Re: Calculating Newtons in Joules and Joules/s


Dennis B wrote:
Randy Poe wrote:
Dennis B wrote:
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?

Let me summarize the situations I have analyzed for you
regarding mass A transferring all of its momentum to
mass B and coming to a halt:

(1) mass A < mass B. Inelastic collision. Some of A's
energy is lost and does not become KE of B.

(2) mass A = mass B. Elastic. All of A's momentum
and KE are transferred to B.

(3) mass A > mass B. Impossible. There is no situation
in which all of A's momentum is transferred to B because
that would require additional energy not present in A.

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".

See #2 above. In that one situation (equal masses) it is indeed
an elastic collision that has the properties you want. I said
that just below.

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.
[snip]
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?

KE of A: 0.5*4*2^2 = 8 Joules.
KE of B: 0.5*8*1^2 = 4 Joules

If the mass is doubled and the velocity is halved, the KE is halved.

Wouldn't it take the same energy to accelerate the 8kg mass to
a velocity of 1m/s?

No. See the expression for KE = 0.5*mv^2.

Assume that we are talking about single atoms
instead of masses comprised of multiple atoms. No energy could be
dissipated as heat.

Why not?

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.

But atom B is not moving at half the velocity.

If it is an elastic collision in which no KE is lost, then
both A and B will be moving after the collision. B won't
be moving at half the velocity.

Working out the equations, I find that atom B will be
moving at 2/3 of the speed that atom A had, and that
atom A will be moving backward at 1/3 of its original speed.

Again, you're trying to insist that you can simultaneously
have

m1*v1 = m2*v2

and

m1*v1^2 = m2*v2^2

when v1 and v2 are nonzero and m1 is not equal to m2. You
just can't. Even though you believe that both these things
SHOULD be true, no amount of belief will make it possible.

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?

When is it exerting this force? You mean at impact?

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?

That's the DEFINITION of the force of impact.

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.

Right. That's the point. C could get a velocity of 1 m/sec by
way of an acceleration of 100 m/sec^2.

Why do you say a = 1 m/sec^2?

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?

mv^2 isn't used to calculate acceleration.

The acceleration would
be dependent upon the energy and momentum of the impacting mass ("mass
A" in this case), would it not.

Yes, but it depends on other things as well.

You make it sound as if the
acceleration were variable. Based upon what?

The stiffness of the material, which will govern how long the
two bodies are in contact. That time is on the order of a msec.
That's the time during which an acceleration happens.

Consider two objects of the same momentum in an elastic
collision against a wall. One hits a steel plate, the other
a pillow. Are the forces the same?

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.

Yes. Necessarily.

The latter assertion would seem to be falsified if we
substitute atoms for the masses.

Only because you think it's possible for such a collision to
happen in which A transfers all of its momentum to C. As I
pointed out, if atom A collides elastically with atom C
of twice the mass, atom C will have 2/3 of the original
speed of A and atom A will rebound at 1/3 of its original
speed. The total momentum of the system is

2m*(2/3 of original) - m*(1/3 of original) = 4/3-1/3 = original

I should point out that atomic
collisios are always perfectly elastic.

I don't know if that's true, but in the case of perfectly elastic
collisions you will NOT get A to transfer all of its momentum
to C if A and C have different masses.

I am beginning to wonder if
there may be a logical insight in the erroneous equations I provided
earlier, wherein I asserted that:


Previously you had expressions here which were at least sometimes
true. The expressions below aren't.

Momentum = 1/2mv^2

No. mv is not 1/2mv^2.

Force = 1/2mv^2

No. ma is not 1/2mv^2.

Kinetic Energy = 1/2mv^2

Yes (for slow objects).

Work = 1/2mv^2

Not necessarily (for reasons previously explained).

There is no "logical insight" to be gained by asserting that
momentum, for instance is something that it isn't.

- Randy

.

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