Recent content by Cryptologica

Oh, ok! Thanks, I think I got it now.

Yes, I know I am really close. However, I cannot get the equation to equal one, without having constants in the numerator other than 1. When I multiply by (18/19) to make it equal to one the numerator never cancels?

Homework Statement The ellipse 18x^2+2x+y^2=1 has its center at the point (b,c) where b=____ and c=____? Homework Equations x^2/a^2 + y^2/b^2 = 1 The Attempt at a Solution 18x^2+2x+y^2=1 18(x^2+(1/9)x)+y^2=1 18(x^2+(1/9)x+(1/324))+y^2= 1+18(1/324) 18(x+(1/18))^2+y^2=19/18...

Oh, and thanks again! :)

Ah, forgot to go back and change in PE, should be 0.000048846/2 = 0.000024423 Substituting accordingly, I got: 900 +/- 1.938e-4 I think that should be correct now.

Recalculated using the Correct mass value and got 0.001938 So, then the velocities should be: 900 +/- 0.001938, correct?

Haha, oooops! I thought it was 13000, but it was suppose to be 1300. Thanks so much for your help and patience, I should be able to take it from here? :)

Ok, calculating ΔPE: [(6.67e-11)(13000)^2]/2 - [(6.67e-11)(13000)^2]/15 = 0.004884, half of that: = 0.002442 Not sure what you mean by "give", however the change in PE should be proportional to to the KE, so we equate the two? 0.002442 = (1/2)(13000)v^2 solving we get: +/-11.26 Though I feel...

Using the formula you gave, the change in PE would be: (Gm^2)/2 - (Gm^2)/15 Right?

PE = GMm/r and M = 2m, so by substitution I came to 2Gm^2/r Alright, I think we are getting somewhere, hopefully? :) Ok, so since they both have the same mass they will gravitate towards each other at the same rate, so we can conclude that they will both have the same velocity and will be...

Alright, so that will yield a positive answer because m_1 is behind CM and m_2 is ahead? (Or whichever is going faster, haven't calculated it out yet). So, I am a bit pedantic so I will just restate what I think I am suppose to do... m_1(900m/s) - m_2(900m/s) = 0 because the masses are equal...

Yes, my described method will only result in one velocity and I need two... Center of mass = MƩmr, correct? A bit rusty...how do I "connect" what I have with C-of-M?

Ok, so I can use: E=1/2mv^2 - GMm/r where G=universal gravity constant M = m_1+m_2 r=distance between them I solve for E, then plug back in and solve for v?

Homework Statement Two identical particles, each of mass 1300 kg, are coasting in free space along the same path. At one instant their separation is 15.0 m and each has precisely the same velocity of 900 m/s. What are their velocities when they are 2.00 m apart? m_1 = m_2 = 1300kg at...

Ok, thanks!