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9

Photons don't directly interact with each other, but if one photon pair produced an e+/e- then the second photon could interact with that pair. The interaction has to conserve the energy of the two photons and conserve their momentum as well of course. But yes they could (and most probably depending on their energy) just pass right "through" each other.


5

There is no such thing as classical motion of an electron in an atom. The quantum states electrons in an atom are in are atomic orbitals, which possess a definite energy, but not a definite position. The Bohr model of the electron, in which electrons are thought of as classical particles orbiting the nucleus, is false. The question whether or not two ...


4

Usually absolutely nothing. Electromagnetism is linear, which means that the result of doing something with two photons is the superposition of the results of doing something with each one individually. By that reasoning, since one photon by itself just goes on its own merry way, then two photons, even if they go near each other, just go along on their ...


2

To leading order, nothing happens in any photon "collision". To higher order there are light-light interactions that involve particle loops, but they don't (can't) depend on the geometry because we can always boost to a frame in which the pair has zero net momentum (even though you can't boost to the frame where a single photon has zero momentum).


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

Andromeda and the Milky Way belong to a group of galaxies called the Local Group. The two galaxies are the largest galaxies in the group, so to a pretty good approximation their interaction can be treated as a two body problem, with the other galaxies in the group producing only minor perturbations to their motion. So as you suspected, it isn't the case ...


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

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 ...


1

You're looking at the net force acting on him at the instant he begins falling. And you are correct that, in the absence of air resistance, the net force that is acting on the man the instant after he begins falling in the same, given by $F_g=mg$. However, there's an old saying that says, "it's not the fall that kills you; it's the sudden stop at the end." ...


1

When you jump from any given height, the force pulling you down is gravity with $F=mg$. This makes you accelerate to faster speeds as you fall farther, obviously. When you hit the ground, you do not experience the same acceleration. Otherwise, it would take just as long to stop falling as it took to get up to that speed. Hitting the ground imparts a much ...



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