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1. ### Question envolves the centre-of-mass and momentum.

Oh i think its because i used KEf = 1/2m1(v1f)^2 + 1/2m2(v2f)^2 Wouldnt that be right?
2. ### Question envolves the centre-of-mass and momentum.

So, I did what you said, and its starting to look like an answer, thanks for that. But now I have: u1x = (u1x)^2 ((1839/1841)^2 + (1840)(2/1840)^2) with the left side being inital and the right side being the final KE's I was wondering though, how would I convert that into a percentage?
3. ### Question envolves the centre-of-mass and momentum.

Ok: u1x = v1x + 1840v2x u1x = v2x - v1x so... v2x= u1x + v1x and then... u1x = v1x + 1840(u1x + v1x) u1x = v1x + 1840u1x + 1840v1x -1839u1x = 1841v1x u1x = -1841v1x/1839 there. What is this even showing me. I have still have 2 unknowns?
4. ### Question envolves the centre-of-mass and momentum.

You did say to combine the equations. But, when you do solve for u1, there are still 2 variables, and you can't find the KEf or KEi without at least one of these, and therefore can't compare the two.
5. ### Question envolves the centre-of-mass and momentum.

So your saying that when you add the equaitons and get: V1i = 1841v2f How can this be used to compare the two KE? since KEi = (1/2)mv^2 KEi = (1/2)(1)(1841v2f)^2 It doesn't give an actual number, and again there ends up with two variables and then KEf would be: KEf =...
6. ### Question envolves the centre-of-mass and momentum.

I have tried combining the two equations, but that only gives me a new equation with two unknown velocities, and I need all three velocities to find the inital and final KE. How would I use an equation with 2 unknown variables to compare the initial and final KE?? Help??
7. ### Question envolves the centre-of-mass and momentum.

Homework Statement An electron collides elastically with a hydrogen atom that is initially at rest. Assume all the motion occurs along a straight line. What fraction of the electron's initial kinetic energy is transferred to the atom? (Take the mass of the hydrogen atom to be 1840 times the...