# Tag Info

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When you push on the ball, it pushes back on you with exactly the same force for the same amount of time. Both of you therefore always experience the same impulse. However, the speed imparted to a object by a specific impulse is inversely proportional to its mass. Both you and the tennis ball will move away from the point of contact, but the tennis ball ...

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"imagine a spherical cow, in vacuum...." Newton's Third Law explains the question. Regarding the follow-up in the comment, the trajectory of your spaceship will not be altered, same reason. You cannot push against the back wall without reacting against something else, and that something else is part of the same solid body (or will be when it hits the front ...

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The angular momentum $L_{A/B}$ of a rigid body $A/B$ about its center of mass is $$L_{A/B} = I_{A/B} \omega_{A/B},$$ where $I_{A/B}$ is the inertia matrix of $A/B$ about its center of mass in the world frame and $\omega_{A/B}$ is the angular velocity of $A/B$. The angular momentum $L_{A/B}^0$ of a rigid body $A/B$ about the origin of the world frame is ...

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Here are two separate ways to address the issue you bring up. One is more mathematical---comparing the relations $mv$ and $\frac{1}{2}mv^2$. The other has more to do with force and energy, which I'm calling physical. Mathematical Let's imagine two objects that are moving in the same direction collide with each other. Just to keep things simple, let's also ...

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Let's take an example with simple numbers : 1+2=3 3+0=3 This can represent the momentum conservation. Now look at the sum of squares : 1*1+2*2=5 3*3+0*0=9 The sum is not conserved because the momentum that was transferred changed differently the result of the squares. In a word, kinetic energy doesn't change linearly with speed (which is obvious since ...

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We know that in an inelastic collision that total momentum of the system before collision equals the total momentum after collision. But total kinetic energy before collision is not equal to total kinetic energy after collision. How is possible given that the formula of momentum is $mv$ and the formula of kinetic energy is $\frac{1}{2} > ... 0 Physicist created momentum as the property of the system that is conserved if on the system act only internal forces. A force is defined internal in the system if, the force itself and its reaction are applied on the system. From the previous hypotesis you can demostrate that momentum is defined as $$\vec{p} = \sum m_i\vec{\upsilon}_i$$ The last example ... 0 An elastic collision means that the over all kinetic energy of the entire system before and after the collision is the same. So the ball can bounce off the wall, and the wall can recoil in such a manner that you have an elastic collision. 3 Momentum is conserved in magnitude and direction. So in order to analyze any situation of momentum conservation, you should always start with $$\sum \mathbf p_{i}=\sum\mathbf p_f$$ where the subscripts denote the initial and final momenta. As to the ball & wall, you are correct that momentum is not conserved if you are only looking at the ball. If you ... 2 It is the momentum of the entire system that is conserved. The fundamental reason for this is that the laws of physics are the same everywhere in space. This argument for momentum conservation is called Noether's Theorem. So where did you go wrong in your original example? Well you assumed that the wall was completely rigid. In reality that isn't actually ... 0 Okay, let's first review the initial setup: 1st particle: mass$m$and initial velocity$v_m>0$in +x direction 2nd particle: mass$Am$with assumption$A>0$, and initial velocity$v_A=0$3rd particle: mass$Bm$with assumption$B>0$, and initial velocity$v_B=0Notation: after each collision, the new velocities will have a prime added to ... 0 Some bits and pieces on angular momentum: Angular momentum is that which is conserved in rotationally invariant systems, just like energy is that which is conserved in time translation invariant systems and momentum is that which is conserved in space translation invariant system. This is the essence of Noether's theorem. The analogue in QFTs are ... 0 If you are looking for quantitative values, look to further than the equations of motion \begin{align} \sum_{i} \left( \vec{F_i} \right) & = m \vec{a}_{cm} \\ \sum_i \left( \vec{\tau}_i + (\vec{r}_i-\vec{r}_{cm}) \times \vec{F}_i \right) & = I_{cm} \vec{\alpha} + \vec{\omega} \times I_{cm} \vec{\omega}\end{align} where the left hand side is ... 0 The angular equivalent of the impulse-momentum theorem states The change in angular momentum of a system is equal to the product of the (average) external torque time the time it is applies In math that is (finite version): $$\langle \vec{\tau}_{ext} \rangle \,\Delta t = \Delta \vec{L} \,,$$ or (infinitesimal version): $$\vec{\tau} \, \mathrm{d}t = ... 1 Linear momentum will be conserved when the Lagrangian that describes your system is unchanged by translations in space. This is a consequence of Noether's theorem, and it's as close as you're going to get to a fundamental explanation for the conservation of momentum. In general a Lagrangian written in the coordinate system of an accelerating observer is not ... 1 Some conservation laws are related to conservation of angular momentum. There is a famous example (from Feynman if I recall correctly), where you assume an infinite flat space and conservation of angular momentum about any point, and then you get conservation of linear momentum for free. Intuitively, to get say the x component of linear momentum is ... 0 The canonical (total) momentum is the sum of the kinetic (mechanical) momentum and the potential momentum. Potential momentum occurs only if the potential energy depends explicitly on velocity. Look at a much more simple case: A particle falls down in constant gravity. The potential energy depends only on height. There is no potential momentum. The canonical ... 0 Eric Angle has it pretty much right. In an inelastic collision some of the kinetic energy is absorbed by the deformation of the material. For example, if two balls of putty collide and stick together, kinetic energy is absorbed by squishing the putty. In a second example, if you shoot a bullet at a log, some of the kinetic energy is absorbed by friction ... 0 One does not need to switch "gravity off" to make a system like that. Replace gravity with a simple string, and cut the string. Physically that's completely equivalent to your problem, as far as I can see. And while this may seem counterintuitive, the angular momentum in a system of a mass rotating at the end of a string is, indeed, conserved, when you cut ... 0 Your visualization is a good one for exploring the "no-center" concept of the universe - that is, if you only count the universe as the boundary of the hyper-sphere. Technically, though, it could be wrong. As you'll find as you look at GR (if you haven't already) is that there are three types of curvature: positive, negative, and flat. A positively-curved ... 0 It's a mistake, it happens, the text says clearly :" ...if the collision is elastic...." (wiki:)The conservation of momentum is expressed by the equation \,\! m_{1}\vec{u}_{1}+m_{2}\vec{u}_{2}=m_{1}\vec{v}_{1} + m_{2}\vec{v}_{2}. If you consider m=1 the velocity of M is simply v_M =\frac {\sqrt{8gl}}{M+1} 2 (1) Since u(\textbf{r}) = u(\textbf{r}+\textbf{R}), we can expand this part in terms of reciprocal lattice vectors, u_k(\textbf{r}) = \sum_\textbf{G}{e^{i\textbf{G}\cdot \textbf{r}}u_\textbf{k-G}}. We can therefore write: \psi_{\textbf k+\textbf K} = e^{i(\textbf k + \textbf K)\cdot \textbf r}\sum_\textbf{G'}{e^{i\textbf{G'}\cdot ... 3 Yes. There are only internal forces of your body. Without external forces, the center of mass of your body cannot change position. As your center of mass did not move, the main body should move in the opposite direction. 1 Tong is alluding to the standard trick in the derivation of Noether's theorem by promoting the (infinitesimal) x-independent parameter \epsilon to become x-dependent, see e.g. this Phys.SE post. 1 It seems like the previous answers were based on the abstract or third-party articles about the abstract, but I gather from this Wired article that the full paper provides more details (behind a paywall, so I can only speak for what the article says). The described tests were on the Cannae drive, the inventor of which believed required slots in the drive to ... 2 From the OP's comments I have a feeling that the real issue is not the virtual particle, but an actual annihilation of a positron and electron into 2 real photons. Like here: And the question is, how can the system of 2 massive electrons accelerate into the speed of light, when accelerating to c requires infinite amount of energy. The answer has two parts: ... 4 Consider the following diagram: This shows a mass m moving past a point P in a straight line. Note that the mass isn't connected to P in any way - it's just moving past in a straight line. The angular momentum of m about P is given by:$$ \vec{L} = \vec{r} \times m\vec{v} $$So the direction of \vec{L} is normal to the screen and the ... 0 Ball A has linear momentum pointing in the direction of its flight path. But it is not conserved, because you have a force acting on it (centripetal force), that as no counter part (unless you specify where the axle is mounted and allow for the mount to move as well). Observe, that you can ascribe an angular momentum to B as well even when it does not move ... 0 For each continuous symmetry, infinitesimal transformations may be expressed, by a bracket involving the conserved charge operator associated to the symmetry :$$\delta_\epsilon \phi(x) = i\epsilon [Q, \phi(x)] \tag{1}$$In our case, we must have :$$\epsilon \theta = i\epsilon [Q_\theta, \phi(x)] \tag{2}$$A solution is then :$$Q_\theta = \theta ... 3 In Feynman diagrams all four vectors conform with special relativity algebra, BUT the internal lines, even though they have the name "photon" are virtual, which means the mass can be different than zero, only the quantum numbers identify them as a photon, quark, electron, etc.. Feynman diagrams are an iconic representation one to one with the integrals and ... 3 it seems curious to me that a system can have a mass, and then accelerate to the speed of of a massless particle There is no paradox here. Special relativity doesn't preserve velocity - in fact the opposite - it allows one to move between frames of reference in which the velocity of objects is relative to the observer. [do the particles] accelerate ... 0 This is a very interesting question. The problem is simpler in kinematics. However, if we view it as a problem in dynamics, invoking forces and Newton's laws, then the question becomes a natural consequence and the answers become rather complicated. The question is one of reference frames, in kinematics. The collision is viewed from the laboratory frame of ... 2 The engine isn't designed to use solar power. The thrust^1\$ is due to the motion of microwaves within a resonant cavity, and the microwaves are generated by a magnetron on the space ship. The power source for the magnetron would presumably be a nuclear generator or some other compact way of storing energy. The original 2006 paper describing the drive is ...

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One of my mentors likes to say, "Nothing resembles a new effect quite so much as a mistake." Conservation of momentum is a fundamental principle of mechanics supported by hundreds of years of experimental evidence since the language needed to discuss it was codified by Newton. Certainly it's the case that electromagnetic radiation carries momentum and can ...

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The first thing that should jump out to anyone is the following excerpt from the abstract of the recent NASA paper: Thrust was observed on both test articles, even though one of the test articles was designed with the expectation that it would not produce thrust. Specifically, one test article contained internal physical modifications that were designed ...

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It is possible to generate thrust using EM radiation such as a laser or microwaves. Discussed in this XKCD blog. However this relies on momentum being transferred from the photons of the EM radiation to the object being propelled. This is not possible if the microwaves are completely sealed within a container as indicated in the article. This paper will ...

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In inelastic collisions, kinetic energy is not conserved, so I'm going to assume you mean a totally elastic collision since you say energy is conserved. O.K, so when the ball hits the wall, the speed of the wall before and after is 0, so that means the kinetic energy of the ball is conserved and thus the magnitude of the velocity is the same before and after ...

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