Garbage can suspended in the air

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In summary: I also think that the example solution is wrong.Same here. I think the correct answer for a 10 kg garbage can (and calling this "weight" in a physics text is not good) is zero. 20 m/s * 0.5 kg/s = 10 Newtons. Double that for a pure elastic collision and you get 20 Newtons, which is not enough to lift a 10 kg object off the ground.The problem says "of weight W". So I'd think it should be W = 10 N, rather than 10 kg.And it also says that W=10 kg. The text of the problem is confusing. Same here. I think the correct answer for a
  • #36
D H said:
I suspect rude man was looking at things from a realistic rather than theoretical perspective. I would expect that if this was tried experimentally, one would see collisions that are much closer to purely inelastic than elastic. When the water hits the can it will bounce off slightly inelastically and immediately collide with incoming water. The end result will be a close to purely inelastic collision with the water dribbling down the sides of the can.

However, the question is asking for the maximum possible height, and to get that you need elastic collisions.

Thanks DH. I guess that's what the question meant by "maximum" height. Meaning a purely elastic collision.

Thanks for your post. I feel better already!
 
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  • #37
rude man said:
That was not my question. My question is why is dp/dt = 2v dm/dt instead of v dm/dt.

In other words, why is the collision of the water with the can bottom considered an elastic vs. an inelastic collision.

The Physics problems in the books are not always realistic. You get higher stationary position for the garbage can if you take the collision elastic. And the problem asked the highest possible position.


ehild
 
  • #38
ehild said:
The Physics problems in the books are not always realistic. You get higher stationary position for the garbage can if you take the collision elastic. And the problem asked the highest possible position.


ehild

What if an elastic collision is not possible?
I hope you get my point. It was a stupidly worded question.
 
  • #39
Then the question would mention the coefficient of restitution or something and not just 'maximum height', it was just a way to get the student thinking about the difference between elastic and inelastic collision and quite clever at that. In real world? Yep, pretty stupid since collision would be more inelastic than elastic.
 
  • #40
Enigman said:
Then the question would mention the coefficient of restitution or something and not just 'maximum height', it was just a way to get the student thinking about the difference between elastic and inelastic collision and quite clever at that. In real world? Yep, pretty stupid since collision would be more inelastic than elastic.

The question should have read, "Assuming an elastic collision between the water and the can, ...".

Since the collision is probably inelastic the "possible height" of 17m or whatever it turned out to compute to would be, ah, impossible.

The whole problem and especially the "hint" were totally screwed up anyway.
 
  • #41
rude man said:
The question should have read, "Assuming an elastic collision between the water and the can, ...".
Too obvious, why give 'em something when you could have them figure it out themselves...

The whole problem and especially the "hint" were totally screwed up anyway.
Amen.
 
  • #42
Now and again I discover how a spoon in the kitchen sink seems pretty efficient at reflecting the incident stream.
 
  • #43
Dimension analysis idea !

I think k = [kg/s]. Now, since velocity (V) is [m/s] -------> KV = [kg/s][m/s] ... which turns out to be: [kg.m/s][1/s] momentum flux dimensions, I think.

the problem does not states if water rebounds with the same velocity that hits... so i asume, stops the momentum one´s it hits the can, so:
forces up = forces down, rigth ?

Force up = KV = K.(V^2 - 2gh)^0.5
force down = W

but i still don't get the clue relation of the book, don´t know why !
 
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