Recent content by mncyapntsi

So I want to subtract the two surfaces, right? I really don't know where to start... I am guessing this would be some sort of triple integral, however I am very confused with the bounds. Any help would be greatly appreciated! Thanks!!

Right! When I do that I get either (v1f) = v, or (v1f) = -(1/3)v. Does that second result sound reasonable?

Oh that's a typo (sorry I don't know proper Latex yet), I meant to cancel all the m's, but did so in the next step: (v)^2 + (1/2)(v)^2 = (vf1)^2 + (1/2)(vf2)^2 (3/2)v^2 = (vf1)^2 + (1/2)(vf2)^2

Oh my goodness - I think I overthought this problem and lost track of what I was effectively doing. Thank you for your patience and help! So if I substitute for vf2 instead... (v)^2 + (1/2)(m)(v)^2 = (vf1)^2 + (1/2)(m)(vf2)^2 (3/2)v^2 = (vf1)^2 + (1/2)(vf2)^2 mv = 2m(vf1) + m(vf2) v = 2(vf1)...

Ah! That's it... (v)^2 + (1/2)(m)(v)^2 = (vf1)^2 + (1/2)(m)(vf2)^2 (3/2)v^2 = (vf1)^2 + (1/2)(vf2)^2 mv = 2m(vf1) + m(vf2) v = 2(vf1) + (vf2) Plugging into KE conservation equation gives me: (3/2)v^2 = (vf1)^2 + (1/2)(vf2))^2 (3/2)(2(vf1) + (vf2))^2 = (vf1)^2 + (1/2)(vf2))^2 (3/2)(4(vf1)^2 +...

Ah yep! forgot about that: I assumed that the initial momentum was 3mv because the momentum of the 2m mass is 2mv and the momentum of the m mass is mv, so the total initial momentum would be 2mv+mv = 3mv. Where am I going wrong? Thanks!

Thanks for catching that! So with that fixed I've got: (v)^2 + (1/2)(m)(v)^2 = (vf1)^2 + (1/2)(m)(vf2)^2 (3/2)v^2 = (vf1)^2 + (1/2)(vf2)^2 3mv = 2m(vf1) + m(vf2) 3v = 2(vf1) + (vf2) vf2 = 3v - 2(vf1) Plugging into KE conservation equation gives me: (3/2)v^2 = (vf1)^2 + (1/2)(3v - 2(vf1))^2...

I know I need to look at the conversation of momentum, as well as the conservation of kinetic energy. However I get stuck with my equations. Any help would be greatly appreciated! I've already got (don't know where I am going wrong): (v)^2 + (1/2)(m)(v)^2 = (vf1)^2 + (1/2)(m)(vf2)^2 (3/2)v^2 =...

Hi! I've been trying to attempt this problem over here but the solutions state that the solution is this below? However, from integrating the density and then plugging it into Gauss's law, I get the exact same thing, except a 15 instead of a 5. Could any please help point out if there is an...

Hello, I am attempting to correctly solve this problem, however I end up with an equation that is slightly different as the one provided in the textbook solution. For question (a) I get the same thing, just instead of cos, I have cos^2 and I can't figure out where I went wrong. My process was...

Oh no! Of course that changes the result haha. Thank you so much for pointing that out :)

Hello! I have done this problem : vf^2 = (4.0x10^5)^2 + 2(6.0x10^12)(5x10^-3) so vf= sqrt((4.0x10^5)^2 + 2(6.0x10^12)(5x10^-3)) I get vf = 4.7 x 10^5 m/s However, the textbook solutions says vf = 8.7x10^5 m/s. Where did I go wrong? Thank you for any help! :)

The solution in my textbook says that for b, us = 0.234. However when I use the formula above I get 0.2364 which I feel like is too far off. Something must have gone wrong... Any help would be much appreciated! Thanks :)

I am so sorry, I got mixed up : by intuition I got that 2 - False! 1- True 2- False 3- True 4- False 5- False I drew out the vectors and their x and y components, and found all this above However---This has been marked incorrect... I really don't understand

Yes, by intuition I wrote True on that - It can be larger than 20N because it is the hypothenuse. However, the pattern 1- True 2- True 3- True 4- False 5- False Has also been marked as wrong by my professor...