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12 votes
Accepted

After how many bounces will a ball's mechanical energy equal zero?

Never, if you treat the problem with the laws of classical mechanics, treat both the ball and the ground as "macroscopically" rigid, and you don't set a threshold $\overline{h}$ under which ...
basics's user avatar
  • 10.6k
6 votes

Noether's theorem by a taste of logic

Noether’s theorem is a mathematical theorem. It connects conservation laws to the properties of Lagrangian functionals. If these properties are contradicted by experiment then this means that a ...
my2cts's user avatar
  • 25.3k
3 votes

After how many bounces will a ball's mechanical energy equal zero?

First approximation: ball and surface are perfectly rigid. This implies the collision is instantaneous, and therefore elastic. The bounce height is always the same. Second approximation: ball and ...
Alex I's user avatar
  • 426
2 votes

Do bodies stick together after an inelastic oblique collision?

Do bodies stick together after an inelastic oblique collision? If the collision is perfectly inelastic, yes. If coefficient of restitution is zero it should mean that after collision the relative ...
Bob D's user avatar
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2 votes
Accepted

Alternative way to compute expectation value of momentum?

Yes, you can work out the expectation value of an operator in that way. The key thing to notice is that the expectation value does not depend on which basis you choose to work with. For example, it ...
Níckolas Alves's user avatar
2 votes
Accepted

Meaning of $d\mathcal{L}=-H$ in analytical mechanics?

I think it is gibberish in your own words. $\frac{\partial}{\partial \dot{q}}$ cannot be replaced by $dt \frac{\partial}{\partial {q}}$ even in the most cavalier approach because $\dot{q} = \frac{dq}{...
John's user avatar
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2 votes

Generalized momentum

Generalized momentum $p^i$ is conserved if the Lagrangian function $\mathscr{L}$ doesn't depend on the generalized coordinate $q^i$. It's readily proved from the Lagrange equation, $$\frac{d}{dt} \...
basics's user avatar
  • 10.6k
1 vote

Why is it less shocking to cross the bump with just one wheel?

Good point. The missing piece is that there is a spring and shock absorber between each wheel and the car. If a wheel is moved to a higher position, the spring exerts a larger upward force on the car. ...
mmesser314's user avatar
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1 vote
Accepted

Direction of impulse

Let's start by asking what is impulse ? Its the change in momentum of the body under consideration i.e. $$\vec{I}=\Delta \vec{p} = \vec{p}_{final}-\vec{p}_{initial}$$ Now , say that the ball's initial ...
CP of Physics 's user avatar
1 vote

Direction of impulse

Since impulse is the difference between two vectors, it will always be along the diagonal of a parallelogram formed by the initial and final momentum-regardless of if there is a change in speed or not....
mike1994's user avatar
  • 978
1 vote

Direction of impulse

Why the Impulse is Along the Bisector: The ball's speed doesn't change, only its direction. This means the magnitude of momentum remains constant, but its direction changes. The change in momentum (...
Ansh Tandon's user avatar
1 vote

Direction of impulse

Impulse is just the change in momentum, which is the difference between the final momentum vector and initial momentum vector. If these two vectors are of the same magnitude, then their difference ...
BioPhysicist's user avatar
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1 vote

Meaning of $d\mathcal{L}=-H$ in analytical mechanics?

Apart from the fact that I am really skeptical about its mathematical validity, your replacement $\frac{\partial}{\partial\dot{q}}\rightarrow\frac{dt}{\partial q}$ makes little sense in the context of ...
paulina's user avatar
  • 1,884
1 vote

How do you solve instantaneous 3 body collisions

How did you solve it for 2 particles? I ask because there's not actually enough information in the question to determine what happens in a collision. Consider a simple example of two equal mass ...
Cort Ammon's user avatar
  • 50.2k
1 vote
Accepted

What's the meaning of the momentum operator?

Disperson relation Before even looking at the Schrödinger eqaution let's look at the wave equation to understand an important concept in physics: the dispersion relation. The wave equation (1D) is ...
AccidentalTaylorExpansion's user avatar
1 vote

Do bodies stick together after an inelastic oblique collision?

The case you are referring to is where the objects are like in this drawing: where the surface of interaction (the red lines) is slippery and the objects are malleable enough to absorb all velocity ...
Jos Bergervoet's user avatar
1 vote

Proper time of two particles being the same when they are under a tree-level interaction

If you want it to be the same then you can set it to be the same. The proper time $\tau$ does not have any particular physical significance. What is important is $d\tau$. So you can set $\tau$ how you ...
Dale's user avatar
  • 103k

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