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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.


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.


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 \begin{equation} ig \sum_i p_i \ , \end{equation} where $g$ is some coupling and $p_i$ the momenta of the particles. This vertex, however, vanishes due to momentum ...


5

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 ...


4

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 $m$ ...


4

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. ...


3

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 ...


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 ...


3

$$ \newcommand{\p}{\partial} \begin{aligned} \p_{\mu}J^{\mu} &= \gamma_{\nu}(\p_{\mu}\p^{\nu}\phi)\gamma^{\mu}\psi+\gamma_{\nu}\p^{\nu}\phi \gamma^{\mu}\p_{\mu}\psi-m\p_{\mu}\phi\gamma^{\mu}\psi - m\phi\gamma^{\mu}\p_{\mu}\psi\\ \end{aligned} $$ Now the first piece can be written as $$\gamma_{\nu}(\p_{\mu}\p^{\nu}\phi)\gamma^{\mu} \psi= ...


2

In short, it continues to rotate simply because it has no reason to stop rotating. Imagine I have a puck and I whack it across an ice-rink. The puck will continue to travel at the same speed until it hits a wall; i.e it will stay the same unless there is a reason not to. This is conservation of linear momentum as shown here. The same thing happens with ...


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 ...


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 ...


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

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 ...


2

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 ...


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 ...


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

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 ...


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 ...


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

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} ...


1

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 ...


1

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 $M$ ...


1

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) ...


1

A force is not an impulse, it's a force. A force can exist without producing any work, stresses in materials are typically generated by forces applied to the same solid that oppose themselves and therefore do not produce any work. Some forces require contact, some forces don't (infinite range, decreasing with distance). As long as you're under the influence ...


1

It's hard to think of a physical system involving a force that acted for zero time. However I think it's useful to consider a collision, perhaps between two billiard balls. When the balls collide they change momentum. We know that the change of momentum is just the impulse, and we know that the impulse is given by: $$ J = \int F(t)\,dt $$ where I've used ...


1

I only expand TwoBs comment to your answer. There is following statement: massless particles with both of helicities $\pm 1$ can't be represented by 4-vector field $A_{\mu}$. The only field (up to equivalence) which represents corresponding particles is $F_{\mu \nu}$. If you decide to represent these particles by $A_{\mu}$, then it won't be 4-vector: $$ ...


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 ...



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