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Newton thought of momentum as "Quantity of motion" - as we can see in the translated version of 'Principia'. Particularly, he defined momentum in the following words: The quantity of motion is the measure of the same, arise from the velocity and quantity of matter conjointly. So yeah, that is the definition of momentum. The question why we defined the ...

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I am only going to leave a brief answer, seeing that the comments are very accurate. The paradox can simply be resolved by considering the elastic nature of all the objects. How so ever instantaneous might the $dt$ or the time of collision seem to the human eye, actually it occurs over a small duration, based on the elasticity of both the objects involved in ...

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Actually A.Zee's book on "QFT in a nutshell" has a very nice explanation on this on chapter I.5, I will breifly sketch it (this is a very rough skeptch), $$Z=\int DA e^{iS(A)} =e^{iW(J)}$$ where W(J) is given by, $$W(J)=-1/2 \int \int d^4xd^yJ(x)D(x-y)J(y)$$ where D(x-y) is the photon propogator and J(x) and J(y) refer to two lumps of matter Plugging in ...

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The momentum operator is not $-i\partial_x$, rather, that is the representation of the momentum operator on the position basis: namely $$\langle x|\hat{p}|\psi\rangle = -i\frac{\partial}{\partial x}\psi(x).$$ Otherwise, the momentum operator is just defined by action on its eigenstates as $\hat{p}|p\rangle = p|p\rangle$. I understand the complex ...

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Yes. It is much easier to think of this in terms of conservation of momentum: Because light (and electromagnetic radiation in general) has momentum, you will have to gain momentum in the opposite direction to conserve total momentum --- just like if you were to throw the flashlight. It is difficult to think of this in terms of forces because we tend to ...

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1)The conservation law of linear momentum does involve the velocity, not just the speed. That said, however, it is ok to use speed in the problem statement. There are no external forces acting on the system bullet-gun. Therefore, if the bullet moves along the positive $x$-axes the gun recoils in opposite sense along the same direction. This is implicitly ...

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Imaging the balls on a string. You are launching N balls per second, at a velocity $u$. This means the distance between the balls is $u/N$. And $N$ balls per second will pass a certain point in space. Now if the car is moving at a velocity $v$ (same direction as $u$), fewer balls per second can hit it - because subsequent balls on the string have further to ...

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For these kinds of system we often define a pair of quantities, one which is characteristic of objects or systems and one which is characteristic of interactions. Examples of these pairs are work (interaction) and energy (system) or impulse (interaction) and momentum (system). There is no commonly applied name for the interaction quantity that pairs with ...

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Keep in mind that the equation $$E^2 = p^2c^2 + m^2c^4$$ is derived from the relations \begin{align} E = \gamma mc^2,\qquad p = \gamma m v. \tag{1} \end{align} Therefore $$p = E\frac{v}{c^2}.\tag{2}$$ Although (1) is only defined for massive particles, it turns out that (2) remains valid when $v=c$, i.e. for massless particles. Indeed, we get $$E= ... 1 Newton (if I recall correctly) typically referred to the concept of inertia, which was an objects resistance to changes in velocity when subjected to external forces. You are right about him not thinking about it as just the speed of the object, because this is where the mass term comes in. Many people think of Newton's second law as being written as F = ... 1 What you're looking for is an intuitive explanation or how you could visualize momentum. You can think of momentum as the quantity/amount of motion or "how much would I not want be in the path of this body." I'm going to try and provide some intuition through a few examples: A car of mass 1000 kg moving at 5 m/s would have the same "quantity/amount of ... 1 I think for 1-dimensional bound states, this is the proof : The expectation value of the momentum operator, \langle \hat{P} \rangle=\langle \psi|\hat{P} \psi\rangle=\int_{-\infty}^{+\infty}\psi^{*}(x)\frac{\hbar}{i}\frac{\partial}{\partial x}\psi(x)dx=\frac{\hbar}{i}\int_{-\infty}^{+\infty}\psi^{*}(x)\frac{\partial}{\partial ... 1 For the forces between elementary particles we have Feynman diagrams, where there exists a mediating particle for the interaction. In the simplest diagrams: for the strong it is the gluon, for the weak it is Zs and Ws and for the electromagnetic it is the photon. Here is Bhabha scattering, where the electron and the positron ( attractive force) are first ... 1 Does the imaginary part have any physical significance? Are we to interpret this as two waves in superposition in the complex plane? In a sense neither the real part nor the imaginary part have physical significance, as these quantities do not directly appear in observables. One way to see this is that any solution \left|\psi \right\rangle to ... 1 No the time taken does not depend of the velocity attained by the first ball(if they are ideally rigid) it rather depends on the elasticity or rigidity of the balls. So for ideally rigid bodies, the time taken to transfer approaches 0. Nothing would happen with an increase in distance between the two balls. See: Is the reaction force for a stone hitting a ... 1 It is correct except for one sign, note that the work done by friction is negative (since you move the block in the opposite direction w.r.t. the friction force) and thus it is equal to$$ -\mu mg d \cos \theta $$with this solving for  v gives you 11.49 m/s. 1 The two laws are the same. To see this break down your rotating object into a sum of point masses. Then consider one of these masses: The angular momentum of our point mass is given by:$$ L = rmv $$so:$$ \frac{dL}{dt} = \frac{d}{dt}(rmv) $$For circular motion r is constant so we get:$$ \frac{dL}{dt} = rm\frac{dv}{dt} = rma  But the second ...

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