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54

Conservation laws don't "propagate". They are inevitable consequences of symmetries of the dynamics by Noether's theorem, and the dynamics propagate with whatever finite speed they do.


28

See https://en.wikipedia.org/wiki/Continuity_equation - I was just writing some text there that I think helps explain. Continuity equations are a stronger, local form of conservation laws. For example, the law of conservation of energy states that energy can neither be created nor destroyed—i.e., the total amount of energy is fixed. But this ...


15

Conservation laws do not propagate instantaneously (or really at all), it really means that some property of the system does not change over time. In the case of momentum conservation, the information about the collision propagates (at most) at the speed of sound in the medium. This is why you see the cars continuing along their path during a collision: ...


5

The Earth's tilt doesn't change so much as it's position around the sun changes. Notice the North Pole sees more sun in summer than in winter cause it's tilted towards the sun in summer but not in winter. The north pole always points in the same direction into space pretty much. It wobbles slightly, and quite slowly, completing a full wobble every ...


5

The creation of particles has to respect a number of conservation laws. For example charge has to be conserved, so it you create an electron with a charge of -1 you have to create a particle with a charge of +1 to balance it out. Likewise lepton number is conserved (in the Standard Model at least). An electron has a lepton number of +1, so if you create an ...


4

You've got it a little backwards - physicists first defined the quantity $m \cdot v$ because it quantified the amount of "motion" an object possessed. They named it "momentum". Modern physics is primarily concerned with the quantity $m \cdot v$ (and the updated versions of that quantity in more recent frameworks of physics) because it is conserved. This ...


4

The answer is basically, angular momentum. The collapsing proto-solar nebula has some angular momentum. Whilst dissipative processes can allow the nebula to collapse along the axis of rotation, there is still the problem of how to shed angular momentum in order to allow gas/dust to orbit closer to the rotation axis. This is just a basic application of ...


4

It does follow from the conservation of momentum. Consider the diagram (from Wikipedia) of a rocket expelling gas of mass $\Delta m$: At $t=0$, the initial momentum is $$ p(t=0)=\left(m+\Delta m\right)V\tag{1} $$ but at $t=\Delta t$, we've lost some mass and gained some velocity, $$ p(t=\Delta t)=m\left(V+\Delta V\right)+\Delta m \left(V-v_e\right) ...


3

Let's look at an example, electromagnetism. The electromagnetic field (the combination of electric and magnetic fields) has momentum. Charges have momentum. The charge feels a force right where it is, a force based on the fields right where it is. This changes the momentum of the particle. The field loses an equal and opposite amount of momentum and also ...


3

Well, since unburned fuel is not distinguishable from dry mass, we could replace the above equation with $$ \Delta v = v_{ex}\ln(\frac{m_i}{m_i - m_b}) - gt $$ where $m_b$ is the mass of the burned fuel. Assuming constant burn rate $r_b$, it would just be $$ \Delta v = v_{ex}\ln(\frac{m_i}{m_i - r_bt}) - gt $$


3

If your fan-boat is in vacuum, they won't move. In the air, they will. Your assumption conflicts with your intuition is because you isolated the system from the air, which should not.


2

There is no doubt that such a system would move, as other people here say, moreover, such a system can be quite practical and is actually used in marshy/shallow water (https://en.wikipedia.org/wiki/Airboat )


2

No, This device would just oscillate (vibrate) When the first two electromagnets switched on the metal ball (which is presumably made out of a ferromagnet) would accelerate towards them but the craft would also accelerate towards the ball (At a lower acceleration as it has more mass). It would then switch to the next electromagnet pair and do the same ...


2

General Question: Why should I use just the friction force rather than the net force to integrate over distance when conserving energy? Answer: In energy conservation problems each way of storing energy generally gets it's own term. In the example problem there is a gravitational potential energy term (GPE), a kinetic energy (KE) term, and a friction ...


2

I was not able to convince him that this propulsion drive cannot work due to conservation of momentum. Am I wrong about that? No, you are not wrong. It's clear that the engine cannot work because of momentum conservation. It's basically just a fixed double pendulum. Why should there be any positive momentum in any direction after one full circle? ...


2

Energy is dissipated in the form of "internal energy", which means that all of the objects kinectic energy is transfered to internal movement of atoms and mollecules of both the object and the surface. When there is a large deformation and no restitution you can argue that some of the energy is stored in some kind of ellastic energy of the mollecular bonds ...


2

Suppose someone suggests that following a perfectly elastic collision, two billiard balls are each traveling twice as fast as they were before (and opposite to their original directions). You can't prove him wrong using conservation of momentum, but you can prove him wrong using conservation of energy. Therefore conservation of energy has implications that ...


2

Notice that you have implicitly chosen to measure angular momentum about the axle of the platform. That means that all the forces exerted by the axle on the platform are applied through the axis for rotation, meaning the torque they exert is $$\text{force} \times \text{lever arm} = F \times 0 = 0\,.$$ And there are no other forces present expect those ...


1

Given the persistence of this kind of thing, maybe determining the misconceptions underlying it are not so easy to determine and resolve. [Given that we're talking about physics students, I guess it's fair to them to sort this out rather than ignore as one might otherwise.] For my money, I'd ask whether within each part of the mechanism Newton's Third Law ...


1

Others have explained that they don't propagate. That's a real puzzler though, and the real question is how does the stuff (energy, charge, whatever) always add up when a transfer takes finite time? The answer to figuring that out was a profound pillar of modern physics: the field contains momentum (etc.). So while an electron is handshaking with another ...


1

It is not enough for it to be moving - it needs to accelerate (or decelerate). An accelerating charged particle will emit radiation and it will lose energy as a result. An excellent example would be the loss of energy of charged particle in synchrotron accelerators. They emit... synchrotron radiation. This is either a boon (e.g. the Diamond light facility ...


1

What is missing: this must be recognized as a relativistic process and be treated accordingly. Why is it a relativistic process: energy transforms into mass or conversely. This is something that cannot be accounted for using only classical mechanics. In older times I would've added "because it needs to account also for relativistic mass", but since using ...


1

In addition to the good answers above, I'd like to add a point about the nuclear reactor losing mass. It's true, it does, albeit at a really tiny rate. The reason that this doesn't contribute to propulsion is that the mass is loss in all directions so the net effect is zero.


1

A helpful yet elementary answer may do the trick, If you are familiar with the Euler-Lagrange equation then it will be straight forward and you can skip ahead a little. If not then you have to accept that there is an equation in physics that generalises classical mechanics called the Euler-Lagrange equation. For a particle moving in one dimension under a ...


1

If you count a $\pi^0$ as a particle, you can create and destroy them individually. This is similar for any particle that is its own antiparticle (like the photon, the $Z^0$ and the Higgs.) You can have an arbitrary number of these.


1

There exist particles which are their own antiparticle, even at the elementary particle level. Photons, and gluons are their own antiparticle and thus no need to come in pairs.


1

Let $t$ be the time from the dropping the fist ball until the collision of the balls. Then, $v_1=gt$ and $v_2=u-gt$. Moreover, $d=ut-\frac{1}{2}gt^2$ and $h-d=\frac{1}{2}gt^2$, so that $ut=d+(h-d)$, which gives $t=\frac{h}{u}$. Because $mv_1=m_1v_2$, we must have $m\left(\frac{gh}{u}\right)=m_1\left(u-\frac{gh}{u}\right)$, or ...


1

Let the balls meet at the height $d$ from the ground then at the time of collision, the velocity of the ball (mass $m$) is given by third equation of motion as follows $$v^2=(0)^2+2g(h-d)$$ $$v^2=2g(h-d)\tag 1$$ & the time taken by ball (mass $m$) to reach at the point of collision is given by second equation of motion as follows ...



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