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I used the center of horizontal momentum frame. I just used the y-moment to find the angle of the velocity component, so I can write the velocity in the horizontal direction. I'll edit the equations to make it clearer

This problem I already solved using another resource (just get the coordinate of the center of mass reach and from it, get to the larger mass. R = (3v02) / (4g)). But I'm having some trouble calculating using moment conservation. Here what I've done so far: $$ 3\vec v_0 = \vec v_1 +2\vec v_2 $$...

"In (1) and (3), what is the initial height of the mass center of each side?" I was trying to calculate from the end of the string, but really from the center of mass it seems a lot simpler. "In (2), you have a factor x (l-x) in the GPE term, but it is (l-x)^2 in (4). Do you see the asymmetry...

You're right! It's as simple as that. But, the logic used to express the initial and final energies do you think is right?

I solved this problem easily using Newton's second law, but I had problems trying to use mechanical energy conservation to solve it. How I solved using Newton's second law: ##\text{(part of the rope that is on the left)}\, m_1=x\rho g,\, \text{(part of the rope that is on the right)}\...

I am trying to prove the following expression below: $$ \int _{0}^{1}p_{l}(x)dx=\frac{p_{l-1}(0)}{l+1} \quad \text{for }l \geq 1 $$ The first thing I did was use the following relation: $$lp_l(x)+p'_{l-1}-xp_l(x)=0$$ Substituting in integral I get: $$\frac{1}{l}\left[ \int_0^1 xp'_l(x)dx...

Sorry, I forgot to organize and correct something here. 1 / | r-r '| is a scalar, in my papers I was disregarding the module, forget that part of the divergence theorem.

it's a scalar (lol I saw the error here, i forgot to fix it here)

Spherical coordinates. Using Cartesian coordinates I solve correctly, i am trying for spherical coordinates.

The integral that I have to solve is as follows: \oint_{s} \frac{1}{|r-r'|}da', \quad\text{ integrating with respect to r '}, integrating with respect to r' Then I apply the divergence theorem, resulting in: \iiint \limits _{v} \nabla \cdot \frac{1}{|r-r'|}dv' =...

Doing R=|r-r'|, i get the expected result: \nabla \frac{1}{|r-r'|} = -\frac{1}{R^2}\hat r=-\frac{(r-r')}{|r-r'|^3} But doing it this way seems extremely wrong, as I seem to be disregarding the module. So I tried to do it by the chain rule, and I got: \nabla...