momentum change

[chiddler]

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A person on a frictionless surface catches a ball.

vs

A person on a frictionless surface gets hit by a ball in a perfectly elastic collision.

The two balls are of same mass velocity and the two persons are identical. Which situation results in a greater change in momentum of the person?

My answer is the next sentence.

In #1, the new momentum is (m of ball + m of person)*v. And in #2, the new momentum is m of person *v.

Is this correct?

 

[MedPR]

A person on a frictionless surface catches a ball.

vs

A person on a frictionless surface gets hit by a ball in a perfectly elastic collision.

The two balls are of same mass velocity and the two persons are identical. Which situation results in a greater change in momentum of the person?

My answer is the next sentence.

In #1, the new momentum is (m of ball + m of person)*v. And in #2, the new momentum is m of person *v.

Is this correct?

Yes, but what is the answer to the question? In which situation does the person experience a greater change in mv?

I think the answer is situation 2. Say the person weighs 50kg, the ball weighs 5kg and the ball has initial velocity of 5m/s. Consider the ball to be thrown from right to left, and also consider this to be the negative direction.

Situation 1:

-(50+5)vf=25
vf=-0.5m/s

Situation 2:

-25=-50v1f + 5v2f, where v1f = final velocity of the man, v2f = final of the ball.
For elastic collisions, KE is conserved, so:

v1i-v2i=-(v1f-v2f)
0-(-5) = -v1f+v2f
v2f=5+v1f

-25=-50v1f + 5(5+v1f)
-25 = -50v1f + 25 + 5v1f
-50 = -45v1f
v1f = 1.11m/s

Though I'm kind of confused where the sign went. :/

 

[chiddler]

This inspired my question. 10 minutes even.

So for what I wrote, I think change in momentum will be equivalent in both cases but change in speed will be different.

I want to look at the question in the video more closely though because in THAT cause, there is no change in mass.

 

[chiddler]

 

[MedPR]

Since this is a thread asking about momentum, I'm going to assume the light somehow converts into kinetic energy and gets this thing spinning. So A or B.

Now to figure out which one. Since black emits no light, I guess that means that it is impossible to excite an electron to a higher energy state (thus no electrons can relax and emit light) so no kinetic energy is created. If you shoot light at the mirrored side, some of the electrons might get excited and by the photoelectric affect some of the energy is converted to KE. I assume by preventing out-gassing they mean that once light goes in it will undergo total internal reflection, so it will serve as a constant source of KE.. So I guess A.

 

[chiddler]

Since this is a thread asking about momentum, I'm going to assume the light somehow converts into kinetic energy and gets this thing spinning. So A or B.

Now to figure out which one. Since black emits no light, I guess that means that it is impossible to excite an electron to a higher energy state (thus no electrons can relax and emit light) so no kinetic energy is created. If you shoot light at the mirrored side, some of the electrons might get excited and by the photoelectric affect some of the energy is converted to KE. I assume by preventing out-gassing they mean that once light goes in it will undergo total internal reflection, so it will serve as a constant source of KE.. So I guess A.

A is right.

But hold on the point was to examine change in momentum. I'm not sure if your reasoning is valid and bringing photoelectric effect into this I think is irrelevant. This has nothing to do with voltage and is only considering momentum.

The question is simply this:

Which causes a greater change in velocity, an absorbed photon on a sheet of metal or a reflected photon on a sheet of metal?

The answer is a reflected causes a greater change in velocity. My question is why? The analogy I was trying to make falls short because photons do not carry mass.

 

[MedPR]

A is right.

But hold on the point was to examine change in momentum. I'm not sure if your reasoning is valid and bringing photoelectric effect into this I think is irrelevant. This has nothing to do with voltage and is only considering momentum.

The question is simply this:

Which causes a greater change in velocity, an absorbed photon on a sheet of metal or a reflected photon on a sheet of metal?

The answer is a reflected causes a greater change in velocity. My question is why? The analogy I was trying to make falls short because photons do not carry mass.

And you're positive the answer has nothing to do with the photoelectric effect? I have no clue then. Whenever I see a photon converting into KE, I think photoelectric. I know nothing else that shares that energy conversion relationship :/

 

[chiddler]

🙁

 

[milski]

Photons don't have a mass but still have a momentum. You cannot use the usual p=mv since it leads to certain contradcitions. The momentum can be derived using the theory of relativity and is E=pc where E is the energy of the photon and c the speed of light.

With that said, the answer should be easier - more photons reflected means more momentum in the direction away from the collision means move momentum for the reflecting surface.

 

[chiddler]

Photons don't have a mass but still have a momentum. You cannot use the usual p=mv since it leads to certain contradcitions. The momentum can be derived using the theory of relativity and is E=pc where E is the energy of the photon and c the speed of light.

With that said, the answer should be easier - more photons reflected means more momentum in the direction away from the collision means move momentum for the reflecting surface.

so p = E/c. But how does this translate into what you wrote last?

 

[milski]

so p = E/c. But how does this translate into what you wrote last?

It shows you that you can treat the photon as a regular particle with some mass for the purposes of conservation of momentum.

Before the collision you had a photon with momentum p and a mirror with p1, total momentum p+p1.

After the collision with a mirrored surface, the photon has a momentum -p which means that the mirror has to have momentum 2p+p1.

After the collision with the black surface, the photon is gone/has no momentum, so the surface has a momentum of only p+p1.

 

[chiddler]

It shows you that you can treat the photon as a regular particle with some mass for the purposes of conservation of momentum.

Before the collision you had a photon with momentum p and a mirror with p1, total momentum p+p1.

After the collision with a mirrored surface, the photon has a momentum -p which means that the mirror has to have momentum 2p+p1.

After the collision with the black surface, the photon is gone/has no momentum, so the surface has a momentum of only p+p1.

oh. very cool. thanks for your help.

wait the implication is that absorption of photons means that the photons have no momentum?

 

[milski]

oh. very cool. thanks for your help.

wait the implication is that absorption of photons means that the photons have no momentum?

Uh, no? After the photon is absorbed all of its momentum is transfered to whatever absorbed it, thus p+p1 for the black surface case.

 

[chiddler]

Uh, no? After the photon is absorbed all of its momentum is transfered to whatever absorbed it, thus p+p1 for the black surface case.

But you wrote that the before and after are equal.

 

[kasho11]

If the first collision is considered inelastic shouldn't momentum be conserved in both cases? So no overall change in momentum?

 

[milski]

But you wrote that the before and after are equal.

The total momentum of the system (photon + surface) is constant.

-----------|photon | surface | system
mirror | p | p1 | p+p1
mirror' | -p | p1+2p | p+p1
black | p | p1 | p+p1
black' | - | p1+p | p+p1

mirror' and black' are after the collision.

 

[chiddler]

The total momentum of the system (photon + surface) is constant.

-----------|photon | surface | system
mirror | p | p1 | p+p1
mirror' | -p | p1+2p | p+p1
black | p | p1 | p+p1
black' | - | p1+p | p+p1

mirror' and black' are after the collision.

Oh! The black absorbs 1x momentum and the reflected absorbs 2x since it shoots the photon back.

Okie dokie thanks.

 

[chiddler]

If the first collision is considered inelastic shouldn't momentum be conserved in both cases? So no overall change in momentum?

I'm not following. Momentum is conserved in both cases, but why would it then follow that no overall change in momentum is thus necessary.

 

[kasho11]

I'm not following. Momentum is conserved in both cases, but why would it then follow that no overall change in momentum is thus necessary.

That is the exact definition of conservation of momentum (no change in momentum)

 

[chiddler]

That is the exact definition of conservation of momentum (no change in momentum)

Oh, duh.

Some of the momentum is transferred to the plates and so the plates move while the photons bounce off with a negative momentum having transferred it to said plates.

 

[pfaction]

Wait, can we go back to the original question the way medPR stated it?

A person on a frictionless surface catches a ball.

vs

A person on a frictionless surface gets hit by a ball in a perfectly elastic collision.

The two balls are of same mass velocity and the two persons are identical. Which situation results in a greater change in momentum of the person?


Aren't these two the same thing? Catching and an elastic collision, both result in the m1+m2 being the same velocity at the end?? momentum is always conserved!

 

[chiddler]

It was my question. Momentum is the same in both cases, but when the person catches the ball, then their speed is lower despite the same momentum as the person who was hit by the ball. This is because the person has more mass attached to him since he caught the ball.

 

[pfaction]

I see. I'm gonna put some numbers to this the same way medPr did.

I got 0.5 using his numbers for the first scenario.
But for the second one I'm having trouble. If KE is conserved in the second scenario I'm getting:
5*5sq = 50V1sq + 5V2sq
125 = 50V1sq+5V2sq
Not sure how to go from here?

 

[chiddler]

I see. I'm gonna put some numbers to this the same way medPr did.

I got 0.5 using his numbers for the first scenario.
But for the second one I'm having trouble. If KE is conserved in the second scenario I'm getting:
5*5sq = 50V1sq + 5V2sq
125 = 50V1sq+5V2sq
Not sure how to go from here?

Here, this helped me a lot and I hope it'll help you too.

Specifically