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0 votes

Could someone give me the explanation for this scenario?

We are interested in the change in the ball's momentum during the collision. The collision takes place in a very short time interval: the average force required to affect a change in momentum so ...
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-1 votes

Calculating momentum change?

let us look at what happens at the wall: the ball has incoming momentum in the positive x-direction, and departing momentum in the negative x-direction ; the momentum for the y-direction is the same ...
1 vote
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Can we say the momentum of a system is always equal to the momentum of the components of the system?

Velocity of the system is better translated as velocity of the centre of mass. We can use the definition of centre of mass: $$\mathbf R{cm} = \frac{M\mathbf R_M + m\mathbf R_m}{M+m}$$ Taking the ...
9 votes

Calculating momentum change?

Sounds to me, you are not familiar with vectors. So I highly suggest to learn about vectors quickly, which is a simple field of mathematics. Let's have a look in some more detail. x-axis points to the ...
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3 votes
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Force Problem with Punching Glove?

When you punch someone/something, you're applying an impact force on it and you yourself are experiencing the same force in return (of same megnitude but in opposite direction in accordance with ...
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6 votes

Calculating momentum change?

I'd recommend that you think first about a simpler case: you throw a ball straight at a wall (in the x-direction, let's say) and it bounces back with equal and opposite velocity. The wall has to exert ...
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7 votes
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Calculating momentum change?

One way to look at this is the following: for any system, you can relate the force the object experiences to its change in momentum by $$ \Delta \vec{p} = \int \vec{F} \, dt $$ In particular, if the ...
7 votes

Exact statement of conservation of momentum

Wikipedia's "in a closed system" and Taylor's "if there are no external forces on a pair of particles" are the same assumption, when applied to a pair of particles. In particular, ...
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0 votes

Why is maximal kinetic energy lost in a perfectly inelastic collision?

But just because the KE is 0 in some frame doesn't mean that it is the least possible in every other frame, does it? Yes, it does. Given two states with the same momentum, the question of which one ...
8 votes
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Adjoint of the Quantum Momentum Operator

Consider a vector space $V$ with an inner product $\langle\cdot,\cdot\rangle:V\times V\rightarrow\mathbb{R}.$ Given an operator $A:V\rightarrow V$, the adjoint is defined as the unique$^1$ operator ...
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Unclear negative sign in rocket propulsion equation

Actually $dm_g = -dm$ means that the mass that the rocket loses in the burned fuel, is the mass of the ejected fuel.The $dm$ is negative as the change is negative(The rocket is losing mass), whereas $...
-1 votes

Unclear negative sign in rocket propulsion equation

The total mass is the mass of the rocket plus mass of the fuel is $$m_t=m_r+m_f=\text{constant}$$ hench $$dm_r+dm_f=0\quad\Rightarrow\\ dm_f=-dm_r$$ f fuel r rocket The rocket equation \begin{align*...
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How does Newton's cradle actually work?

As @Steeven remarked already: Ad #1 and #2: $\vec p = const.$ You may want to review the video Stack of balls from this perspective, to change perspective. Though it are inelastic collisions, ...
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1 vote

Why is a slap shot in ice hockey more powerful than a wrist shot?

Force is not the same. Nor is it constant through either stroke. Consequently we can't apply any high school physics approaches that rely on constant forces. If we had a very good way of measuring the ...
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2 votes

Why is a slap shot in ice hockey more powerful than a wrist shot?

In fact it’s mostly the other way around. $dT$ is the time the puck is in contact with the blade of the stick and it’s roughly the same (to a factor of 2 depending on the players) for both shots. In ...
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0 votes

Why do the surfaces in contact of two colliding balls deform?

If a ball traveling 50m/s hits a stationary ball of equal mass then the traveling ball can only go as fast as the ball it collides with. When they collide one can't instantaneously accelerate or ...
1 vote

Unclear negative sign in rocket propulsion equation

$m$ is the mass of the rocket and $dm$ is the change in the mass of the rocket after burning a small amount of fuel. The negative sign means that the rocket is losing mass. In other words, the initial ...
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0 votes

Why did NASA need to observationally confirm whether DART successfully redirected Dimorphos?

This article confirms John Doty's answer that the biggest unknown was indeed how much ejecta the DART impact would throw out from Dimorphos: Researchers estimate that this spray of rubble meant ...
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2 votes
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Pressure due to a single photon

I don't think that $A$ is well-defined concept for a photon, and I'm not too sure about $\Delta t$ either. What does make sense is to interpret your expression as the mean pressure exerted per photon ...
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0 votes

Pressure due to a single photon

Too many variables: $$ f = c/\lambda $$ and by dimensional analysis: $$ \Delta t \propto \lambda/c $$ and $$ A \propto \lambda^2 $$ so $$ P\propto \frac{hc/\lambda}{c\lambda^2\lambda/c} = \frac{hc}{\...
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1 vote

4-momentum of photon

let speed of light c=1(by choosing the right units We do not neccessarily need the proper time to define the four velocity and hence, the four momentum. Any scalar λ which is independent of the frame ...
1 vote
Accepted

Momentum and Center of Mass of a Rocket split into two

The rocket had some horizontal momentum $4mv$ just before exploding. You have forgotten to consider this term in your equation for the conservation of momentum. Assuming the rocket to be a simple ...
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0 votes

Physical Significance of Frequency of matter waves

I think I got the answer after thinking a lot. Our teacher had said that for electrons, E=hν was actually wrong, rather we should consider, E=0.5mv² , or, E=p²/2m so, this means if we use de Broglie's ...
3 votes

Is the operator $P=-i\hbar\frac{d}{dx}$ self-adjoint given the Hilbert space of the problem of particle in a box?

The correct answer to your question has already been posted, but I wanted to supplement that by noting a particular misconception: If we want to apply this idea to the problem of a quantum particle ...
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6 votes
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Is the operator $P=-i\hbar\frac{d}{dx}$ self-adjoint given the Hilbert space of the problem of particle in a box?

You forgot to include the differentiability condition for the domain of $P$ and $P^\dagger$. Indeed, suppose our Hilbert space is $L^2(I)$ with $I:=[0,2\pi]$.$^\ddagger$ Define the domain of $P:=-i\...
1 vote

Physical Significance of Frequency of matter waves

I think you would do best to ignore this comment in your book; Personally, I even consider it wrong. We routinely talk about the energy of particles and that is a precisely defined quantity with no ...
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1 vote

Physical Significance of Frequency of matter waves

The wavelength $\lambda$ is physically significant in the sense that you can directly measure it. You can perform the double slit experiment and that will directly tell you the wavelength of the ...
0 votes

Difference between 'go over speed bump with one side' and 'both sides at the same time'

If both wheels hit the bump, the center line of the front of the car is raised to the height of the bump. If one wheel hits, the center line is raised half as much. The time it takes for one or two ...
2 votes
Accepted

Difference between 'go over speed bump with one side' and 'both sides at the same time'

When driving on one side, the center of mass of the car is raised to a lower height. This means the (one front and one rear tire) bumps you experience should be smaller when driving on one side, as ...
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2 votes
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Does a box of Pendulum's weigh more if they are swinging?

The double pendulum apparatus has an apparent weight equal to the sum of the weights of the box and two pendulums when at rest, and weighs more than that at the bottom of the swing, but weighs less ...
0 votes

If the particle moves with group wave, what $\lambda$ in De Broglie equation should we use?

Well, then you have to operate on momentum and wavelength ranges, like : $$ \Delta p \Delta \lambda = h \tag 1$$ And this a lot reassembles uncertainty principle (lower bound), except for the $\hbar$ ...
1 vote

If the particle moves with group wave, what $\lambda$ in De Broglie equation should we use?

In general that equation is only for particles that have a specific $\lambda$, or even a specific momentum. In general a particle need not be in a state which has those properties well defined. So at ...
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2 votes
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Where is Conservation of Momentum derived from?

Conservation of momentum follows immediately from Newton's 2nd and 3rd law. For simplicity, I will assume a 2-particle system but the logic can be easily generalised to many particles. For the purpose ...
1 vote

Why do we drop the renormalization term in momentum Klein-Gordon Field Theory?

During quantization, classical quantities are typically replaced by normal ordered operators to ensure correct action on the vacuum state, cf. e.g. my Phys.SE answer here. Nevertheless, in OP's case ...
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5 votes
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Why do we drop the renormalization term in momentum Klein-Gordon Field Theory?

The normal ordering procedure in flat spacetime is nothing but the subtraction of the (infinite) contribution of the Minkowski vacuum. As a matter of fact, an added finite renormalization term could ...
0 votes

Why do we drop the renormalization term in momentum Klein-Gordon Field Theory?

That term has nothing to do with renormalization, it’s just a result of ordering ambiguity in transitioning from classical physics to quantum physics.
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