Garbage can suspended in the air

[rude man]

That's still wrong. 8.2 Newtons is the weight that can be suspended at 17 meters. It's about 15 meters for the 10 Newton garbage can.

That's more like it.

So the can's weight is not 10 kg, it's 10 N. An 80 kg man weighs 80*9.81 N. Really confusing!


I have one question: the assumption is that the change in momentum per unit time is 2v dm/dt. Why not v dm/dt? I would expect the water v to be zero once it impacts on the can's bottom, not reverse with equal |v|.

 

[ehild]

That's more like it.

So the can's weight is not 10 kg, it's 10 N. An 80 kg man weighs 80*9.81 N. Really confusing!

When I started school, (a very long time ago) we used kg for weight. And then (in my country) we said kg-weight for the weight and kg for mass. And then we used kilopond for the force equal to the weight of 1 kg mass, which was equivalent to Newton.

I have one question: the assumption is that the change in momentum per unit time is 2v dm/dt. Why not v dm/dt? I would expect the water v to be zero once it impacts on the can's bottom, not reverse with equal |v|.

Why should it be zero?

The maximum possible height is needed. If the change of momentum is less, the force of the geyser is equal to the weight at higher speed, at lower height.

ehild

 

[D H]

i have one question: The assumption is that the change in momentum per unit time is 2v dm/dt. Why not v dm/dt? I would expect the water v to be zero once it impacts on the can's bottom, not reverse with equal |v|.

why should it be zero?

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.

 

[rude man]

When I started school, (a very long time ago) we used kg for weight. And then (in my country) we said kg-weight for the weight and kg for mass. And then we used kilopond for the force equal to the weight of 1 kg mass, which was equivalent to Newton.



Why should it be zero?

The maximum possible height is needed. If the change of momentum is less, the force of the geyser is equal to the weight at higher speed, at lower height.

ehild

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.

 

[D H]

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

Because the problem is to find the maximum height at which the garbage can rides, not the height at which the garbage can will ride realistically. Elastic collisions will yield the largest possible height.

 

[rude man]

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!

 

[ehild]

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

 

[rude man]

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.

 

[Enigman]

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.

 

[rude man]

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.

 

[Enigman]

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.

 

[NascentOxygen]

Now and again I discover how a spoon in the kitchen sink seems pretty efficient at reflecting the incident stream.

 

[aamaritoo]

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 !