Re: Calculating Newtons in Joules and Joules/s



Randy Poe wrote:
Dennis B wrote:
Randy Poe wrote:
Dennis B wrote:
Randy Poe wrote:
Dennis B wrote:
Since the 2kg mass and the 4kg mass have the same momentum, wouldn't
both of the 3 kg masses have the same momentum as well?

Momentum and energy are separately conserved. Due to
those requirements, your hypothetical momentum transfers
can't both happen.

If the 3 kg mass were to get all of the momentum of the
2 kg mass, it would be moving at 2/3 of the velocity and
would therefore have (3/2)*(4/9) = 2/3 of the KE that the
2 kg mass originally had. That is possible in the right
inelastic collision, with the remaining 1/3 being lost to
heat.

What does the (4/9) represent?

(2/3)^2.

KE1/KE2 = (m1/m2)*(v1/v2)^2 = (3/2)*(2/3)^2

I found it more convenient to analyze this problem in terms
of ratios rather than work out the actual velocities.


I will have to study this in detail.



If the 3 kg mass were to get all of the momentum of the
4 kg mass, it would be moving at 4/3 of the velocity and
would have (3/4)*(16/9) = 4/3 of the KE that the 4 kg mass
originally had. That is impossible.

What does the (16/9) represent?

(4/3)^2



It is not possible for a 4 kg mass to transfer all of its momentum
to a 3 kg mass in a collision.

- Randy

Make it two 4 kilogram masses instead of two 3 kg masses.

Originally you had a 2 kg and 4 kg mass with the same
momentum, each of which is striking a 3 kg mass. There
were no "two 3 kg masses". So what is the situation now?
What are the moving masses and what mass are they
striking?

And make the
collision totally inelastic, so that all of the momentum is
transferred.

"Totally inelastic" means that the objects stick together
after the collision, both moving together as one mass.
Not that one comes to a halt and the other gets all
the momentum.


Yes. I realise. That is why I corrected myself in a later message, and
said that the collision should perhaps be a perfectly elastic collision

In a perfectly elastic collision, kinetic energy is conserved. No
energy is lost to friction or deformation.

wherein the momentum of the two masses is completely transferred.

No, that is not necessarily what happens in an elastic collision. In
general, both masses have momentum after the collision.

In
other words, one mass continues on where the other left off. This would
be a form of perfectly elastic collision, would it not?

No.

Somehow I sense
you are going to try to tell me that such a collision is not possible

It is not possible if it violates conservation of energy. I showed
you it *is* possible for a smaller mass to transfer all of its
momentum to a larger mass, but not vice versa. The first
requires some KE to be lost. The second requires some
KE to be created.

(although I know that it is).

Yes, you "know" that it is possible to violate energy conservation.


Yet, what of the "executive's toy"made of hnging balls? When the
swinging ball strikes the others, it tops and the momentum is
transferred to the outermost ball on the opposite side. Is that not an
elastic collision?

But the universe thinks otherwise.

Your expectations, based on your intuitions, continue to
be at odds with the laws of physics and experimental
results. Yet when told such things, you continue to insist
that your beliefs must be right and the universe is wrong.


Perhaps I do not understand what a transfer of momentum is...or even
what it means to say that the momentum of two differing masses is the
same. What precisely do such descriptions mean. In other words, what
physial property is considered to be the same (based upon the measured
effect).

For purposes of this discussion, mv suffices as a definition
of momentum. So the physical property conserved is the
vector sum of mv over all the masses.

I would presume that if a 4 kg mass travelling at 2 m/s (mass
A) collided with another 4 kg mass having identical properties (mass B)
in a perfectly elastic collision of the sort

You shouldn't use the term "elastic collision" unless you mean
what we mean by it. You want another type of collision, use
a different term.


What would you call the collision observed in the "executive's toy".

However, in the case of an elastic collision of equal masses,
it will be of the type you want. It's just with unequal masses
where your intuition is leading you astray.

I described above (wherein
one mass continues where the other left off), mass B would have the
same velocity as mass A. Correct?

Yes. m*v before the collision = m*v after. Since the masses are
equal, so is the velocity.

If mass A collided with an 8 kg mass
(mass C) having identical material properties, I would expect mass C to
have half the velocity as mass A (1 m/s).

Again, yes. And it would have 1/2 the kinetic energy as well.
So this would be an inelastic collision. The other half is
lost in heat or deformation somewhere.

This does not make sense to me. Why would there be a reduction of
energy? Wouldn't it take the same energy to accelerate the 8kg mass to
a velocity of 1m/s? Assume that we are talking about single atoms
instead of masses comprised of multiple atoms. No energy could be
dissipated as heat. If an atom collided with an atom with twice the
mass (atom B), atom B should have the same energy as atom A although
atom B would be moing at half the velocity.


Consequently, I would expect
the force exerted by mass A to be the same for both mass B and mass C,

This I can't comment on until you tell me what force you mean.

The force exerted on what? Do you mean the force of impact?


The force exerted by mass A upon mass B or C. What other force could
there be? The force of acceleration as mass B or C accelerate as a
result of the impact? Would that not be a measure of the force of
impact? After all, we measure force by it's effects, such as the
acceleration of a mass.

since:

ma = F

Then:

8kg * 1m/s^2 = 8N

What makes you think a = 1 m/s^2? It could be 100 m/s^2.

a is the rate of change of v. If mass C accelerates from 0 to
1 m/sec in 0.01 sec, the rate of change is delta-v/delta-t
= 1/0.01 = 100 m/sec^2.


100 m/sec^2 is NOT the same as an acceleraton of 1m/s^2. Perhaps I
confused with the anerror in my math. Note that the force calculations
were erroneous. I was multiplying mass time velocity (mv) and not mv^2.
Perhaps I should reserve my USENET activities until I have fully
awakened. I've noticed that I am prone to such errors until the evening
hours (presuming I keep my sleep scedule on track).

The acceleration certainly won't be 1 m/s^2, as the time
of acceleration is certainly much shorter than 1 second.


I see what you are saying. Yet, how could mass A induce an acceleration
greater than what would be calculated from mv^2? The acceleration would
be dependent upon the energy and momentum of the impacting mass ("mass
A" in this case), would it not. You make it sound as if the
acceleration were variable. Based upon what?

Note by the way that we had to make two different assumptions
in these two collisions. For A to transfer its momentum to B,
we have a perfectly elastic collision. For A to transfer its
momentum to C, we have an inelastic collision where half of
A's energy is going somewhere else other than C.

Not necessarily. The latter assertion would seem to be falsified if we
substitute atoms for the masses. I should point out that atomic
collisios are always perfectly elastic. I am beginning to wonder if
there may be a logical insight in the erroneous equations I provided
earlier, wherein I asserted that:

Momentum = 1/2mv^2
Force = 1/2mv^2
Kinetic Energy = 1/2mv^2
Work = 1/2mv^2

The reason for such an assertion being that then and only then do the
numbers stay the same regardless of the mass in the examples I
provided. I believe that my (perhaps erroneous) intuition is based upon
the expectation that momentum and force are ultimately measures of
energy (which seems to be shared by others). Therefore, to say that
momentum is the same when energy is not conflicts with my (possibly
erroneous) idea of momentum. I would expect, as explained previously,
that if the momentum were the same even when mass and velocity are not,
that the moving masses would exert the same effects. Yet...this is
obviously not the case using the standard equations.

As a more realistic solution, I ask, would it be possible for physics
to function properly using only a measure of energy? I ask becuse then
when we say that the two differing masses moving at different
velocities have the same value (of energy), they truly will exert the
same effects (thus, clearly having the same value). For example, a 1kg
mass travelling at 2 m/s (having a kinetic energy of 2J) will exert the
same effects as a 2 kg mass travelling at 1 m/s (which also has a
kinetic energy of 2J). This not only makes physics more intuitive, it
makes it simpler and easier to learn. Alas, what is the point of
specififying that momentum is the same if it does not exert the same
effects? To point out that momentum is the same when energy is not
seems to me an observation that a numerologist (or fortune teller)
would make in response to a synchronicity of numbers.

-Dennis B

.

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