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You have the right ideas, but you don't dare to put them in maths ! Momentum conservation is vectorial. Here, you have a 2D system, so let's write it with 2 component vectors. I will denote $\vec{P}_{i/f}$ the initial and final momentum respectively. So the momentum conservation yields $\vec{P}_i = \vec{P}_f$ I choose to represent the $x$-axis as the ...

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Momentum conservation says $\vec{p}_A = \vec{p}_B + \vec{p}_C$, we can split this in components: $$p_{A,x} = p_{B,x} + p_{C,x} \\ p_{A,y} = p_{B,y} + p_{C,y}$$ Some of these momentums are $0$. Which one? For the block B - the only one with a motion not parallel to one of the axes' - you have to use trigonometry. To get the velocity, once you have the ...

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Frankly, until you bring along an accelerometer as well as tracking the bus's speed thru corners (I'm assuming you mean you sway during a turn, not just going down the road), I'm going to remain skeptical of your claim. Centripetal force is centripetal force. Now, if you happen to counteract turning force unconsciously, it may be that you feel more ...

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It is not immediately obvious, but the block has calculable angular momentum at the point just before impact. the block has velocity $v$ tangential to the disk's center of rotation which is a distance $r$ away., and so has angular velocity $\omega=v/r$. the block also has calculable moment of inertia around that center, $I=mr^2$. Then, it is simply ...

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I suppose you know the mass and extent of the disk. Let's just ignore the mass of the stick. (We don't have to do that, but it makes everything simpler). We can then just use conservation of energy: You can calculate the kinetic energy of the block. Once it stops, this energy will be transferred completely to the rotational energy of the disk. Using the ...

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You have already got "practical" answers, so I intend to answer form another point of view. There is a quite famous theorem due to Stone and von Neumann, later improved by Mackay, and finally by Dixmier and Nelson, roughly speaking establishing the following result within the most elementary version. (Another version of the theorem focuses on the unitary ...

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A perfectly inelastic collision (also known as a plastic collision) occurs when the maximum amount of kinetic energy of a system is lost. In a perfectly inelastic collision, i.e., a zero coefficient of restitution, the colliding particles stick together. In such a collision, kinetic energy is lost by bonding the two bodies together. This bonding energy ...

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Lionel's answer affords good intuition. For me, another notion that really motivates all this is the notion of symmetry and the proof (in a particular sense) that for every symmetry we can see in nature, there must be a conserved quantity. This idea is embodied in Noether's theorem. At high school it wouldn't be reasonable to expect you to have the ...

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One way to think of it is "flux", i.e., the amount of stuff flowing or moving through somewhere. If I wanted to quantify the rate at which fish are escaping from a pen, I could say that $x\,\mathrm{kg}$ fish swim past the opening in the pen in $1s$, or that the flux is $x\, \frac{\mathrm{kg}}{\mathrm{s}}$. Faster fish or heavier fish would contribute more to ...

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Firstly, impulse is the change of momentum. Force is the RATE of change of momentum. The equation you have given is for force, not impulse. So, Force=change in (mass x velocity) / time of impact However, impulse, force* time= change in momentum. For example, if a car suddenly crashes with a wall, the passenger will continue to move in a straight line ...

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The electron is stuck to the atom, which isn't going anywhere. This means that $\langle p \rangle = 0$. The uncertainty $\Delta p$ measures the RMS fluctuation of the momentum: $$\Delta p^2 = \langle p^2 - \langle p \rangle^2 \rangle = \langle p^2 \rangle$$ Since $E = p^2 / 2m$, this means that $$\langle E \rangle = \frac{\Delta p^2}{2m}$$ The ...

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I) At least three different quantities in physics are customary called an action and denoted with the letter $S$: The off-shell action $S[q;t_i,t_f]$, The (Dirichlet) on-shell action $S(q_f,t_f;q_i,t_i)$, and Hamilton's principal function $S(q,\alpha, t).$ For their definitions and how they are related, see e.g. this Phys.SE answer. II) OP's ...

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The "center-of-charge" is part of a more general concept that is used quite often in physics: Multipole expansion. The general idea of multipole expansion is the following: If you view a charge (or mass) distribution from a large distance, then most of its internal structure is irrelevant to you. Instead, it suffices to do all calculations based on a few, ...

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Mass and charge are not so similar for the charge having "center of charge". The notion of "center of mass" appears in many applications when number of bodies move. In this situation, the movement can be splitted into movement of center of mass and individual movements of bodies relative to the center of mass. This occurs because of dual role of mass: it ...

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Of course you can define such a quantity, but the question is: does it mean anything physically? Contrary to what has been stated in some of the answers/comments, this quantity is not comparable to a "normalized" dipole moment. A dipole is a system of two charges equal in magnitude but opposite in sign. The corresponding dipole moment, which is of great ...

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The reason why you arrive at different solutions is that the assumptions in the assignment are inconsistent. One way to show this is to show in isothermal flow if ideal gas in straight duct of constant cross-section with no friction the gas has to have the same pressure everywhere; different pressures at the entrance and the exit $P_1, P_2$ are not possible. ...

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If we have some coordinates $q_i$ and some momenta $p_i$, then a generator of a transformation is defined as a function $g(q_i, p_i)$. By definition, this generates the transformation $$q_i \to q_i + \epsilon \frac{\partial g}{\partial p_i}$$ $$p_i \to p_i + \epsilon \frac{\partial g}{\partial q_i}$$ So if we want the generator of translations, we want ...

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Dirac argues from symmetry in his Principles of QM: In a 1-D system, $\hat{q}$ and $\hat{p}$ are both observables, with eigenvalues extending from $-\infty$ to $+\infty$, and are connected by the commutation relation $[\hat{q},\hat{p}]=i \hbar$. Since one can interchange $\hat{q}$ and $\hat{p}$ in these equations if $i$ is replaced by $-i$, it follows ...

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Say that we are handed a 1D quantum mechanical system, which satisfies the canonical commutation relation $$\tag{1} [\hat{Q},\hat{P}]~=~ i\hbar~{\bf 1},$$ and handed some choice of eigenstates $|q\rangle$ and $|p\rangle$ for every value of $q,p\in\mathbb{R}$. The eigenstates satisfy $$\tag{2} \hat{Q} \mid q \rangle ~=~q\mid q \rangle, \qquad ... 1 A force affects the motion of the center of mass only (call it point C). The rotational motion is defined by the total torque applied on the center of mass. If the applied force (F_x,F_y) is at location (r_x, r_y) relative to the center of mass then$$ \begin{aligned} F_x &= m \ddot{x}_C \\ F_y &= m \ddot{y}_C \\ r_x F_y - r_y F_x ...

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When you hit anything with a force, it causes an acceleration in the direction of the force. So if you hit the square at the corner like so: It would start moving upwards. However, it also experiences a torque about its center of mass. This torque can be calculated as the force multiplied by the perpendicular distance between the center of mass and the ...

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You may use the following notation for hypersurfaces in four dimensions : $d\sigma_\mu = \epsilon_{\mu\alpha\beta\gamma}dx^\alpha dx^\beta dx^\gamma$ For instance $d\sigma_0= d^3x$ The expression of the momentum-energy is then : $P_\nu = \int d\sigma^\mu \Theta_{\mu\nu}$ The same kind of expression could be used with the charge : $Q = \int ... 2 You don't actually need$\hat{p}$to do this. You can start from the fact that$\hat{x}$, when applied to a position eigenfunction, has to produce the corresponding position eigenvalue ($x_0$) times that same function: $$\hat{x}\phi_{x_0}(p) = x_0\phi_{x_0}(p)$$ In the momentum representation, position eigenfunctions take the form$\phi_{x_0}(p) = ...

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Assume one dimensional and $\hbar=1$ By \begin{array} \hat \hat{p} |p \rangle &= p | p \rangle \\ \langle x | \hat{p} |p \rangle &= p \langle x | p \rangle \\ -i \frac{\partial}{\partial x} \langle x| p \rangle &= p \langle x| p \rangle \,\,\,\, (1) \end{array} Because you already knew $\hat{p}=-i\frac{\partial}{\partial x}$, hence ...

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Here's how you do it. First, notice that for any state $|\psi\rangle$, we have \begin{align} \langle p|[\hat x, \hat p]|\psi\rangle &= \langle p|\hat x\hat p-\hat p\hat x|\psi\rangle \\ &= \langle p|\hat x\hat p|\psi\rangle - \langle p|\hat p \hat x|\psi\rangle \\ &= \langle p|\hat x \hat p|\psi\rangle - p\langle p|\hat x|\psi\rangle ...

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This should get you going in the right direction. I am using different variables so you can understand the concepts behind impacts and collisions (or explosions). Two rigid bodies are attached at common point A located a distance $\vec{r}_{A1}$ and $\vec{r}_{A2}$ from their respective centers of mass. With the CM velocities $\vec{v}_{1}$ and ...

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The answer is simple. The optimum strategy to slow (accelerate) the system is to remove equal amounts of mass from each small mass at the same time and in the same (opposite) direction as the current motion of the pair of point masses connected by the rigid rod. By symmetry the momentum change is then all in translational momentum in the direction ...

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