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12

First, something we need to get out of the way: Kinetic energy as $\frac{1}{2} m v^2$ is not a precise formula; it is merely a good approximation for anything that is traveling slowly compared to the speed of light. In fact, more precisely, the energy is \begin{equation} E = m\, c^2\, \frac{1}{\sqrt{1-v^2/c^2}} \approx mc^2 + \frac{1}{2} m v^2 + ...


6

Your lecturer got the eigenvalue using the fact that the operator $\hat{p}$ is Hermitian so you can do this: \begin{align} \langle p| \hat{p} &= \left( \hat{p}^\dagger |p\rangle\right)^\dagger\\ &= \left( \hat{p} |p\rangle\right)^\dagger\\ &= \left( p |p\rangle\right)^\dagger\\ &= \langle p| p \end{align} I think it ...


5

That formula for momentum is only true for massive particles. Here's what is always true: A particle with a mass $m$ ($\geq 0$) can have an arbitrary momentum $p$ (in some direction, with magnitude $\geq 0$). The energy of such a particle is $$ E = \sqrt{m^2c^4 + p^2c^2}$$ The velocity of a particle is equal to $$ v = \frac{pc^2}{E} $$ When $m = 0$, $E ...


5

The problem does not mention any radii, but if we did know the radius of each sphere, would it be possible to skip conservation-of–linear momentum calculations altogether The accepted answer has confused you: You can simplify things by considering these conservation laws in the center of mass frame. There the total momentum is zero, therefore ...


4

In both cases and at all times, the force from the (wall/tire) on the hammer equals the force from the hammer on the (wall/tire) : total momentum must be conserved. However, in the first case, the initial energy is dissipated in the wall (as heat and/or damage), so at the end the hammer is stopped. In the second case the initial energy is stored as ...


4

Inertia is an intrinsic characteristic of the object related to its mass. Inertia tells you how much force it will take to cause a particular acceleration on the object. Momentum is a function of an object's mass and velocity. Momentum is a measure of the kinetic energy of the object. A massive object can have any momentum (at least as long as its velocity ...


4

Momentum: The resistance of an object to a change in its state of motion. That sounds like a fishy definition of momentum to me. A slightly better definition, at least at your level, is that momentum represents the "amount of motion" an object has. Granted, "amount of motion" is a very vague term, but it stands to reason that if "amount of motion" were ...


4

A few quick clarifications: a particle cannot just annihilate. It disappears when it interacts with something else. The obvious example of this is an electron and positron annihilating to turn into two photons. Also, the total energy of a particle (this applies to electrons, positrons and photons) is given by: $$ E^2 = p^2c^2 + m^2c^4 $$ where $p$ is the ...


4

Yes, $F=ma$, but also $v=at$. That means that, as you fall for a longer time, your speed will increase. After 1 second, you are going at $9.8 m/s$ or $35 km/h$, about the speed of Usain Bolt. After 10 seconds you would reach $98 m/sec$ or $350 km/h$. For a free-falling human, the air resistance actually limits you to about $200 km/h$. When you hit the ...


3

You are right. If $q$ is a generalized coordinate then $\dot{q}$ is the generalized velocity and hence the generalized momentum is \begin{equation} p = \frac{\partial L}{\partial\dot{q}} \end{equation} Therefore, your sequence looks correct. Further, equations (20) and (21) of the article you have referenced also tell that the $p_{\theta_i}$ are indeed ...


3

Find the nearest box-spring mattress. With your hand, execute a slow motion "impact" between your hand and the mattress. You should notice that when your hand is not touching the mattress, there is, of course, no force between your hand and the mattress. As your hand begins to touch the mattress, you should feel a very light force. As your hand presses ...


2

The thrust coming from a rocket engine is exerted on the engine bell, and it is directed along its axis of symmetry. It's not completely clear how you're modelling your ship but it is probably more realistic to apply the force to the "thruster fire" block, whatever that is. It's important to note, though, that if applying the force to the engine bell and ...


2

If the collision is not perfectly along the line connecting the centers of mass of the pucks, they will exert torques on each other as well as forces. The angular momentum of the pair will be conserved, so if the incoming puck was not spinning, the pucks will exit the collision spinning in opposite directions. If the surface they slide on is frictionless, ...


2

Your analysis will need to be a combination of theory and experiment. You are not maximizing impulse alone: you need to think about drag as well. So you need to think about the shape of your rocket as well as the volume of water and initial pressure. The drag on the rocket is probably well modelled by the ram pressure equation ...


2

Force is a concept which describes, and can be used to manipulate, real phenomena that exist regardless of the existence of the human race. Newton's second law (Acceleration = Force / Mass) is a definition of force. Mass certainly exists, as do velocity and its time derivative, acceleration. There is no reason to suppose that force does not exist. Force ...


2

Without defining what a thing is, it makes little sense to discuss the ontology of a thing. Does an apple exist? First, one must say what an apple is; once we agree on that, it's straightforward to show (by example) that apples exist. Given a definition of force, force certainly does exist; we can point to time derivatives of momentum that we observe, and ...


1

This seems like a list question which aren't very good for SE. Here is a proof of your statement: $m_1v_1 = m_2v_2$ $\dfrac{m_1}{m_2} = \dfrac {v_2}{v_1}$ Let us assume that $m_1$ < $m_2$ $\dfrac{(m_1)^2}{(m_2)^2} = \dfrac {(v_2)^2}{(v_1)^2} < 1$ $v_1$ must be greater than $v_2$ The only case the energy of a higher velocity is beaten is shown by ...


1

If you consider things classically (for the moment forgetting about virtual particles as mediators of the force) things get more clear. For instantaneous forces (which do not exist in nature), momentum conservation comes from the fact, that the forces in nature fulfil Newtons axiom actio = reactio, meaning, that for two particles, that interact we have the ...


1

I will bring an example from classical electrodynamics. In EM(electromagnetism) you have to consider that the fields(electrical and magnetic) also have energy and momentum. A classical example is to apply the third law of Newton(each action has an equal counter-action) to two moving charges. Then you'll conclude that the third law does not hold-thus the ...


1

You didn't cancel out the mass $m$ properly. $$-mv_0\cos a+0=mv_1\cos b+0\\ -v_0\cos a=v_1\cos b\\ \frac{-v_0\cos a}{\cos b}=v_1$$ The incident angle is equal to the exit angle in such collision, $a=b$. The above reduces to: $$v_1=\frac{-v_0\cos a}{\cos b}=\frac{-v_0\cos b}{\cos b}=-v_0$$ And here you see what you probably expected. In the perpendicular ...


1

Hint: The center of mass of pieces A and B moves in the same path as the intact shell would. (This arises from the conservation of momentum.) Edit: The gravitational force is only acting along the vertical direction. So there is net acceleration only in the vertical direction. Looking at the horizontal one, if we neglect things like air resistance (which ...


1

If we did know the radius of each sphere, would it be possible to skip conservation-of–linear momentum calculations altogether and use a single conservation-of-energy equation... The radius of a ball is always irrelevant to the outcome of a collision, what counts is the ratio of the masses. Two equations are always necessary to determine the outcome ...


1

As CuriousOne inexplicably said in the comments, but not as a formal answer, you should use this equation: $$\vec{L}=m{\vec{r}}\times{\vec{v}}$$ This is the standard equation for angular momentum in vector form. Once you have your angular momentum vector, you can get the individual components. You can see how to take a cross product here. If you need to ...



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