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The problem does not mention any radii, but if we did know the radius of each sphere, would it be possible to skip conservation-of–linear momentum calculations altogether The accepted answer has confused you: You can simplify things by considering these conservation laws in the center of mass frame. There the total momentum is zero, therefore ...


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That formula for momentum is only true for massive particles. Here's what is always true: A particle with a mass $m$ ($\geq 0$) can have an arbitrary momentum $p$ (in some direction, with magnitude $\geq 0$). The energy of such a particle is $$ E = \sqrt{m^2c^4 + p^2c^2}$$ The velocity of a particle is equal to $$ v = \frac{pc^2}{E} $$ When $m = 0$, $E ...


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The cases, assuming the spheres are not rigid: Collision is perfectly elastic=they will only touch Inelastic collision=particles can get displaced because of the compression caused Plastic collision= spheres would stick to each other, the particles will be displaced. Thanks


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They will not touch. This prevented due to the Pauli exclusion principle. Also, the electrical repulsion of the electron prevents this. Momentum is conserved, however.


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No, the (elementary solution for the position representation of the) wavefunction of a free particle, $\psi(x) = \mathrm{e}^{\mathrm{i}px}$ is not an "explicit function of both" position and momentum. It is a function of position - the momentum of the plane wave is fixed, and the momentum space wave function of this is its Fourier transform $\tilde{\psi}(k) ...


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In an elastic collision involving two objects both momentum and energy are conserved. You can simplify things by considering these conservation laws in the center of mass frame. There the total momentum is zero, therefore one object will have the opposite momentum of the other. Since energy is the square of the momentum divided by twice the mass, this means ...


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Answer to the title of your question : "NO!"... First of all,in general, in most of the mechanics questions, you should make it clear that what is your system and what conservation laws are applicable on it. Like if you have a system on which some constant net external force is acting throughout the time duration you are analyzing the system, you can not ...


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The first thing to note is that the angle of incidence and refraction are equal only for elastic collisions. In such a case, the kinetic energy and momentum of the ball-plane system will be conserved. The motion of the plane is a factor on how the ball moves. Let us assume here that the plane has a mass significantly greater than the ball. The angle of ...


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.... We are told that all the subsequent collisions involving the balls and floor are elastic. We are asked to determine the maximum height to which the small sphere will rise on the rebound. The problem does not mention any radii, but if we did know the radius of each of the spheres, would it be valid to bypass conservation of linear ...


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From my readings; the key to conservation of momentum appears to be based on defining a closed system to see if any mass crosses the boundaries of the system.


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Check the principle of relativity of Galileo (there are no absolute velocities, only relative ones). Both scenarios are the same and the only difference is how you choose to a stationary frame. One usually assumes that the surface/wall doesn't move at all, given the eventual very high mass (compared to the ball), to simplify the analysis.


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The question is so ambiguous that it allows a resounding yes. This is because "balance" is not defined, neither are the dimensions and the material used for the pencil, nor the location of where the "balancing" is to happen. The material and the shape of the surface to balance the pencil on, are not specified, nor the length of time it should stay balanced. ...


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Let me be the firs to answer 'Yes' (more or less). As the saying goes: In theory there is no difference between theory and practice. In practice there is. What I'm getting at is that there will always be differences between theory and practice, and that it is up to the physicist to decide which assumptions/simplications are suitable and which ones are ...


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I'm not sure I have fully grasped what you are asking. The equation is for calculating the momentum of the proton in the inertial frame of the observer i.e. the frame velocity is zero by definition. The only thing moving is the proton, at a speed $v$. If, as in your question 2, you have a different frame then the speed of the proton in that frame is given ...


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No. The weight of the pencil is roughly 1 Newton, and the area is about 500 square picometer (5 * 10-22) which means the pressure on the tip is around 2 ZettaPascal. That's quite a bit more than what graphite (or diamond) can withstand (that's masured in GigaPascal)


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No. Firstly, the point of the pencil is generally not sharp enough to have just one single atom. People attempt to make that kind of tip in STM's. Even if you somehow did manage to get it sharp enough, graphite is so soft the the weight of the pencil will crush the tip. It won't stay a single atom wide. So there is no way to balance the pencil on a single ...


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TL;DR: there are many factors that prevent a pencil remaining perfectly balanced. The most important of these is the uncertainty principle that will make the pencil fall over in less than four seconds. For details, read on... Short answer: NO. The first photon of light that hits it would disturb your perfect equilibrium. The moon's tidal forces (which are ...


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No. To balance perfectly, the pencil would have to be perfectly upright and perfectly still. The uncertainty principle limits how well you can do both at the same time. Momentum and position form a conjugate pair. $\Delta x \Delta p \geq \hbar$. Angular momentum and angular position form one too. $\Delta L \Delta \Theta \geq \hbar$ This doesn't guarantee ...


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Isn't it just a change in the reference frame and therefore the forces are equal? Update: Suppose your initial conditions: the ball ($m_1$) hits the man ($m_2$). Its velocity is equal to $u_1$. As you've said: $\m_1(v_1-u_1)/t$=$\-m_2(v_2-u_2)/t$ Now consider that an observer is moving at the velocity $u_1$. The ball looks static and the man looks like he's ...


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Without going into wave equations, lets just say the segments of a string does not only move along the vertical direction. There are horizontal movements as well, although to a much smaller amplitude. A segment is being pulled towards the source horizontally when departing the equilibrium position and pulled back when heading back from maximum amplitude. ...


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According to quantum mechanics, $p=\frac{h}{2\pi}k$,where $k$ is wave vector and $h$ is Planck's constant. As we know $k={2\pi\over \lambda}$, where $\lambda$ is the wavelength of the wave. So momentum and wavelength are associated to each other. Moreover we can view as the motion of the every particles as well as wave and particle.This is known as duality ...


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I did not get my copy of Srednicki out but from what you have written... Srednicki is referencing the method of steepest descent. Although these notes look to be better than wikipedia. Another page that is directly applicable to the quantum field theory case is here. In short, exponential integrals may be estimated by the saddle points of the integrand. ...


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Absolutely. If Compton scattering occurred in first order in $e$, the only contributing diagram would be the obvious one. Say we're in a frame with the electron initially at rest and an incoming photon in the $z$ direction. Then the electron 4-momentum is $$p^\mu_{\text{in}} = (m,0,0,0)$$ while the photon 4-momentum is $$k^\mu_{\text{in}} = ...


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$\newcommand{ket}{\left| #1 \right>}$ $\newcommand{bra}{\left< #1 \right|}$ $\newcommand{\bk}[3]{\left< #1| #2 |#3\right>}$ In a one dimensional problem $\langle \hat p \rangle$ is always zero. $$\langle \hat p \rangle = \bk{\psi}{\hat p}{\psi}=\int \mathrm{d}x \,\psi^* \hat p \, \psi \propto \int \mathrm{d}x \, \psi^* \psi' \overset{(1)}{=}-\int ...


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A hint on this could be the fact that a superposition of stationary states of different energies is NOT a stationary state, because you can not express the wave function as the product of a single time-dependent exponential tiames a spatial function.


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I'll make an example, to make things clear. Take a two body system, in which the particles are seperated by a constant distance $d$ and have mass $m_1 = m_2 = m$. This is a holonomic constraint, since $$ | \vec{r}_1 - \vec{r}_2 | = d $$ with the particle-positions $\vec{r}_1$ and $\vec{r}_2$. This system is therefore reduced to 5 degrees of freedom (6 minus ...


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I think this should help clear things up. Suppose you take a rod at rest and apply a force $F$ perpendicular to the rod at a distance $r$ away from its center of mass for a short time $\delta t$ - short enough that the orientation of the rod does not change much during the time the force is applied. The rod's linear momentum will become $$F\delta t$$ (from ...


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Without specifying the domains of the involved operators all the discussion has not much sense. Let me say that, if $A :D(A) \to H$ with $D(A)\subset H$ a linear dense subspace of the Hilbert space $H$, $A$ is self-adjoint if $D(A)=D(A^\dagger)$ and $A=A^\dagger$. Notice that consequently (I stress that the converse is false) self-adjointness implies ...


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It was an incorrect statement, as it is explained here by Griffiths himself. Anyways, the mathematical explanation is straightforward: given a self-adjoint operator $A$ with domain $D(A)$, any sufficiently regular real function $f(A)$ of it (and the square is perfectly ok for $-\Delta=p^2$) is self-adjoint on some domain $D(f(A))$ by the spectral theorem ...


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Acceleration of the center of mass is always $F/m$, so if force and mass are the same, the center of mass will accelerate the same way, regardless of the point where the force acts. After the same time of experiencing the same force, the body in the rotating case has greater kinetic energy than in the non-rotating case. This is due to greater work done by ...


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I'm writing this assuming you're talking about something like a disk that cracks and the pieces move without interacting with each other. You need to conserve energy: $m_0v_0^2 = m_1v_1^2 + m_2v_2^2 $ $I_0w_0^2 = I_1w_1^2 + I_2w_2^2 $ where $v_0$, $v_1$, and $v_2$ are the magnitudes of the velocities $v = \sqrt{v_x^2 + v_y^2}$ You also need to ...


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Choosing any two points in the trajectory exposes one problem: if the two points are along the width of the circle, both momentum vectors are going to be parallel and the difference is a vector that doesn't points inwards but is tangent to the trajectory. In case the two points are the same, but one is placed "one lap" later, both vectors are going to be the ...


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Let's solve this by considering the time-reversed process: two spinning objects collide inelastically. We know that linear and angular momentum are conserved: $$ m_1\vec{v}_1 + m_2\vec{v}_2 = m_0\vec{v}_0 $$ $$ m_1\vec{r}_1\times\vec{v}_1 + m_2\vec{r}_2\times\vec{v}_2 = m_0\vec{r}_0\times\vec{v}_0 $$ The latter can also be written (assuming you have the ...


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I'm not sure about the “pick it up with a laser” part, but let's simplify by assuming that we can hit the golf ball from below the ground with photons. Let's further assume that the mass of the ball is $m_{ball} = 46\,\mathrm g$ and we want to hurl it five yards away ($\approx 4.6\,\mathrm m$). If the ground is level and we neglect air drag and wind, the ...


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Hurl it several yards away Rather than a simple yes/no, let's figure out some numbers - as usual I work with "round" numbers. Mass of golf ball = 50 gram "several yards" = 5 meters Projectile launched at 45° with initial velocity $v$ travels for a time $$t = 2\frac{v}{\sqrt{2}g}$$ during which time it will travel a distance $$x = v_x t = ...


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If the ball was completely reflective, yes. The concept was discussed e.g. here http://en.wikipedia.org/wiki/Solar_sail edit: e.g.- radiation pressure is what keeps Sun from collapse


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What's the difference between where you are and space? Atmosphere (ie air pressure), temperature, gravity, radiation. Think about how each of those affects what happens to someone after they pull the trigger of a gun, on earth. Atmosphere: has no significant impact on the effects of recoil. It will help slow the bullet down but that's not relevant to ...


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As the other answers have stated, the primary oversight in the original question is the mass of the astronaut/cosmonaut holding the firearm. However, your original number for the mass of the projectile is off by an order of magnitude. Therefore, the original calculation - as well as some of the other samples provided afterward - are all still an order of ...


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The problem (read error) with your calculation is when you assume the speed of the bullet to $$v_{bullet}=800 m/s$$ This is the speed the bullet gets if the gun is being held relatively still by a person standing on fixed ground. The speed of the bullet comes from the impulse of the high pressure in the piston after the "explosion". Since the impulse is ...


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For most guns, you can roughly hold them in place while fired. That is, the repulsion will not only "hit" the gun's mass but the astronaut's mass too, not allowing the gun to gain such high speed. With your numbers this leaves at most $$ v \approx 0.11~\text{ms}^{-1} = 0.38~\text{kmh}^{-1} $$ for an astronaut + spacesuit + gun with $m=225~\text{kg}$, if no ...


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You've calculated the speed of a remote-triggered gun after it fires the bullet, true. However, there's actually nothing about space in your calculation, as @ACuriousMind noted. In theory, a gun fired on Earth could fly off just as fast, at least for a second. What you should use is not $m_\mathrm{gun}$ but $m_\mathrm{gun} + m_\mathrm{person}$. The gun never ...


1

$\newcommand{ket}[1]{\left| #1 \right>}$ $\newcommand{bra}[1]{\left< #1 \right|}$ $\newcommand{bk}[2]{\left< #1 \big| #2 \right> }$ Though I am not sure 100% if what I am going to do is legitimate I would suggest the following (I am about 90% sure that it is legitimate): The confusion arises because the author has used $x'$ for two distinct ...


1

In my opinion, your explanation's weak point lies with the intuitive jump you make by take the limit of $\Delta t$. You may need some mathematics or a diagram to support this intuition. For example, drawing a diagram showing the velocity vector and the small change in velocity vector, showing that in order to keep the magnitude of velocity constant, then the ...


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When you compute the final velocity of the parcel you have forgotten that it's no longer traveling at 37 degrees - that was the angle at the end of the chute. While it drops, the horizontal component of velocity doesn't change - it is still $3.4\cdot \cos 37° = 2.71 m/s$. With that, you should be able to solve this.


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I) Many of OP's questions on how the Lagrangian formalism works is already addressed in e.g. this Phys.SE post and links therein. For instance the question about the total time derivative in the EL equations is discussed in my answer. II) In this answer, we would like to explain mathematically the various definitions in the Lagrangian formalism (of ...


-1

Why can't the particle borrow some of its momentum from the field or lend some of its momentum to the field thereby increasing or reducing its velocity? Because the particle is the field. We make an electron (and a positron) out of electromagnetic waves in gamma-gamma pair production. In atomic orbitals electrons "exist as standing waves". We can diffract ...


1

It's acutally possible. The phenomenon is called Bremsstrahlung (its direct translations would be something like "stopping radiation"). It can be described purely with classical theories like special relativty and electromagnetism. If a charged particle is accelerating, it "borrows" some of his momentum and energy to the EM-field which is then radiated as ...


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Because the equations for linear and angular motion are very symmetrical In Newtonian mechanics, linear momentum is a vector while angular momentum is pseudo-vector which hints at its true nature as a higher rank tensor object. In relativistic mechanics, four-momentum is a four-vector while angular momentum is a (rank 2) four-tensor. So, the ...



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