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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 ...


8

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 ...


7

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 ...


5

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 ...


5

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 ...


4

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 ...


4

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 ...


3

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 ...


3

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 ...


2

Assuming no air friction, you compute the initial velocity needed from conservation of energy: $$ \frac12 mv^2=mgh $$ The impulse needed is $mv=F\Delta t$. The product of these ($F, \Delta t$) is constant - shorter time implies higher force. The above assumes the time of impact is short enough not to affect the over all time (otherwise you need to solve ...


2

The external force acts only for the small time when the cue has been struck. Once it moves, there is no force. This means that the ball is moving with zero external force, which means according to Newton's second law, the velocity of the ball is the same. here the act of friction is of less importance as it requires in a billiard play. So the center of mass ...


2

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 ...


2

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 ...


2

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 ...


2

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 ...


2

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 ...


2

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 ...


2

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 ...


2

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: ...


1

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.


1

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.


1

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.


1

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 ...


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 ...


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 ...


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 ...


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 ...


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 ...


1

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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