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20

It is a standard exercise in quantum electrodynamics to find the angular dependence of the differential cross section. Which more or less means how probable it is for the photons to scatter at a certain angle, given the energy of the incident particles. So assuming the spins of the electron-positron pair is averaged, and that you don't care about the photon ...


14

We do. The LHC accelerates two protons, each with 3.5 TeV of energy, giving a total of 7 TeV in the CoM frame (The energies are from the initial phase of the previous LHC run. Later in the run this was increased to 8 TeV and the combination of the two dataset was what discovered the Higgs boson. The energies are roughly doubling now for Run II, to 13 TeV). ...


5

Many modern particle accelerators do accelerate both particles towards each other. LEP accelerated electrons and positrons in opposite directions in the same chamber, and the Tevatron did the same for protons and antiprotons. The LHC is a proton-proton collider, and so it has two stacked rings that accelerate protons in different directions. For the BaBar ...


5

The problem is equivalent to 4 spheres colliding simultaneously, where top sphere center is at $60^o$ relative to the $x'x$ axes (same goes for bottom sphere): We'll name them: sphere A (dark blue), and spheres 1, 2, and 3. During the collision the spheres will behave like springs with an infinite hook constant. The forces on the spheres will be ...


4

Here is how you deal this this problem as a system of equations. For each contact pair assign a normal direction $\hat{n}_k$ and and impulse $J_k$. The possible contacts are AB, AC, and AD. We can introduce symmetries and simplifications later. The initial velocity if body A is $v_A$ along the horizontal axis, and after the collision it is $v_A + \Delta ...


3

.... We are told that all the subsequent collisions involving the balls and floor are elastic. We are asked to determine the maximum height to which the small sphere will rise on the rebound. The problem does not mention any radii, but if we did know the radius of each of the spheres, would it be valid to bypass conservation of linear ...


3

Let a sphere be dropped from height $h$ to a fixed horizontal plane. If $u$ be the velocity of the sphere just before striking the plane, then $$u^2 = 2gh \implies u = \sqrt{2gh}$$ If $v$ be the vertically upward velocity with which the sphere rebounds to a height $H$, then $$v = \sqrt{2gH}$$. Since both $u$ & $v$ are perpendicular to the horizontal ...


2

the speed of both should be according to the conservation of energy If both balls have the same speed after the collision, the collision is inelastic, i.e., kinetic energy is not conserved. If the balls are identical, then conservation of momentum requires that $$\mathbf v'_1 + \mathbf v'_2 = \mathbf v_1 + \mathbf v_2 = 150 \mathrm{\frac{m}{s}}$$ If ...


2

When you compute the final velocity of the parcel you have forgotten that it's no longer traveling at 37 degrees - that was the angle at the end of the chute. While it drops, the horizontal component of velocity doesn't change - it is still $3.4\cdot \cos 37° = 2.71 m/s$. With that, you should be able to solve this.


2

I was thinking about building my virtual elastic bodies as systems of "nodes", each with a certain mass, interconnected by springs. You just described Finite Element Modeling - the cornerstone of mechanical engineering, and the method used for making sure that that bridge won't collapse when an 18 wheeler passes over it. This is an extremely well ...


1

The problem is ambiguous and has infinite solutions. The reason is that now you cannot ignore the relative forces between the A and B,C and D. For instance, if you solve the problem as non simultaneous collisions you will get a different answer depending on the order of the collisions. Suppose that A collides first with C, then A will stop, C will move at ...


1

The ball will rise again to height 'h' if the surface is absolutely smooth and there is no air friction (elastic collision). The ball will rise to 'eh' if the collision is inelastic. This is because some fraction of initial velocity is lost due to rough surface etc. The ratio i.e COR is 1 when it is the former case and < 1 when it is the latter. COR ...


1

It will help if you study this diagram of what a vacuum tube is If a cathode is heated, it is found that electrons from the cathode become increasingly active and as the temperature increases they can actually leave the cathode and enter the surrounding space. When an electron leaves the cathode it leaves behind a positive charge, equal but ...


1

Suppose you and a friend shoot pool balls dead-on, which then bounce off of each other. Right before the collision, you count up the kinetic energy, and you measure it to be $E_1$. Right after the collision, you count up the kinetic energy, and you measure it to be $E_2$. There are clearly three possible cases: $E_1 = E_2$: This is an "elastic" collision. ...



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