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7

Let me rephrase those precise equations in the language of finite-dimensional linear algebra. You have a vector $A$ and two bases $\beta=\{e_i\}_i$ and $\beta'=\{e_i'\}_i$. This means you can write the components of $A$ with respect to $\beta$ as $$ A_i=e_i·A=\sum_j\delta_{ij}e_j·A $$ and the components with respect to $\beta'$ as $$ ...


4

Probably the best way to think about this is to say that $$p = mv\\ F=\frac{dp}{dt}=m\frac{dv}{dt}+v\frac{dm}{dt}$$ (Using the usual product rule for differentiation - thanks @ja72 for the suggestion). If velocity is constant the first term vanishes and your result follows.


4

This is because $H'=UHU^{-1}$ for a certain unitary operator $U$, therefore $\psi$ is an eigenvector of $H$ with an eigenvalue if an only if $U\psi$ is eigenvector of $H'$ with the same eigenvalue. Thus the two operators have the same point spectrum. $U = e^{i \lambda X/\hbar}$. From $[X,P]= i \hbar I$ one finds $$e^{-i \lambda X/\hbar}Pe^{i \lambda ...


4

Your gut feel is correct. Both are exactly the same. Look at the acceleration in both scenarios. 35 mph to 0 in the time it takes for the cars to fold up and stop. Everybody gets this wrong. Good question.


4

If you rely on Newton's second law, the definition of mass turns out to be circular or very intricate as also the notion of (undefined) force appears therein. A better approach consists of starting from the experimental fact that momentum is conserved. In a very theoretical picture you can deal with as it follows. You have a set of bodies and you already ...


4

Why is the vector |S⟩ represented as Ψ for both bases when working out the components for the quantum mechanics case above? The first of the final two equations is simply an expression for the sifting property of the delta 'function'. $$f(x) = \int dx' f(x')\delta(x - x') $$ Let's back up just a bit and write the state (ket) as a weighted 'sum' of ...


3

For Newton's Laws to hold, mass must not vary. Wherever you read otherwise was mistaken. Take a look at this answer. It contains a description of why mass must be constant in Newton's Laws in the context of the rocket equation ... but the analysis applies generally. Newton's Laws are not valid for variable mass systems.


2

As a complement to another answer, I'll demonstrate moving from the coordinate representation (wave function) to the momentum representation. Recalling that $$\Psi(x,t) = \int dp \frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}}\Phi(p,t)$$ and putting this expression into the (coordinate representation of the) TDSE, we have $$i\hbar\frac{\partial}{\partial ...


2

Yes, there are more unknowns than equations. You do not have sufficient information to solve for the requested quantities. Someone might be playing a prank on you! In reality, each ball and each paddle would have a specific finite stiffness, and one could use this information along with some clever math to determine the final velocities of all the bodies. ...


2

Courtesy of its spin the electron has a magnetic dipole moment. That means if we place it in a magnetic field the two states aligned with and against the magnetic field have different energies. The magnitude of the energy difference depends on the strength of the field and the size of the magnetic dipole moment, which in turn depends on the spin. So by ...


2

The two equations are the same to first order, which is all that is important. If I were writing down the equation for the total momentum P(t+dt) myself, I would probably jot down the first equation (that of Morin) since I would be thinking of the instantaneous velocity of the rocket at time t rather than at time t+dt. But, again, the distinction is not ...


2

The collision doesn't happen at a single point in space - rather the colliding objects exert a forces on each other over a distance as they approach and the recede. Consider a tennis ball hitting a racket - the ball and the strings of the racket deform and we get an increasing elastic restoring forces until the two objects at at their closest approach. ...


1

Work (or energy) is transferred from one particle to another, but the net effect is no change overall. How? Consider a collision force acting between two particles over a small time frame. During that time frame the particles move, and the work done on one particle is ${\rm d}W_1 = \boldsymbol{F} \cdot {\rm d}\boldsymbol{x}_1$. Since an equal and opposite ...


1

However, I think that both these answers are circular in nature, as Newton didn't derive mass $m$ in terms of force $F$ , he derived $F$ in terms of $m$. Newton's 2nd law does not "derive $F$ in terms of $m$"; it states if force acting on the body $\mathbf F$, mass of the body $m$ and acceleration of the body $\mathbf a$ are determined independently, ...


1

For two particles to influence each other you need some sort of interaction. For (macroscopic) mass this is clearly Coulomb-interaction. Two atoms can not be at the same place, because their cores repell each other. If you look at smaller scales, strong and weak interaction might add their part. Photons have no charge, no color-charge and don't interact ...


1

Assume that you want the rocket to move to the left. In the combustion chamber there is a chemical reaction which leads to the increase in the kinetic energy of the atoms/molecules and these molecules move in all directions. As momentum is conserved then the increase in momentum of the molecules moving to the right is balanced by the increase in momentum of ...


1

Even a chemical reaction would maintain conservation of momentum. So, the blocked side (rocket) moves forward, and the open side (exhaust gas) moves backward. The chemical reaction is needed to produce the gas. The speed/amount of exhaust alone is enough to compute overall forward thrust of rocket. If you open the mouth of an inflated balloon, it moves while ...


1

As the net external force acting on the rocket-fuel system is 0 N, the system's linear momentum is constant, the same before and after the burning and ejection of some fuel. Note that linear momentum is a vector quantity with a magnitude equal to mass x speed. A change in one direction of part of the system results in a change of equal magnitude in the ...


1

I do not get why system such as the rocket in space are defined as "variable mass" since the mass of the system is not varying. This depends entirely on where one draws the system boundaries. One possible boundary is the rocket plus all of the exhaust gases it has released. The center of mass of the rocket + exhaust gas cloud system moves per the ...


1

For a 2D planar simulation with zero friction do the following Definitions Each body has 3 degrees of freedom. These are $(x_1,y_1,\theta_1)$ and $(x_2,y_2,\theta_2)$ defined at the center of mass. Each body has mass and mass moment of inertia. These are $m_1$, $m_2$ and $Iz_1$, $Iz_2$. The contact is at point A with coordinates $(x_A,y_A)$ and with ...


1

Yes and no. depends on the direction considered. Momentum is a vector quantity. Since there is no force in the x direction, momentum is conserved in that direction but not in the vertical direction because gravity (an external force for the system of Wedge and block) is acting. Yes, for your 2nd comment above. You can use COLM (conservation of linear ...


1

The safest way to start with is the representation-free Schrodinger equation, $$ i\hbar\frac{\partial}{\partial t}|\Psi(t)⟩= \hat{H}|\Psi(t)⟩. $$ Referring to your case, we take the separable Hamiltonian: $H=\frac{p^2}{2m}+V$ so that $$ i\hbar\frac{\partial}{\partial t}|\Psi(t)⟩= \left(\frac{\hat{p}^2}{2m}+V\right)|\Psi(t)⟩. $$ Now is the time the ...


1

The answer is two arcs. One arc with a constant gee loading in one direction and then flipping to the opposite direction. This is called the bang-bang method, and it is no very smooth, but the gee forces never exceed the specified maximum. Given a path $y(x)$ the instantaneous radius of curvature at each x is $$ \rho = \frac{ \left(1+ \left(\frac{{\rm ...


1

We consider friction to an impulsive force, in cases when normal force is impulsive. Here's how:We know that $f=\mu N$(only during slipping motion, for no slipping frictional force is equal to applied force RESISTING friction). Since friction is proportional to normal reaction, it will be impulsive only when normal force is impulsive.Thus, if in a situation ...


1

Take the inverse Fourier transform of your last equation, $\psi(x,t)=\frac{1}{\sqrt{2\pi\hbar}}\int e^{i p x/\hbar}\phi(p,t) dp$, to see that the "coefficients" of $\psi(x,t)$ are not the same in the two representations: in the $x$ representation, the coefficients are $\psi$ and in the $p$ representation, the coefficients are $\phi$.


1

One could not defend a book you are most probably misquoting. I strongly suspect the book says "the mass distribution is not constant", that is M is constant but the distribution and number of constituent $m_i$s may vary, i.e. they may split or agregate, a common feature in astrophysics. You are confusing yourself with symbols and definitions and proofs. ...


1

The external force acts only for the small time when the cue has been struck. Once it moves, there is no force. This means that the ball is moving with zero external force, which means according to Newton's second law, the velocity of the ball is the same. here the act of friction is of less importance as it requires in a billiard play. So the center of mass ...



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