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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 ...

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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 ...

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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, ...

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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 ...

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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 ...

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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 $\vec ... 4 It Depends on Your Model How closely do you want to model reality? The truth is, most collisions are some mixture of inelastic and elastic. (That is, momentum is transferred, but not always "cleanly," some of that energy gets transferred into deforming the objects.) You can see this sort of thing if you watch slow-motion videos of things striking other ... 4 If the total kinetic energy before the collision equals the total kinetic energy after the collision, the collision is elastic. Otherwise, it isn't elastic. given the mass, the velocity, and the 'angle' the two objects are going two be when they collide - how can I know if I need to compute an elastic or an inelastic collision? The mass, velocity ... 3 There is not such thing as a "partially elastic" collision. Classical collisions between particles can be separated into two categories: elastic and inelastic. Elastic collisions are defined as collisions in which no energy leaves the system (i.e.$E_i = E_f$). All other collisions are inelastic, as some energy is lost ($E_i > E_f$). A perfectly inelastic ... 3 I will make my comments into an answer: If we are talking of the classical domain, energy is conserved but can change forms. Inelastic scattering has a strict definition and is usually used describing two body scattering processes. In two body impact situations conservation of input energy in the output products would require that the target has some ... 3 It helps to think of the extreme cases here. If the bat was not moving (like a bunt), the resulting bounce does depend on incoming speed. Now, make the incoming ball really slow and the batter hitting fast. The result does not depend on incoming speed. The true answer is the superposition of these two cases. 3 There are several things that can happen, depending on the nature of the obstacle. The simplest case is if, classically, we think that the obstacle is not too much smaller than the wavelength of the light. Then, the light wave can be reflected, absorbed, or diffracted. Most people are familiar with reflection, the light more or less bounces off the ... 3 I understand that the inner product of two 4-vectors is conserved under the Lorentz transformations Yes,$p_1.p_2$is a Lorentz invariant So that the absolute value of the four momentum is the same in any reference frame. It is not correct to speak about the "absolute value" of a (quadri)vector. Which is conserved in a Lorentz transformation ... 2 In special relativity, if you add two velocities, you have to use the formula $$v = (v_1+v_2)\left(1+\frac{v_1v_2}{c^2}\right)^{-1} \text{ .}$$ So you cannot simply add two velocities together. Usually, velocity is not a good variable to work with in special relativity. It's much easier to use four-momentum conservation, which is simply given by $$p = p_1 ... 2 Normally you solve an elastic collision with just momentum and energy conservation, because you really don't know what happens at impact. The formulas are given in this answer. For equal masses in one dimension the velocities are exchanged. It turns out your interaction time and acceleration multiply to get the correct velocity change, but since you ... 2 I'll try to answer in the context of cricket; baseball shall follow as a special case. (Cricket is richer in terms of the role of the batsman, as compared to baseball!) The objective is to transfer momentum to the ball, which can be done in two broad ways (this isn't the standard terminology, this is how I see it)- 1 Guiding - Here, you use the momentum ... 1 To parameterize the degree of inelasticity you use the "coefficient of restitution" which is 1 for elastic processes and 0 for inelastic processes. This is described by$$ \text{coef. of restitution} = c_R = \frac{\text{final relative speed}}{\text{initial relative speed}} = \frac{v_2 - v_1}{u_1 - u_2} \,. \tag{*} $$This also tells you how to compute the ... 1 Perfectly elastic and perfectly inelastic collisions are just limiting cases on a scale of how much kinetic energy is retained. As noted in @Nathan's answer, if you work in the center-of-mass frame, a perfectly inelastic collision results in 0% of the kinetic energy retained, while perfectly elastic collisions have 100% of kinetic energy retained. So, you ... 1 Luminosity is necessary in order to turn number of interactions to crossections, because theories provide crossections to compare with experiments. Experiments measure number of interactions. A well known crossection, as is Bhabha scattering, substituted on the right will give the luminosity to be used in the other observed interactions in the experiment. ... 1 Here's a parallel answer to Luboš's but purely classical. Start by noting that the momentum vector of a plane wave with wavelength \lambda is:$$ \vec{p} = \frac{2\pi}{\lambda}$$In some elastic scattering experiment, e.g. X-ray or some other diffraction measurement, we have something like: where$\vec{p}_{in}\$ is the momentum of the incoming wave ...

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In general, the elasticity of a collision is dependent on the properties of the colliding objects. In a perfectly elastic collision, no kinetic energy is dissipated, which means the collision creates no heat, no sound, etc. In a perfectly inelastic collision, the maximum possible amount of kinetic energy is dissipated as heat, sound, etc. This corresponds ...

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The direction the ball will take depends on the angular momentum. The velocity with which the ball moves or bounces backwards but the chief determinant is the spinning effect of the incoming ball.

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