# Does momentum "get transferred" during collisions?

#### Pmb

PHF Hall of Fame
Momentum and energy exist in the same way that the number three exists. We just make physics easier to speak of we don't explain everything in exact way.

For example: "height" isn't a thing which I can put into a box, throw it on the grill with some barbeque sauce over it, etc. Yet I can measure my height easily. I can estimate the height of he highest point on a house.

You can think of energy and momentum the same way that you think of money. If you have \$7,874 in cash and you put it into your bank account, is that money still yours? Where is it?

When a capacitor is said to be charged and has a certain amount of energy "in it" what really happened is that energy was required to do work to assemble the charges in the capacitor. There is a one to one relationship between the charge in the cap and the energy required to charge it. So once one is given the other is known. And the amount of energy required to charge the cap is the exact amount of energy "stored" in the cap.

Someone's gonna dump on me now because I said stored, right?

#### studiot

PMB
Someone's gonna dump on me now because I said stored, right?
No one is gonna dump on you because we have all read (and agree with) your reference which states that the theory of energy as a movable fluid is wrong but very convenient.

But no one seems to want to discuss the nitty gritty of energy and momentum either.

I posted this in post 4 but no one has been prepared to try their hand at it.

Consider the following:

A rifle bullet of mass 0.1kg is fired at block of wood, of mass 1kg and situated on a long frictionless surface. The bullet impacts at 440m/s.
The impact embeds the bullet in the block and the two speed off at what velocity?

What is the kinetic energy before and after the impact?
So we have Momentum before = M1V1 + M2V2 = (0.1)(440) + (1)(0) = 44 kgm/s

Momentum after = M3V3 = 44 kgm/s by conservation of momentum.

Thus V3 = 40m/s

Kinetic energy before = (0.5)(0.1)(440)^2 + (0.5)(1(0) = 9680J

Kinetic energy after = (0.5)(1.1)(40)^2 = 880J

So kinetic energy is not conserved ? What is happening?

Alternatively

If energy is conserved

Kinetic energy after = Kinetic energy before = 9680J

Then velocity after = sqrt ((2)(9680)/1.1) = 132m/s

Substantially different from the momentum conservation result.

This simple example shows why the subject has hidden depths I will leave all you wise heads to discuss.

#### Pmb

PHF Hall of Fame
In my case I don't mind discussing it but I'm burned out a bit on physics. Its taking all of my mental capacity to do a solid and thorough review of QM. After that I'm wiped out for the most part.

#### Pmb

PHF Hall of Fame
If energy does not exist and is a bookkeeping concept...why is it accurate for a physicist to describe energy as being absorbed?

....thats what my Oxford textbook on IB physics does.
In a sense it is. Its like money again. When a wire transfer takes place between banks one bank looses money while the other gains or "absorbs" it even though its not a physical thing. In quantum mechanics there's a quantity called a probability current. One region of space has a decrease in probability the particle being detected there while the probability increases elsewhere.

Its the way language is constructed.

As far as absorbing energy please show me where you read that. Something can absorb energy in the forum of thermal energy or EM radiation but its eak motiioon of atoms being changed.

Did you read taht webpage I wrote on all of this?

As you study physics you are absorbing knowledge!

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