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19

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 ...


18

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 ...


17

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 ...


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 ...


15

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 ...


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 ...


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

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

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 ...


9

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 ...


9

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 ...


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 ...


7

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 $$ ...


6

The rotating surfaces of the spheres would just slide over each other at the instant of contact: no forces perpendicular to the line connecting the centers of the two spheres would exist (i.e. no torque would exist). They would undergo a perfectly elastic collision (no loss of energy, thus no friction), thus conserving angular (they just keep spinning) and ...


6

99% of the speed of light generates a Lorentz factor of only $$ \gamma = \left[ 1 - (.99)^2 \right]^{-1/2} \approx 7 $$ which means that you have only about 14 times the mass of a down-quark to make additional particles. The PDG puts the bare mass of the down quark in the neighborhood of 5 MeV, so $14 \times 5\,\mathrm{MeV} = 70\,\mathrm{MeV}$ isn't enough ...


6

Yes the I must have the ^-1 exponent, otherwise the unit would not end up in $s^-1$ (the unit for angular velocity). $\hat n$ is the unit vector in the direction of exit after collision. Moment of inertia of a 2D or 3D object is the same as long as they have the same cross section from the perspective of the dimension you want to ignore (for example ...


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

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. ...


5

Head on collision is not required for elastic collision. Or the collision you described above can be an elastic collision. To be an elastic collision, the momentum and kinetic energy should both be conserved, that is: assume the velocity for sphere 1 and 2 are $\vec{v_1}$ and $\vec{v_2}$ accordingly, and the direction of neither of them is along the ...


5

For completeness, here's an another solution (though not so elegant as Shanth's solution). The equations of motion (from the 2nd Newton's Law) of the two point masses $m_1$ and $m_2$, respectively, are: $$G\frac{m_1 m_2}{(r_2-r_1)^2}=m_1 \ddot{r_1}\Leftrightarrow \ddot{r_1}=G\frac{m_2}{(r_2-r_1)^2}$$ $$-G\frac{m_1 m_2}{(r_2-r_1)^2}=m_2 ...


5

As a clarification the energy per nucleon is always lower: for example, currently in the LHC the proton top energy is 3.5 TeV. Now the Pb energy is 3.5 TeV times Z so the energy per nucleon is 3.5*Z/A and A is greater than Z for every nucleus (except the proton where it is equal to one). But the goal of ion-ion collision is not to increase the total energy ...


5

The hypothetical collision has been mathematically modeled, and the results of those models are consistent with what see. The impacting body (referred to as "Theia") would have hit at a low velocity and relatively shallow angle. It would certainly have affected the Earth's orbit, but not enough to knock it out of the Solar System or into the Sun. Knocking ...


5

A subtle problem you seem to overlook is that the proton-proton cross section is very small, about 0.07 barns (a barn is $10^{-28}$ square meters) at the LHC energies and not dramatically different at your lower "fusion energies". It means that at the LHC, much like at your dream machine, most of the protons simply don't hit their partners. It is not really ...


5

The initial and final momentum are not the same because the ball is not an isolated system. The wall exerts a force on it. In principle the ball and the wall (and the planet it's connected to!) form an isolated system with a conserved momentum, but you'd have to take into account how much the wall moves after the collision. The change of momentum is final ...


5

As far as I know, nobody has ever done this, at least not at what we currently consider high energy. (Electron-electron collisions happen at low energy all the time, of course.) I doubt that anything interesting would happen, primarily because electrons are mutually repulsive, and they have a low mass. That means two colliding electrons would just bounce ...


5

Am I right to say that some of the kinetic energy can be converted to angular momentum[?] No, angular momentum is a conserved quantity. In any isolated interaction you get out exactly as much as you put in. But you may have intended to ask Can a ball that is not spinning when I toss it at the ground come off with spin? to which the answer is ...


5

The Kepler orbit of the Earth around the Sun is determined by two constants: the specific orbital energy $E$ and the specific relative angular momentum $h$: $$ \begin{align} E &= \frac{1}{2}v_{r,\oplus}^2 + \frac{1}{2}v_{T,\oplus}^2 - \frac{\mu}{r}= -\frac{\mu}{2a},\\ h^2 &= r^2\,v^2_{T,\oplus} = \mu a(1-e^2), \end{align} $$ where $\mu = G(M_\odot + ...


5

The explanation can be found in the author manuscript of the article at this HAL preprint of the original journal article (Phys. Rev. Lett. 110 no. 17 (2013), 174302). It is my understanding that, for larger times, the number of cracks is determined by minimizing the sum of stretching energy and fracture energy. You can also read the Physics Focus piece ...



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