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Originally Posted By: paul

the mass has fallen for a distance and now
has a acceleration of 40 m/s it strikes a
inelastic part of the earth , how much force
does the earth feel from the collision?


You can't know from just that information.

F=ma

What's the acceleration while it's applying a force to the earth? Who knows. Depends how soft it is. Depends how long it spends applying that force before completely stopping.

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Originally Posted By: paul

the bungee jumper feels a smaller force for a longer time.


Exactly. The force depends how long it takes to stop. Sudden hard stop = high force, slow cushioned stop = low force.


How about you either explain where you got

F = m v

from or stop using it. I've given you two very good reasons why it's wrong. Maybe you just used it by mistake once and are clinging to it to save face? Whatever the reason, back it up with something, or throw it out.

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paul Offline OP
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the 100kg mass has fallen for a distance and now
has a acceleration of 40 m/s/s it strikes a
inelastic part of the earth , how much force
does the earth feel from the collision?


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Originally Posted By: paul
the 100kg mass has fallen for a distance and now
has a acceleration of 40 m/s/s it strikes a
inelastic part of the earth , how much force
does the earth feel from the collision?


You mean 40m/s/s is the acceleration it experiences while stopping, not while falling? Then it's easy:

F=ma
F=100kg * 40m/s^2
F=4000N

boom.

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paul Offline OP
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No , I mean it has a acceleration of 40 m/s/s as it strikes
the earth.

we can say that the collision is inelastic so when the two
collide they do not deform , the 100kg mass transfers its force to the earth.

why didnt you say it strikes the earth with a 8000N force

its the same mass and the same acceleration as in the example.
the 100kg mass can not tell if its hitting the earth or being turned around in the pipe.


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Originally Posted By: paul

we can say that the collision is inelastic so when the two
collide they do not deform , the 100kg mass transfers its force to the earth.

That's not what 'inelastic' usually means. But if they really don't deform then the ball stops instantly. Stopping instantly applies an infinite force, which of course isn't possible. Perfectly undeformable bodies aren't possible either so it's OK.


Quote:

its the same mass and the same acceleration as in the example.

No. In the pipe the acceleration wasn't 40m/s^2. Just work through my calculations in bold in a recent message, I think that sums up my entire argument pretty clearly. Point to any specific lines you don't like.


I'm about to give up Paul. You used to be pretty competent with physics, but all your skills seem to have gone to the dogs with this tube thing, and getting progressively worse.

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I tried to use the velocity that the mass has as it strikes.

you demanded acceleration as it strikes.

then I used acceleration as it strikes.

then you say that is wrong , let me ask you the question
and you plug in whatever you require.

the 100kg mass is traveling at a speed of 40 m/s
the 100kg mass strikes the earth , what force will the mass apply to the earth?

4000N or 8000N


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Originally Posted By: paul

you demanded acceleration as it strikes.

then I used acceleration as it strikes.


You just changed the units. Doesn't mean it's the actual acceleration.


Quote:

the 100kg mass is traveling at a speed of 40 m/s
the 100kg mass strikes the earth , what force will the mass apply to the earth?

4000N or 8000N


We can't know without knowing its acceleration as it's stopping.

In the pipe case I worked out the acceleration using:

a = change in velocity / time taken
Right??

We both agreed that initial velocity is -40m/s and final velocity is +40m/s.

Change in velocity is (+40m/s) - (-40m/s) = +80m/s

Time taken is specified by you as 1s.

a = 80m/s / 1s
a = 80m/s^2


What's wrong with that?
Don't like my 'change in velocity'? How would you do it?

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Quote:
We can't know without knowing its acceleration as it's stopping.


use 1 second to stop the mass.

Quote:
We both agreed that initial velocity is -40m/s and final velocity is +40m/s.

Change in velocity is (+40m/s) - (-40m/s) = +80m/s

Time taken is specified by you as 1s.

a = 80m/s / 1s
a = 80m/s^2


What's wrong with that?
Don't like my 'change in velocity'? How would you do it?


nothing has a speed of 80 m/s

the mass has a speed of 40 m/s
the pipe has a speed of 4 m/s

you cant use 80 m/s as a speed or 80 m/s/s as a acceleration.
the distance the mass travels through the 1st turn is only 40 meters.

the turn does not stretch to 80 meters because the mass moves through it.

the mass will not increase its speed to 80 m/s in 1 second as it passes through the 1st turn.

you cannot use a change in direction as a force.





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Originally Posted By: paul
Quote:
We can't know without knowing its acceleration as it's stopping.


use 1 second to stop the mass.




OK so it's slightly different from the tube. Here the mass stops in 1s rather than reversing in 1s.

a=change in v / time
change in v = 0m/s - 40m/s
|change in v| = 40m/s

time = 1s

a = 40m/s / 1s
a = 40m/s^2

F=ma
F=100kg * 40m/s^2
F=4000N

Good enough? I got a bit sloppy with the + and - signs, so maybe it's -4000N but who cares.

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I think its amazing how we cannot seem to get a agreeement
on this , I think things have been made too confusing or too
hard to figure out using todays math.

are they teaching that momentum and a change in momentum is a force where you come from?

because you constantly use changes in direction as forces.

this is our main problem , I have never encountered momentum being used as a force so that is why I have trouble with your results.


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So you agree that a 4000N force will be placed on the earth
by the 100kg mass.

now suppose we move the earth and put the pipe there , the pipe would feel the same 4000N force?

if so then how could the mass moving through the turn apply a 8000N force to the pipe?

have you ever considered that the 4000N force is distributed evenly to the turn?



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Originally Posted By: paul

because you constantly use changes in direction as forces.

Not intentionally. Show me where I did and I'll correct my mistake.

Quote:

this is our main problem , I have never encountered momentum being used as a force so that is why I have trouble with your results.


Who said anything about momentum? We've dropped that concept.

I'm just using a couple of equations again and again for all sorts of purposes:

F=ma
a = (change in velocity) / (time taken)
(change in velocity) = (final velocity) - (initial velocity)

Well that's pretty much it really!!! Google them if you don't believe me.

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Originally Posted By: paul

now suppose we move the earth and put the pipe there , the pipe would feel the same 4000N force?

Of course.

Quote:

if so then how could the mass moving through the turn apply a 8000N force to the pipe?

Because the turning mass isn't just stopping, it's turning around. It's a whole different situation.

Quote:

have you ever considered that the 4000N force is distributed evenly to the turn?

Of course, that's been a fundamental assumption throughout. Sure it doesn't have to, but why complicate things with varying forces.

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Quote:
Who said anything about momentum? We've dropped that concept.

then how are you getting 8000N from 4000N if your not using
a change in momentum or a change in velocity to get the
8000N?


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Quote:
have you ever considered that the 4000N force is distributed evenly to the turn?

Of course, that's been a fundamental assumption throughout. Sure it doesn't have to, but why complicate things with varying forces.


so are you saying that the mass moving through the 1st turn only applies a force of 4000N to the pipe?

because that is where we dissagree , you say it has a force of 8000N applied I say it has a force of 4000N applied.

which of these do you agree on?


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Originally Posted By: paul
then how are you getting 8000N from 4000N if your not using
a change in momentum or a change in velocity to get the
8000N?


I'm using a change in velocity, not a change in momentum.

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Originally Posted By: paul

because that is where we dissagree , you say it has a force of 8000N applied I say it has a force of 4000N applied.


I say 8000N and I've shown my calculations of how I got that number.

Why do you say 4000N? Where's your calculations? Not the ones using F=mv.

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Quote:
Change in velocity is (+40m/s) - (-40m/s) = +80m/s

Time taken is specified by you as 1s.

a = 80m/s / 1s
a = 80m/s^2


your using the change in velocity above to get the 80 m/s/s
acceleration.

F=ma = 80m/s/s * 100kg = 8000N

change in velocity is not a force

and change in velocity cannot be used to accelerate the pipe.

my calculations?

F=ma

the 100kg mass applies a force of 4000N for 1 second.
since force is determined by m*a
and acceleration is the change in velocity over time

so force = d/(dt)*mv

F=40m/(40m*1s)*mv
F=40m/40 * mv
F=1*mv
F=1*100kg*40m/s
F=1*4000
F=4000N

that 4000N force is Applied to the pipe
and it is applied to the mass.

so the mass still has +4000N
and the pipe has recieved -4000N

you were getting 8000N but the only way you can get
a 8000N force is to calculate the force by using a
200kg mass , if you keep the time and the velocity the same.

F=40m/(40m*1)*mv
F=40m/40 * mv
F=1*mv
F=1*200kg*40m/s
F=1*8000
F=8000N







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Originally Posted By: paul

your using the change in velocity above to get the 80 m/s/s
acceleration.

F=ma = 80m/s/s * 100kg = 8000N

change in velocity is not a force


Of course not, if it was, I'd have written:

F = change in velocity
F = 80m/s

which is nonsense.


Change in velocity becomes a force after you divide it by the time it took and multiply it by the mass who's velocity is changing. That's Newton's 2nd law again, F=ma.

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