Momentum Definition and 1000 Threads

I am very confused when textbooks say the direction of Angular velocity is perpendicular ot radius and theta for that matter direction is in perpendicular direction. I know this comes from cross product rule but what is the meaning of Angular velocity and Angular momentum directing in upward...

$$\langle p | W | p' \rangle = \int \langle p | x \rangle \langle x W | x' \rangle \langle x' p' \rangle dx dx'$$ $$\langle p | W | p' \rangle = \int \langle p | x \rangle \delta(x-x') W(x) \langle x' | p' \rangle dx dx'$$ $$\langle p | W | p' \rangle = \int \langle p | x' \rangle W(x') \langle...

Hi all - related to a question I asked some time ago: If one introduces a momentum cutoff, the result in the most basic case is Lorentz violation. That is, some form of preferred frame must be introduced. I'm wondering what this does to the vacuum state? That is, how does one keep the vacuum...

Here's the question ^ My first thought to solving this is to use Heisenberg's uncertainty principle. $$\Delta x \Delta p = \frac{h}{4\pi}$$ Now, we approximate ##\Delta x = \frac{L}{2}##. Then, plug and chug we end up with:$$p =\frac{h}{2\pi L}$$ I thought this was it, especially because this...

To show that when ##[J^2, H]=0 ## the propagator vanishes unless ##j_1 = j_2## , I did (##\hbar =1##) $$ K(j_1, m_1, j_2 m_2; t) = [jm, e^{-iHt}]= e^{iHt} (e^{iHt} jm e^{-iHt}) - e^{-iHt} jm $$ $$ = e^{iHt}[jm_H - jm] $$ So we have $$ \langle j_1 m_1 | [jm, e^{-iHt} ] | j_2 m_2 \rangle $$ $$ =...

I calculate in this way : Angular Momentum = I W = [ ( 1/12 ML^2 + m(L/2)^2 ] (V/ L/2) = [ 1/12 ML^2 + 1/4 mL^2 ] 2V/L = 2VL/4 [ M/3 + M] but can not find a matching answer. Why?

Let's imagine an ideal scenario where you're lifting your own weight in its entirety. Let's say a woman weighing 100 lbs. Suppose she's doing an idealized handstand and pushup from that position. So she's lifting 100 lbs. Let's say ideally all of the forces are on her arms only. Do these forces...

mball = 2 kg, mputty = 0.05 kg, L = 0.5 m, v = 3m/s a) Moment of inertia : I = (2mball + mputty ). ¼ L^2 = 0.253125 kg.m^2 Linitial = Lfinal => mputty. v. r = I.ω => ω = (4.mputty.v.r) / I = 0.148 rad/s b) K initial = 1/2 m v^2 = 0.225 J K final = 1/2 Iω^2 = 2.85.10^(-3) J => Kfinal /...

Here we are talking about non-relativistic quantum physics. So we all know kinetic energy T = E - V = \frac{1}{2}mv^2 in classical physics. Here V is the potential energy of the particle and E is the total energy. Now what I am seeing is that this exact same relation is being used in quantum...

My proposed solution: When the student stops at the end, suppose the carriage is moving at speed u. 0 = (M+2m)u - m(v - u) ==> u = mv/ M+3m After jumping out, the total momentum of the Carriage + collector system is 0 - mu = -m^2v/ M+3m. By conservation of momentum for the Carriage +...

In quantum mechanics hydrogen atom problem ##L=\sqrt{l(l+1)}\hbar##. What that means physically when ##L=0##?

To find the mass in other pan, i need to find the force caused by beads on the pan ∴ KEinitial + PEinitial = KEfinal + PEfinal 0 + mgh = ½ mv^2 => v = 3.13 m/s ∴ The change in momentum : p2 - p1 = m ( v2-v1) = m( v - (-v)) = 2mv ∴ F = Δp / Δt = n. m. v How can i apply the rate of 100...

80km/h = 22.2 m/s Through momentum: 1940(v_f) = 540 (22.2) + (1400)(-22.2) => v_f = -9.84 m/s I figured the work that the energy lost in a collision is equal to the work done to crumple the cars. So W = K_i - Kf = [1/2 (540)(22.2)^2 + 1/2(1400)(-22.2)^2] - 1/2(1940)(-9.84)^ 2 = 384110 J At...

Maybe a silly question but on the above question using the conservation of momentum: momentum before firing (0) = momentum after firing (55*35)+(M*2.5) If I re-range the above it's M = -(55*35)/2.5 = -770kg. I can I reconcile that minus sign (basically get rid of it)? Thanks

Hello! If we have a transition between 2 ro-vibrational levels of the same electronic state of a diatomic molecule the selection rules require for the changes in the rotational quantum number J that ##\Delta J = \pm 1##. Why can't we have ##\Delta J = 0##? The photon carries one unit of angular...

I know that the eigenstates of momentum operator are given by exp(ikx) To construct a real-valued and normalized wavefunction out of these eigenstates, I have, psi(x) = [exp(ikx) + exp(-ikx)]/ sqrt(2) But my trouble is, how do I find the expectation value of momentum operator <p> using this...

Hello everybody at the forum I'm from Ukraine, I have Chemistry degree, and last year I began to self studying Quantum Mechanics. I'm reading this article: R. Garcia, A. Zozulya, and J. Stickney, “MATLAB codes for teaching quantum physics: Part 1,” [Online]. Available...

1. Hello, so the difficulty I am having with this problem is that is seems relatively straightforward. I have tried to solving it by assuming that this is a collision in which momentum is conserved. Therefore, I found the total momentum before the collision and used this to resolve it must be...

I try to use relativistic energy equation: $$E'=\gamma mc^2$$ But, I use $$\gamma=\frac{1}{\sqrt{(1-(\frac{v'}{c})^2}}$$ then I use Lorentz velocity transformation. $$v'=\frac{v-u}{1-\frac{uv}{c^2}}$$ At the end, I end up with messy equation for E' but still have light speed c in the terms. How...

Textbook solution: ##v## is the instantaneous velocity, $$P(t)=(M+b t) v$$ Then $$impulse = \Delta P = (M+b t) v = \int^{t}_{0} F dt'$$ Thus $$v=\frac{F t}{(M + bt)}$$ What I did instead was: Let ##M## be the instantaneous mass, and ##M_0## be the initial mass, then $$M=M_{0} + b t$$...

Equation for rocket motion: $$\frac{d \mathbf{P}}{d t} = M\frac{d \mathbf{v}}{d t} - \mathbf{u}\frac{d M}{d t}$$ But $$\mathbf{F}=\frac{d \mathbf{P}}{d t}=M\frac{d \mathbf{v}}{d t}$$ So $$M\frac{d \mathbf{v}}{d t} = M\frac{d \mathbf{v}}{d t} - \mathbf{u}\frac{d M}{d t}$$ And $$-...

For the sake of this question, I am primarily concerned with the position wave function. So, from my understanding, the wave function seems to 'collapse' to a few states apon measurement. We know this because, if the same particle is measured again shortly after this, it will generally remain in...

Hello! I just started reading some molecular physics and I am a bit confused about the electron angular momentum in diatomic molecules. Let's say we have just 2 protons and an electron for simplicity and we are in the Born-Oppenheimer approximation, so we assume that the nuclei are fixed in...

Take for example earth. Earth has angular momentum about its own axis. However, if we ignore the orbital portion, the angular momentum of the Earth relative to the sun's axis is the same. Another example is the spinning bike wheel/person holding it in a chair. It has angular momentum about its...

The textbook says that by squaring and subtracting the expressions you can eliminate u. E² - p² = y² (mc²)² - y²m²u²

My attempt: ##E^2 = p^2c^2 + m_0^2c^4## ##2E dE = 2pc^2 dp ## ##\frac{dE}{E} = \frac{pc^2}{E^2}dp=\frac{p^2c^2}{E^2}## % (dp/p = 1%) ##=\frac{E^2-m_0^2c^4}{E^2}## % ##=1-\frac{m_0^2c^4}{E^2}## % ##=1-\frac{m_0^2c^4}{m^2c^4}## % ##=1-\frac{1}{\gamma ^2}## % ##=\frac{v^2}{c^2} ##% =0.81...

I have trouble solving this problem any help would be appreciated.Problem statement ##J=\frac{mr^2}{2}## a) Determine the motion of yoyos for ##n=1,2,3## The case for ##n=1## is simple, however, I am having trouble with ##n=2## and ##n=3##. for ##n=2## I started by drawing all the forces...

Unfortunately, I couldn't arrive to the correct answer ($$=0.28mL^2 \omega^2$$ ) and will be happy to understand what am I doing wrong. **My attempt:** Using $$ E_k = \frac{1}{2} I \omega^2 $$ I obtain that the difference I need to calculate is $$ \frac{1}{2} (2mL^2)(0.8\omega)^2 +...

In the classic example of a person holding a spinning bike wheel, as they flip the wheel over, angular momentum is conserved by the person/chair spinning with 2x the angular momentum of the initial wheel. Not questioning that. However, I thought ang momentum is always conserved about a...

Why, in lagrangian mechanics, do we calculate: ##\frac{d}{dt}\frac{\partial T}{\partial \dot{q}}## to get the (generalised) momentum change in time instead of ##\frac{d T}{dq}##? (T - kinetic energy; q - generalised coordinate; p - generalised momentum; for simplicity I assumed that no external...

I honeslty don't quite know how to start. It seems like the Hooke's coefficent k is independent of the answer to this problem. I would also appreciate any clue of expressing the condition when "balls will collide again". The fact that all balls can keep moving make this rather difficult. It...

L = mvr = mr (dr/dt) = 2m*r*(dr/dt)/2 = 2m*(dA/dt) So, A = (L/2m)T so, ## L = \frac{2 \pi a b m}{T}## Now, ##T^2 = \frac{4 \pi^2}{GM} a^3## So from all these, I get ##L = \sqrt{ \frac{GM m^2 b^2}{a}}## But answer given is ##L = \sqrt{ \frac{2GM m^2 ab}{a+b}}## (This, they have derived from...

I know that in QM, there is LS coupling. So the interaction is there. But is such an interaction possible in macroscopic objects like a planet?

A stationary observer sees a particle moving north at velocity v very close to the speed of light. Then the observer accelerates eastward to velocity v. What is its new total velocity of the particle toward the north-west relative to the observer? I ask because while the particles total...

Question 7.6 Official solution It seems that the solution uses the conservation of angular momentum to solve this question (τ=0). But the problem is that the frame is set on the centre of mass of the guy, which is non inertial. I would like to know why it is correct to do it this way. My...

I don't understand the reasoning of this question's answer. The answer is velocity = 0 (option A). A while ago, I was told that, since the magnets were held at-rest (before being let go), they must have no velocity after the collision. What about the velocity which they had just before the...

So I've managed to confuse myself on this problem :) Since the problem says we can assume ##m_p << m_b##, I'm assuming that the velocity of the bowling ball will be unchanged, such that ##\vec v_{b,i} = \vec v_{b,f} = -v_{b,0} \hat i## I started out using the energy-momentum principle, ##(\vec...

List of relevant equations: Angular Momentum = L (vector) = r(vector) x p(vector) Angular velocity of rotating object = w(vector), direction found using right hand rule. Torque = T(vector) = dL(vector)/dt I have a few questions about torque and angular momentum direction and...

Question: If we place the frame of reference on an accelerating point, does the total rotational momentum still remain the same? I attempted to solve this question by manipulating the equations as shown below. $$\text{Define that }\vec r_i=\vec R+\vec r_i'\text{, where r is the position vector...

First I calculated the momentum of m1. Since m2 was at rest after the collision, all its momentum was transferred, so m1 has a momentum of 158 i hat. L=r x p, so its 916 k hat. This would also be the change in L because it was initially 0 when m1 had no velocity, so I know this is the net...

First I found the moment of inertia, I=1.8(5.5^2+3.9^2+4.9^2) =125.046 Then I tried to find the rotation rate using the equation L=rotation rate*I rotation rate=3773/125.046=30.173 But the answer is suppose to be 21.263?

I know how to get to the answer but that's what is confusing me. To find final velocity I multiply the acceleration by the time the object fell. Then multiply the velocity by the mass to get momentum. Now the angular momentum is r x p. Since the initial angular momentum was 0, this was also...

So I first tried to find L using torque, Torque=d/dt*L, and took the integral of this. Ended up with 23.28484t Now I square the equation L=rotation rate*I to get L^2=rotation acceleration *I^2 Angular acceleration=L^2/I^2 I feel like I am doing something wrong though, this doesn't give the...

m1v1+m2v2=m1vs'+m2v2' , if car hits small fluffy object m2, initially v2=0, and v1'=v2' ... so m1v1=[m1+m2](v2') but why not energy? Why is there a KElost? .5m1v1^2+.5m2v2^2=.5m1v1'^2+.5m2v'2^2 +KElost , and again v2=0, v1'=v2' .5m1v1^2=.5[m1+m2]v2'^2+KElost using consv of momentum...

So to cut to the chase, I missed my class' lesson on momentum - have tried to catch up, quite successfully but am baffled about this question. I know the conservation of momentum etc. but after trying for ages it's just not happening this question so any help would be much appreciated, Oscar.

Please scroll-sown to Question 52: https://www.undergraduate.study.cam.ac.uk/files/publications/engineering_s1_qp_2017.pdf The correct answer is 'B'. This is the working I did: F = (change in momentum) / (change in time) change in momentum = mv - mu, where v = final velocity and u = initial...

So we know that the initial intertia of the merry go round is 250 kg m^2 and its angular speed is 10 rpm. MGRs angular momentum would be L=Iw=250(10)=2500kg m^2 rpm. We know the mass if the child is 25kg, and the child's linear velocity is 6m/s. We convert linear to angular w= v/r = 6/2 =...

Hi, I have the following problem: A homogeneous disc with M = 1.78 kg and R = 0.547 m is lying down at rest on a perfectly polished surface. The disc is kept in place by an axis O although it can turn freely around it. A particle with m = 0.311 kg and v = 103 m/s, normal to the disc's surface at...

I know that the force must be a central force and that under central forces, angular momentum is conserved. But I am unable to mathematically show if the angular and linear momentum are constants. Radial Momentum ##p=m\dot r = ma\dot \theta=ma\omega## Angular Momentum ##L=mr^2\dot\theta =...