I have 2 cubes where one mass is greater than the other. $M$ and $m$, where $M>m$, and there is a hill that is symmetrical on both sides, and has a friction factor of $k$ between the object and the surface.
If we place both objects on one side of the hill, point 1, that has no slope, so it is flat, and give both of the masses the same velocities in the direction of the top of the hill so they finally land on the other side, on the flat area, point 2 (symmetrical to the point 1).
How can I prove that the speed of the heavier cube is smaller than the lighter one in the point 2? (If I'm correct) I tried equating $E_{kinetic}$ and $E_{potential}$... I see there is work from the friction, but whenever I try anything I cross out the mass.
Heres a diagram: http://www.wolframalpha.com/input/?i=sin%5E2(x)%2C+0%3Cx%3Cpi&x=0&y=0
Where point 1 is at (0, 0), and point 2 at (pi, 0).
What I have so far: where $g$ is constant of gravity and $k$ friction factor, and $d$ distance.
$E_{kinetic.start} - W_{friction} = E_{kinetic.end}$ $M(\frac{v_0^2}{2}-gkd)=M(\frac{v_1^2}{2})$
I could change $M$ with $m$ and nothing would change, they'd both have same speeds...
So it's obvious that the masses cancel out and the energy is proportionally lower at both masses, the distance for friction is the same and starting velocities are the same, therefore the velocity should be the same for both bodies at the end. This confuses me as I believe the smaller body should be faster.
(Cant noone help me with this? Im now doubting the solution i've provided first is right. I cannot prove otherwise.)