# Tag Info

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 ...

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 ...

10

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 ...

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 ...

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

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

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 ...

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 $\mu = G(M_\odot + ... 5 The explanation can be found in the author manuscript of the article at this HAL preprint of the original journal article (Phys. Rev. Lett. 110 no. 17 (2013), 174302). It is my understanding that, for larger times, the number of cracks is determined by minimizing the sum of stretching energy and fracture energy. You can also read the Physics Focus piece ... 5 You're confusing the acceleration of your car with the acceleration in a collision. You actually have to look at it "backwards" from what you've described above. That is, in the collision you don't do a$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 ... 4 In the ideal case where the collision is instantaneous and there is only a single point of contact, the forces experienced by each object can only be along the line that connects the centers and passes through the contact point. Actually, the "force" will be infinite, but it will impart a finite impulse (i.e. change in momentum) during the infinitesimal ... 4 Here are real events relating to the last page of the pdf link you gave: Fig.1 This bubble chamber picture shows some electromagnetic events such as pair creation or materialization of high energy photon into an electron-positron pair (green tracks), the Compton effect (red tracks), the emission of electromagnetic radiation by accelerating charges ... 4 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 ... 4 That depends strongly on specifics of the crash, and where the other occupants of the car are. Let's assume you slam straight into a brick wall. If you sit in the back, there is nothing to hold you back at the time of the crash. You will slam hard into the seat in front of you (if you happen to have been sitting normally), and you might bruise or break ... 4 This is pretty basic physics: We know the following formulae $$F=ma$$ $$a={v_f^2-v_i^2\over2\Delta d}$$ In both cases, the final velocity is$0$. Assuming you have the same room,$\Delta d$, to decelerate in a crash, $$F=m{v^2\over2\Delta d}$$ Due to the square of the velocity, if you increase the impact speed by a factor of 2, you increase the impact ... 3 The two vehicles experience a force of the same magnitude due to Newton's third law: If object$A$exerts a force$\mathbf F_{AB}$on object$B$, then object$B$will exert a fore$\mathbf F_{BA}$on object$A$and $$\mathbf F_{BA} = -\mathbf F_{AB}$$ However, what you're probably thinking about is that motion of the car is more drastically affected by ... 3 What is missing in your question, (and maybe not emphasized properly in the book), is the domain of application of the statement :"measure". Here are individual electrons in a bubble chamber interacting with a magnetic field and turning into helical paths. It shows an electron and positron pair generated by a photon interacting with a nucleus in the ... 3 Mosquitos, windshields and trains are NOT perfectly rigid bodies.. In fact, the mosquito will suffer a completely inelastic collision with the windshield and will get squished when it hits. So the mosquito center of mass will gradually decrease from 4m/s to 0m/s and then to -300m/s and all of this will happen over the time frame of the squishing of the ... 3 Right before the collision, the mosquito is moving at 4 m/s, right after the collision it is moving at -300 m/s. In between, it will be quickly accelerated, so yes, it will eventually be not moving for an instant. But the train and the mosquito velocities are only equal after the collision, not while it is happening. So there is nothing in theory that ... 3 Taking the case of point particles and "contact" collisions seriously actually causes trouble even in the two dimensional case: the instantaneous forces are necessarily infinite even if the impulses remain finite. The solution to that problem--to recognise that all real particles interact via fields over non-zero distances--solves the three particle problem ... 3 As the collision is not known to be elastic or inelastic. We just go with checking options , as you did. a),d) are easily eliminated . But now b),c) gives in problem. Now we see that after collision the bodies must separate out. $$0\le\text{coefficient of restitution }(e)\le1$$ Otherwise$e$will go negative. Now we can see in c)$e=-1/2$but in b) ... 3 According to the National Highway Traffic Safety Administration the safest place to sit is in the centre of the back seat. I couldn't find anywhere they detail what research they used to come to this conclusion, but it seems reasonable on the grounds that it is the point in the car farthest away from anything that might intrude into the car body. You should ... 3 First, figure out how much energy is lost by the two balls as they fall to the ground. Now, the first ball reverses its momentum upon hitting the ground. Now, you have the one ball going toward the ground with speed$v$and the other ball going upward at speed$v$. What happens when two balls collide elastically with a head-on collision of speed$v$in ... 3 Both will exert the same impulse on your body, since this is equal and opposite to the impulse exerted on the bullet, which was stipulated to be identical in both cases. Impulse is force x time. The difference will be that the lighter gun will push you with a higher force for a shorter time. This will make the impact feel sharper, which can make it hurt ... 3 Another way to think about Newton's second law (and the way he originally defined it) is$F=\dfrac{d\rho}{dt}$, where$\rho=mv$is momentum and$\dfrac{d\rho}{dt}$is the rate of change of momentum. I think you meant to say that the obstacle will exert a force on you - and that is correct. If you could calculate your change in velocity, and the amount of ... 2 To clarify, it sounds like you are suggesting that: 1) There is an instant where the mosquito's velocity is 0, and: 2) Since the mosquito is stuck to the train's windshield... It follows that that the train must also have an instant where its velocity is 0. Of course, however, this is not really what theory predicts... The discrepancy comes from the fact ... 2 What would happen if the scientists would use leptons instead of hadrons? They would go in the opposite direction - electrons are the opposite charge to protons. The LHC doesn't use electrons because protons are 2000x heavier so you get a lot more energy in the collision. Especially: What would happen if they would collide electrons? ... 2 The very basic reason LEP stopped going to higher energies ( it reached over 200GeV center of mass, at the last stage, LEPII) and the tunnel was used for LHC is synchrotron radiation . Note that radiated power is proportional to 1/m^4 It is not possible to feed a circular beam of electrons the energy needed to raise it to higher energies at the radius ... 2 Nope. Initially, your object has no net force on it, and thus zero acceleration. Once it hits the wall, the wall exerts a force on it (and it exerts an equal/opposite reaction force back) which decelerates it according to$F=ma$Note that$F=ma\$, expanded, means "net force on a body is equal to its mass times its acceleration at any instant". It makes no ...

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