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101

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 ...


48

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 ...


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 ...


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 ...


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

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 ...


7

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, ...


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

What acts here is called impulse (of a Force) Suppose balls A, B are made of stainless steel and (m = 0.1 Kg r = 0.03 m) they collide. B is at rest (v = 0): Ball A will exert on b the Impulse of a Force $J$ and its velocity, momentum and KE will increase: $$J = [F . t] = \Delta p$$ If you know exactly of what steel the balls are made you can calculate the ...


6

Classically we would add the speeds to get $1.8c$, which is obviously not allowed. In relativity you simply use the relativistic velocity addition formula: $$V = \frac{u+v}{1+uv/c^2}$$ Where $u$ and $v$ are the velocities of the particles as seen from some reference frame, and $V$ is the velocity of one particle in the rest frame of the other, i.e., the ...


6

If you collide two ideal billiard balls, then that would be what you would call a perfectly elastic collision. If you have a large dense collection of billiard balls, and you slam a new one into the collection, then there are a whole lot of elastic collisions, transfering energy and momentum in many ways that you would be hard-pressed to calculate exactly. ...


6

You're confusing the acceleration of your car with the acceleration in a collision. You actually have to look at it "backwards" from what you've described above. That is, in the collision you don't do a $F = ma$ calculation where $a$ is the acceleration of your gas pedal. Instead in the collision you have a force $F$ resulting from the collision and you ...


6

So how is momentum conserved in inelastic collisions? It is basic law of physics that momentum is always conserved - there is no known exception. Kinetic energy does not need to be conserved, because it can turn into other forms of energy - potential energy, internal energy - "heat". Momentum can also turn into other form of momentum - momentum of the ...


6

Regardless of the physical undefinability of "painfulness", I'd like to plug some numbers in a particular scenario: Let's have a momentum of $p = 1000 $m$\cdot$kg/s, A 0.25kg bullet would be fatal, moving at $v = 1000/0.25 = 4000$m/s, while a 2000kg car moves at $v=0.5$m/s, So at least in this scenario and particularly for relatively low momentum ...


6

Backspin! Those shots in which the cue ball "draws" backwards after hitting the target ball involve backspin. Without backspin, the cue ball cannot reverse direction. Consider what happens when the cue ball is not spinning at all when it hits the target ball. The cue ball will come to a dead stop if it hits the target ball straight on. Think of Newton's ...


6

Microscopically, i.e. in the quantum theory the scattering with radiation is a collision of particles with photons such as $$ e^- + \gamma \to e^- + \gamma$$ The momentum vectors of the particles above are $$ \vec p_1+\vec p_2= \vec p_3 + \vec p_4$$ where the identity holds due to momentum conservation. But in general $\vec p_1\neq \vec p_3$ and $\vec ...


5

Unless I'm missing an easy way to do this problem it seems a surprisingly hard one. This diagram shows the problem (I've exaggerated the altitude of the satellite to make the diagram clearer): The satellites are in circular orbits (dotted line) at a distance $r$ from the centre of the Earth, so their orbital velocity as (as you say): $$ v = ...


5

While a single LHC particle wouldn't be doing much harm, being hit by the LHC beam would be certainly deadly and it would damage the machine badly. Any dense matter that comes into the LHC beam will instantly act as a beam dump. We have a very good idea about what happens in the LHC beam dump, see e.g. ...


5

I'm not sure that there is a good way to figure out exactly what happened unless there is video of the crash, however both statements can be true. There will have been some loss of energy in the collision due to overcoming the kinetic friction of your car (getting it moving from being stopped) as well as the energy absorbed by permanently deforming the ...


5

A simple version of this is bremsstrahlung, i.e. an electron that decelerates and produces electromagnetic radiation / photons. By your reasoning the energy of the electron should only be able to go into other electrons: maybe it should radiate other electrons, maybe a single electron shouldn't lose energy as it travels. But the electron can transfer some ...


5

Collisions can be elastic or inelastic. Elastic collisions are collisions where the incoming and outgoing kinetic energies are the same and only the angles change The same holds true classically, example: billiard balls ; and in the elementary particle framework. example: An electron hitting an electron has a probability of scattering elastically. ...


5

First of all, if the collision is elastic, the distribution of momentum in between the components is completely determined by momentum and energy conservation! This statement is most obvious in the center-of-mass frame where the total momentum is zero and the two objects are moving in opposite directions. The momentum conservation (the total momentum is ...


5

Let's turn this around. In an inelastic collision, some of the energy, instead of remaining with the center of mass of the objects colliding, is dissipated as heat - an increase in the random motion of the atoms and molecules of the objects colliding. At the (sub) atomic level, the two particles involved in a simple collision are the same two objects that ...


5

Possibly your confusion arises from not considering that KE is a scalar, whilst momentum is a vector. Yes, there is of course a connection between KE and momentum: $K = p^2/2m$ (for non-relativistic bodies). Thus, for two equal-mass particles heading directly towards each other at equal speeds $v$, their velocities are $\pm {\bf v}$ and their momenta $\pm ...


4

I'm 27 and since I was about 15 I had the same doubt you do. Only a couple of years ago I realized why momentum is always conserved in a collision, whereas the same is not enforced for energy. (They must have told me this at some point -- or points --, but I guess sometimes I just don't pay much attention) First of all, let's make it empirically clear that ...


4

Is there no relation between the momentum and kinetic energy? Are they not linked with each other? No, they're not. Conservation of momentum only says that $$m_1 v_{1_-} + m_2 v_{2_-} = m_1 v_{1_+} + m_2 v_{2_+}$$ I used subscripts 1 and 2 denote the particles involved in the collision and the subscripts - and + denote the pre- and post-collision ...


4

To stop instantly, you would need infinite deceleration. This in turn, requires infinite force, as demonstrable with this equation: $$\vec F=m\vec a$$ So when you hit a wall, you do not instantly stop (e.g. the trunk of the car will still move because the car is getting crushed). In a case of a change in momentum, $m\vec v$, we can use the following equation ...



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