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You will agree that the velocity of A depends on the elasticity of the collision. Now, momentum is conserved regardless of the elasticity of the collision. Therefore, you cannot expect momentum conservation alone to determine the outgoing velocity of A. In detail, the total outgoing momentum depends on two unknowns, i.e. A's velocity and B and C's common ...

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Way back in time, scientists studied the following equation intensely because it was so new: $$F=m*a$$ By experiment, they discovered that if you apply a force to a mass and it's free to move, then it will accelerate according to that equation. Back then, it was considered a LAW. It was immutable. It was a "law of physics". It wasn't derivable. Never in ...

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$\hat{X} = i\partial_{k_x}$ in momentum space is wrong. The left hand side is an operator, the right hand side is the representation of an operator when acting on functions, taken with respect to the scalar product with the basis $|\ p\rangle$. In particular we have $$\langle x\,|\,\hat{x}\,|\,\phi\rangle = x\phi(x)$$ and $$\langle p\,|\,\hat{x}\,|\,\... 1 You must look at all forms of energy. Just before the explosion, the projectile has gravitational potential energy GPE, kinetic energy KE, and also chemical potential energy CPE stored in the dynamite. Just after the explosion the 3 fragments all have the same GPE as before. The CPE has disappeared in the explosion. As Jim says, we must assume that it ... 3 As indicated in the answer to your previous question, since as E(=30\:\mathrm{eV}) is higher than V_2(=20\:\mathrm{eV}), that particle is not bound, it's not an eigenstate of the system's Schrödinger equation (it's a scattered state). Its wave function for x \to +\infty would something like:$$\psi=c_1e^{-ikx}+c_2e^{+ikx}$$... where both complex parts ... 0 The thing is, the book claimed the catapult had no recoil during launch thanks to its fixed counterweight. Is this physically correct? Of course not. The projectile has been given some momentum, so the launcher must recoil by conservation of momentum. You can't dodge that. In this case the problem with the proposed mechanism is that after the (initially ... 4 Update 1: 1) Note added in proof: The photon stress-energy densities obtained below more or less heuristically are identical to those obtained in more rigorous approaches from the electromagnetic stress-energy density tensor. 2) The physical reason why the stress-energy argument retrieves the detailed balance result in the OP, but is inequivalent to simply ... -3 This problem is equivalent to Einstein's original thought experiment with the center of mass of a tubular spacecraft in an inertial reference frame emitting a photon from one end and reabsorbing it at the opposite end. This is the same thought experiment convinced an audience of Newtonian physicists that E=mc^2 by arguments related to center of mass. ... 1 When the man throws the ball, both the ball and the man get equal momentum in the opposite directions. Since the collusion is elastic, i.e: no loss in energy, the ball rebounds with momentum of the same magnitude but in the opposite direction. At this point, both the ball and the man have momentum in the same direction with equal mangitude. When the man ... 0 Straight after the man throws the ball, his velocity is determined by conservation of linear momentum, that is, the momentum of the man recoiling is equal to the momentum of the ball leaving his hand: Mv_1=mv After the ball bounces elastically off the wall, it returns towards the man at opposite velocity (equal in magnitude, opposite direction), so ... 1 When a man in frictionless surface throws the ball in forward direction, by conservation of linear momentum he gets pushed back (exactly the case in space where astronaut throws something back to move foreward).Here,when man throws the ball, the momentum of ball and man are exactly equal and their velocities are in opposite direction. But you need to note ... 2 @ChesterMiller: answer is good, but I would just simplify it. What you say is that the total momentum is the momentum of the contents of the control volume plus the sum of that which traverses the boundaries. If you take the change in momentum per unit time, you have momentum flux, which equals force. 3 The forces acting on the fluid are changing its momentum. There are forces acting on the stationary portions of the control volume boundary, and there are also forces being applied to force fluid into, and acting to prevent it from flowing out of the control volume. The total rate of change of momentum in the control volume at any instant of time is equal ... 0 First, the authors positioned their reference frame at the center of mass, if you calculate the momentums that way, the total momentum will be zero, at least instantaneously. But that is not the point. In addition to the effects of other objects in our own solar system, mentioned in the other answer, remember that the Sun travels around the milky way ... 1 I don't know what numbers you are using, or how much precision you are expecting, but the problem could be your assumption that the earth - sun system is an isolated system. First there is earth's moon that (if not taken into account) might induce errors. Then there is the effect of other planets, especially jupiter. All of these must be considered if you ... 1 Let's first address the general question--do "waves" of any kind have (or can they have) inertia? I suppose here by "inertia" you mean "resistance to changes in velocity." This is certainly true of waves in, say, water--you've certainly felt resistance to your hand if you sweep it through water to make a wave; the destructive force of a tsunami is a more ... -4 TDSE TISE Here you have the time dependent and independent Schrodinger wave equations, respectively. These relate to the energy of particles, but the trident symbol, Psi is representative of the actual wave equation I believe you are referring to. While De Broglie and schrodinger and others like them do describe particles as behaving like waves, they ... 1 The velocity you calculate by linear momentum is the velocity of centre of mass of the body under study. Angular momentum should be conserved at a point where the impulsive forces act. 3 Although this is almost an engineering question, there are some simple physical constraints that mean the answer is an emphatic no. At staging, significant change in momentum / trajectory of the payload-bearing stages can only arise from high impulse transfer between the payload bearing and jettisoned stage; this statement is a reformulation of conservation ... 2 I have asked a professor about this and he gave me the answer. After replacing \mathbf{k} by -i\nabla in H(h\mathbf{k}), we are actually getting a new Hamiltonian that acts on envelop of wave functions. To make this answer relatively complete, I will briefly introduce the main steps focusing on only one band. Suppose the band we are interested in ... 2 I would call this quantity "first moment of mass" or just "moment of mass". Have a look at this wikipedia article to read about the general concept of moments in physics. As pointed out in the comments to your question, the first moment of mass is closely related to the center of mass (CoM). For a collection of particles with masses m_i and positions \... 4 When exploring deep questions in physics, like you are, it is important to remember that nature appears to obey the laws of physics. Empirical studies such as science cannot prove ontologically that nature does obey the laws of physics, or obey laws at all. For an extreme test case, consider the concept of idealism. In idealism, one claims that there is ... 4 We know or reasonably assume that momentum and energy are conserved because of two reasons mainly: Mathematical plausibility: If we assume that nature follows mathematical descriptions, then e.g. Noether's theorem makes it a necessity that momentum is conserved. Otherwise, something with the mathematical description would be wrong. And so far, in the vast ... 12 The answer by WetSavanna... is complete but I want to particularly address the part I hope it is clear that I'm not trying to suggest that I don't trust these laws to be true but rather that I'd like to know how we know they are true. Physics theories are mathematical models that fit current observations/data and are predictive of new ones. Prediction ... 32 We know through experimental observation. That is the beginning and end of the subject of physics, at least the part of it the tells it apart from, say mathematics. Conservation of momentum is simply an inductively reasoned hypothesis to summarize certain patterns in experimental data. You are alluding to the conservation of momentum's being "explained" ... 2 Let's consider the frame in which initially both the masses are at rest to be the frame O. In frame O, momentum conservation is trivially followed because of the symmetry of the problem. For the energy conservation, we require that M = m \sqrt{1-v^2}, where m is the initial rest mass of each of the particles and M is the final rest mass of each of ... 2 If there is no gravity in your simulation, your thoughts are on the right lines. The amount of centripetal force (=thrust) required to maintain a space-craft of mass m in circular orbit of radius r at speed v is F=mv^2/r. Thrust has to be directed towards the centre of the circle and has to be maintained constantly - so unlike orbiting a planet using ... 3 If you shoot light out the backside of your craft then indeed you would move towards the front. But then you'd use up some energy in order to go because the light that leaves would carry energy as well as momentum. The energy of your craft would decrease and thus the mass of your craft would go down. You'd arrive at your destination with a less massive craft ... 1 You have proved with your analysis that v_2 cannot be zero. m_2 must be moving with some non-zero velocity, either in the same direction as m_1, in which case v_2 must be smaller than v_1 (or m_1 will never catch up to m_2) or m_2 must be moving in the opposite direction to m_1. With the information given, there are 4 unknowns: v_1, v_2, ... 1 Momentum (p) is "really" mv, even for light and EM fields. This can be proven by the use of E = mc^2. The momentum for a photon (EM) is p = mv. Where the mass is given by m = E/c^2 and v = c. Substituting these into the equation, one obtains, p = E/c. Although this equation "looks" different from p = mv, because it was derived using the ... 4 A eigenstate of a crystal hamiltonian can be written as a Bloch function in space representation$$ \psi(\mathbf{r}) = e^{i\mathbf{k}\mathbf{r}} u_\mathbf{k}(\mathbf{r}) $$u is periodic with respect to the unit cell. The momentum is now given by$$ \langle \psi|\hat{\mathbf{p}} |\psi\rangle = -i\hbar \int e^{-i\mathbf{k}\mathbf{r}}u_\mathbf{k}^*(\mathbf{...

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