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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 ...


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

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

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 ...


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

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

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 ...


4

In particle physics there exists elastic scattering for all interactions: change of direction but not of energies. When a photon penetrates into a medium composed of particles whose sizes are much smaller than the wavelength of the incident photon, the scattering process, also known as Rayleigh scattering, is also elastic. In this scattering process, ...


4

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 ...


4

Pairs of charged particles and/or objects attract via the $Q_1Q_2/R^2$ Coulomb's law. This is a classical approximation that quantifies how their velocities are changing when the objects are large or distances are much longer than the Compton wavelength etc. When the particles get really close, there are new effects that are neglected by the laws of ...


4

Total momentum is always conserved, in both elastic and inelastic collisions, but total kinetic energy is only conserved in elastic collisions. This example seems to be a completely inelastic collision, because at the end the objects merge. There is a formula to calculate the final velocity $v$ of two object with speed $u_1$ and $u_2$ and mass $m_1$ and ...


4

This is an example in which one needs to be careful to make the distinction between net force, which may vanish, and one of the individual, nonzero interaction forces that contributed to the net force The fact that your knuckles are not accelerating does not mean that the contact force with the person's face is zero, it means that the net force on your ...


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 ...


4

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 ...


4

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. ...


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

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

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

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

Most gamma rays from $pp$ collisions come from neutral pions ($p+p\to p+p+\pi^0$), you'd first have to do some relativistic momentum & energy conservation to determine the energy of the neutral pion. It's easiest if you consider the two subsequent reactions: $$ p+p\to p+\Delta^+ \\ \Delta^+\to p+\pi^{0} $$ (it's up to you to figure out the kinematics). ...


3

In a typical introductory physics class, the terms elastic and inelastic refer to whether the macroscopic kinetic energy is the same before and after the collision. By this I mean the large-scale motion of the objects that you typically consider in collisions, like carts or balls, not the detailed particle motion/potential. You are correct that the total ...


2

Without friction, the forces during the collision (glancing or head-on) are applied exclusively through their centres of mass. (Illustration available on Wikipedia.) The torque is given by $\tau=\mathbf r \times \mathbf F$ - but if the forces are applied through the centre of mass, then $\mathbf r$ and $\mathbf F$ are parallel, and hence $\tau=0$. Without ...


2

This problem has a recursive flavor that we'll not try to avoid. Conservation of momentum tells us that $$m v_0 + (p+n-1)m v(n-1) = (p+n)m v(n).$$ Imposing the boundary condition $v(0)=0$ we find $$v(n) = \frac{n}{n+p}v_0$$ as claimed. Let $a_n$ be the time at which the $n$th bullet strike occurs. We have $a_1=x_0/v_0$ and $$\begin{equation*} ...



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