# Re Calculating Newtons in Joules and Jouless by TaylorRandle

VIEWS: 0 PAGES: 5

• pg 1
```									                           Re: Calculating Newtons in Joules and Joules/s

Re: Calculating Newtons in Joules and Joules/s

Source: http://sci.tech−archive.net/Archive/sci.physics/2006−07/msg00705.html

• From: "Dennis B" <Utopian@xxxxxxxx>
• Date: 6 Jul 2006 22:50:12 −0700

Dennis B wrote:

Randy Poe wrote:

Dennis B wrote:

PD wrote:

Dennis B wrote:

Randy Poe wrote:

Dennis B
wrote:

You
are
blind
if
you
cannot
see
that
force
is
energy.

Saying it
over and
over won't
make it so.

Force
can
be

Re: Calculating Newtons in Joules and Joules/s                                  1
Re: Calculating Newtons in Joules and Joules/s
measured
in
J/m.

Yes, and
energy
can't.

The
distinctions
between
momentum,
energy,
work,
and
force
are
that
they
are
different
measurements
of
one
thing:
Energy.

Incorrect.

For
everything
is
energy.
You
cannot
argue
with
that.

Yes I can.

− Randy

I don't have the time or
energy at the present

Re: Calculating Newtons in Joules and Joules/s                            2
Re: Calculating Newtons in Joules and Joules/s
moment to respond to all
of the messages in detail.
Not that I don't intend to. In
the interim,
perhaps following evidence
will prove to you that
momentum, force,
work, and energy are all
unified:

Momentum = mv
Force = (delta)mv
(Kinetic) Energy = 1/2mv^2
work = 1/2mv^2

In other words, momenum,
force, kinetic energy, and
work are all
measurements of mv...

Perhaps now you will begin
to understand my
perspective?

A couple of comments:

These formulas you have here are NOT
universal definitions, despite
what you may have read. They work
sometimes, under some circumstances,
for some kinds of things.

Just because m and v appear in all of them
does NOT mean that they are
the same thing. For example, if momentum
and energy were the same
thing, then if the momentum for an object A
happened to be the same as
the momentum for object B, then you would
expect the energies to be the
same, too. But they're not. Let's take an
example:
Object A: m = 2 kg, v = 4 m/s
Object B: m = 4 kg, v = 2 m/s
Here momentum of A = (2 kg)(4 m/s) = 8
kg*m/s
And momentum of B = (4 kg)(2 m/s) = 8
kg*m/s. They have the same
momentum. So if momentum and energy are
the same thing, then if these

Re: Calculating Newtons in Joules and Joules/s                            3
Re: Calculating Newtons in Joules and Joules/s
two momenta for A and B are equal, then so
should be the energies.
Here energy of A = 1/2(2 kg)(4 m/s)^2 = 16
J
But energy of B = 1/2(4 kg)(2 m/s)^2 = 8 J.
Not the same.
So the momenta of A and B are equal, but
the energies of A and B are
NOT equal.

−Dennis

I have a question for you: If the 2kg and 4kg masses have the
same
momentum, what would happen if they each collided with a
3 kg mass
(each having identical properties) and transferred their
momentum?

In collisions, typically one mass doesn't "transfer it's momentum"
to another. Most of the time both masses end up with some
momentum and KE after the collision.

The amount of momentum and KE transferred will be
different for the 2 kg and 4 kg masses if you assume
perfectly elastic or perfectly inelastic collisions.

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?

Re: Calculating Newtons in Joules and Joules/s                                        4
Re: Calculating Newtons in Joules and Joules/s

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?

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. And make the
collision totally inelastic, so that all of the momentum is
transferred. Then what would the results of a transfer of momentum be?
I don't see why the velocities would be any different. Although the 2kg
and 4kg masses differ, the 2kg mass travels at a greater velocity to
make up for the difference so that the momentum of the two is the same.
Therefore, I would expect that the 2kg mass should have just as much of
an impact as the 4 kg mass.

−Dennis

Perhaps I should ask what the results of a perfectly elastic collision
would be as opposed to an inelastic collision. The reason being that
while studying momentum I was astonished to find that in an inelastic
collision, the colliding masses supposedly stick together, presumably
such that both then travel together. Yet, I ask...is it not possible
for the mass at rest to absorb all of the energy of the colliding mass
and accelerate to the same velocity of the colliding mass which, having
given up all of it's kinetic energy, then effectively absorbs the zero
velocity "momentum" of the mass at rest such that it stays put while
the other mass carries on where the other left off. That would be a
true transfer of momentum. An inelastic collision wherein both masses
continue onward together after the collision does NOT represent a
complete transfer of momentum in my opinion, since the colliding mass
retains some of it's momentum.

−Dennis

.

Re: Calculating Newtons in Joules and Joules/s                                       5

```
To top