amp1098's Profile
amp1098
April 6, 2016
6
Physics, Materials Science, Fabrication, Physics Education
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Notice how on the PVA graph the velocities all converge. When the first ball hits, it transfers kinetic energy to the next one, and then to the next one, ad nauseam. I'll divide the total system into two subsystems: the moving mass and the stationary mass. The moving mass has a velocity of ( 5m/s, 0m/s ), and the stationary mass has a velocity of ( 0m/s, 0m/s ). So, what causes the resulting system, the moving mass plus the stationary mass, to have a lower velocity than the moving mass? The answer lies in momentum. Momentum is defined as "P = mv", with "m" and "v" representing mass and velocity, respectively. In any closed system, momentum must be conserved, since no energy is being added or subtracted. The moving mass has a mass of 1 (I assume this is in kilograms) and is moving, in the "x" direction, 5m/s. The momentum for this system is therefore 5 kg/m/s . The stationary system has a momentum of 0 kg/m/s, since it isn't going anywhere. When the moving mass hits the stationary mass, the velocity drops to 0.556 m/s, or around 0.6 m/s. Going back to P=mv, you can see that, in order for P to remain the same, one of the variables has to change if the other one changes. The relationship between "m" and "v" is known as "Inversely Proportional"; when one goes up, the other goes down. Putting in the data, we get: 5 kg/m/s = 10kg * .5m/s. Wait... 0.5m/s? That isn't equal to 0.556m/s. It is close, but not exact. This error is due to the simulation itself, I think. If any of this is wrong ( I am very prone to mistakes), please say so; I do not wish to misconstrue anyone reading this.
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Notice how on the PVA graph the velocities all converge. When the first ball hits, it transfers kinetic energy to the next one, and then to the next one, ad nauseam. I'll divide the total system into two subsystems: the moving mass and the stationary mass. The moving mass has a velocity of ( 5m/s, 0m/s ), and the stationary mass has a velocity of ( 0m/s, 0m/s ). So, what causes the resulting system, the moving mass plus the stationary mass, to have a lower velocity than the moving mass? The answer lies in momentum. Momentum is defined as "P = mv", with "m" and "v" representing mass and velocity, respectively. In any closed system, momentum must be conserved, since no energy is being added or subtracted. The moving mass has a mass of 1 (I assume this is in kilograms) and is moving, in the "x" direction, 5m/s. The momentum for this system is therefore 5 kg/m/s . The stationary system has a momentum of 0 kg/m/s, since it isn't going anywhere. When the moving mass hits the stationary mass, the velocity drops to 0.556 m/s, or around 0.6 m/s. Going back to P=mv, you can see that, in order for P to remain the same, one of the variables has to change if the other one changes. The relationship between "m" and "v" is known as "Inversely Proportional"; when one goes up, the other goes down. Putting in the data, we get: 5 kg/m/s = 10kg * .5m/s. Wait... 0.5m/s? That isn't equal to 0.556m/s. It is close, but not exact. This error is due to the simulation itself, I think. If any of this is wrong ( I am very prone to mistakes), please say so; I do not wish to misconstrue anyone reading this.