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New answers tagged energy-conservation

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Kinetic energy of the billiard ball will be maximum when the ball will hit the billiard ball in straight line with a given velocity.$$mv=mu+MU...........1$$And,$$mv^2=mu^2+MU^2............2$$Now the kinetic energy $\frac{1}{2}MU^2$ is a fraction of the kinetic energy $\frac{1}{2}mv^2$$\frac{1}{2}MU^2=k\frac{1}{2}mv^2............3$$From 2 and 3 find$k$in ... 0 The way I would approach this problem is through the impact parameter,$b$. You can find the definition on Wikipedia. The first thing to note is that for$b > r + R$, there is no collision. Here$r$and$R$are the radii of the two balls. For$b$less than this limit, you can use geometry to determine the direction of$\mathbf{U}$. This is because the ... 3 Here's a general overview of how to approach this: Since the only external forces are vertical (gravity pulling the balls down, normal force of the surface holding the balls up), we can use conservation of momentum in the plane. Similarly, there is no external torque rotating things in the plane, so that component of the angular momentum is conserved. And ... 1 If you impact the second body its axis of percussion it will purely rotate. By carefully choosing the inertial properties of the two objects you can make the first object stop translating in the process. See this post for more details on a particle to rod impact. 2 To calculate for a situation like this, consider the Law of Conservation of Momentum: Pi = Pf In the case of the billiards: KEi = 1/2 mu1^2 + 1/2 Mu2 (u1, u2 = initial velocity) KEf = 1/2 mv1^2 + 1/2 Mv2^2 (v1, v2 = final velocity) Based on this law, initial kinetic energy and final kinetic energy in an elastic collision are equal. KEi = KEf Hope ... 0 This is not an answer, but is there any contradicción with Noether theorem to say that Kinetic energy gives diferent value according to system of reference? 3 We don't know! If we were to observe a situation where energy conservation did not appear to work, that would be a major puzzle. As you say, either we would have to discover some alternative contribution to the energy that we had been neglecting, or we would have to give up on energy conservation. A priori it is not obvious which one of those two resolutions ... 7 Noether's theorem states that to every continuous symmetry of a physical system there is an associated, conserved quantity. The conserved quantity associated with time translation invariance (i.e. it doesn't matter if you perform an experiment now or tomorrow, provided you set it up the same way) is what we call energy. Therefore, somewhat tautologically, ... 1 A "collision course" is a very fuzzy concept: if you are "barely going to hit" you are on a collision course but don't need a lot of deflection. However, let's assume for a moment a stationary earth, a meteorite of mass$m$at distance$D$, heading for earth of radius$R$with velocity$v$. The equations you need are conservation of angular momentum and ... 0 The center of mass moves as though all forces act there - so the two cases will result in the same (linear) velocity of the center of mass. However, in the second case you also get rotation. No conservation of energy is violated because it you apply the force for the same (short time), the distance moved in the second case is larger (because the rod starts ... 1 For what it is worth, your formulation of the spherical wave business is flawed. See here for example. These have the form $$u(r, t) = \frac{A}{r} e^{i(\omega t \pm k r)},$$ where the symbols have their usual meaning. So, intensity, which goes as the amplitude squared, will go as $$I = \left(\frac{A}{r}\right)^2$$ falling off as the square of the distance, ... 1 There is a confusion between the terminology "perpetual" , which means "continuously", devices that almost move forever, and a machine that can produce energy. As the other answers point out energy is conserved and if it looks as if energy is provided from nothing a closer analysis shows the mistake, as in the drinking bird perpetual setup. In the case of ... 3 You might be able to get it to work for quite a while, depending on your skill as an engineer. But there is a critical difference between a well-engineered machine that runs for a while, and a perpetual motion machine that runs forever without input. The latter is impossible. 4 The key here is the antimagnetic strip, quite aside from whether or not such a device can be built. When you insert the anti-magnetic strip, you must change the shape of the magnetic field. You must force the magnetic field to "leave" the high permeability ball. The same magnetic induction$|\vec{B}|$in a high permeability$\mu$material represents a lower ... 4 If the magnet is strong enough to pull the ball up the bottom slope, it will be too strong to let the ball fall. Even worse, as you have drawn the diagram the magnet is pulling the ball down the lower slope when it is toward the left end. Anywhere to the left of where the perpendicular from the magnet to the ramp, the magnet is pulling more right than up. ... -1 Too many abrupt changes in direction. Also, is this spring powered? If so, I would think it to be simpler by using an electromagnet to apply and cut voltage. More moving parts mean more opportunities of failure. 1 I interpret your question as What type of equation is the equation $$\int_{t_0}^{t} \vec{F}(\vec{x}(t')) \cdot d\vec{x}(t') = \frac{1}{2} m ((\dot{\vec x}(t))^2 - (\dot{\vec x}_{0})^2) \; \; \;\; (1)$$ where$t$is variable and$\dot{\vec x}_{0}$fixed? The answer is, it is a first order integro-differential equation, not a mere ordinary differential ... 1 Intro I'm the original asker of this question (9 months ago); thanks to the comments and answers I've gotten here, I think I've pieced together an answer that I'm happy with. Short forms used in this answer: CoLM = Conservation of Linear Momentum CoAM = Conservation of Angular Momentum KEB = the Kinetic Energy Balance CoE = Conservation of Energy ... 0 "If a photon hits an electron". If only it were that simple! The electrons (in this case) are bound in atoms (or ions). If the photon has the "right" energy, then the probability of the atom making a transition from one state to another may become appreciable and the transition may occur. If the photon does not have the "right" energy - and you can think of ... 1 When we lift those three weights from Y to X, we can use the reversbile machine. So, we ease the machine B, because machine A takes those weights. Is that a right picture? There are two weights with mass$M$and$3M$. Initially, both weights are at the same height$h_0$which we can freely set to zero:$h_0 = 0$. Now, machine B lowers mass$M$to ... 0 A "fish in a pond" analogy of the expanding universe is better: If we were all fish in a pond, imagine what it would be like if water was being constantly added to the pond. Space is not expanding, rather new space is being added. In the rubber sheet analogy, there is no new real estate being added and hence it suggests some kind of elastic tension, which ... 0 What is "nothing"? Can somebody please get this experimentalist a couple of liters of "nothing", so that he can do some basic physical measurements on "nothing"? Moreover, can someone point me to a paper where the total energy of the universe has been measured experimentally? 1 No. There are more conservation laws: conservation of angular momentum, electric charge, color charge, weak isospin. Without the charge conservation you cannot explain why electrons do not decay into neutrinos. 0 Not actually just energy conservation and momentum conservation but there are some other rules also that are fundamental in physics. If all the rules that govern the universe are given and the data of universe at any one instant of time is given then a computer that calculates all the simulation at a very good speed can tell all the events that will occur in ... 1 I don't think energy conservation and momentum conservation can explain charge conservation in particle physics. 0 No. Consider the radioactive decay of an atom. The energy and momentum is conserved but the direction of emission is not known. We do know that on average, i.e. if we take many identical atoms and observe their decay, that the direction of emission is isotropic. However, for a single event we just can't seem to predict this direction beforehand! These kinds ... 1 Both are correct. The elastic energy of solid is stored in the chemical bound. On the other hand, it also contributes, although by very very tinny amount, to the total mass of the system due to$E=mc^2$. So combining both facts, we arrived at the conclusion that the chemical bound has mass, even though the mass of the chemical bound is about only$10^{-9}$... 1 BECAUSE WE MEASURE WITH ATOMS (or atomic properties) a change in the measured value can be either a true change of the observed quantity or a change in the properties of the ruler (the atoms). I'll revert the paradigm the space expands to this one: the atoms are shrinking THEN I have simple answers : If the universe is expanding, what is it ... 1 The modern mind picture of electrons bound to a nucleus is that of an orbital rather than of an orbit as you seem to be thinking. However, your ideas are still somewhat meaningful in this modern picture in that region of electron delocalisation is more tightly confined around the nucleus for lower energy orbitals than it is for higher energy ones. You might ... 0 First of all what you need to understand about quantum physics is that it's a theory of probability, not realist classical probability, but it is still a probability theory. The second thing you need to understand is that realism is wrong. Conservation of energy in quantum physics simply means that the Hamiltonian is not time dependent. That's it. From this ... 0 Of course, one usual intepretation of quantum tunneling is that the particle will borrow some energy from the vaccum in order to pass an unsurmountable barrier otherwise and then restitute it asap after crossing the barrier. As many others have said, this is a valid interpretation. I am not sure it is necessary though. In fact, in quantum tunneling, what ... 0 You are right to be confused because these processes do not conserve energy when taken individually. If you are lucky a couple of particles may appear close to you and you can collect their energy for free. However, the inverse is also equally likely. Imagine you spend some energy and store it in a couple of particles, then there is a chance that these will ... 0 Temporary violations of conservation of energy is allowed in quantum mechanics. A system can make a transition to a state that violates conservation energy by an amount$\Delta E$as long as it stays in that state for a time shorter than$\Delta t$where$\Delta E\Delta t \leq \hbar/2$. This is a form of Heisenberg's uncertainty principle. 4 One mistake you are making is equating a constant power$\frac{\delta E}{\delta t}$with a constant force. One does not imply the other. I'm going to switch to a more standard notation and use$K$for kinetic energy. And let's say$U=\text{const}$to keep things simple in our system. Finally, let's imagine there's only this one force$F$on this ... 4 "applying a constant force, F with a rocket engine that I supply a constant amount of energy to" Heere's your problem! Rockets are not as simple as you think. Yes, you could apply a constant force (thrust). And yes, this would consume fuel at a constant rate, but as @garyp points out, this does not mean you are adding to the kinetic energy of the ... 2 Conservation of energy only applies to closed (isolated) systems, and your example, with an external source of energy, is definitely not of that kind. 4 @soumyadeep is on the right track, but wrong. The object starts at rest. When the "external force" is removed, it starts to slide down the slope, picking up speed. When it reaches the point where the force of friction plus the force of the spring equal the force of gravity, it doesn't just stop - it stops accelerating. Thus, to get equality you need to ... 14 1. If the universe is expanding, what is it expanding into? The universe isn't expanding into anything. Space-time isn't curving into a higher-dimensional space. So what do we mean by "curved" and "expanding", words usually having a meaning only for objects in space? The answer is it is just an analogy. Mathematicians have found properties of space an ant ... 1 Work done on a system increases the energy of the system. But you have to be clear about what system you are talking about. For the system consisting of the Earth and your other object, there is no external force, and no work done on the system. The total energy is constant; it just changes from potential energy to kinetic energy. If your system is the ... -1 In a closed system there is no where for the energy to go. There is no way for the gravitational force to change the overall energy. I would simply change "the work done equals the energy change, BUT not when the work is done by a conservative/internal force" to "The work done equals the kinetic energy change". This just means that you can convert ... 2 Energy is conserved so it can't be created or destroyed. All we can do is change energy from one form to another. In your example we are changing the potential energy of the mass$m$into kinetic energy. The increase in kinetic energy must be equal to the decrease otherwise energy wouldn't have been conserved. By an external force I assume you mean some ... 5 Good question! Many astronomers think that the moons of Mars, Phobos and Deimos, are captured asteroids. Others object precisely because of the issues that you raised. Capture is not easy. Sans a collision, capture is impossible in the Newtonian two body problem. A hyperbolic trajectory stays hyperbolic. On the other hand capture in the multi body problem ... 2 According to the question, distance$d$is measured from the pivot down: [...] a peg located a distance$d$straight down from the pivot point You have computed it from the bottom of the swing up. I expect that$d = \frac{2L}{3}$is one of the answers in your list... Your analysis is otherwise correct. One word of advice: ALWAYS draw a diagram. If you ... 1 Let's look at what happens if we put numbers into your equations to see if there actually is a violation of the conservation of energy. Using numbers for Earth from wikipedia we have a velocity of$29.78*10^3\frac{m}{s}$for orbital velocity and a mass of$5.972*10^{24}kg\$. For say the Voyager 2 we have mass of about 730 kg and a speed of ...

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I read that the KE a free-falling ball acquires is not originated by the attracting body but that energy was actually stored in the ball when it had been lifted to the height it dropped from. In this way, it was said, gravity is subject to the conservation of energy principle and cannot change the total energy of an object. Consider now the ...

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Energy is in fact conserved, even in gravitational slingshots. After the slingshot, the velocity of the spacecraft may indeed change, which means its kinetic energy will also change. If this happens, the energy increase (or decrease) will be made up by a commensurate decrease (or increase) in the kinetic energy of the planet. In plain English: The planet ...

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Cory, here's a different way of thinking about gravity assists that may help: First is my short answer for readers in a hurry: What is really going on is a giant game of pool, with fast-moving planets acting as massive cue balls that impart some of their energy when they whack into tiny spacecraft. Since you can't bounce a spacecraft directly off the ...

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Gravity assists don't change speed in the two body problem. An object approaching a lone gravitating body will enter and leave the vicinity of that body with exactly the same speed. All that a lone gravitating body can do is change the direction in which the object is heading. The body that provides the assist needs to be moving with respect to the target ...

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But once it's moving in the vacuum of space, with no gravity or magnetic field nearby, could it spin nearly forever (aka billions of years) producing a magnetic field, No. A rotating magnet creates a changing magnetic field. Similar to an oscillating electric field, it will radiate electromagnetic energy. This energy will come from the rotational ...

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... if the momentum of the system is conserved, the velocity of the centre of mass of the system should remain the same. True. ... the velocity of the centre of mass of the system has to be different after the collision for the kinetic energy to be different. False. So, how can there be a change in kinetic energy of the ...

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