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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 ...

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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 ...

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 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 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 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 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 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 ... 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 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 ... 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 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 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 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 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 ... 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 $$\begin{equation*} ... 2 They do not need to necessarily collide like balls. I guess the picture in your book is illustrative. Conservation laws apply to any kind of interaction between them. Note that details of the collision are not even provided in the question but you still can calculate the answer. The detailed theory of photon-electron interactions is called Quantum ... 2 I worked on a physics engine written in C# that does just this. Here are my notes on this topic. Objects have both translational and rotational momentum. When two objects collide, the overall algorithm goes like this: 1> Find the total momentum of both objects. Calculate the translational and rotational momentum, the vector sum of this is the total ... 2 Suppose the collision between particles A and B lasts some time \tau, and during the collision the force between the particles is some complicated function of time, F(t). Consider particle A. The total impulse on it is the force times time, or more generally the integral of the force over the collision time:$$ I_a = \int_0^\tau F_a(t)dt$$And the ... 2 When the two balls arrive at the ground, they both have speed$V=\sqrt{gh}$and are travelling downward. Break the interaction down into two stages. First,$m_2$collides with the ground. Then,$m_1$and$m_2$collide (they cannot collide before$m_2$collides with the ground because they have the same velocity). After this,$m_1\$ necessarily has positive ...

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