New answers tagged

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With problems like this you need to decide what reference frame to choose. The obvious choice is the centre of mass frame because this is the inertial frame in which the total momentum is zero. Conservation of momentum then assures us that the total momentum will remain zero. To see how this helps consider your example of the five electrons and one ...


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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 ...


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I) There are already several good answers. OP is asking about the momentum of the non-relativistic string in the transversal model, 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; $Y$ is the ...


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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 ...


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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 ...


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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.


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A simplified answer is obtained by ignoring all possible effects and dependencies other that the mass of the earth (M), its angular velocity ($w$), the mass of the impacting object($m$,) and its velocity ($v$). The rotational energy of the earth is $I w^2/2$, the kinetic energy of the object is $mv^2/2$. To stop the rotation and then make it go in ...


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But the acceleration is not a partial derivative! Its a total derivative, $\frac{\mathrm dp}{\mathrm dt}$, with a $\mathrm d$ instead of a $\partial$. Anyway, I guess you might want to read about the Hamilton-Jacobi equation.


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Momentum and position are conjugate variables in classical mechanucs, which means they satisfy the Poisson bracket relationship. When quantum mechanics was invented the Poison bracket relation was replaced by the operator commutation relationship which results in the relation under consideration.


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These relations are found in every book on QM, but the usual notation is $$ X|x\rangle=x|x\rangle $$ and $$ P|p\rangle=p|p\rangle $$ To go from these equations to the ones you've written, you just have to project them into the position basis $|x'\rangle$ (and use $\langle x'|x\rangle=\delta(x-x')$ and $\langle x'|p\rangle\sim\exp[ipx]$). Edit Important: ...


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Ab initio the momentum operators can be constructed using de Broglie Plane waves In one dimension, using the plane wave solution of the Schrodinger equation,the wave function Psi = exp. i (kx -wt) , if one takes the partial derivative w.r. to x of the wave function delta/delta x (Psi) = ik. Psi and using de-Broglie relation p = hbar . ...


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Momentum is the generator of spacial translations, even in classical physics. Anyway, you can find a derivation here or in Sakurai's book Modern Quantum Mechanics. They are more or less the same and go like this: The translation operator is the operator $T( a)$ such that $$T( a) \mid x \rangle = \mid x+a\rangle$$ From the definition it follows that the ...


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Theoretical physics is about constructing a mathematical model which we hope describes the phenomena it's being modeled for and hence helps predicting stuff. In classical physics this mathematical model is based simply on the real numbers (at least locally) because of the nice behaviour of things. In quantum mechanics it's not the case. Experiments started ...


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That $\hat{P} = - i \hbar \partial_x$ generates translations comes from a straight-forward computation: if $\psi$ is continuously differentiable, and $\Psi$ as well as its derivative are square integrable, then you can prove that \begin{align} i \frac{\mathrm{d}}{\mathrm{d} y} \bigl ( \psi(x - y) \bigr ) \big \vert_{y = 0} = - i \partial_x \psi(x) ...


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Historically, you probably want to start with the de Broglie relations (i.e. $p = \hbar k$), which are just a wild guess. This immediately pops out the form of $p$ as an operator if the wavefunction is a plane wave. Mathematically, $p$ should be defined as the generator of translations (or equivalently the conserved quantity corresponding to translational ...


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You are absolutely right in everything you said. The momentum is non zero only if the wave has a longitudinal mode, which is in fact the realistic case. Moreover when this is the case, the wave equation is not that simple. Let me try show this. Longitudinal Mode Let us assume that when in equilibrium the string, of density $\mu$, is along with the $x\equiv ...


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A fake derivation We can rather easily compute a horizontal velocity for the string fi we assume that the total velocity vector is everywhere normal to the string (this assumption is not always valid, see below). The following picture then illustrates the computation: Take two infinitesimally separated points $x$ and $x+\mathrm{d}x$ and let the wave ...


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Quick answer My question is how does the presence of nonzero $J(x)$ results in a non-trivial spacetime dependent value of $\langle 0|\phi(x)|0\rangle$? The equation $\phi(x)=\mathrm e^{-iPx}\phi(0)\mathrm e^{iPx}$ works both for $J=0$ and $J\neq 0$. Therefore, $$ \langle \phi(x)\rangle_J= {}_J\langle 0|\mathrm e^{-iPx}\phi(0)\mathrm e^{iPx}|0\rangle_J ...


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Newton was not the first to devise or discover the concept of momentum. This goes at least as far back as Fr. Jean Buridan (ca. 1295 - ca. 1358), professor of natural philosophy and rector of the University of Paris, who was the first to introduce the impetus theory of projectile motion. See this translated excerpt from his Quaestiones super octo physicrum ...


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Finding the missing equations Coordinate transforms just complicate the issue. The heart of the matter is that in n dimensions you have n degrees of freedom for the velocities of the COM of each sphere, and you only have n momentum conservation equations plus one energy conservation equation. That means you need an additional n-1 equations to solve the ...


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Both states $\Psi_{k,\sigma}$ and $\Psi_{k',\sigma'}$ are meant to be states of the same particle species i.e. they have the same values of the squared mass $k^2$. The inner product of one-particle states from different species $s$ is zero which one might indicate by additional $s,s'$ labels and a Kronecker symbol $\delta_{s,s'}$. Weinberg claims about the ...


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Use the work-energy theorem to find the speed of the fragment that you were travelling in after the collision. Find the net momentum North and the net momentum West after the collision. Find the magnitude of the momentum and its direction after the collision which will be the magnitude of the momentum and the direction of the car and occupants of mass 1600 ...


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Newton's third law is a statement that momentum is conserved, so it is equivalent to the law of conservation of momentum. Conservation of momentum follows from a fundamental symmetry (of the action) called space shift symmetry and as far as we know this applies to all our physical theories. So Newton's law is still valid but has to be treated with some care ...


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The linear momentum of the system is always conserved when there is no external force acting on it, so you should always consider it as a whole. And as said in the comments, it doesn't matter which point you chose for calculating angular momentum.


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Mandelstam $t = (p_1 -p_1')^2$ corresponds to the square of the momentum transfered between the two scattering particles, in an elastic scattering process. If you look at the scattering in the centre of mass frame, like you suggest, then clearly they cannot transfer any energy between them, since that would vilolate conservation of momentum.


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This is one of those three part dynamics questions. For the first part you need to use energy conservation to work out the horizontal speed of the person just before hitting the pole. The second part is the application of the conservation of angular momentum about the pole's pivot point when the person grabs hole of the pole. Note that the collision between ...


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This is a quite subtle problem. You have to be careful about three different situations. A ball can be thrown with velocity (relative to the ground): a) $v_0-v_e$. b) $v(t)-v_e$, where $v(t)$ is the velocity of the car just after the ball is thrown. c) $v(t)-v_e$, where $v(t)$ is the velocity of the car just before the ball is thrown. You actually stated ...


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All of your math is correct, but your result conflicts with our intuition as well as the reality of how rockets work. Here is why: You assume that "The ball will be moving at $v_\mathrm{ball}=v_0-v_e$". This approximation is valid only in the limit where $m$ is much smaller than $M$. (i.e. the limit of a continuous stream of tiny balls, like a normal ...


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All of these resources are saying the same thing, but you have to pay extremely close attention to the definitions of their differential operators. Specifically, in the brown.edu link, they define the divergence of a tensor $\mathbf{A}$ as $$ \nabla \cdot \mathbf{A} = \frac{\partial A_{ij}}{\partial x_i} $$ with summation over the first index of ...


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It's because you have to replace the erroneous $\gamma$ by $\gamma^2$ (and similarly $m$ by $m^2$) in the inner product and because $$\gamma^2 m^2 c^2 - \gamma^2 m^2 v^2 = \gamma^2 m^2(c^2-v^2)=\dots$$ and $$\gamma^2 = \frac{1}{1-v^2/c^2} =\frac{c^2}{c^2-v^2} $$ and $c^2-v^2$ from the explicit factor cancels against the denominator of $\gamma^2$, while ...


1

I will try my hand in simplicity... Momentum is the measure of motion. So, it is a measure of how much stuff is moving, and how fast that stuff is moving. What we call mass at the very basic level is "amount of stuff". Very interestingly, more stuff is heavier (pulled more strongly by the Earth - gravitational mass) and harder to move around (change ...


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You are studying the very basics of mechanics. In that case, as @dmckee♦ quoted, we will discuss it on the fundamental level. Momentum is the characteristic property of a moving body. There are many dynamical variables associated with the motion of an object, like the displacement, velocity, acceleration etc. But, these quantities are just only variables ...


1

I would suggest not using "relativistic mass". The strength of the concept is that it preserve non-relativistic formulae such as: $F = ma$ and $\frac{dp}{dt} = F$. which was valuable in 1916, but is dated in 2016. It's more productive to work in Minkowski space, where rest mass: $mc^2 = \sqrt{E^2-(pc)^2} $ is a four-scalar (the same in all frames), and ...


0

"when it collides directly" is to be interpreted as a "head-on" collision? So the balls go off along the line of the original velocity direction. The question tells you that the direction of $A$ is reversed so the direction of $B$ must be in the opposite direction ie in the original direction of $A$'s velocity otherwise momentum cannot be conserved.


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The point is that you cannot "make the gun unmovable". Total initial momentum is $0$, so, by conservation of momentum, we will have: $$m_{(gun)} v_{(gun)} + m_{(bullet)} v_{(bullet)} = 0$$ So that $$v_{(gun)} = - \frac{m_{(bullet)}}{m_{(gun)}} v_{(bullet)}$$ If you want $v_{(gun)}$ to be exactly $0$ (with $v_{(bullet)}\neq 0$, of course) you have to ...


26

When you say we make the gun unmovable what this really means is that you are fixing the gun to the Earth. So now when you fire the gun the momentum of the bullet must be equal and opposite to the momentum of the gun + the Earth. So when you fire the gun you change the velocity of the Earth very slightly. However the Earth is so much more massive than the ...


1

That the bomb breaks apart due to the explosive forces which are internal to the system, has nothing to do with the trajectory of the center of mass. As user Sammy Gerbil points out correctly, the initial trajectory of the bomb was parabolic and even if the bomb exploded into million fragments of different masses, the center of mass would continue on it's ...


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UPDATE : The explosion itself conserves linear momentum, regardless of how small the fragments are. If we ignore gravity and air resistance and all other external forces, there is no change in total momentum. This is because the internal forces all occur in equal and opposite pairs (Newton's 3rd Law). If we take the external forces into account, then ...


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It will stay the same, if we neglect the variation due to gravity (every external force is going to change the momentum). If we assume a uniform distribution of the shrapnels' mass (same size for all shrapnels), the shrapnels going in the direction the bomb was originally going will have, on average, higher velocity. With a great simplification, we can say ...


0

Consider two objects, 1 and 2, colliding for some short time interval $\delta t$. During $\delta t$ let's ignore all forces except the contact force that 1 exerts on 2, $\vec{F}_{12}$ and the contact force that 2 exerts on 1, $\vec{F}_{21}$. As long as the objects touch each other, both of these forces exist, and by the principle of Newton's 3rd Law we know ...


1

I come to think that when a mass collides with another, both of them should always have equal velocities post-collision. To paraphrase Feynmann, no matter how beautiful you may believe your reasoning to be about this, if it doesn't agree with experiment, it is wrong. If the two objects 'stick together' after the collision (the collision is totally ...


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Often when two object collide it is often represented as an instantaneous impulse exchange. However in reality this happens continuously. Namely both objects are not completely rigid and will deform during the collision, storing energy in the elastic deformation (like a spring) and dissipating energy with any inelastic deformation. During such a collision ...


1

What is conserved during a collision is not velocity, but momentum (mass times velocity). If the collision is elastic, kinetic energy is also conserved: https://en.wikipedia.org/wiki/Elastic_collision. If the collision is inleastic, only momentum will be conserved: https://en.wikipedia.org/wiki/Inelastic_collision The resulting equations (which you can ...


1

Yes, there is an integral, which comes from the LSZ reduction formula, $$ \langle f|i\rangle\sim \int \mathrm dx\ \mathrm e^{ikx}\square_x G(x) $$ where $x=(x_1,x_2\cdots,x_n)$, $k=(k_1,k_2,\cdots,k_n)$ and $G$ is the $n$-point function. If you go to momentum space you'll get that integrand depends on $x$ only through exponentials, and therefore there is a ...


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There is no need of integrating over anything to obtain the 4-momentum conservation. Indeed, if you think in terms of perturbation theory (Feynman diagrams), each vertex conserve momentum, so that the diagram itself automatically conserves momentum.


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The experiment certainly does produce a very general complex superposition of momentum eigenstates. The spread is not "small" in any way – virtually all allowed (by conservation laws etc.) final states are represented in the superposition for any initial state. We detect particles of particular momenta in the final states because the detectors (e.g. at the ...


0

To calculate the minimum energy needed for the reaction the products are assumed to be stationary, i.e. the momentums are zero. With $ \pmb p^2 = E^2 - \vec p^2 = E^2 = m^2$ follows: $$({p_1^\mu}' + {p_2^\mu}' + {p_3^\mu}' + {\bar p_4^\mu}')^2 = (E_1 + E_2 + E_3 + E_4)^2 - (\vec p_1 + \vec p_2 + \vec p_3 + \vec p_4)^2 = (E_1 + E_2 + E_3 + E_4)^2 = (m_p + ...


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but still we have momentum telling us that both blocks must rise to the same height That's not true here. In the first case where the bullet is embedded, the final velocity of the bullet and the block must be identical. Since initial momentum of the two shots were the same, then the final momentum will be the same as well. Because they are connected, ...


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The loss of energy (efficiency<1) manifests itself in two ways Friction. When doing free body diagrams add frictional forces at the sliding contacts and find which coefficient of friction $\mu$ gives you the efficiency values you expect. Structural damping (hysteresis). This is more difficult to put in the equations of motion, because they assume ...


0

First of all, the transmission of force between any two pairs of mating gears happens due to surface contact between the two and hence through friction. Next, any loss is a dissipation in energy. Though I am not that well-versed in vibration analysis, I know that an periodic dissipation of energy is accounted in the force-equation using a damping ...



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