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All the laws of Physics are derived from the law of coservation of energy,also the law of conservation of momentum and Newton`third law of motion...


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Pressure from combustion of gunpowder will far exceed pressure of air trapped when muzzle is touching the wall. The bullet will still be driven into the wall, but not quite as far, since it will have had to expend a small amount of energy against the trapped air ahead of it. Not sure whether you were asking about my credentials or fundamental concepts. At ...


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Newton's third law states that if object A acts on object B with force $\mathbf{F}_{AB}$, then object B must act on object A with force: $$\mathbf{F}_{BA}=-\mathbf{F}_{AB}$$ When expressed in terms of A and B's momentum, the same equation can be written as: $$\frac{\mathrm{d} \mathbf{p}_A}{\mathrm{d}t} = -\frac{\mathrm{d} \mathbf{p}_B}{\mathrm{d} t}$$ ...


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It has always struck me as odd that Newton's 3rd law was never explained in any detail in my physics degree lectures. I have quoted it glibly for decades, but it was only when a 10 year old asked me WHY? that I thought about how to understand this law. In the macroscopic world, we apply a force and there is an "equal" reaction force. What's actually going ...


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I suspect you are meant to treat the collisions between spheres 1 and 2 and between spheres 2 and 3 as separate collisions. Solve 1 and 2 first. There will always be a solution where 1 continues with unchanged speed - reject this solution, it would mean that 1 passed through 2 without colliding at all. Then, take your solution for the velocity of 2 and solve ...


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You always have to satisfy the momentum equations, which is only the linear momentum equation for this one dimensional case: $$m_1v(t_1)-m_1v(t_0) = \int_{t_0}^{t_1} F dt$$ Assume the collision is completely elastic and all is conservative, so no plastic deformation, drag or any kind of damping. Then the only force which acts is the gravity: $$ F = ...


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The only thing that the device "knows" when it is hit, is the force with which it gets hit, and the duration of that hit. Transfer of momentum $m\Delta v = F\Delta t$. So what matters is the momentum of the hammer's head - or more specifically, the momentum that you are able to transfer. Ultimately it comes down to giving the most momentum to the head of the ...


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If you consider your photon as a point object, it cannot bend its own path. It will always travel on the ridge it creates, speaking in terms of curvature of space. The other idea is possible. Two photons having a momentum, attract each other, trapping each other, like a positronium (typical example for this behavior). In the model of relativity this is ...


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I believe you are all using things you know little about to make your claim seem reliable. For example, from this simple strain of comments I have heard anything from black holes to the expansion of the universe. These things, although interesting, have nothing to do with whether or not a stone will halt after being thrown into space; the forces you have ...


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As I indicated in my comment on Rod's answer, some very powerful lasers do exist - and while their photons don't have much momentum, they do "pack a mean punch". In fact, laser ablation (where a laser beam produces significant local heating and material is ejected at high speed) may just produce the phenomenon needed. Let's take the example of the LLNL ...


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It is Minkowski spacetime. Both $\bar{p}=\left(\begin{array}{c}E\\{\vec p}\end{array}\right)$ and $\bar{x}=\left(\begin{array}{c}t\\ \vec x\end{array}\right)$ are invariant 4-vectors in spacetime. A 4-vector is composed of a time component and a spacial component (with 3 sub-components related to ordinary 3-space). By invariant we mean that their magnitudes ...


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There are already a few answers that explain the mathematics behind it, but since you've said that your "math knowledge is quite limited" I'll try and break it down into simpler terms. You're already familiar with a 3D vector dot product, and it seems your confusion arises from dot products of a four-vector. Now what they didn't teach you when they taught ...


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In Minkowski spacetime you assign four coordinates to your events: $x=(x^0,x^1,x^2,x^3)$ - in this notation $x^0 =ct$ and the other three coordinates are the spatial ones. Suppose these are the coordinates of a point in spacetime reached by a light ray which started at the origin, then you have that: $(x^0)^2=c^2 t^2 = \sum_{i=1}^3 (x^i)^2 \Rightarrow s^2 = ...


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That's the definition of the dot product in Minkowski space-time. To be clear, any space-time is endowed with a metric. Standard ${\mathbb R}^3$ that you may be familiar with has a metric $\delta_{ij} = \text{diag}(1,1,1)$, $i=1,2,3$. Given two vector $\vec{v} = (v^1,v^2,v^3)$ or $v^i$ for short and similarly $w^i$, the dot product is defined as $$ \vec{v} ...


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Supposing we could shield ourselves with a perfectly nonabsorbing, reflective shield so that light would perfectly elastically bounce off us, thus preventing high power beams from incinerating us as Anna V's answer validly argues they would. Then the "fundamental" answer to your question is "because light has zero rest mass"; to explain further: the ...


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A ray of light is a geometrical line describing the propagation of an electromagnetic wave. The electromagnetic wave is composed of zillions of photons each with a tiny momentum. The momentum is not large enough to sense an impact, it is pico newtons even for a laser beam. Lasers can have very high energy and momentum, but like knives, they cut soft tissue ...


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But the only way I can see the EM field changing is if the top particle accelerates to the left, under the action of F, producing its own counter electric field towards the right. However this won't work as the mass M of the top particle is assumed to be large so that its acceleration, and thus its induced electric field, is negligible. Density of ...


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Two things. First of all, in general, if you make simplifying assumptions that aren't actually true, then why would you expect actual momentum conservation? You should only expect your calculation to yield approximate momentum conservation. Secondly (and more importantly), is that if you are applying a constant force $f$ to this system, then you should ...


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Assuming no air resistance and no rolling resistance, the cart's momentum can only be transferred to the rain. As BowlOfRed says, assume that rain leaving the cart at any one instant does so with the same horizontal velocity component as the cart has at that moment. And assume that before then, its horizontal velocity component was zero. You'll also need ...


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Conservation of linear momentum, for a physical system whose particles are initially at rest in a given inertial reference frame is equivalent to the fact that the center of mass of the system remains fixed at its initial position. You see that this constraint is quite week for a system made of a large number of particles as a human body. Starting form a ...


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Linear momentum is always conserved. If you swing your hand there is always a change such that linear momentum is conserved. If you swung your hand in space there would be some motion. At least until the electrostatic forces in your body brought you to rest. On Earth, the friction between your body and the ground and the air drag wouldn't allow the ...


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If the (entire) system does not change if you displace it in space, then the total momentum (of all particles) will be conserved. Momentum is the generator of translation, which is a specific case of Noether's theorem. To relate it to your example of living creatures: imagine cat on very slippery ice and neglect air resistance and friction. If you slide ...


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Speed can be understood physically, as the distance traveled within a certain amount of time, it makes sense to me. By contrast, I can't attribute a physical explication to linear momentum. How can I understand it physically? Why do we multiply mass by speed? 'speed' (or velocity) is the measure of 'motion' and is related to KE, a unitary mass ...


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Since momentum is a vector quantity, the negative sign indicates directionality. In the case of changes in momentum, $$ \Delta p<0\Rightarrow p_2< p_1 $$ which typically means one of two things: the object lost mass without changing velocity ($m_2<m_1$) the object slowed down without changing mass ($v_2<v_1$) Both points come from the ...


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You cannot understand this notion without considering the equations of motion. The force $\mathbf{F}$ during interval $dt$ changes a quantity $m\mathbf{v}$ by an addendum $\mathbf{F}dt$: $$d(m\mathbf{v})=\mathbf{F}dt$$ That's it.


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The idea of momentum is the idea of "quantity of motion". You want to be able to formulate in some sense the idea of "how much motion there's in this particle?" and you can think about the simplest model for that like that: the faster the particle moves intuitively more motion there is, so the quantity of motion should be proportional to the velocity. Also, ...


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Since $v=r\omega$ Where $v$ is velocity and $r$ is radius and $\omega$ is angular velocity So $\omega=v/r$ This equation shows that if $r$ decreases $\omega$ increases


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Start with the force felt while holding weights and spinning with arms at full extension. Ask if it is easier or harder than when not spinning. Here you are forcing the weights to move from a straight line and to go in a circle, the force has to be felt all the time to keep pulling the weights into a circle. To make the circle smaller requires even more ...


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Because the conserved angular momentum is the product of radius and angular velocity, thus in order to remain conserved (a compatibility condition) the velocity has to increase when radius decreases. it is similar in a way to the lever principle (based on conservation of energy) Did not notice the intention of the question (updating with other answer) ...


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Suppose total kinetic energy were conserved (i.e. if you can ignore potential energy changes), then because it's a nonlinear fuunction of the velocities, the sum of the kinetic energies of the particles that make up a composite body does not equal the kinetic energy of the center the center of mass of the object. In practice, what this means is that you ...


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I'm 27 and since I was about 15 I had the same doubt you do. Only a couple of years ago I realized why momentum is always conserved in a collision, whereas the same is not enforced for energy. (They must have told me this at some point -- or points --, but I guess sometimes I just don't pay much attention) First of all, let's make it empirically clear that ...


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Possibly your confusion arises from not considering that KE is a scalar, whilst momentum is a vector. Yes, there is of course a connection between KE and momentum: $K = p^2/2m$ (for non-relativistic bodies). Thus, for two equal-mass particles heading directly towards each other at equal speeds $v$, their velocities are $\pm {\bf v}$ and their momenta $\pm ...


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Is there no relation between the momentum and kinetic energy? Are they not linked with each other? No, they're not. Conservation of momentum only says that $$m_1 v_{1_-} + m_2 v_{2_-} = m_1 v_{1_+} + m_2 v_{2_+}$$ I used subscripts 1 and 2 denote the particles involved in the collision and the subscripts - and + denote the pre- and post-collision ...


0

Individual momenta are not conserved, only the total momentum is. It is not entirely determined with the kinetic energy, there are some additional "degrees of freedom" or "dimensions", loosely speaking. The total energy is conserved too, you just have to consider a potential energy of interaction during (even elastic) collision. Heat released in a body is ...


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Work done by a force in a system does not depend on whether the force is internal or external. In your case, the internal forces are doing work. The net work done by these internal forces is $W=\Delta K =K$, where '$K$' is the final kinetic energy of the system.


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The work done was expended in moving all the components of the bomb away from it, largely dissipated as heat. Picture the process in microscopic slow motion: Chemicals transform from a solid to a gas + heat (which will increase the gas pressure some more). This isn't really "work" in the classical mechanics form (yet). Gas pressure will exceed the limits ...


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You didn't say how big is the mirror. Assume the photon is in the visible domain, and the mirrors are classical object, i.e. big. Such objects will remain insensitive to the linear momentum lost by the photon, because the recoil velocity of the mirror is practically zero. The reflection of the photon will be elastic, no energy imparted to the mirror, ...


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This part is incorrect: When photons are reflected, they impart a small amount of their momentum/energy into the reflector and are red shifted. Momentum is always conserved. If in the frame of reference you choose, the photon accelerates and increases the momentum of the mirror, then the reflected photon will have less momentum (or less eneergy). In ...


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The canonical momentum $p$ is just a conjugate variable of position in classical mechanics, for we have the relation $p=\frac{\partial L}{\partial \dot{r}}$. When making the transition to quantum mechanics: we need substitute $p$ by an operator $-ih\nabla$ in the Hamiltonian; similarly, we need substitute $r$ by $i\hbar \nabla_p$ in momentum representation. ...


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The relation between Kinetic Energy and momentum is derived in the following way. I hope this answers your question.


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Try this method for solving the majority of Momentum/Impulse Problems with these two simple equations. Michel Van Biezen on Youtube teaches this method. Sure beats m1v1 + m2v2 (initial) = m1v1 + m2v2 (final). Given: Girl- Mass of 45.5kg; Velocity +1.47m/s Plank- Mass of 140kg Questions: QA Find the VELOCITY of Plank (that is girl + plank) on ice. QB ...


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The simplest way to get this intuitively is to consider a rocket where the exhaust gasses escape in two opposite directions. So, there is a nozzle at one side and also the opposite side. In this case, the rocket will go nowhere. The chemical reactions produce gasses at high temperature and pressure and they then accelerate and escape in both directions. If ...


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The best common example is often shown in HS physics classes. A student in a low friction chair holds a CO2 fire extinguisher and points it in a safe direction. When they pull the trigger and release CO2 gas moving very fast from the nozzle of the fire extinguisher that rapidly moving gas leaving the system (chair student and gas bottle) causes the system to ...


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Gyroscopic force, rake, trail etc. have been identifiers for theories why bikes don't fall when moving. Camber would be a better one. The bike can stay upright by camber so leaning is the natural expression of bicycling. To understand camber, watch a bicycle fall over. The wheels are held at the road by the camber force. A force balance on the bike will ...


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A model I encountered as a kid, back when the Space Race was in full swing, is still the simplest explanation I've found: Think about an inflated balloon with its neck closed. It doesn't go anywhere, because the pressure in all directions is equal. (Which is what keeps it inflated, too.) Now open the neck. What makes the balloon fly around isn't actually ...


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I'm a little confused why you ask about GR at the very end of your question. If your question is simply how a rocket is able to accelerate in space without having anything to "push" off, then we can tackle your question pretty well with classical mechanics. Newton's Third Law has its limitations at times, but for this question it will work just fine. ...


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I am wondering how the rockets could thrust in the empty space and move in the opposite direction. In very simplistic terms the rocket motor thrusts against the closed end of the nozzle. Once the gas leaves the nozzle it no longer has any interaction with the rocket - there is no need for it to 'hit' anything else.


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Newton's third law is pretty near to the mark. All of the phenomena you cite stem from the principle of conservation of momentum in an isolated system, itself ultimately a result (through Noether's theorem) of the fact the physical description of that isolated system is unchanged if we shift the spatial origin of our co-ordinate system. So, if you're in ...



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