Recent content by mncyapntsi

  1. mncyapntsi

    Confused about polar integrals and setting up bounds

    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!!
  2. mncyapntsi

    1D collision, varying masses but same initial velocity

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

    1D collision, varying masses but same initial velocity

    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
  4. mncyapntsi

    1D collision, varying masses but same initial velocity

    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)...
  5. mncyapntsi

    1D collision, varying masses but same initial velocity

    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 +...
  6. mncyapntsi

    1D collision, varying masses but same initial velocity

    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!
  7. mncyapntsi

    1D collision, varying masses but same initial velocity

    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...
  8. mncyapntsi

    1D collision, varying masses but same initial velocity

    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 =...
  9. mncyapntsi

    E-field of solid sphere with non-uniform charge density

    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...
  10. mncyapntsi

    Centripetal acceleration along a latitude of Earth

    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...
  11. mncyapntsi

    1-D Motion, calculating final velocity

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

    1-D Motion, calculating final velocity

    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! :)
  13. mncyapntsi

    Banked curves, coefficient of static friction

    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 :)
  14. mncyapntsi

    Object at rest hanging from two ropes

    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
  15. mncyapntsi

    Object at rest hanging from two ropes

    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...
Back
Top