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8

Notice that $\psi(x)$ is defined on a circle of circumference $a$. Multiplying $x$ on this circle is really multiplying a periodic extension of $x$, i.e., the sawtooth function $x - a\lfloor x/a\rfloor$, where $\lfloor y\rfloor$ means the largest integer not greater than $y$. So, the commutator of the position and momentum operators involves the derivative ...


5

This is what happens if one cares not for the subtlety that quantum mechanical operators are typically only defined on subspaces of the full Hilbert space. Let's set $a=1$ for convenience. The operator $p =-\mathrm{i}\hbar\partial_x$ acting on wavefunctions with periodic boundary conditions defined on $D(p) = \{\psi\in L^2([0,1])\mid \psi(0)=\psi(1)\land ...


5

It is not about momentum. The total momentum is conserved: the fingernail flies in one direction, and you (plus the nail clipper) suffer a recoil to the opposite direction. The thing is that your piece of fingernail has a very small mass compared to your body mass, and the firing speed is actually not so high, so the recoil is negligible. About why the ...


4

The point is that your equation (1) does not make any sense. On the left side, you have operator acting on a vector in some abstract vector space. On the right side you have the position representation of momentum operator acting again on the same abstract vector. Those objects live in different spaces, you cannot just multiply them (actually you can if you ...


3

The answer may be deeper than you expect. The operator $K$ that the OP defined --- here it will be called $-i\nabla_x$ --- is the generator of space translations. And the operator $X$ --- here called $x$ --- is the generator of momentum translations. If you exponentiate them, to form the corresponding one-parameter unitary groups of translation, you get ...


3

Well I would say your first approach is absolutely correct and will result in right answer if the calculations are done correctly (I have not done the calculations) but your second approach is faulty. Problem with second approach In the second approach you have used the conservation of mechanical energy it seems. The problem is that conserving mechanical ...


2

Your first part was correct; but for the second part, you have to equate the energy of A plus the stored energy in the spring to the energy of B (because you start with no energy in the spring, and all the energy as kinetic energy in B). So the expression for the stored energy is $E_\mathrm{spring}=\frac12 k x^2 = \frac12 m_b v_b^2 - \frac12 m_a v_a^2$. ...


2

"For sake of simplicity lets take both objects rigid particles." Well, we can't both do this and answer your question. In such a model, we really just imagine that the particles experience an infinite force for an infinitesimally small amount of time (a dirac delta function). So in this model, momentum is mysteriously acquired instantly. In reality, the ...


2

We don't take conservation of momentum as an assumption, and neither do we take Newtonian mechanics as an assumption. Instead the fundamental assumption we make is that the systems we study in classical mechanics obey the principle of least action. It is hard to overstate just how important this principle is. From it we obtain Lagrangian mechanics and ...


2

Go this this site and find the section entitled "skater's spiral". You will get a nice detailed analysis that will show you that since the velocity is at right angles to the string, no work is done on the bead, so the speed is constant. But the radius of curvature of the path is changing as the string gets shorter, so the tension in the string changes.


2

The x and y coordinates of the mass can be expressed parametrically in terms of the wrap angle $\theta$ (of the string around the cylinder) as follows: $$x=R\sin\theta+(L-R\theta)\cos\theta$$ $$y=-R\cos\theta+(L-R\theta)sin\theta$$ where x and y are measured from the center of the cylinder, R is the radius of the cylinder, and L is the initial length of ...


2

This is essentially the same question as https://astronomy.stackexchange.com/questions/13302/ In a perfectly elastic collision, both momentum and kinetic energy are conserved. The initial momentum is $\text{m1} \text{v1}$ and the initial kinetic energy is $\frac{\text{m1} \text{v1}^2}{2}$, since m2 is at rest. Let u1 and u2 be the velocities of the ...


2

In Galilean relativity (pretty sure that is what you mean/need), you just add up the velocities. So $ V_{p_1} =v/2+x$ and $V_{p_2}=v/2+y=v/2-x$


2

OP's ket first equation$^1$ $$\tag{1} \hat{p}|x\rangle ~=~+i\hbar\frac{\partial |x\rangle}{\partial x}$$ is explained in eq. (7) of my Phys.SE answer here. In short, eq. (1) is consistent with the corresponding bra equation $$\langle x |\hat{p} ~=~-i\hbar \frac{\partial \langle x |}{\partial x} $$ via Hermitian conjugation. The bra equation in turn is ...


2

In a collision momentum is conserved if there are no external forces. enefgy is also cionserved but in the examples kinetic energy is an important parameter. Elastic collisions are ones when kinetic energy is conserved. In non-elastic or inelastic collisions kinetic energy is not conserved and some kinetic energy can be converted into heat, sound and in ...


2

Momentum, energy, angular momentum, and charge are conserved locally, globally, and universally. One must remember that conservation locally (within a defined system) does not mean constancy. Constancy occurs only when the system is closed/isolated from the rest of the universe. Conservation means that these quantities cannot spontaneously change. Let's ...


1

The small object exerts a force in the opposite direction to the normal force on the cart


1

If a ball of say radius $R$ rolls without slipping it has both linear ($p$) and rotational momentum ($L$): $$p=mv$$ $$L=I\omega$$ Where $m$ is mass of ball, $I$ is inertial moment of ball, $v$ is translational (linear) speed and $\omega$ is angular speed. For rolling without slipping the following condition also holds: $$v=\omega R$$ The ball will have ...


1

Note that pendulum motion is only sinusoidal for small angular displacements; as you increase the amount of swing the harmonic approximation fails. Lagrangian mechanics gives you a handle on all of the cases.


1

If you added up the momentum of all the molecules of gas (vectorially), the combined momentum will be equal in magnitude to that of the rocket, and in the opposite direction. In other words, the velocities of the molecules will be biased away from the rocket.


1

Momentum and inertia are closely related properties. Newton's first law states that an object will continue in a straight line with constant momentum if no net external forces act on it. When you apply the brakes, the road applies a net external force on the car-plus-wheels. This force will cause the car to slow down, and the road to speed up (conservation ...


1

I would say the answer is rather simple. If you do various experiments to measure the position (or momentum) of a particle, then QM only gives you consistent and accurate predictions when the mathematical symbols $x$ and $p$ are associated with the measured position or momentum. However, position and momentum are not the only canonically conjugate operators ...


1

This assumes a smooth surface collision. The component of velocity (momentum) along the surface of the wall cannot change because there is no friction and hence no forces along that direction. Because the mass of the wall is assumed to be much greater that the mass of the ball and the collision is assumed to be elastic the normal component of velocity of the ...


1

Exactly. In a 2D problem, it's usually a good idea to break the components into two dimensions based on the environment. In this case, the wall makes the best split, let's say that x is the direction of motion along the wall while y is perpendicular. In this simple situation, the force on the ball can only act in the Y direction. Which has the following ...


1

Force is the rate of change of momentum, $$ \vec F = \frac{d\vec p}{dt} = m\frac{d\vec v}{dt} = m\vec a. $$ If you have an external force (like friction), you are exchanging momentum with the outside world. In this case, conservation of momentum demands that the momentum in your system change. Only with zero net external force can you observe conservation ...


1

Two things are required to prove this. First, you haven't really defined your state $|p\rangle$ in its entirety since you haven't defined what normalization you are using for this state. In the formula above, it seems to me that you are using the normalization $$ \langle {\bf p} | {\bf p}' \rangle = \frac{2 E_p }{ 2\pi } \delta^3 ( {\bf p} - {\bf p'} ) $$ ...


1

As another hint to your problem, you'll also want to consider conservation of energy (there's a big clue in your problem that conservation of energy is important --there is no friction between the cart and the path). To start you off, here are the important conservation of energy equations for your problem: $$E_k = \frac{1}{2}mv^2= ...


1

It relies on conservation of energy and momentum and the equation for energy in special relativity: $E^2 = (pc)^2 + (mc^2)^2$. Here you go. Energy of photon: $E_\gamma = \hbar\omega = p_\gamma c$, where $p_\gamma$ is the momentum of the photon. Assume the electron is initially at rest, so it's energy is simply $m_ec^2$. By conservation of energy, the ...


1

$F = ma$ and the only forces on the cart are gravity and the reaction force of the surface that balance. So the cart does not accelerate: it keeps moving at constant velocity.



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