21

Momentum is conserved only if there is no net external force on the system. Consider the snowball and the tree as the system. In your case, the earth provides an external force on the tree, so the momentum of the snowball/tree system is not conserved. If the tree is "suspended" (not attached to the ground) momentum would be conserved, but the ...


9

Momentum of the body will be transferred to earth, let's exaggerate the numbers in your favor and assume your snowball is $10kg$ and you managed to throw it with $v_s = 200\frac{m}{s}$ where subscript "s" stands for snowball. Mass of the earth is approximately $5.97\cdot10^{24}kg$. For simplicity assume it was an elastic collision $$m_sv_s = m_ev_e$...


6

The tree is attached to the earth. The momentum from the snowball is transferred to the tree and then distributed troughout the whole earth. Because a tree is a solid object this transfer happens almost instantaneously. The earth is so large that this tiny amount of momentum won't be noticable at all.


4

If you are asking about frictional forces that would slow the rotation of earth around the sun, then since there is no air in space (space is almost a vacuum), then there is no friction so the earth cannot be slowed down as such. For the same reason, its rate of rotation on its own axis will also not be hindered. Also, the rotation of earth cannot be slowed ...


3

The true fact is that conservation of total momentum is a consequence of translational invariace in Lagrangian mechanics. I.e. the fact that the Lagrangian is translationally invarint implies the conservation of total momentum. However in Newtonian mechanics, under the assumptions that in an inertial system and for an isolated system of material points, (a) ...


3

Two complex scalar fields $\phi_{1}$ and $\phi_{2}$ can be rewritten as four real fields, in terms of their real and imaginary parts, $$\Phi=\sqrt{2}\left[\begin{array}{c} \Re\{\phi_{1}\} \\ \Im\{\phi_{1}\} \\ \Re\{\phi_{2}\} \\ \Im\{\phi_{2}\} \end{array}\right].$$ For the free theory, the Lagrange density is actually equal to $${\cal L}=\frac{1}{2}\partial^...


3

Your $Q$ still is four independent quantities - one for each independent choice of the translation direction $a^\nu$. Noether's theorem states that for a one-parameter continuous quasi-symmetry \begin{align} x^\mu & \mapsto x^\mu + \epsilon \delta x^\mu \\ \phi^a & \mapsto \phi^a + \epsilon \delta \phi^a \end{align} where the theory is quasi-...


3

Let give a bigger picture. Assume the snowball appears (due to atmospheric processes) at some height "h". From this moment, gravity accelerates it against earth (discard wind forces). In the same way, earth is accelerated towards the snowball. Gravitational forces are the same for snowball and earth in modulus but opposite sense, same for momentum, ...


3

In an axisymmetric system, the gravitational force on a particle is going to be radial. It therefore exerts zero torque about the origin, so angular momentum about the origin is conserved: $$L = mr^2\dot{\varphi} = \text{const.} $$ Now differentiate both sides with respect to time to get $$ 2mr\dot{r}\dot{\varphi} + mr^2\ddot{\varphi} = 0,$$ which upon ...


2

I think that to understand your misunderstanding we have to consider the conditions for conservation of momentum: momentum is conserved so long as there is no net external force acting on the system in question. Considering the simple case in a vacuum, even friction between balls or between the strings and the post is acting. There may even be other forces ...


2

The block is on a horizontal surface. The normal force acts in a vertical direction. Since you are considering motion in the horizontal direction, there is no net force.


2

If $\gamma \rightarrow\pi^+ + \pi^-$ is valid process, then the conservation of total 4-momentum equation $$\tilde \gamma = \tilde \pi_+ + \tilde \pi_-$$ expresses a lightlike vector as the sum of two future-timelike vectors. UPDATE: To address your last question in any reference frame, the energy $𝑝^0$ seems to be always positive. Why is that so? The ...


2

As with photon non-decay to $e^+e^-$, the trouble is that the massive system always has a rest frame, but the photon does not. In a comment, you write I believe here's a problem in your statements. The quantity that is conserved is the 4-momentum, not the 3-momentum. Thus, there is no contradiction in having no 3-momentum conservation. Here you are mixing ...


2

The photon is an elementary particle of mass zero. If it could decay into two pions, it would violate Lorenz invariance, because four vector invariant mass does not change, that is why it is called invariant. The photon should have had at least the summed mass of the two pions. The length of the energy-momentum 4-vector is given by The length of this 4-...


2

I did not draw in the image another direction of pressure acting upwards from below because that was not the image in my mind but the one that I have drawn is. Your image in your mind is that pressure is a force which does not act upwards. That means that an object will experience an unbalanced downward force due to the atmospheric pressure of $14\text{ psi}...


2

I don't think this is a bad question at all - I don't understand the down votes. Issues like this used to confuse me too, and I don't feel that any of the comments or answers so far have really got to the bottom of the issue. Let's think about your setup a bit. You have a Newton's cradle toy in a vacuum, so it's isolated from air resistance. But you still ...


2

Your assumption that both bodies come to rest is false. Recall that in an elastic collision with two balls $A$ and $B$ with initial velocities $v_{A,\;i}$ and $v_{B,\;i}$, the final velocities are given by: $$v_{A,\;f} = \dfrac{m_A - m_B}{m_A+m_B} v_{A,\;i} + \dfrac{2 m_B}{m_A+m_B}v_{B,\;i}$$ $$v_{B,\;f} = \dfrac{2m_A}{m_A+m_B} v_{A,\;i} - \dfrac{m_A-m_B}{...


1

The ground applies a frictional/constraint force , so there is a net external force on the system and hence, Conservation of momentum can't be used on the snowball and tree as a system.


1

The chance in angular momentum is given by:$$\frac{dL}{dt}=\tau,\tau=\hat{r}\times F$$ In an ideal rocket, the main engine does not apply torque to the system, since the thrust is parallel to the position vector from which it acts (here it'll be the distance from the CoM). So we simply have $$\tau _{tot}=\tau_{thrusters}$$ so it wouldn't matter when we use ...


1

tl;dr: In principle, yes at the amplitude level, but no at the rate level. In the Standard Model, no, not at all. I assume you are talking about something like "photon decay in vacuum" because you say "lone photon." So I won't consider photon conversion due to interactions with some other particle, for which there are more possibilities. ...


1

The first one is for the amputated diagrams, which do not contain any external legs. The second one is the full matrix element plus the delta for the conservation of momentum. The delta for the conservation of momentum is not fundamental, since it can be added at any moment when evaluating cross sections. From my experience, i'd rather put the conservation ...


1

If I assume that your diagram is to be read from left to right, then it seems that two quarks respectively coming from the two colliding protons, annihilate. However, for quarks to annihilate, one needs to be an anti-quark. So what one would expect instead is a gluon exchange between the two quarks with enough energy to try and kick the quarks out of the ...


1

Momentum is conserved if there is no NET external force acting on the system. The normal force on the block is balanced out by the force of gravity on the block. You may say, "what about the force of gravity acting on the bullet -- does that cause an external force?" It does, however, if you consider the speed of the bullet right before the ...


1

Let's simplify a bit more. Suppose there are only two balls, and they are made of wet clay. One balls swings toward the other and sticks - totally inelastic. You have probably already done the math for collisions like this. Momentum is conserved, kinetic energy is not. Both balls have the same velocity afterward. The outcome is the velocity is $0.5$ m/s. The ...


1

Atmospheric pressure comes from air molecules bouncing of the object. As the molecues are moving in random directions, the average force they exert is perpendicular to the surface they are hitting. The total force on an object at rest in the air is therefore it's weight together with the sum of the pressure forces. This sum is called buoyancy, and is ...


1

Suppose you are given a general collision problem in 1-dimension with two known masses and their initial velocities. Conservation of total momentum is an equation with two unknowns, and thus can't alone determine the final velocities. You need additional information... like a final velocity... or a relation between the final velocities. For a totally ...


1

Solving for velocities in 1D an elastic collision for two objects $A$ and $B$ boils down to solving the following system of equations, $$ \left\{ m_Av_A + m_Bv_B = m_Av_A' + m_Bv_B'\quad \text{cons. of momentum} \atop \dfrac1 2m_Av_A^2 + \dfrac1 2m_Bv_B^2 = \dfrac1 2m_Av_A'^2 + \dfrac1 2m_Bv_B'^2\quad \text{cons. of KE} \right.$$ The above system yields the ...


1

The form of the E.o.M almost implies straight away there's a conserved quantity: i.e. it looks as if you can write it of the form $\frac{d}{dt}(...) = 0$. In the case you presented, $2 \dot{r} \dot{\phi} + r \ddot{\phi} = 0$, it should be easy to see that this can be written as $$ \frac{1}{r} \frac{d}{dt} \big(r^2 \dot{\phi} \big) =0 \ ,$$ which is exactly ...


1

I remember finding this confusing. Sometimes they call conserved quantities 'integrals of the motion', which I really don't think is a good name. Let us differentiate $$ mr^2\dot\phi \to \frac{d}{dt}mr^2\dot\phi=2mr\dot r\dot \phi+m r^2\ddot \phi=mr(2\dot r\dot \phi+r \ddot \phi) $$ Its called an integral of the equation of motion because if you ...


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