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14

Elastic collisions do happen at the LHC. The TOTEM experiment measures the differential cross section (rate as a function of angle) for proton-proton elastic scattering at the LHC. Here is their latest result. They don't publish an estimate of the elastic cross section, but according to their data it must be at least 25 mb (millibarns) (my first version of ...


8

Anything that is not forbidden must happen. That's an important statement to keep in mind when approaching quantum physics. It doesn't mean that anything that can happen always happens, but it must happen at some time or another just like someone eventually has to win the lottery. That said, some protons do go through the LHC, ram into each other and ...


6

elementary particles (e.g. protons) Protons aren't elementary particles, they're made of partons (quarks and gluons) in "soup". Below, $\lambda$ is the wavelength corresponding to the energy of the interaction via the usual de Broglie relation and $r_p$ is the radius of the proton. At low energy with $\lambda >> r_p$ the interactions are just ...


5

The problem does not mention any radii, but if we did know the radius of each sphere, would it be possible to skip conservation-of–linear momentum calculations altogether The accepted answer has confused you: You can simplify things by considering these conservation laws in the center of mass frame. There the total momentum is zero, therefore ...


4

In both cases and at all times, the force from the (wall/tire) on the hammer equals the force from the hammer on the (wall/tire) : total momentum must be conserved. However, in the first case, the initial energy is dissipated in the wall (as heat and/or damage), so at the end the hammer is stopped. In the second case the initial energy is stored as ...


4

Yes, $F=ma$, but also $v=at$. That means that, as you fall for a longer time, your speed will increase. After 1 second, you are going at $9.8 m/s$ or $35 km/h$, about the speed of Usain Bolt. After 10 seconds you would reach $98 m/sec$ or $350 km/h$. For a free-falling human, the air resistance actually limits you to about $200 km/h$. When you hit the ...


3

Find the nearest box-spring mattress. With your hand, execute a slow motion "impact" between your hand and the mattress. You should notice that when your hand is not touching the mattress, there is, of course, no force between your hand and the mattress. As your hand begins to touch the mattress, you should feel a very light force. As your hand presses ...


2

Here's how. The ball gains a velocity $v$ due to gravity before hitting the ground. So each time it hits the ground its velocity is changed from $v$ to $-v$ (taking down as positive) during the collision, then returning again with $v$. The force $F_1= m\frac {dv}{dt}$ is experienced by the ball due to the collision, however this force is felt after the ...


2

If the collision is not perfectly along the line connecting the centers of mass of the pucks, they will exert torques on each other as well as forces. The angular momentum of the pair will be conserved, so if the incoming puck was not spinning, the pucks will exit the collision spinning in opposite directions. If the surface they slide on is frictionless, ...


2

Remember that the rest mass of a proton is about 940MeV. So, a proton at 0.6c is pretty darn energetic. I will leave a precise energy up to the reader. However, there are a number of experiments in the physics literature of smashing protons into various things. At low (1MeV-ish) energies, you are looking mainly at classic Rutherford scattering cross ...


1

You didn't cancel out the mass $m$ properly. $$-mv_0\cos a+0=mv_1\cos b+0\\ -v_0\cos a=v_1\cos b\\ \frac{-v_0\cos a}{\cos b}=v_1$$ The incident angle is equal to the exit angle in such collision, $a=b$. The above reduces to: $$v_1=\frac{-v_0\cos a}{\cos b}=\frac{-v_0\cos b}{\cos b}=-v_0$$ And here you see what you probably expected. In the perpendicular ...


1

If we did know the radius of each sphere, would it be possible to skip conservation-of–linear momentum calculations altogether and use a single conservation-of-energy equation... The radius of a ball is always irrelevant to the outcome of a collision, what counts is the ratio of the masses. Two equations are always necessary to determine the outcome ...


1

As the Earth isn't a closed system with regard to other objects in the Solar System, including atomic particles, its momentum is affected by collisions. But if the collisions were inelastic (if the atomic particles were absorbed into elements of the crust and the atmosphere), the momentum of the atomic particles and the momentum of the Earth involved in the ...



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