# Tag Info

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Alpha particles are only significantly scattered by nuclei. Electrons are so much lighter than an alpha particle that it is hard for the alpha particle to transfer much momentum to them. But nuclei are small. The radius of a nucleus is of the order of $10^{-5}$ times the radius of an atom, so the cross-sectional area of the nucleus is of order $10^{-10}$ ...

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If $\Delta S = r cos\theta$ then $dl=ds=rd\theta$ and $F_g=mg$ $$W_{g}=\int_0^\pi mgrcos\theta d\theta$$ If you're taking the angle from the center of the circle (which you are, since you said that $\Delta S = r cos\theta$, then the initial position of the ball is $-R$, since displacement is a vector quantity (and the final ...

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Maybe we will need to deflect an oddly shaped asteroid heading towards Earth someday?

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So I came up with a graphical solution to this kind of problem. It might help you understand the process of collisions, without giving you a direct answer. Consider an Cartesian coordinate system xy for measuring momentum. Draw the initial momentum vectors $\vec{A} = m_A \vec{v}_A$ and $\vec{B} = m_B \vec{v}_B$. Draw a circle with the two vectors as ...

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Rotate the digram so the line connecting the circles is horizontal at the moment that they touch - since you know dx and dy, you just take the arc tangent. Now you move the frame of reference so the point where the two balls meet is stationary. The actual speed of the center of mass is just the vector mean of the velocities of the two balls (if they have ...

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Just solve the second order differential equation obtained from using Newton's Laws i.e. $$F=-\frac{k}{r^3}$$ or $$m\frac{d^2r}{dt^2}=-\frac{k}{r^3}$$ If you solve this differential equation, then your equation for the path will be of the radius as a function of time. The equation will be a non-central conic. HINT (TO SOLVE THE DIFFERENTIAL EQUATION): ...

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In relativistic QFT for instance, the Hamiltonian does not respect Lorenz invariance. That is not to say that the symmetry is not present, in the Hamiltonian. Of course it is, the Lagrangian and Hamiltonian are equivalent ways of analysing the same thing. But the Hamiltonian does not have Lorenz invariance autocratically built into it, as it singles out ...

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An important thing to recall about alpha particles is that at energies up for a few tens of MeV they range out in very short distances (less than a milimeter in many cases). Multiple scattering can be expected to generate non-trivial scattering angles only over larger ranges than the penetration depth of all alpha-decay alphas and a great many alphas that ...

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Let us say you have two frames of reference; frame $F$ and frame $F'$ such that $F'$ is moving at velocity $v$ in the positive $x$ direction of $F$. Given a space time event that occurs at $(ct,x,y,z)$ in frame $F$ the Lorentz transform helps us to find the space-time coordinates $(ct',x',y',z')$ of that event in frame $F'$. If, however, you know the event ...

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