The goal of this challenge was to solve for the additional weight on one can, and being allowed to gather the velocity of the cart's and the mass of both cart's without the addition of the can.
In order to solve for the unknown additional weight on one of our two carts, we used the stuck to unstuck formula as our carts were sprung from a joined position. This is
stuck=unstuck
m(a+b)*v(ab)=mava+mbvb.
Our initial given's were the mass of each cart on its own (0.5Kg), and we solved for speed using pasco capstone software.
In order to do this we took the formula m(a+b)*v(ab)=mava+mbvb and added the pertinent info resulting in:
m(500+ (500+x)*v(ab)=mava+mbvb. The initial velocity was 0, and cart a's final velocity was -55, while cart b's final velocity was 0.24. This resulted in the further change to the formula (1000+x)*(0)=(500)(-55) + (500 + x)(0.24). This further resulted in 0=-278+220+0.24x. Continuing on this series of equations 0=-158+0.27x, which simplifies to 158=0.24x, and when we divide both sides by 0.24, it results in .658Kg =x.
