Recent content by TheGreatDeadOne

Old question, and I forgot about it (It's not worth reviving an old thread). Going through some papers I found these notes and tried to redo it. PS: At the time I already had solved this problem in another way before making this older post (it's in the first part of the description), I just got...

Yes, but considering it that way, there would be no way to solve it via momentum conservation. In the question we have (it's in the Homework Statement): "The pieces continue moving in the same plane as the entire projectile and reach the ground together.", so the components are equal in both...

Momentum conserves only in the ##\hat x## direction. What I wrote later there was just me wasting ink from my pen. So $$m\vec v_{0} = \frac{1}{3}m\vec v_{1} + \frac{2}{3}m\vec v_{2} \longrightarrow mv_{0} \cos{\theta} = \frac{m}{3}v_{1x}\cos{\theta} + \frac{2m}{3}v_{2x}\sin{\theta}$$ The...

It's not really an explosion - the question uses that term - because then we'd be adding energy to the system. It's more like a decoupling of the mass ##m##, which is why I think it's the same ##v_{0x}## for the other particles.

"Where does the factor of 3 in the denominator come from and why did it disappear in the next equation?" Typing error.

I've already solved this problem using another resource (just get the coordinate of the range of the center of mass and from there, get it for the larger mass ##R_{2}=(3v_{0}^{2})/(4g))##: Range CM: $$R_{(CM)} = \frac{v_{0}^2 sin{2\theta}}{2g}=\frac{v_{0}^{2}}{2}$$ then: $$ R_{(CM)}=...

I'll try to solve it tomorrow and post it here. Corrected some spelling errors (typed on cell phone)

Ohhhhh I focused a lot on the radial (fixed) part that I forgot about the azimuthal and polar dependence. edit: misspelling

Ok, edited. The vectors within the module would be the separation vector ## R = |\vec{r} - \vec{r'}|##.

Yes, they are vectors, I thought it was implicit in the question. But I still don't understand your point in saying "mistakes". That integral at the beginning is given in the problem.

Solving the integral is the easiest part. Using spherical coordinates: $$ \oint_{s} \frac{1}{|\vec{r}-\vec{r'}|}da' = \int_{0}^{\pi}\int_{0}^{2\pi} \frac{1}{|\vec{r}-\vec{r'}|}r_{0}^2 \hat r \sin{\theta}d\theta d\phi$$ then: $$I = \dfrac{1}{|\vec{r}-\vec{r'}|}r_{0}^2(1+1)(2\pi)\hat...

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?