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109

Things are not empty space. Our classical intuition fails at the quantum level. Matter does not pass through other matter mainly due to the Pauli exclusion principle and due to the electromagnetic repulsion of the electrons. The closer you bring two atoms, i.e. the more the areas of non-zero expectation for their electrons overlap, the stronger will the ...


47

Amazingly this actually happened to a Russian scientist called Anatoli Bugorski (WARNING: this is pretty gruesome). The beam basically just killed all the tissue it passed through. The symptoms were the relatively mundane ones expected from tissue death. The LHC has a much, much greater energy than the one that struck Bugorski, so it would cause a lot more ...


33

A charged particle will create charge separation (ionization) along its path. This will cause harmful chemical reactions to occur in the body, including DNA damage. The effects of these chemical reactions depend on their amount. The body can heal from a low amount on its own, while a high amount will cause radiation sickness and probably death. This can be ...


29

Look at it this way: Suppose you are in a train travelling at 10 m/s. Somebody inside the train throws a ball at you in the opposite direction at 10 m/s. You feel the pain belonging to your first experiment. However, somebody looking at this experiment from outside the train would say that the ball is standing still and you are travelling towards the ball ...


28

This is actually a really good question. (And I'm not one of these people who insists that there's no such thing as a dumb question; I just think we shouldn't be embarrassed to ask dumb questions. Anyway, this isn't a dumb question.) As you may know, collisions between two protons (like those the LHC usually does) can produce many different types of ...


22

I don't think any of the other answers have made the following point clear enough, so I am going to give it a try. Both scenarios are very similar before the collision, but they differ greatly afterwards... From a stationary reference, you see the cars driving towards each other at 50mph, but of course if you choose a reference frame moving with the first ...


21

While everyone agrees that jumping in a falling elevator doesn't help much, I think it is very instructive to do the calculation. General Remarks The general nature of the problem is the following: while jumping, the human injects muscle energy into the system. Of course, the human doesn't want to gain even more energy himself, instead he hopes to transfer ...


18

Contrary to what is stated in many textbooks, energy-momentum conservation alone cannot explain the behavior of Newton’s cradle. For N balls we have two equations and N final velocities to calculate. Hence, conservations laws can do the job only for N=2. This means that if we want to give an explanation of the cradle behavior based on conservation laws, we ...


18

First of all -- it wouldn't be called "the Large Hadron Collider", right? Looks like one would rather call it something like "Large Electron-Positron Collider". In that case one definitely would need another abbreviation for it. Something like "LEP" instead of "LHC"... Now, guess what was there in the same tunnel before? Edit: since my shenanigan got ...


16

You should be able to use energy conservation to write down the velocities of the bodies as a function of time. $$ \textrm{Energy conservation (KE = PE): } \frac{p^2}{2}\left( \frac{1}{m} + \frac{1}{M} \right) = GMm\left(\frac{1}{r} - \frac{1}{r_0}\right) $$ And $$ \frac{dr}{dt} = -(v + V) = -p\left( \frac{1}{m} + \frac{1}{M} \right) $$ Momentum ...


13

First, you state a few things that aren't quite right in your question. While the view that's generally talked about is that Phobos and Deimos are likely captured asteroids, dynamically it's a pretty difficult problem (you generally need a third (in this case fourth?) body to take away the extra energy, and it's hard to get a circular orbit around the ...


13

Yes, unfortunately. Because of the equivalence of inertial reference frames, the the physical laws are the same in both reference frames. However, another possibility, which is non abelian, is that instead of feeling the same amount of pain, you could be feeling the opposite amount of pleasure. It depends if pain (X) are fermions or bosons, that is, if ...


13

Nothing happens obviously, when one high energy particle penetrates flesh as cosmic rays continuously impinge on us and some have the energies of the LHC. The cosmic rays reaching us are mainly muons and the damage they do is with electromagnetic scatters/ionisations in their path. The mean energy of muons reaching sea level is about 4 GeV. Muons, being ...


12

It conserves both energy and momentum in the collision at the same time. By design, when the balls collide the strings that hold them up are vertical (assuming balls are only swung from one side). This means there are no horizontal forces from the string on the balls so linear momentum in the direction of swing must be conserved in the collision. Energy is ...


12

There are two points in answering this question: Design: The design of the collider would have to be different. Electrons/positrons in a cyclotron radiate synchrotron radiation when they are accelerated (which itself is a useful device). To get above a few GeV, researchers use linear accelerators, such as SLAC. The proposed International Linear Collider is ...


11

As an addition to already posted answers and while realising that experiments on Mythbusters don't really have the required rigour of physics experiments, the Mythbusters have tested this theory and concluded that: The jumping power of a human being cannot cancel out the falling velocity of the elevator. The best speculative advice from an elevator ...


11

The anti-particle corresponding to a neutron is an anti neutron! The neutron is made up of one up quark and two down quarks. The anti-neutron is made up of an anti-up quark and two anti-down quarks. Both have zero charge because the charges of the quarks within them balance out. You are correct that elementary particles with no charge are often their own ...


11

These collisions don't produce significant amount of light in the visible range, so the easy answer is "no". They also take place in a vacuum, inside a beampipe which is itself buried in a detector apparatus that is ten meters plus on a side and packed full of stuff with no room for a human. That said, there are several ways in which a high energy ...


10

Perhaps the simplest and most intuitive approach is to regularize the hard wall potential $$V_0(x)~=~\left\{ \begin{array}{rcl} 0 &\text{for}& x<0 \cr\cr \infty &\text{for}& x>0\end{array}\right. $$ as $$ \lim_{\varepsilon \to 0^+} V_{\varepsilon}(x) ~=~V_0(x).$$ For instance, one could choose the regularized potential as $$ ...


10

To figure this out, you need to know about momentum ($p$). That's a combination of how fast something is moving ($v$, for velocity) and how much it weighs ($m$, for mass). You'll also need to understand algebra, which is just using a letter to mean some number you don't know yet. $$ p = m\cdot v $$ Momentum is conserved, which means the momentum from both ...


10

The reason that jumping can make a relatively large difference is that the kinetic energy is proportional to the square of the velocity. Thus relatively small changes to the velocity can result in relatively large changes to the kinetic energy. In addition, the velocity which a human can achieve in jumping is a substantial percentage of the velocity of fatal ...


10

What acts here is called impulse Suppose balls A, B are made of stainless steel and (m = 0.1 Kg r = 0.03 m) A has v = 4 m/s, B is at rest (v = 0, p = 0) they collide in 1-D. Since the mass is the same, A will stop dead and give B all its velocity and momentum p = (v = 4 * m = 0.1 Kg) = 0.4 Kg m/s: $$J = [F . t] = \Delta p = 0.1*4 -[0] = 0.4 Kg*m/s$$ If you ...


9

Yes, when you fire a pistol the hammer hits the bullet with a relatively small initial kinetic energy but the kinetic energy of the hammer and bullet after the collision is considerably higher. This may seem a silly example, but I think it actually highlights the important principle involved. In general when two bodies undergo an inelastic collision part of ...


8

If it were possible for one object to pass through another object, then it would be possible for one part of an object to pass through a different part of the same object. Therefore the question asked here is equivalent to the question of why matter is stable. See this question on mathoverflow. That question was more about the stability of individual atoms, ...


8

Without having heard this argument before I would guess that the plan it to reduce the degree to which the head rattles around. Most of the brain damage (short term and long term) associate with a punch comes from the brain bouncing off the skull a few times as the head whips back and forth. Minimize the motion, minimize the damage. By leaning in you get a ...


8

A simple counterexample: Imagine two particles with opposite direction and equal speed. The center of mass does not move, yet the kinetic energy of the system is non-zero. Now let both particles come to rest (by friction, hitting a wall, whatever). The kinetic energy is now zero, and total momentum has been conserved, while energy is not. The crucial ...


8

Let's take everything out of our scenario other than you and the ball. No baseball stadium, no Earth, no spherical cows, NOTHING in the entire universe but you and the ball. (Nope, not even microwave background radiation) Now the question has changed. Now you need to ask whether it's possible to decide whether you're moving towards the ball or vice versa.


7

While people normally quote Newton's Second law as $\vec F = m \vec a$, it is better written as $$ \vec F = \frac{d\vec p}{dt} $$ Force is a rate of change in momentum. This means that the average force applied when an object undergoes some discrete change in its momentum is $$ F_{\text{avg}} = \frac{\Delta p }{\Delta t} $$ The change in your momentum ...


7

Ayush: Isn't the question telling that the bullet always loses 1/n th of its velocity no matter which plank? Based on the answer provided, it seems the writer wanted you to assume that the energy loss per plank is constant. This is not the same as the bullet losing $1/n^\text{th}$ of its velocity per plank (however, the fact that the question does not ...


7

It all depends on your definition of visible. Elementary particle collisions have been made visible since the time of cloud chambers and bubble chambers. A good site for bubble chamber pictures exists . Unfortunately proton scattering is not as photogenic as scattering by other particles so I was unable to find a photo of a proton proton scatter in a ...



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