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6

They're both saying the same thing: the relativistic momentum is given by $$\mathbf{p}=\gamma(v)\,m\mathbf{v}$$ The confusion, it seems, is that you are using Feynman's $m=\gamma m_0$ as equivalent to the $m$ in Resnick & Halliday's text; the actual correlation is Feynman's $m_0$ to Resnick's $m$--both of these terms are the (invariant) rest mass. ...

4

If the initial momentum is $mv$, then when the particle bounces off the wall, it is going in the opposite direction, so its new momentum is $-mv$. The difference is $2mv$ because $mv - (-mv) = 2mv$. By the way, this is not really how it works; molecules will speed up or slow down when they hit the wall. If the molecule is moving slowly, it's likely to speed ...

3

The equation is just the kinetic and rest energy, it does not include potential energy. But potential energy in relativity is not the proper concept. The linked question has some useful answers, but I think your true question is about how to learn to do things relativistically that you used to do non relativistically. And since the other answer so far takes ...

2

Yes, indeed the out-of-plane component of the wave vector of a surface plasmon is imaginary. A purely imaginary wave vector means the wave does not radiate in that direction, but instead is evanescent. (That's what you get if you plug in a purely imaginary $k_z = -i\alpha$ into the formula $$E(z, t) = e^{i (k_z z - \omega t)} = e^{-\alpha z} e^{-i\omega ... 2 Imaginary wavevectors are possible and, as ptomato's answer correctly points out betoken evanescence. I'd like to add a few words to his answer that might help clear up your confusion. Imaginary wavenumbers always betoken Evanescence. Sometimes the vague term "nearfield" is used to connote something not propagating. Evanescence is NOT dissipative; this is ... 2 Imagine this situation: at time t=0, we have a infinite long straight wire with current zero, and a charged particle q with zero velocity. at time t=T, we make the current to be I, thus we have a  \mathbf{B} field, and  \mathbf{A} field. during this process,  \mathbf{A} is build up from zero to some value, therefore we have induced electric field ... 2 You cannot change your linear or angular momentum in open space at all. You need something to transmit it to. if you swing your legs your body will rotate in the opposite direction while you swing, and stop when you stop swinging. If you are out of fuel there is no way to accelerate. Only by releasing mass you could change momentum, as Bender well shows you, ... 2 So, I suppose that Φ(k) is the probability density of the momentum. Is this true? Almost. \Phi(k) is the probability amplitude for the momentum of the particle. The probability density is obtained as usual by squaring the amplitude, giving |\Phi(k)|^2. For a free particle, all values of momentum are always allowed, which enables the superposition ... 2 In a collision it's often the case that it's hard to measure exactly how long the collision lasts and exactly how the force between the objects changes during the collision. Squishy objects like nerf balls will collide relatively slowly while hard objects like billard balls will have a short collision time. However there is a well defined quantity called ... 2 In classical mechanics there is no distinction between free and bound as far as this relation is concerned. In relativistic quantum mechanics (i.e. QFT), a particle that satisfies this relations is said to be "on-shell" or a physical observable asymptotically free particle. It is certainly not satisfied for virtual particles, but they are as their name ... 1 The law states this: Two objects masses and with velocities m_1, m_2, \vec{v}_1 and \vec{v}_2 collide with contact normal \vec{n}. The final velocities are \vec{v}_1^\star and \vec{v}_2^\star such that the coefficient of restitution \epsilon is defined by$$\vec{n}\cdot \left( \vec{v}_2^\star - \vec{v}_1^\star \right) = -\epsilon \;\left( ...

1

$N.s$ is the unit of Momentum and Impulse. Let's consider, what the quantities itself are so that you might be able to correlate them with their units. Speaking colloquially, Momentum is a measure of strength and a measure of how difficult it is to stop an object, and Impulse is the measure of how much the force $changes$ the momentum of that object. ...

1

An elastic collision is defined as one which conserves energy. When you jump against a wall, most of your kinetic energy is dissipated as heat into your tissue as your legs and muscles absorb the impact. Therefore, energy is not conserved so by definition this is an inelastic collision.

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I didn't redo your calculations and I assume that they are correct, which actually doesn't play any role in what I'll describe now. Notice that in the second scenario the 2kg ball will inevitably start to move. By keeping it still you change the reference frame one more time, which invalidates the use of conservation laws. You cannot use the conservation of ...

1

You need to keep the amount of time the same as well. The change in momentum of a system is equal to impulse delivered to it, which is just the time-integral of the force: $$\Delta \vec{p} = \vec{J} = \int_{t_1}^{t_2} \vec{F} \, dt = \vec{F} \Delta t \text{ (if \vec{F} is constant with respect to t.)}$$ So just having $\vec{F}$ constant isn't enough; ...

1

I use this arrangement in my introductory classes when time allows, but not for momentum. I call it the "work-energy mini-lab". (It's a mini-lab because I don't expect a detailed write-up and guide the class through some of the harder analysis.) Some things to note. By measuring how far the mass hanger has to drop you get the distance over which the ...

1

Static magnetic fields by themselves have no momentum, you need an electric field and a magnetic field to have momentum. Also, the momentum comes from the total fields. So even if you think of two magnetic fields $\vec B_1$ and $\vec B_2$ and two electric fields $\vec E_1$ and $\vec E_2$ the momentum density is $\epsilon_0\left(\vec E_1+\vec ... 1 The way to define momentum in a special relativistic context is the following: Start with the trajectory of the particle parametrized by its proper time$x^{\alpha} (\tau)$; define the four-momentum by$p^\alpha = m \frac{\mathrm{d}x^{\alpha}}{\mathrm{d}\tau}$, where$m$is the mass of the particle (note that I'm only using one mass, not distinguishing ... 1 Am I missing something here? Yes. What you're missing is "the mass of a body is a measure of its energy-content". Read Einstein's original paper, and take note of this: "If a body gives off the energy L in the form of radiation, its mass diminishes by L/c²". Next, imagine your body is a massless photon in a gedanken mirror-box. It isn't actually at rest ... 1 In quantum mechanics observables are represented by (some classes) of self-adjoint operators on some Hilbert space. Saying that you can precisely measure a quantity given by the operator$A$means that your state can be one of the eigenstates of that operator. Likewise, if you want to precisely measure two quantities$A, B$together your state needs to be an ... 1 Imagine the rocket before and after throwing a small ("infinitessimal") amount of fuel out its exhaust. You apply the momentum conservation notion by equating the increase in the rocket's forwards momentum with the momentum of the fuel thrown backwards. The easiest inertial frame to do one's analysis in is that of rocket immediately before the increment ... 1$E = pc$is only true for massless particles. For massive particles you have the mass-shell relation:$E^2 = m^2c^4+p^2c^2$After you use$E=T+mc^2$and you can find$p\$

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