New answers tagged

7

Without touching on electromagnetism, I'd like to bring up this construction from mechanics (it's in the Feynman lectures). Consider two equal particles approaching each other with equal speed. A----> <----B You can argue from first principles that if they stick together they will not be moving afterwards -- any argument you could make ...


0

First of all, I shall assume that the system is kept on a horizontal plane so as to simplify the calculations. Secondly, I shall assume that the line joining the mass $m$ and $M$ is perpendicular to the direction of motion of mass $3m$. Thirdly, I shall assume that the spheres to be point objects (in other words, you can consider it to be a perfect head-on ...


10

As the wiki article you quote states, momentum is defined as the product of the velocity times the mass of an object. Classical mechanics developed theoretically on the lines explained by WetSavanna in the other answer, the conservation of momentum and energy being cornerstones of the theory. Classical mechanics is a very successful theory, and ...


44

Momentum / energy are the conserved Noether charges that correspond, by dint of Noether's Theorem to the invariance of the Lagrangian description of a system with respect to translation. Whenever a physical system's Lagrangian is invariant under a continuous transformation (e.g. shift of spatial / temporal origin, rotation of co-ordinates), there must be a ...


0

Most people believe that you need mass to transmit a force, or even to be able make one, but the more fundamental concept happens not the force but the momentum. The momentum is the capacity to interact with another entity and transfer it some change in speed. In the case of the photon, it is massless but however has momentum and energy. The energy and ...


0

This is newton second law: $$ \mathbf F = \frac{\Delta\mathbf p}{\Delta t} $$ As you can see, its variation of momentum that brings force. So, its the transfer of momentum. Granted, photons are massless, but they do have momentum. There are two ways of a photon transfer momentum to a solar sail: The photon is absorbed (also heats up the sail), or, the ...


2

They are trying to explain the resonant cavity thrusters , controversial propelling engines. It seems to me that the paper you quote confuses photons with light waves, attributing a real space wave nature to the photon. From their fig1: Our reasoning is that when light waves combined with opposite phases, the photons do not vanish for nothing but ...


2

I am not supporting or seconding any finding in the article. They appear to be just incoherent chatting. However the pressure excerted by a electromagnetic wave is $P_{rad}=\frac{I} {c}$ where I is the intensity of light and c is light velocity. In case of totally reflecting surfaces this will be doubled. Here you can see that the pressure is not of one ...


1

force is change in impulse over time, so saying that impulse is lower than static friction does not make sense, in the same way that saying that some speed is larger than some acceleration, they are different things. Impulse from a moving object is transferred to the one at rest through a force, that results in an exchange in momentum. During the ...


4

I am answering the question formulated after the "edit" in a newer version of the text because that one seems well-defined. Indeed, a situation with a uniform field $\vec E$ may be said to be "uniform" or translationally invariant in space. Noether's theorem says that this "uniformity" (spatial translational invariance) implies the existence of a conserved ...


0

I am not completely sure, but I would say, it depends what is the exact phase structure of the entire beam. If I was trying to approach the problem, I would use wave formalism and analyze the phase profile vs time. Good question.


2

The momentum of the photon is $p=h\nu/c$, so it only depends on its frequency, not its phase. At constant frequency, all photons will have the same $p$ regardless of phase.


1

To be able to solve this problem you should assume that all interactions are elastic so kinetic energy is conserved. Do the problem in two stages and consider each stage as a one dimensional interaction: The interaction of mass $m$ and mass $2m$ along the line of the string joining them and this should give you the velocity of mass $2m$. The interaction ...


2

The solution is actually strange for me. It assumes that the distance traveled by each car (in the CM frame) during the acceleration (crumpling) is also equal to the crumple length of each car, but is not always the case. Following the assumption of the solution: In the CM frame, the heavier car moves slower than the lighter car. So that at the moment that ...


3

Regarding total momentum conservation, the point is that in non-inertial reference frames inertial forces are present acting on every physical object. Momentum conservation is valid in the absence of external forces. However, if these forces are directed along a fixed axis, say $e_x$, or are always linear combinations of a pair of orthogonal unit vectors, ...


0

First off I would recommend a very large boot. "Size billion" comes to mind. Second I would query "in what direction are you planning to kick said ball with said billion sized boot, fine sir?" Since direction is not specified in this...ahem...matter...ahem...I find the issue at hand confounding indeed. Still...we all could use a few more hours in our day ...


0

This is similar to Can humans control rotation of the Earth? and How can you find the impact necessary to change the direction of Earth's spin? Suppose the ball is kicked against the direction in which the Earth is rotating. This increases the speed of rotation of the Earth, reducing the rotation period (ie the length of 1 day). The angular momentum ...


1

Your impulse cannot be converted to pressure because you are considering only one collision. Impulse actually has units of force times time. And there are a few trillion collisions per second on any normal container, so you have to spread the impulses through time to get the proper force. The easiest way is probably to estimate the average time between two ...


4

We shall consider the man to be at the middle of the box initially. If the man starts walking left, the box shifts towards right (assuming no friction) such that the centre of mass of the man box combined system remains on the tip. If the man reaches the end of the box and starts jumping, the box begins to oscillate about the centre of mass and lf cases are ...


2

The short answer to your question: Yes, they can. But in the particular example you are considering, they don't. As mentioned by Jahan, it is the gravity that gives the system (man+box) a net non-zero momentum. A rather interesting point to worry about in this scenario is 'who gives the system a net non-zero angular momentum?' Since gravity acts through the ...


1

The net force isn't zero. There is an external force, namely gravity. If you did this experiment in outer space, where there truly was no external force, you wouldn't see the box topple over.


3

Well, you know that the eigenfunction of $\hat p$ is $\exp(ipx)$, so let's try to find what the expectation value of $x$ is, to begin with: $$\langle p | x | p\rangle = \int \exp(-ipx) x \exp(ipx) \, dx = \int x\, dx$$ and this integral doesn't exist. Then it shouldn't come as a surprise that trying to take the time derivative gives nonsense. The underlying ...


0

Update: Comments by Robin pointed out my confusion. Consider the following: $$ [x, p^2] = x p p - p p x = x p p - p x p + p x p - p p x = [x,p] p + p [x,p] = 2 i h p $$ If you plug this in the initial expression, you will get exactly what you would expect from this observable: $p/m$. But formally correct manipulations that you performed provide a ...


21

TL;DR: every time you use momentum conservation. One way to see this is to take a close look at Newton's cradle: Image is published under GNU Free Documentation License You can start with Newton's second law: $$\mathbf{F}=m\mathbf{a}=m\frac{d\mathbf{V}}{dt}$$ By calculating the scalar product with the velocity vector on both sides of the equation we ...


13

The reason why one often thinks that all the familiar phenomena can be explained just on the basis of the second and the first law of Newton is that it is not clearly emphasized (mainly in school textbooks) that Newton's second law for a system of particles can take the form of $F_\textrm{external}=\dfrac{\mathrm d}{\mathrm dt}(p_\textrm{system})$ only when ...


0

Newton's third law emanates from the fact that the momentum of an isolated system is always conserved viz. $$\mathrm d\mathbf p_1 +\mathrm d\mathbf p_2 ~=~0 \;.$$ From this, it can be inferred that $$\int_{t_\mathrm i}^{t_\mathrm f}~ \mathbf F_{21}~\mathrm dt ~=~ - \int_{t_\mathrm i}^{t_\mathrm f}~ \mathbf F_{12}~\mathrm dt\tag 1$$ It could be that the ...


0

This is a tricky question. I think you are all right and this problem, because using momentum, is valid for both elastic and inelastic problem. the tricky part is, particle A cannot penetrate particle B. If, they collide, particle A moves at -97.5m/s and particle B moves with -2m/s, do you think that is possible? So if particle A initial velocity is ...


1

I) There are already several good answers. OP is asking about the momentum of the non-relativistic string with only transverse displacements, which in textbooks usually is given as $$ {\cal L}_T ~:=~\frac{\rho}{2} \dot{\eta}^2 - \frac{\tau}{2} \eta^{\prime 2}. \tag{1}$$ II) Let us fix notation: $\rho$ is the 1D mass density; $\tau$ is the string tension; $...


1

You have the wavefunction $\Psi(x)$ of the particle given in position representation. In this representation, $\int_a^b |\Psi(x)|^2 dx$ gives you the probability to find the particle somewhere in the interval $[a, b]$. You can convert this function into the equivalent momentum representation (by taking the Fourier transform of $\Psi(x)$), in which the ...


0

The compression pulse that propagates through the metal spheres of Newton's cradle are not ordinary sound waves. They are approximate solitons (a nonlinear wave form that balances dispersion against nonlinearity). It is this property of soliton pulses that is responsible for the observed behavior. More Details: Newton's cradle is a physical manifestation ...


2

The simple answer is that not only momentum but also energy needs to be conserved. This puts constraints on the number of balls that can be activated in the cradle. Note that this does not always give a unique solution either. But it enforces that $ n $ balls to $ n $ balls is a "stable" solution.



Top 50 recent answers are included