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Since the game is just an approximation of real-world physics, you have some flexibility, but generally, these kinds of games simulate a perfectly elastic collision (in which total kinetic energy is preserved) and the angle of incidence = the angle of reflection. Graphically this looks like . Because your breakout paddle (unlike the diagram) is probably ...

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by putting $p=\gamma mv$ and then get a value for $m$ (which will be 0 for a photon) and therefore rendering the equation to $E=0$ First, let's write this out in full (in 1D) $$p = \frac{m v}{\sqrt{1 - \frac{v^2}{c^2}}}$$ Then, solve for $m$ $$m = p\frac{\sqrt{1 - \frac{v^2}{c^2}}}{v}$$ Now, holding $p$ constant, see that the limit of $m$ as $v ... 5 You can understand the expression by attempting the limit for$v\rightarrow c$and$m\rightarrow0$. Notice that$\gamma\rightarrow\infty$when$v\rightarrow c$. Therefore$m v \gamma$is an undetermination of the form$0\cdot\infty$. From your expressions, you cannot say that$E=0$. The limit does not exist, and this implies that this expression is not valid ... 6 That's because the relation$p=\gamma mv$doesn't hold universally. As you just showed yourself, using this relation for a photon would lead to a contradiction because the energy of a photon isn't zero. A heuristic way of seeing why this relation won't hold for a photon is by recognizing that $$p=\gamma mv =m\frac{d x}{d\tau}$$ but a photon doesn't ... 0 As soon as the ball is released from the rod, it starts going in a tangential direction. Since there are no external forces on the system, the linear momentum of the system (ball + motor) must be conserved and the motor starts moving in the opposite direction of the ball with some velocity (depending upon the masses of ball and motor), so that the linear ... 1 I'll provide an answer in non-relativistic quantum mechanics. The short answer is that momentum and mass commute, so a particle can have a well-defined momentum and mass simultaneously. But really, mass isn't considered an operator in quantum mechanics; it's a parameter, a number. So for some system, it is presumed that the mass is known always. There's no ... 0 Well, the two are constrained by $$p_{\mu}p^{\mu} = -m^2$$ or $$-\mathcal{E}_{\vec{p}}^2 + \vec{p}^2 = -m^2$$ so I would say that yes they can. Let us consider the types of interactions people make in the LHC at CERN. I am not an experimentalist, but from my undergrad particle physics course, I seem to remember that linear-momentum can be measured (at ... 0 Let us take an electron's track in a bubble chamber where there is also a magnetic field. We can measure the momentum of the electron, the change due to ionisation, and its position as it goes through the spiral and finally know its final (x,y,z) at rest, and 0 momentum. Even though we are dealing with an elementary particle we are still, with our ... 8 Yes. Actually photons exert pressure on any surfaces exposed to them. For example, photons emitted by the Sun exert pressure of$9.08 \mu N/m^2$on the Earth. Reference: https://en.wikipedia.org/wiki/Radiation_pressure 6 Yes - you can even propel spaceships with it - Solar Sail Although Solar radiation pressure at the Earth is around 9E-6N/m2 while the thrust from a Saturn V rocket is 34 MN, so you would need a solar sail something like 2000Km on a side to get the same acceleration 1 I can give you one example. In a semiconductor reverse-biased p-n junction, a potential barrier exists that prevents electrons from crossing the junction. There is an energetically-forbidden region in the vicinity of the junction. The wave functions of electron states in both the valence and conduction bands are real exponential in this region. ... 1 In the dynamical systems jargon, this is the usual Lyapunov stability. If additionally the system reaches the equilibrium state, then the equilibrium is called asymptotically stable. For the formal definition, see this link: http://www.scholarpedia.org/article/Stability#Definitions:_Stability_of_an_Equilibrium It is also worth to remark that a equilibrium ... 2 If there is oscillation, the equilibrium is stable in at least one dimension. The situation where momentum keeps increasing could apply to a driven mass oscillating about a stable equilibrium. It would also apply to a non-driven mass at a saddle point. Consider that the mass begins oscillating approximately, but not exactly, along the concave up curve in ... 0 I am not sure of what you mean but in mechanic it's called unstable equilibrium. Check this http://en.wikipedia.org/wiki/Mechanical_equilibrium to get illustration on differents equilibrium states 1 According to this article (the choice of article has no significance other than it came up first in my Google search): Remington's 12-gauge 2 3/4-inch Premier Magnum turkey load has 1 1/2 ounces of shot and a 1260 fps muzzle velocity Converting these figures to metric,$m_{\text{shot}} = 0.0425$kg and$v_{\text{shot}} = 384$m/sec, so the momentum of ... 1 The current density formula you consider is not right, because$\frac{d(\psi\psi^*)}{dt}$is always 0. The current density is defined as$\bf{j}=(i\hbar/2m)(\psi\nabla\psi^*-\psi^*\nabla\psi)$. In your one dimensional case change$\nabla$to$\partial/\partial x$, you will get your answer. A more physical definition of current density operator is this: ... 0 Oh, I think I've figured it out... I sort of realised this just as I posted my question : From my earlier intros to quantum mechanics I know that p = mv =$\hbar k$. By simple substitution I can then obtain a final answer of : j =$v\lvert{A}\rvert^{2}$I'll wait for someone to confirm my line of thinking. 0 You can compute the velocity using this approach: $$\gamma = \frac{E_\text{kin}+E_\text{rest}}{E_\text{rest}}; \quad\beta=\sqrt{1-\frac{1}{\gamma^2}}; \quad v=\beta c.$$ where the total energy for each of the two particles is half of the mass of the particle that you want to obtain. This is the sloppy-theoretical-minimum energy: it requires the two ... 1 Yes you do need the exact velocity when it hits, and the exact velocities when they shoot off. Theres a fair bit of friction involved, therefore it can be tricky to calculate the velocity. Also momentum is conserved along a direction. Did you take that into account? 1 Well that's a result using differentiation and derivation. Have you studied calculus? If not, there is a simple way to look at it. $$\frac{\Delta p}{\Delta t} = \frac{\Delta (mv)}{\Delta t}$$ (Yes there should be a$\Delta t$in the denominator, too.) Now, what does$\Delta(mv)$mean? It represents the change in the quantity$mv$. For current situations, ... 1 You are right, there is a$\Delta$missing in front of the$t$.$\Delta v = v_2 - v_1$. If the mass is not changing, then$\Delta (mv) = mv_2 - mv_1 = m(v_2 - v_1) = m\Delta v$. Hope that helps. The equation that includes$\frac{\Delta m}{\Delta t}$is not Newton's second law. The second law is valid only for systems of constant mass. An equation like ... 3 1) Yes indeed, the absence of a$\Delta$in the second expression is just a typo. 2) The last expression is derived assuming that mass is a constant. If it helps, just set the mass equal to 4, or something. If we want to know how the quantity$ 4v $changes, we really only need to know how the quantity$v$changes. Suppose$v$changes from$v_1$to$v_2$. ... 1 Yes. It should be: $$\frac{dp}{dt}=\frac{d(mv)}{dt}$$ I'm using$d$instead of$\Delta$because I am thinking about the limit where the changes in$p$and$t$are very small. Then these are called infinitessimal changes, and denoted by a$d$. Usually, when one considers simple problems in Newtonian mechanics, what one does is study a given object with a ... 0 I've been wondering a lot about this too. So far I haven't got any straight answers, and I've asked around a lot. Angular acceleration is easy - if you have either a moment of inertia (2D) or an inertia tensor (3D+) I, then the angular acceleration due to a force is simply$\alpha=I^{-1}\tau$for torque$\tau=\vec{F}\textrm{x}\vec{r}\$, with a force ...

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To prove the conservation of quantities, you need to be able to compute the motion of the system so that you can directly compute these quantities from the time dependent coordinates and verify that they do not change with time. However I don't think that the motion of this system is integrable: it looks like a multiple oscillator and it is very prone to ...

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$$K.E.= \frac{1}{2} p v$$ since: $$K.E.=\frac{1}{2} m v^2$$ and $$p=mv$$

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