New answers tagged

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To calculate the minimum energy needed for the reaction the products are assumed to be stationary, i.e. the momentums are zero. With $ \pmb p^2 = E^2 - \vec p^2 = E^2 = m^2$ follows: $$({p_1^\mu}' + {p_2^\mu}' + {p_3^\mu}' + {\bar p_4^\mu}')^2 = (E_1 + E_2 + E_3 + E_4)^2 - (\vec p_1 + \vec p_2 + \vec p_3 + \vec p_4)^2 = (E_1 + E_2 + E_3 + E_4)^2 = (m_p + ...


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but still we have momentum telling us that both blocks must rise to the same height That's not true here. In the first case where the bullet is embedded, the final velocity of the bullet and the block must be identical. Since initial momentum of the two shots were the same, then the final momentum will be the same as well. Because they are connected, ...


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The loss of energy (efficiency<1) manifests itself in two ways Friction. When doing free body diagrams add frictional forces at the sliding contacts and find which coefficient of friction $\mu$ gives you the efficiency values you expect. Structural damping (hysteresis). This is more difficult to put in the equations of motion, because they assume ...


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First of all, the transmission of force between any two pairs of mating gears happens due to surface contact between the two and hence through friction. Next, any loss is a dissipation in energy. Though I am not that well-versed in vibration analysis, I know that an periodic dissipation of energy is accounted in the force-equation using a damping ...


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Assuming no air friction, you compute the initial velocity needed from conservation of energy: $$ \frac12 mv^2=mgh $$ The impulse needed is $mv=F\Delta t$. The product of these ($F, \Delta t$) is constant - shorter time implies higher force. The above assumes the time of impact is short enough not to affect the over all time (otherwise you need to solve ...


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Supposing these vehicles are very solid, have both of them crash into identical parked train wagons, from the back. If after the crashes both vehicles are still intact and not moving any more, then you will see both train wagons moving away at the same speed. The extra supposition (solid vehicles still intact after the crash) is there to assume no momentum ...


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The external force acts only for the small time when the cue has been struck. Once it moves, there is no force. This means that the ball is moving with zero external force, which means according to Newton's second law, the velocity of the ball is the same. here the act of friction is of less importance as it requires in a billiard play. So the center of mass ...


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Just a note, you can rewrite (3) as $$\vec{F} = \frac{d\vec{P}}{dt} = \frac{dm}{dt}\vec{v} + m\frac{d\vec{v}}{dt}$$ If you have a system where the total mass is not changing, then $\frac{dm}{dt}=0$, which brings us back to the form we're more commonly used to.


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One could not defend a book you are most probably misquoting. I strongly suspect the book says "the mass distribution is not constant", that is M is constant but the distribution and number of constituent $m_i$s may vary, i.e. they may split or agregate, a common feature in astrophysics. You are confusing yourself with symbols and definitions and proofs. ...


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Well if we neglect the hidden momentum the conservation law of momentum in electromagnetism is simple: The momentum can be stored in static fields ($D\times B$); the mechanical momentum ($mv$) + electromagnetic momentum ($D\times B$) $= constant$. The similar formula is valid for angular momentum (where it is not hidden momentum) See Feynman's Lectures ...


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This is because $H'=UHU^{-1}$ for a certain unitary operator $U$, therefore $\psi$ is an eigenvector of $H$ with an eigenvalue if an only if $U\psi$ is eigenvector of $H'$ with the same eigenvalue. Thus the two operators have the same point spectrum. $U = e^{i \lambda X/\hbar}$. From $[X,P]= i \hbar I$ one finds $$e^{-i \lambda X/\hbar}Pe^{i \lambda ...


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From the equation $H$ is a linear function of $H_0$. In that case, the eigen values of $H_0$ are eigen values of $H$ also. That's why the eigen values remain invariant under such a transformation.


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Probably the best way to think about this is to say that $$p = mv\\ F=\frac{dp}{dt}=m\frac{dv}{dt}+v\frac{dm}{dt}$$ (Using the usual product rule for differentiation - thanks @ja72 for the suggestion). If velocity is constant the first term vanishes and your result follows.


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Considering momentum: $$m_1u_1+m_2u_2=m_1v_1+m_2v_2$$ Therefore, $$m_1u_1-m_1v_1=m_2v_2-m_2u_2$$ Factorising gives us: $$m_1(u_1-v_1)=m_2(v_2-u_2)$$ Allowing us to rearrange to: $$\frac{m_{1}}{m_{2}}=\frac{v_{2}-u_{2}}{u_{1}-v_{1}}$$ Using the fact that you say the velocities can be interchanged, we obtain a final answer of: ...


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We consider friction to an impulsive force, in cases when normal force is impulsive. Here's how:We know that $f=\mu N$(only during slipping motion, for no slipping frictional force is equal to applied force RESISTING friction). Since friction is proportional to normal reaction, it will be impulsive only when normal force is impulsive.Thus, if in a situation ...


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The answer is two arcs. One arc with a constant gee loading in one direction and then flipping to the opposite direction. This is called the bang-bang method, and it is no very smooth, but the gee forces never exceed the specified maximum. Given a path $y(x)$ the instantaneous radius of curvature at each x is $$ \rho = \frac{ \left(1+ \left(\frac{{\rm ...


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For Newton's Laws to hold, mass must not vary. Wherever you read otherwise was mistaken. Take a look at this answer. It contains a description of why mass must be constant in Newton's Laws in the context of the rocket equation ... but the analysis applies generally. Newton's Laws are not valid for variable mass systems.


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Courtesy of its spin the electron has a magnetic dipole moment. That means if we place it in a magnetic field the two states aligned with and against the magnetic field have different energies. The magnitude of the energy difference depends on the strength of the field and the size of the magnetic dipole moment, which in turn depends on the spin. So by ...


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No, isn't correct, because saying that the acceleration $a$ you feel is $F \over m$ implies that you're using the law of motion: $$ F = ma $$ Which is valid in classical mechanics, not in special relativity. The correct relation con be derived formally by a least action principle In special relativity, the law we deduce has to be invariant under Lorentz ...


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Yes, there are more unknowns than equations. You do not have sufficient information to solve for the requested quantities. Someone might be playing a prank on you! In reality, each ball and each paddle would have a specific finite stiffness, and one could use this information along with some clever math to determine the final velocities of all the bodies. ...


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No, but Newton's third law of motion implies the conservation of momentum. In other words, Newton's third law is a special case of the more general law, which is the conservation of momentum.


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As a complement to another answer, I'll demonstrate moving from the coordinate representation (wave function) to the momentum representation. Recalling that $$\Psi(x,t) = \int dp \frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}}\Phi(p,t)$$ and putting this expression into the (coordinate representation of the) TDSE, we have $$i\hbar\frac{\partial}{\partial ...


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Why is the vector |S⟩ represented as Ψ for both bases when working out the components for the quantum mechanics case above? The first of the final two equations is simply an expression for the sifting property of the delta 'function'. $$f(x) = \int dx' f(x')\delta(x - x') $$ Let's back up just a bit and write the state (ket) as a weighted 'sum' of ...


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Take the inverse Fourier transform of your last equation, $\psi(x,t)=\frac{1}{\sqrt{2\pi\hbar}}\int e^{i p x/\hbar}\phi(p,t) dp$, to see that the "coefficients" of $\psi(x,t)$ are not the same in the two representations: in the $x$ representation, the coefficients are $\psi$ and in the $p$ representation, the coefficients are $\phi$.


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Let me rephrase those precise equations in the language of finite-dimensional linear algebra. You have a vector $A$ and two bases $\beta=\{e_i\}_i$ and $\beta'=\{e_i'\}_i$. This means you can write the components of $A$ with respect to $\beta$ as $$ A_i=e_i·A=\sum_j\delta_{ij}e_j·A $$ and the components with respect to $\beta'$ as $$ ...


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The safest way to start with is the representation-free Schrodinger equation, $$ i\hbar\frac{\partial}{\partial t}|\Psi(t)⟩= \hat{H}|\Psi(t)⟩. $$ Referring to your case, we take the separable Hamiltonian: $H=\frac{p^2}{2m}+V$ so that $$ i\hbar\frac{\partial}{\partial t}|\Psi(t)⟩= \left(\frac{\hat{p}^2}{2m}+V\right)|\Psi(t)⟩. $$ Now is the time the ...


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Yes and no. depends on the direction considered. Momentum is a vector quantity. Since there is no force in the x direction, momentum is conserved in that direction but not in the vertical direction because gravity (an external force for the system of Wedge and block) is acting. Yes, for your 2nd comment above. You can use COLM (conservation of linear ...


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For a 2D planar simulation with zero friction do the following Definitions Each body has 3 degrees of freedom. These are $(x_1,y_1,\theta_1)$ and $(x_2,y_2,\theta_2)$ defined at the center of mass. Each body has mass and mass moment of inertia. These are $m_1$, $m_2$ and $Iz_1$, $Iz_2$. The contact is at point A with coordinates $(x_A,y_A)$ and with ...


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No. Momentum is still conserved. In particular, the component of momentum parallel to the ground is conserved. So if the ball is going to the right before hitting the ground, it will continue going to the right after. The formula you refer to is for one-dimensional collisions. That applies only if the elements are arranged so that there actually is ...


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You are making it more complicated than it needs to be. If your mass is $M$, your momentum is $2M$, so the momentum of the slab is $-2M$ and the velocity is $\frac {-2M}{5M}=-0.4$. You don't know where the CM is because you don't know where the person is. If the person starts in the center, the CM is at $\frac x2$


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Your assertion that if we place a charged object (such as a sphere) in the "donut hole" region of the inductor, this object will be accelerated back and forth through the "donut hole" by the changing electric field created by induction. is completely incorrect. There is never any magnetic field outside the solenoid, which means that there is never any ...


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Think of pushing a wagon full of sand down the road with a constant applied force. $F=ma$ handles that just fine. But now open a hole in the wagon so the sand slowly drains out as you push. If you still push with a constant force the mass of the wagon and remaining sand in it decreases with time and of course you'll get it going faster in the end. You ...


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I do not get why system such as the rocket in space are defined as "variable mass" since the mass of the system is not varying. This depends entirely on where one draws the system boundaries. One possible boundary is the rocket plus all of the exhaust gases it has released. The center of mass of the rocket + exhaust gas cloud system moves per the ...


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I do not get why system such as the rocket in space are defined as "variable mass" since the mass of the system is not varying. It is because the "system" refers to particles that are still in the rocket, no to the whole system including the expelled exhaust gas. This is often done when the point of the analysis is in describing and understanding ...


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$F=ma$ is true only when the mass of the system is fixed. However, in your case where the spaceship is moving across space, the system is losing mass due to the ejection of fuel out of itself. For such cases, where the total mass of system is varying, we cannot use this version of Newton's 2nd Law. That version of the law that is written down in your ...


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The two equations are the same to first order, which is all that is important. If I were writing down the equation for the total momentum P(t+dt) myself, I would probably jot down the first equation (that of Morin) since I would be thinking of the instantaneous velocity of the rocket at time t rather than at time t+dt. But, again, the distinction is not ...


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Work (or energy) is transferred from one particle to another, but the net effect is no change overall. How? Consider a collision force acting between two particles over a small time frame. During that time frame the particles move, and the work done on one particle is ${\rm d}W_1 = \boldsymbol{F} \cdot {\rm d}\boldsymbol{x}_1$. Since an equal and opposite ...


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The collision doesn't happen at a single point in space - rather the colliding objects exert a forces on each other over a distance as they approach and the recede. Consider a tennis ball hitting a racket - the ball and the strings of the racket deform and we get an increasing elastic restoring forces until the two objects at at their closest approach. ...


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For two particles to influence each other you need some sort of interaction. For (macroscopic) mass this is clearly Coulomb-interaction. Two atoms can not be at the same place, because their cores repell each other. If you look at smaller scales, strong and weak interaction might add their part. Photons have no charge, no color-charge and don't interact ...



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