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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

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 hydrodynamics, conservation means that what flows into the control volume is equivalent to the flow out of the control volume. With respect to momentum, we mean precisely that any change in momentum of the fluid within a control volume is due to the net flow of fluid into the volume and the action of external forces on the fluid within the volume ...


4

I will try to give a short introduction into the ideas of scientific truth as I understand them. In mathematics, the world is beautifully simple. We have axioms that the set to be true, and from these we can deduce a plethora of statements to be undoubtedly true - given that the axioms are true. There may be undecidable statements about which we cannot say ...


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

Let's assume that an experiment is performed that seems to indicate a violation of the conservation of energy principle. Now, I suppose that it's logically possible that the experiment actually and unambiguously falsifies the principle in which case we must conclude that the principle is approximate and we must seek a deeper principle to guide our ...


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

Newton's third law. For the force the magnet exerts on the metal, there must be an equal an opposite force on the magnet exerted by the metal. Since both form one system (metal + truck + magnet), the net force on the system is zero, and it won't move.


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

You may doubt that energy is conserved, but it is a direct consequence of Noether's theorem together with the assumption of time translation invariance, and this latter assumption is perhaps a bit more palatable/fundamental. That is, it is mathematically true that if the outcome of an experiment doesn't depend on when we perform it, the quantity we call ...


1

An other way to see the argument of the answer of @fuenfundachtzig , is that, concerning $SU(3)$ representations, there is an equivalence between $(3*3)_\text{antisymmetrised}$ representation ("red * green") and $3^*$ representation ("antiblue"). Why ? Well, thanks to the completely anti-symmetric Levi_Civita symbol. Using objects upon which act the ...


1

It works if you assign colors like this: one red up, one green up, down is blue, $X$ takes red and green which are equivalent to antiblue ("yellow"), thus color is conserved. I didn't take into account the last fact which explains my confusion.


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 ...


1

Emmy Noether proved both the theorem and its converse. Look for the book "The Noether Theorems" for a precise and discussed formulation of her statements, as well as a translation of the original paper. It seems there is a link to the pdf in the princeton math website (I don't know about copyright issues, however).


1

This is a general property of (pseudo)Riemannian geometry. I do not think there is anything specifically physical about it beyond the geometry. It holds even if $\varphi$ in not Lorentz invariant. In (pseudo)Riemannian geometry the covariant derivative $\nabla_i$ replaces partials $\partial_i$. The Laplace-Beltrami operator $$\Delta \triangleq ...


1

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 ...


1

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.


1

The continuity equation (without sources) is usually written as follows $$\partial_t \rho + \nabla \cdot \mathbf{j} = 0$$ If you identify $\rho$ as the mass density, integrate over some volume $V$ and use the divergence theorem you get the result that you mention in your question. Namely, the change in mass in $V$ equals the amount of mass flowing through ...



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