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The pattern does not continue. In fact it doesn't always hold as written. In particular: $\sum_i m_i = \sum_j m_j$ only if rest mass is conserved, which it is not when, say, you have particles and antiparticles being created by and annihilating into photons. $\sum_i m_i \vec{v}_i = \sum_j m_j \vec{v}_j$ only if all momentum in the system is of the form ...

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The difference in force to stop the trains you are talking about here is the difference in force is needed to bring the train to a stop within a particular distance Let me tell you what I mean. When you try to stop the train, you'll obviously be dragged in front of the train. Say the dragging causes an uniform force (due to the friction from the ground) ...

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First of all you should note that Newton's law says when $F$ acts on a mass $m$, then that mass will move with acceleration $a$. Here, we should apply the laws of collision and by using the conservation of momentum, find out what your velocity will be after the collision. Before collision we have: $p_{tot}=mv$ and after collision $p_{tot}'=mv'+MV$ where $M$ ...

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The spatial wavefunction is $Y_L^m(\theta,\phi)$. When exchanging the two particles, the spatial wavefunction becomes to $Y_L^m(\pi-\theta,\pi+\phi)$. Mathematically, we have $Y_L^m(\pi-\theta,\pi+\phi)=(-1)^L Y_L^m(\theta,\phi)$. If $L$ is odd, the spatial part is antisymmetric, otherwise symmetric. BTW, you may post this question as a comment of the ...

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$\nabla T=0$ is not a conservation law. You are not considering the energy of the gravitational field in this way. If you do the calculation you find something like $\partial_{\nu} (\sqrt{-g} T^{\mu \nu}) \neq 0$, so you can't define a conserved charge like $P^{\mu}= \int \sqrt{-g} d^4x T^{\mu0}$. To account for the gravitational scalar field you ...

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Consider the following results: From the definition of scalar product of four vectors, $$\tag{1}(p_1 p_2)^2 \equiv (p_{1\mu}p_2^\mu )^2 = (E_1E_2 - \textbf{p}_1 \cdot \textbf{p}_2 )^2.$$ The usual dispersion relations: $$\tag{2} E_i = \sqrt{ | \textbf{p}_i |^2 + m_i^2}.$$ The velocity $\textbf{v}_i$ in terms of momentum and energy: $$\tag{3} ... 0 The net force on raindrop plus wagon is zero. Consider a single rain drop. Let the momentum in the direction of travel of the combined wagon/raindrop system be p. Now p = p_wagon_before + p_raindrop_before where p_wagon_before & p_raindrop_before are the momentum of the wagon and the raindrop before the drop hits the wagon. We have then: ... 0 This is the opposite of the rocket problem. In a rocket, acceleration occurs becaue mass is thrown out the back end. F = d/dt(mv) = m.dv/dt +v.dm/dt. If F= 0, then m.(-dv/dt) = v.dm/dt <-- note the negative term with the acceration. In a "typical" problem mass does not tend to change significantly, but in the rocket this mass term is highly ... 6 I don't think that there would be any more diagrams. Having a total derivative term in the Lagrangian leads to derivative interaction vertex, which after symmetrising gives you something like $$ig \sum_i p_i \ ,$$ where g is some coupling and p_i the momenta of the particles. This vertex, however, vanishes due to momentum ... 0 But if the velocity of the wagon changes, the net force can't be zero, right? Only true if the mass is constant (it's not if a wagon is filling up with water). If mass and velocity both change, you can't say anything about the force. Record the experiment and play the video backwards. You will see a wagon moving backwards. The wagon is spraying water ... 0 Note that in your suggested motion after the collision, momentum would not be conserved. The Law of Conservation of Momentum is kind of a big deal in Physics, and especially useful when analyzing collisions. It states that for a closed system, momentum is conserved. It's also sometimes restated that for a closed system experiencing a collision, the momentum ... 0 Because in your example the action is not in the same direction than the velocity. The action and reaction are normal to the wall of the table, so the billiard ball only feels a reaction force (and thus a change in its velocity) in that direction. The component of the velocity of the ball parallel to the wall is not affected. 1 The force from the ball on the wall is exactly equal and opposite to the force of the wall on the ball. Both forces however are perpendicular to the wall (and must be assuming the wall is frictionless) and not necessarily perpendicular to the ball's initial direction of motion. Being perpendicular to the wall the force on the ball has absolutely no effect on ... 1 I think we are supposed to assume that the buoyancy force of the balloon is equal to the weight force of the balloon, ladder, and climber. If this is the case, the system is in equilibrium with its environment, with no net forces to the environment. You could look at it as a center of mass problem. We assume the ladder has negligible mass. When the climber ... 45 To deal with this type of problem, you must be careful to define exactly what system you are dealing with, and then not change that system part way through the problem. This definition allows you to be very clear about whether the "system" has any external forces acting, and thus whether the momentum of the system is constant or not. In this case, you seem ... 32 When the raindrops hit the wagon's surface, they aren't moving relative to the tracks. Friction is required to accelerate the raindrops to the wagon's speed. By Newton's third law, there must therefore be a reaction force on the wagon surface by the raindrops. 9 The mass of the cart is changing! This is the variable-mass system, which says,$$ F_{ext}+v_{rel}\frac{dm}{dt}=m\frac{dv}{dt} $$where v_{rel} is the relative velocity of the escaping/entering mass. In your case, there are no external forces so,$$ v_{rel}\frac{dm}{dt}=m\frac{dv}{dt} $$So the change of velocity comes from the change in the mass. 2 *elastic collision occur between atomic particles? inelastic collision occur between ordinary objects? perfectly inelastic collision occur during shooting? super elastic collision occur during explosion?* as John Rehnnie has explained, an elementary particle as the term implies, is not made of other particles, has no lattices ... 1 The same question in the OP has been asked and replied in this question (already linked in one comment here). Probably, if you had made the example of the glass, your post would not have been misunderstood. I have just observed a simple fact there: if a glass can take, support only 100N if we exert a greater force, it will break and offer no support, the ... 1 This follows on from my answer to your previous question: Factors on which Coefficient of restitution depend. The coefficient of restitution of a collision depends on the available degrees of freedom for energy to be lost. If you take your example of the collision of atomic particles, let's say two electrons, then there isn't anywhere for the initial ... 0 A purely elastic collision is a theoretical construct, which helps to simplify some situations. In all collisions, some energy is lost as heat, and thus the collision is said to be inelastic. Perfectly elastic collision does not occur between atomic particles Your problem here is that you are assuming atomic collisions can be represented by classical ... 1 Short answer: No. Momentum-energy conservation arises by dint of Noether's theorem, which says that if a system's Lagrangian is invariant with respect to a continuous transformation, there is one conserved quantity, called the "Noether Charge" for each such transformation (technically: for each linearly independent tangent vector in the Lie algebra of the ... 2 In as far as I know, the universe is actually gaining energy. And as far as we can see, the universe is expanding as a product of pressure, in the direction of the difference between opposing pressures, blah blah.... So would momentum be conserved under this rule? I suppose, if the exact same amount of energy were being "destroyed" as "created" within the ... 1 Your scenario of ball-girl-rail is not completely clear to me - but I can tell you this: both Newton's laws, and conservation of energy, are quite generally applicable. When the girl pushes against a light object (like the ball), then that object will accelerate away from her. If she maintains a certain force for a certain time, she will impart the same ... 1 Note that force is just the momentum per unit of time: F=\frac{dP}{dt} I.e. the force is the speed of momentum transmission. When two bodies interact, they will exchange momentum. The quantity of momentum that will be transmitted from one body will be the same quantity, received by another. This is depicted by following picture which is the third ... 2 Conservation of momentum works here like everywhere else. When A (with mass m_A) throws a ball with mass m_b with velocity v, then v_A=-v_b\frac{m_b}{m_A} so that after the ball is thrown, the net momentum is zero; note that the ball will not be moving towards B at velocity v but instead at v-v_A since A started moving backwards... When B ... 1 Would A move backwards at the same speed? no.. Momentum will be conserved,not speed and since A has greater mass than the ball so his speed will be less.. if A throws the ball with speed 'u1',then his speed(in the backward direction) will be m1u1/M1 where m1 is the mass of the ball and M1 is the mass of the person A Now, what would happen when B catches the ... 2 What is the difference that leads to conservation of kinetic energy in elastic collision ? The difference is only in the properties of the material of a body. If it is elastic (happy ball) it can deform itself (thus absorbing KE) and then recover the original shape, giving back roughly the same amount of KE, which is considered as temporarily stored ... 2 The simple answer is that in an elastic collision (for objects >> in mass than typical molecules) energy moves from kinetic to potential then back to kinetic as long as the "elastic limits" of the materials are not exceeded. In other words, as long as they act like springs. In non-elastic collision the energy goes mostly from kinetic of the colliding masses ... 0 yes, you assumption is correct, for an isolated system, conervation of linear momentum is equivalent to the velocity of the center of mass being constant. The term with variable mass from another answer is incorrect. You can only have variable mass in a non-isolated sysrtem. 1 The linear momentum of a system is given by \vec{p} = m \vec{v}. If you differentiate this with respect to time in an inertial frame, you have:$$ {d\vec{p} \over dt} = m {d\vec{v} \over dt} + {dm\over dt} \vec{v} $$If \vec{v} is constant with time, this becomes$$ {d\vec{p} \over dt} = {dm\over dt} \vec{v} $$Which means, for d\vec{p}/dt to be ... 1 Unfortunately you got answers that are wrong in many aspects. 1) I read all the previous answers concerning the 3rd law and I have seen that it is definitely not universal The third law is universal in classical mechanics. You stop to talk about forces in quantum field theory, but that is another issue. 2)Conservation of momentum is a universal ... 2 The laws of physics are discovered through a mixture of heuristics, modelling and inference. In case of momentum, the story goes like this: It is possible to 'transfer motion' from one body to another. However, experiment shows that it is not velocity that is conserved during such transfers, but another 'quantity of motion'. We give that quantity the name ... 7 If classical mechanics were valid at cosmological levels, the answer would be yes. But general relativity is what describes the dynamics at this larger scale, and it is not generically possible in GR (in an arbitrary spacetime) to define conservation of momentum or conservation of mass-energy. 1 There is an indigenous group of people in South America, the Aimaras. When they mention something that happened in the past, they do not signal to their back, but to their front. To their perception, the past is in front of them, because they can see it. And the future is behind just for the opposite reason. At least in western countries, our perceptions is ... 3 Why is mine wrong? What is the intuition? To see what happens, use finite differences to write$$(M - \Delta m) \Delta v = v_{rel}\Delta m$$which leads to$$M \Delta v - \Delta m \Delta v = v_{rel} \Delta m$$Dividing through by \Delta t yields$$M \frac{\Delta v}{\Delta t} - \Delta m\frac{ \Delta v}{\Delta t} = v_{rel} \frac{\Delta m}{\Delta ...

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Newton's 2nd Law states that the 'the change in momentum (p) of a body is proportional to the force (F) exerted on the body'. Mathematically, we write: $$F_{ext}=\frac{dp}{dt}$$ Where a body of mass $m$ and velocity $v$ has momentum $p=mv$. For a body of constant mass $m$, this becomes: $$F_{ext}=\frac{dp}{dt}=m\frac{dv}{dt}=ma$$ Where the mass $m$ ...

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Newton's second law relates force to acceleration: $$F = Ma \tag{1}$$ but force can also be expressed as the rate of change of momentum: $$F = \frac{dp}{dt} \tag{2}$$ In the case of a rocket it's the rate of change of momentum of the exhaust gas has that produces the force. Momentum is given by $p = mv$ (note $M$ is the mass of the rocket and $m$ is ...

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As someone who has spent some time playing with gyros, I am very familiar (though on a non-mathematical level) with these objects. I have six battery powered high precision Gyros found here: http://www.gyroscope.com/d.asp?product=SUPER2 I have been intensely fascinated with the way they behave when spinning at maximum speed. I feel hesitant to comment ...

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Relation between Forces and Potential Energy Can you explain why can't we define potential energy corresponding to a non-conservative internal force? In order to examine the relation between two terms we must consider the definitions of each term: a) forces: 1) internal forces are those that act inside a body (note that in engineering also a ...

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Rather than beating your student over the head with facts, try to approach the problem the way scientists did in the first place, by following the scientific method. Your student should come up with a hypothesis, and use known theory to make a prediction (calculate the momentum transfer in some idealized model), and then build a model to test the prediction. ...

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You don't decide yourself to make $dv$ negative or positive, you only find out what it is! For example, take the equation $x < 0$. Well that means $x$ is negative, so (by your logic) I should replace the equation with $-x < 0$, but now it says $x$ is positive?? This is in essence what you're doing, so I hope you can see the flaw. The way derivatives ...

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edit - issue with the second method - the way I see it that if you change the sign of $dv$ you also need to change the sign of $v$. Yes normally, in a problem like this we use integration with $dx$, but in the middle I would normaly use $\delta x$ when we talk about small quantities. The change to $dx$ when integrating with infinitessimally small step ...

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Work: $$W = \int_\gamma\mathbf F\cdot\mathbf{dr} = \int_{t_0}^{t_1}\mathbf F(\mathbf r(t))\cdot\mathbf{r}'(t)dt$$ If it is conservative then $\mathbf F = -\nabla U$. Therefore: $$W = \int_{t_0}^{t_1}\mathbf F(\mathbf r(t))\cdot\mathbf{r}'(t)dt = \int_{t_0}^{t_1} -\nabla U(\mathbf r(t))\cdot\mathbf{r}'(t)dt = -\int_{t_0}^{t_1}\left[U(\mathbf ... 0 The relationship between the potential energy and what you call "conservative" force is:$${\bf F} = \nabla U = {\bf n} \frac{\partial U}{{\partial\bf n}}\tag{1} $$where {\bf n} is the direction normal to the equipotential surface passing through some point P. Then, the mechanical work dW done by the force {\bf F} along WHATEVER path dL that ... 0 If you have an example in mind I could point out what is the problem. But you can prove mathematically that if a force is not conservative then it cannot be written as the gradient of a potential. And the reverse is true too. You can show mathematically as a theoren that is a force is conservative, it can be written as the gradient of some potential. 1 Potential Energy is negative of the work done by conservative forces. The reason why we define potential energy is so that we can get the result:: Work done on a body = change in potential energy of the body (if only conservative forces act on the body) this is proved by the very definition of potential energy. However let us say we were to define ... 1 before I answer your question, let me put forward another scenario. Imagine a mass less rod with two massive spherical objects (of mass m) is placed at either of its ends. Further imagine that the rod rotates about an axis. Let's define the axis to be the center of the rod. Let's choose the reference point at a distance x from the axis and on the rod. ... -1 About your reference point, torque acting on earth is not zero. But total torque of sun and earth is zero. That's why total angular momentum of earth and sun remains conserved. 11 You are a point in the circle. The torque is:$$ \mathbf\tau = \mathbf r\times\mathbf F  Where $\mathbf r$ is the position of Earth and $\mathbf F$ is the force (radially towards the sun). Notice when your reference point is somewhere in Earth's orbit, as you said, and your object is earth, the force will not be parallel to the position. Therefore, the ...

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