Here's the question that was on the exam I took earlier this afternoon:
A ball going 3 m/s collides with a ball ball going towards it at 3 m/s. After they collide, they each move apart at 6 m/s. Does this violate anything about conservation of momentum, energy or both?
A ball going 3 m/s collides with a ball ball going towards it at 3 m/s. After they collide, they each move apart at 6 m/s. Does this violate anything about conservation of momentum, energy or both?
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It doesn't violate the conservation of momentum. Do the calculations and you'll see:
Total momentum before = 0
Total momentum after = 0
So the momentum is conservered (that is: "before" = "after" = 0).
Regarding the kinetic energy, you can easily show:
KE before = ½m(3meters/s)² + ½m(3meters/s)² = 9m(meters²/s²)
KE after = ½m(6meters/s)² + ½m(6meters/s)² = 36m(meters²/s²)
So clearly, there is more KE after the collision than before.
HOWEVER: This doesn't technically violate any conservation law either, because there is no law that says the KE has to be conserved during a collision; and in fact the KE usually ISN'T conserved in real collisions (it usually decreases).
But a collision fitting this problem's description--in which the KE _increases_ -- is perfectly plausible, provided there is some source of potiential energy within the balls that can be converted into KE during the collision. For example, imagine the balls are coated with a layer of gunpowder that gives a little POP during the collision. The chemical PE in the gunpowder can provide the necessary KE to increase the balls' speeds.
(Still, it's my guess that your teacher is expecting you to say, "the conservation of energy is violated," since in most collisions involving balls, the KE decreases rather than increases, and in this case the opposite happens. Frankly, your teacher is wrong if that's the answer he/she expects. It would be more correct to say that such a collision would (in the absence of gunpowder or other energy source) actually violate the 2nd Law of Thermodynamics.)
Total momentum before = 0
Total momentum after = 0
So the momentum is conservered (that is: "before" = "after" = 0).
Regarding the kinetic energy, you can easily show:
KE before = ½m(3meters/s)² + ½m(3meters/s)² = 9m(meters²/s²)
KE after = ½m(6meters/s)² + ½m(6meters/s)² = 36m(meters²/s²)
So clearly, there is more KE after the collision than before.
HOWEVER: This doesn't technically violate any conservation law either, because there is no law that says the KE has to be conserved during a collision; and in fact the KE usually ISN'T conserved in real collisions (it usually decreases).
But a collision fitting this problem's description--in which the KE _increases_ -- is perfectly plausible, provided there is some source of potiential energy within the balls that can be converted into KE during the collision. For example, imagine the balls are coated with a layer of gunpowder that gives a little POP during the collision. The chemical PE in the gunpowder can provide the necessary KE to increase the balls' speeds.
(Still, it's my guess that your teacher is expecting you to say, "the conservation of energy is violated," since in most collisions involving balls, the KE decreases rather than increases, and in this case the opposite happens. Frankly, your teacher is wrong if that's the answer he/she expects. It would be more correct to say that such a collision would (in the absence of gunpowder or other energy source) actually violate the 2nd Law of Thermodynamics.)
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You would be multiplying the kinetic energy by 2 so yes it would violate a few rules
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As stated it violates Conservation of Energy.
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Both.
Calculate and you'll see.
Calculate and you'll see.