Recent content by Tea_Aficionado

I think what Lnewqban is getting at is that we should split the force due to gravity into "x" (direction parallel to incline) and "y" (direction perpendicular to incline). I noticed in some of your earlier threads that you err in the pattern of thinking that mg is always perpendicular to motion...

In my humble opinion, I'd suggest checking your setup for acceleration. What other forces are there acting upon the "x" (parallel to the plane) direction?

EDIT for clarity: I solved the question, just asking for if the explanations make sense and if the mechanical energy is considered to be conserved before and after the collision due to reasons listed below the photo. I hope this image is readable (grr, scanner is janky). I'm guessing the...

ah sorry haru, I didn't fully process that until I carefully read through it. Thank you!

OMG, thank you so much! Read through it all. That makes sense -- that we are assuming that mechanical energy is conserved due to ideality. I'll still post my second thread -- just for reference sake ig. But again, thank you all so much! :bow::oldbiggrin::oldbiggrin::oldbiggrin:

Right. I think I'll make another thread about the similar problem that I kept on referring to -- maybe it'll help illustrate my confusion. But once again, thank you all so much for helping me throughout this thread! I understand that it's pretty mentally taxing to try to fill the gaps in my...

I understand that generally energy is always conserved based on the law of conservation of energy, but if we define the system to be just that of the block and the spring, wouldn't the block/spring system lose energy due to deformation? Wiki states that in "such a collision, kinetic energy is...

Huh, okay. So we assume collisions onto ideal springs conserve energy before and after collision?

Wouldn't it be $$mgdeltax+mgh=\frac{1}{2}kxmax^2 $$ where ##h=0.4##, and ##deltax = 0.098... oh, no, we're saying that xmax + h = hnew we don't use deltax at all so then, $$mgxmax+mgh=\frac{1}{2}kxmax^2 $$ and solving for xmax we get exactly 0.1m Should we always assume for these SHM...

1. It'd just be h + delta x right? 2. The assumption that gravitational potential energy before the collision is completely converted into potential spring energy after the collision? Oh I also just realized. Maybe by "maximum" it means max. in general considering favorable conditions...

Would you say that this is presumed true for all ideal springs? Or is it just a presumption we're supposed to make in this problem only? On a side note, thanks to everyone for chiming in! I woke up today and I was pleasantly surprised. Thanks y'all! (not end of thread though lol)

But the energy transferred is different for the mass stuck on the spring (inelastic) vs. the mass bouncing away (elastic) right? Or is it that whether it's inelastic or elastic doesn't depend on whether or not it sticks or bounces? How do we know that the problem is an elastic collision?

Oh, that's true. I guessed that it would be inelastic because I did another problem before that had the collision be inelastic. That would explain why we include h in the total height... but at the same time aren't we assuming that the mass sticks onto the spring because the motion is sustained...

We know that the Ug is converted to KE and Us. I thought that since the system loses energy after the collision that we shouldn't use the equation hnew= delta x + h. I thought instead that maybe the h we should use is xmax, because that's when there is maximum Ug and there is no other energy...