the ground) after the swing and the hit.

Again, this is only when one of the objects is much more massive than the other and so its speed barely changes.
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Post by ME »

*The barge velocity gets added twice? If so, why?

A ball leaves with -40 f/s, hitting (full elastic) a moving barge which moves +20 f/s.
Relative to that barge (thus becomming 0 f/s for easy calculus) the ball hits at -60 f/s. It bounces and becomes +60 f/s.
To get back to our normal frame of reference we have to lift the relativity of the barge and add +20 f/s: The ball goes +80 f/s.
It hits a panel +80, bounces with -80. Hits the barge with -80, but relative -100 and bounces relative +100, normalized to 120.

or, see: https://en.wikipedia.org/wiki/Elastic_c ... _Newtonian
v1 = ( u1*(m1-M2)+2*M2*v2 ) / (M2+m1)
When m1 is very small compared to M2 then we can leave out the (m1)
v1 = ( - u1*M+2*M2*v2 ) / (M) = 2*v2 - u1

...or/and, what Furcurequs said.

For a continuous collision it's nice to add that as the speed goes up and the distance smaller the frequency will ramp up - I think exponentially.
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Post by Mark »

Thank you, gentlemen, for taking the time to explain it to me. I appreciate it. As soon as I read the explanation I realized that I had forgotten to take into account the relative masses.

It's like I'm starting to get Alzheimer's, but I'm only 60 years old! May be the new medication that my doctor put me on, I'll have to check that out.

Regardless, I am seeing a trend in my postings that I do not like. Maybe it's time for me to step away from all this.
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Post by Mark »

Kirk, I apologize for my unwarranted posts.
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re: I think the secret is here

Post by ME »

Mark, if you ask me: You don't need to apologize for a perfectly good question, even if you think it's below your own expected standards (which we don't know about anyway).
Unless you have some better alternative to this forum then please stay, and use it to keep your grey-matter (and ours) busy.
Besides, you still have some greenie-credits: so do your worst.

(I'll not mention we all have our own embarrassing "hmm-I-could-have-done-that-a-little-better"-moments)
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Re: I think the secret is here

Post by Kirk »

Mark wrote:
Kirk wrote:The 80 fps is to a stationary observer on the bank.
Are you saying that the barge velocity gets added twice? If so, why?
The ball is moving 40fps when it encounters a target moving towards the thrower at 20fps. The ball sees 60fps elastic compression. When it rebounds the surface it leaves is moving 20 fps and the ball accelerates 60fps off that for a total of 80 fps relative to the thrower.
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re: I think the secret is here

Post by Tarsier79 »

Ha ha Mark. You had me second guessing about that too.
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