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88

Things are not empty space. Our classical intuition fails at the quantum level. Matter does not pass through other matter mainly due to the Pauli exclusion principle and due to the electromagnetic repulsion of the electrons. The closer you bring two atoms, i.e. the more the areas of non-zero expectation for their electrons overlap, the stronger will the ...


10

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 $$ ...


10

To figure this out, you need to know about momentum ($p$). That's a combination of how fast something is moving ($v$, for velocity) and how much it weighs ($m$, for mass). You'll also need to understand algebra, which is just using a letter to mean some number you don't know yet. $$ p = m\cdot v $$ Momentum is conserved, which means the momentum from both ...


9

Yes, when you fire a pistol the hammer hits the bullet with a relatively small initial kinetic energy but the kinetic energy of the hammer and bullet after the collision is considerably higher. This may seem a silly example, but I think it actually highlights the important principle involved. In general when two bodies undergo an inelastic collision part of ...


7

Ayush: Isn't the question telling that the bullet always loses 1/n th of its velocity no matter which plank? Based on the answer provided, it seems the writer wanted you to assume that the energy loss per plank is constant. This is not the same as the bullet losing $1/n^\text{th}$ of its velocity per plank (however, the fact that the question does not ...


7

If it were possible for one object to pass through another object, then it would be possible for one part of an object to pass through a different part of the same object. Therefore the question asked here is equivalent to the question of why matter is stable. See this question on mathoverflow. That question was more about the stability of individual atoms, ...


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

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

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. ...


6

Backspin! Those shots in which the cue ball "draws" backwards after hitting the target ball involve backspin. Without backspin, the cue ball cannot reverse direction. Consider what happens when the cue ball is not spinning at all when it hits the target ball. The cue ball will come to a dead stop if it hits the target ball straight on. Think of Newton's ...


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

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 ...


5

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. ...


5

I'm not sure that there is a good way to figure out exactly what happened unless there is video of the crash, however both statements can be true. There will have been some loss of energy in the collision due to overcoming the kinetic friction of your car (getting it moving from being stopped) as well as the energy absorbed by permanently deforming the ...


5

Classically we would add the speeds to get $1.8c$, which is obviously not allowed. In relativity you simply use the relativistic velocity addition formula: $$V = \frac{u+v}{1+uv/c^2}$$ Where $u$ and $v$ are the velocities of the particles as seen from some reference frame, and $V$ is the velocity of one particle in the rest frame of the other, i.e., the ...


5

First of all, if the collision is elastic, the distribution of momentum in between the components is completely determined by momentum and energy conservation! This statement is most obvious in the center-of-mass frame where the total momentum is zero and the two objects are moving in opposite directions. The momentum conservation (the total momentum is ...


5

Microscopically, i.e. in the quantum theory the scattering with radiation is a collision of particles with photons such as $$ e^- + \gamma \to e^- + \gamma$$ The momentum vectors of the particles above are $$ \vec p_1+\vec p_2= \vec p_3 + \vec p_4$$ where the identity holds due to momentum conservation. But in general $\vec p_1\neq \vec p_3$ and $\vec ...


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

If the total kinetic energy before the collision equals the total kinetic energy after the collision, the collision is elastic. Otherwise, it isn't elastic. given the mass, the velocity, and the 'angle' the two objects are going two be when they collide - how can I know if I need to compute an elastic or an inelastic collision? The mass, velocity ...


4

It Depends on Your Model How closely do you want to model reality? The truth is, most collisions are some mixture of inelastic and elastic. (That is, momentum is transferred, but not always "cleanly," some of that energy gets transferred into deforming the objects.) You can see this sort of thing if you watch slow-motion videos of things striking other ...


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

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

2 dimensional collision can be reduced to a 1-dimensional problem in the case of spheres--see here. The $\pm$ you encounter when solving the kinetic energy is likely because there are two solutions and the equations are satisfied by either one. One solution is simply where the particles pass right through eachother, which you can discard.


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 ...



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