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Here I present a probable answer to my question and I would really appreciate any feedback of yours. In a few words, the provided mathematical proof tells us that the eccentric masses $2m$ (internal parts of the system $M$) tangential acceleration can be converted to the isolated system ($M$) linear acceleration in agreement with Newton's 3rd law and the ...

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When you say "rectilinear," I assume you mean "collinear." I want to give a little input on this. Newton's third law does not require the internal forces to be collinear, only equal and opposite. You can envision then a system of two particle such as this one, where both particles are at rest: The CG is halfway between each particle. ...

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I answered your earlier question. The answer here is similar. If you want to compare quantum to classical in a limit, you cannot do it piece by piece. You need to look at it as part of a coherent and self-consistent limit. In this case, trying to separately look at the limit of $\hat{p}$ in the $\hbar \rightarrow 0$ limit (see related comment in my ...

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You are willfully misunderstanding the $\hbar/S \to 0$ classical limit. ℏ is dimensionful, so choosing enormous units to measure it with, like MKSA units to measure moving trains, makes it look small. The "proper", enormously subtle, classical limit runs on the above dimensionless ratio comparing the characteristic action quantity S of the system ...

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If you have a system of particles, indeed the Newton third law pairs act on different particles. For a given particle in the system the net force does not have to be zero. Internal forces in the system can accelerate the component particles and can change the total kinetic energy of the system. Hovewer, for any change in momentum produced by a force $F_{ij}$ ...

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The generator of $x$-translations is more precisely the $x$-derivative $$\frac{\partial}{\partial x}~=~i\frac{\hat{p}}{\hbar}~=~i\hat{k},$$ the wavenumber operator, which is independent of $\hbar$.

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When a photon bounces of a mass, there are two things to remember: the momentum transfer is orthogonal to the mass's four momentum. there is always a frame in which the energy transfer is zero. In the rest-frame of the mass: $$k^{\mu} = (k,0,0,k)$$ $$p^{\mu} =(M,0,0,0)$$ and for 180 degree scattering: $$k'^{\mu} = (k',0,0,-k')$$ $$p'^{\mu} =(\sqrt{M^... 3 Imagine that the object with mass M gains momentum p = \frac{2h}{\lambda} and now has energy E = \frac{p^2}{2M}. Let's get the new wavelength from the new photon energy, taking off the energy stolen by the object. \frac{hc}{\lambda'} = E' = \frac{hc}{\lambda} - \frac{p^2}{2M } = \frac{1}{2}pc - \frac{p^{2}}{2M} = \frac{1}{2}pc \left[1 - \frac{p}{Mc} \... 0 The momentum is neither of them, becuase you missed and square root in the \gamma. The momentum is $$p_{x} = \frac{1}{\sqrt{1 - \left(\frac{U}{c}\right)^2}} M U_{x}$$ The reason of why it is that why is that it's just a convinient definition. The Lorentz factor \gamma = \frac{1}{\sqrt{1 - \left(\frac{U}{c}\right)... 2 a) The conservation of momentum and energy is a consequence of Newton's 3rd law; however, the 3rd law by itself cannot determine whether an isolated system should accelerate or not (by utilizing internal forces). Your (a) is not correct. Mathematically, when object A interacts with object B then Newton’s 3rd law says \vec F_{AB} = -\vec F_{BA} where \vec ... 1 Let us consider this relative to the S frame. You seem to have a good understanding of the motion before the collision. You are confused here as to why both balls do not make 2D motion. It is reasonable to think so as well as you are correct. Think of it this way. The vertical momentum of the system before the collision is zero. There is only a horizontal ... 0 The momentum is conserved in both frames! the velocity changes , if you use a different frame. The same ist true for energy . it is conserved in both frames, but the measure of the energy changes when you change the frame of observation. 0 mc^2 is the rest energy and KE is all the energy due to motion. Add them and you get the total energy = all energy due to motion + all energy at rest. This bears some similarity to the classic equation E = KE + PE. 1 It may be helpful to think of an analogy, such as an arm-wrestling or a tug-of-war match. During much of the match, both combatants may be applying equal force, and there's no motion (in actual practice it's hard to be consistent, so there will be some fluctuation back and forth, but we can simplify this). But eventually one of them will weaken more than the ... 3 In the case where the table holds, we have a lot of information: The book is not accelerating The book has mass m The only vertical forces on the book are gravity and the normal force from the table. Given the lack of acceleration, we know the total vertical forces on the book must sum to zero and F_N = mg. The weight of the book is entirely supported ... 2 There is always some amount of material deformation when two objects are held against each other by a force. In this case material deformation of the table exceeds its structural integrity, so it breaks. Now your action reaction pair is the block falling towards the Earth and the Earth falling towards the block. Since Earth's mass is so much greater than the ... 13 The law still applies, but we have to be careful with the forces. If the book has a mass m, we cannot simply say that the book applies a force of F=mg. If the table broke, then there was some maximum force it could withstand, which was less than mg. The key to remember is that the sum of the forces on an object equals 0 only if that object is not ... 0 Any real world position measurement result also implicitly includes a momentum measurement. Why? Any measurement of x results in some psi(x), its Fourier transform is psi(p), both are measurement results and quantum states, they are simply represented in different bases and have inversely related widths, neither width can be zero. If the position measurement ... 1 For M=\mathbb{R}^m, starting from the canonical commutation relations (CCRs), we have the Stone-von Neumann theorem, which proves the existence of the standard Schrödinger position representation (i.e. without the 1-form \omega) up to unitary equivalence, cf. e.g. this Phys.SE post. However conjugating with unitary multiplication operators naturally ... 2 Look at the discussions of units, especially at the bottom of page 194. If you adopt conventional SI units, then one would say "The mom-energy of a particle is a 4-vector whose magnitude is proportional to its mass". If you adopt geometric units (where "the value of spacetime displacement and proper time for that displacement are the same"... 1 The name "moment" comes from the multipole expansion, which can be formulated several (equivalent) ways. One, for electrostatics, is as follows. Suppose you have a charge distribution \rho(\vec r), contained in some volume ||r|| \lt ||R||, then outside of ||R||, you can expand the potential into multipoles:$$ \Phi(\vec r)=\frac 1 {4\pi\...

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You see to be confusing moment and momentum. They are two different concepts.

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Your approach looks reasonable. Since the ball always keeps some proportion of its kinetic energy, it will bounce an infinite number of times. You might expect the time for each bounce to approach 0 as the bounces get lower and faster, but that's apparently not the case, as the problem states that each rebound requires a fixed amount of time (which isn't ...

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It is sure that with an infinite number of bounces and a finite duration for each bounce, the total time will become infinite. So we can conclude that the model is not good, at least to predict the total duration ! To have a finite total duration, you could introduce a rebound duration which decreases according to the collision speed with the ground: it is a ...

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The contact forces are not required to go through the combined COM. Since the net load is zero, its location does not affect the nature of the problem. The simplest situation to consider is a single contact point where two equal and opposite forces act, along the contact normal $\hat{n}$. Often in contact modeling we consider an exchange in momentum, ...

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If you are looking for a mathematical approach, We know that the Center of Mass of the two particles is given by: $$R_C= \frac{ M_1 r_1 + M_2 r_2}{M1+M2}$$ Now for these particles to end up at the same location(collide), we can say that $r_1=r_2=r$. The center of mass of a system is fixed. It does not move unless an external force acts on it. So it is still ...

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The assertion sounds a bit weird - by saying that the vector lies along the worldline and has magnitude equal to the mass, you've defined it by giving a direction and magnitude. There is no other way to construct that vector, frame-independent or not, because only one vector fulfills the condition. Showing that the condition is fulfilled is simple. The ...

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First of all, when a body is been dropped, the initial velocity (U)=0. Then using the 3rd equation of motion, V^2=U^2+2as. This third equation of motion tends to change when a body is been thrown upward or downward. Therefore we have our equation as v^2=u^2+2gh. And from our question, h= 2.5m, g(acceleration due to gravity)= 10ms-1, U=0, V=? From these we ...

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If we have an operator whose representation in coordinate space is proportional to $\frac{\partial^2{}}{\partial{x}^2}$, and we need that the expectation value must be positive, we need the negative sign. Indeed, integrating py parts: \int \psi^*(x)\frac{\partial^2{}}{\partial{x}^2}\psi(x) {\mathrm d}x=-\int \left(\frac{\partial{}}{\partial{x}}\psi^*(x)\... 0 To be positive, has to be not the "operator", but the eigenvalues of such operator. If you solve Schrodinger equation : -(h²/2m)d²/dx² f = Ef You find f=\exp i\sqrt k x With k= 2mE/h² Hermitianity of p garantuees that only real E can be eigenvalue; the square root tells us that E has to be positive; furthermore, every E positiva has its ... 0 As @SolubleFish already said, the Hamiltonian operator is derived by H = \frac{p^2}{2m} + V(x) Now, if you want to explain why \hat p = -i\hbar \partial_x, you could probably start by saying that you need the position and momentum operator to be canonical conjugates (this is important for the Heisenberg uncertainty principle and other things): \begin{... 2 Temperature is a thermodynamic variable , i.e. it depends on a large number of particles because it can be measured in matter, and thermodynamics as a theory developed long before the particle nature of matter was studied. When Newtonian mechanics was developed, it was shown that the theory of thermodynamics emerged from statistical mechanics . In ... 2 Consider \,\mathbf f\boldsymbol{=}\left(f_1,f_2,f_3\right)\, to be a 3-vector representing a force applied on a particle of rest mass \,m_{0}\, moving with velocity 3-vector \,\mathbf u\boldsymbol{=}\left(u_1,u_2,u_3\right). If this force is preserving the rest mass \,m_{0}\, of the particle then we can ''build'' a Lorentz 4-vector \,\mathbf F\,... 1 At low speeds the spacial components of Minkowski's four-vectors become practically equal to the quantities Newton cared about. In particular, if (F_0,F_1,F_2,F_3) is four-force then Newton's force is (F_1,F_2,F_3)\,. In short: you can ignore the time component. Same for momentum. Power is neither a four-vector nor a three-vector. 0 The reaction force on the rocket engine depends on the change in momentum of dm which is caused by the engine, but not that caused by gravity. 0 I think it is because the force \vec{F} is proportional to the mass of the system: most often, it is the gravitational force \vec{F}=m\vec{g} and therefore on the mass dm, \vec{dF}=dm\vec{g} which leads to a second order term, in dmdt, for the impulse. 3 This Hamiltonian operator is derived from the classical one H = \frac{p^2}{2m} + V(x) where p and x are replaced by the corresponding operators : \begin{align} \hat x \psi(x) &= x\psi(x) \\ \hat p \psi(x) &= -i\hbar \partial_x\psi(x) \end{align} The - signs comes from the i in the expression for \hat p, so your question boils down to : ... 2 You can see how the operator acts on a plain wave \psi_k(x)=e^{ikx}:-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} e^{ikx} = -\frac{\hbar^2}{2m}(ik)^2e^{ikx} = \frac{\hbar^2k^2}{2m}e^{ikx}$$As you can see, the imaginary unit comes down twice when calculating the derivative and compensates the minus sign in the operator, so the eigenvalue is ... 0 I'm not sure I understand your specific question, but if you're wondering why there's a negative sign for the KE term which is conventionally positive, note that you have written down the Hamiltonian operator. In QM the Hamiltonian operator is used to give you the total energy by acting on the wavefunction of the system (in this case). However it is not the ... 1 The main reason I have ran into is so that the wider front forks on a motorcycle do not hit the fuel tank which is generally located close to the front forks. Also most motorcycles have a "triple tree" type front end which has to have more frame clearance than the typical bicycle steering stem needs. 0 I will discuss first the case of a force exerted between two objects by a cable being reeled in. From there I will move on the the case of gravity. Let it be granted that F=ma holds good. And I need to make this explicit: let it be granted that F=ma will obtain the same at every location of space, and independent of orientation in space. Let there be two ... 0 There are a lot of factors at play that will dictate what happens in practice, some of which have been touched upon in the previous answers. An additional one that is important is the exact nature of the interaction between the blade and the stock- how large the patch of contact is, and how long it is applied for. The stock will not instantaneously shift ... 3 If two objects didn't have equal and opposite gravitational forces that would break conservation of momentum. Change in momentum is defined as \Delta p = F \Delta t. Conservation of momentum for two objects in a closed system goes as p_{1i} + p_{2i} = p_{1f} + p_{2f} So therefore F_1 \Delta t = - F_2\Delta t; F_1 = -F_2 If, for example, a skydiver ... 3 For an object to appear "flashed", I am guessing it would move at the same speed as say an airplane propeller, since they also appear "flashed". Such propellers have on average a rotational speed of \approx 2500rpm \approx 43rps and the length of their blades are on average 1.5m. This would give each blade (at the tip of it) a speed ... 0 If I understand well, what you are trying to simulate is essentially a simplified version of what is computed by a molecular dynamics software with periodic boundary conditions (although you were a bit vague as to how your boundary conditions exactly work). As a first advice, I recommend that you check out such softwares (LAMMPS, gromacs, etc…) to see if ... 0 First, you need to be careful in how you apply the conservation of momentum. In an inelastic collision (where both objects stick together) you need to look at the total momentum before the collision and the total momentum after the collision. This is not m_a * v_a = m_b * v_b as you suggested above. Rather, the correct equation is:$$m_a * v_i + 0 = (m_a +...

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Lets start from the beginning: Physics theories are mathematical models that fit existing measurements and observation and, very important predict correctly new measurements and observations. In order to pick from the plethora of mathematical solutions those that can be used to consistently fit the data extra axiomatic statements are used, as strong as the ...

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How do you move on Earth? You push the Earth in one direction and you accelerate in the other direction. How do you move in space? You don't have an Earth handy, so you carry whatever you will push with yourself. The thing you push in a rocket is called "rocket propellant". You can't carry much, that's why you push it really hard. And you can't ...

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I believe you already gave the explanation: in the combustion chamber, the pressure is very high and pushes back. While the pressure at the outlet of the nozzle is much lower. The fact that gas escapes into a vacuum does not remove the pressure difference.

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Force comes from Energy, via Work being done. It is the transference of Energy, which normally we cannot see and do not know exactly what "energy" is, only observe its effects on other things. One of those effects we call force. Others can be "transfer of heat". When the energy transfer reaches a threshold we can observe change, often we ...

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