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6

The answer to your title question is "neither". An object that moves has both momentum and kinetic energy, but it acquired these because a force was applied for a certain length of time. Now $F \cdot \Delta t$ has the same dimensions as momentum (and in fact, $F \Delta t = m \Delta v$ so there is a direct relationship between impulse which is $force * ...


4

While people normally quote Newton's Second law as $\vec F = m \vec a$, it is better written as $$ \vec F = \frac{d\vec p}{dt} $$ Force is a rate of change in momentum. This means that the average force applied when an object undergoes some discrete change in its momentum is $$ F_{\text{avg}} = \frac{\Delta p }{\Delta t} $$ The change in your momentum ...


3

Yes. There are only internal forces of your body. Without external forces, the center of mass of your body cannot change position. As your center of mass did not move, the main body should move in the opposite direction.


3

Since the photon reflects, its momentum changes: $p_{ph}'=-p_{ph}$. But total momentum of the system is conserved: $p_m+p_{ph}=p_m'+p_{ph}'$. Thus, the mirror will change its momentum. But, if the mirror has large mass, then it'll get very small energy from the collision. For zero-mass particle (photon) falling onto the mirror with mass $m_2$, the energy of ...


3

Momentum is conserved in magnitude and direction. So in order to analyze any situation of momentum conservation, you should always start with $$ \sum \mathbf p_{i}=\sum\mathbf p_f $$ where the subscripts denote the initial and final momenta. As to the ball & wall, you are correct that momentum is not conserved if you are only looking at the ball. If you ...


3

I'm with Floris that the correct answer is "neither", but I'd like to to take the explanation in a slightly different direction. The notion that something is "moving an object" carries an unstated assumption that the natural state of an object is to be at rest. That is one of those things that is both obvious and wrong.1 The natural state of an object is ...


2

It is the momentum of the entire system that is conserved. The fundamental reason for this is that the laws of physics are the same everywhere in space. This argument for momentum conservation is called Noether's Theorem. So where did you go wrong in your original example? Well you assumed that the wall was completely rigid. In reality that isn't actually ...


2

Why does speed have nothing to do with inertia? It depends on what you mean by "inertia". With one exception, physicists don't use this word much. That one exception is "moment of inertia". The problem with "inertia" is that sometimes it means "mass", other times, "momentum". Mass in classical mechanics doesn't change with speed. Momentum very much ...


2

The operator $d/dx$ isn't a "function" and Dirac surely never claims so. It's an operator, something that changes one function to another. By a function, we mean something that maps one number to another. Functions of $x$, like $f(x)$, may also be connected with operators on the space of (wave) functions. The wave function $\psi(x)$ is mapped to ...


2

You have to interpret $|\frac{d}{dq} \psi\rangle$. Knowing that decomposition of the basis $|q'\rangle$ gives : $$|\psi\rangle = \int dq' \psi(q') |q'\rangle \tag{1}$$ You have : $$|\frac{d}{dq}\psi\rangle = \int dq' \frac {d\psi(q')}{dq'} |q'\rangle\tag{2}$$ So, applying it to $|\psi\rangle = |q"\rangle = \int dq' \delta(q"-q') |q'\rangle$, you get : ...


2

Welcome to the exchange. When you ask "would it be easier to rotate", there are two different answers. If you mean, would it be easier to accelerate the wheel to some predetermined speed, the answer is yes. If you mean would it be easier to maintain that speed, the answer is no. For a rotating body, you need to learn about angular momentum, and angular ...


2

There is a small error in your math by the way - the factor of $4$ should be a factor of $1$ - check the denominator of the quadratic equation. Why is this so unreasonable? At a glance, it looks like the momentum of the sail is a bit larger than what it started out with. If I make the reasonable assumptions that: The sail is non-relativistic ...


1

Force is the rate of change of momentum, or alternatively mass times acceleration. So an object moving at a constant speed isn't experiencing any force. We only get a force when we try and slow the object down or speed it up. So as it stands your question can't be answered - or the answer is zero, which I doubt is very helpful. I'm guessing you're really ...


1

Feynman makes a point of stating explicitly, in vol. 1 of his Lectures on Physics, that $F = \frac{d(mv)}{dt}$ is not the definition of force. In section 12-1 he states If we have discovered a fundamental law, which asserts that the force is equal to the mass times the acceleration, and then define the force to be the mass times the acceleration, we ...


1

Momentum is $p = m v$, so it does have to do something with speed in the sense that it is proportional to it. However, you can have a different momentum with the same speed and vice versa. Take a bullet that has a mass of 3 g and a velocity of 370 m/s. Then the momentum would be about 1 kg m/s. You get the same momentum if you take a pack of flour (1 kg) ...


1

Can you explain why speed has nothing to do with inertia? It is not clear if you think it has something to do or not. In a very faint way they are related, because inertia is that fictitious force that once a body has been accelerated and has gained KE and velocity, keeps it going forever if nothing stops it. Momentum is the quantity of motion that a ...


1

I propose the following order-of-magnitude, very rough line of reasoning. The rate of momentum escaping from the aperture is $$\frac{dp}{dt}=\rho\pi(D_2/2)^2 v^2,$$ all in the horizontal direction. I assume that a fraction $(1-\cos\theta)$ of the horizontal momentum will be lost by the fluid since the change of direction of its motion. This fraction of ...


1

There is no law of physics that requires an external force to be present when some double time derivative of a parameter in a system changes. What you were taught in high school, that is $$F = m \cdot \dot v$$ and $$M = I \cdot \dot \omega$$ are actually the simplified version of $$F = \frac{d}{dt}m \cdot v = m \cdot \dot v + \dot m \cdot v$$ and $$M ...


1

While in some contexts, $$ \frac{d\textbf{p}}{dt}dt = d\textbf{p} $$ is correct and is mathematically rigorous, there is a straightforward way to derive impulse as the change in momentum. Consider a function $f(x)$ where $f:\mathbb{R}^m \to \mathbb{R}^n$ and $f \in C^1$. Suppose its derivative is $g = f'(x)$. We can consider integrating $g(x)$ over some ...


1

Linear momentum will be conserved when the Lagrangian that describes your system is unchanged by translations in space. This is a consequence of Noether's theorem, and it's as close as you're going to get to a fundamental explanation for the conservation of momentum. In general a Lagrangian written in the coordinate system of an accelerating observer is not ...


1

when low mass object hits high mass object it is reflected gaining opposite velocity almost the same as initial velocity. If I jump onto the wall why my body is not reflected? I know that collision is not fully elastic but it should be at least similar. Human body is not elastic: it cannot be deformed/ compressed in any way and then return to ...


1

Maximum penetration will be achieved if there is a small (on the order of several centimeter) gap between the muzzle and wall. This is because, with muzzle against wall, the wall impedes the escape of air and combustion gases. Somewhat analogous to a muffler lowering the effective horsepower of an engine. Beyond a "small gap for exhaust gases to escape," ...


1

'The mirror is given a momentum twice that of the incoming photon. As a mirror is typically quit heavy, lets say one gram. Its kinetic energy due to momentum it received will be extremely small. However, the photon will actually change its energy by the same amount, thus its wavelength changes, but not much.


1

The angular momentum $L_{A/B}$ of a rigid body $A/B$ about its center of mass is $$L_{A/B} = I_{A/B} \omega_{A/B},$$ where $I_{A/B}$ is the inertia matrix of $A/B$ about its center of mass in the world frame and $\omega_{A/B}$ is the angular velocity of $A/B$. The angular momentum $L_{A/B}^0$ of a rigid body $A/B$ about the origin of the world frame is ...


1

The rocket is in free fall along with the book. The nearest gravitating bodies are very far away, so whatever meager acceleration they cause will be almost exactly the same on the rocket and on the book. Suppose you instead went on a very close flyby of a neutron star. Now the book will fall rapidly, and away from the rocket's center of mass. The only way ...



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