Bouncy Collision - 1 Moving / 1 Stationary
A cart is moving at a constant velocity towards another cart which is stationary. Each cart has a magnet attached to the end which will allow the carts to "bounce" off of each other displaying this type of collision.
A cart is moving at a constant velocity towards another cart which is stationary. Each cart has a magnet attached to the end which will allow the carts to "bounce" off of each other displaying this type of collision.
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Cart A (Moving): Position (m) vs. Time (s)
Cart B (Stationary): Position (m) vs. Time (s) |
Cart A (Moving): Velocity (m/s) vs. Time (s)
Cart B (Stationary): Velocity (m/s) vs. Time (s) |
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Cart A (Moving): Momentum (kg*m/s) vs. Time (s)
Cart B (Stationary): Momentum (kg*m/s) vs. Time (s) |
Both Carts: Momentum (kg*m/s) vs. Time (s)
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Initial Momentum = Mass A * Initial Velocity A + Mass B * Initial Velocity B = 0.250 kg * 0.666 m/s + 0.250 kg * 0 m/s = 0.1665 kgm/s
Final Momentum = (Mass A + Mass B) * Final Velocity A and B = 0.250 kg * 0 m/s + 0.250 kg * 0.661 m/s = 0.1653 kgm/s
Percent Difference = |0.1665 - .1653|/0.1659 = 0.72%
We have a 0.72% difference which is extremely small especially since the numbers we are dealing with are small themselves. The small difference could be due to many factors, friction probably being the most common. Through the data we are able to see that no impulse was added or taken away from the system, and momentum must be conserved. In result, no momentum theoretically should be gained or lost in this experiment.
Final Momentum = (Mass A + Mass B) * Final Velocity A and B = 0.250 kg * 0 m/s + 0.250 kg * 0.661 m/s = 0.1653 kgm/s
Percent Difference = |0.1665 - .1653|/0.1659 = 0.72%
We have a 0.72% difference which is extremely small especially since the numbers we are dealing with are small themselves. The small difference could be due to many factors, friction probably being the most common. Through the data we are able to see that no impulse was added or taken away from the system, and momentum must be conserved. In result, no momentum theoretically should be gained or lost in this experiment.
Bouncy Collision - Both Moving
Both carts are moving at a constant velocity towards one another. Each cart has a magnet attached to the end which will allow the carts to "bounce" off of each other displaying this type of collision.
Both carts are moving at a constant velocity towards one another. Each cart has a magnet attached to the end which will allow the carts to "bounce" off of each other displaying this type of collision.
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Cart A (Moving): Position (m) vs. Time (s)
Cart B (Moving): Position (m) vs. Time (s) |
Cart A (Moving): Velocity (m/s) vs. Time (s)
Cart B (Moving): Velocity (m/s) vs. Time (s) |
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Cart A (Moving): Momentum (kg*m/s) vs. Time (s)
Cart B (Moving): Momentum (kg*m/s) vs. Time (s) |
Both Carts: Momentum (kg*m/s) vs. Time (s)
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Initial Momentum = Mass A * Initial Velocity A + Mass B * Initial Velocity B = 0.250 kg * 0.8181 m/s + 0.250 kg * 0.7159 m/s = 0.3835 kgm/s
Final Momentum = (Mass A + Mass B) * Final Velocity A and B = 0.250 kg * -0.6666 m/s + 0.250 kg * -0.4982 m/s = 0.2912 kgm/s
Percent Difference = |0.3835 - .2912|/0.3374 = 27.36%
There is a 27.36% difference between our two momentum values, which is quite large this time. We are dealing with small numbers, so a small amount of error can result in a large percent difference, but this is most likely due to having both carts moving and different friction factoring into the forces. Additionally we can see through the position time graph that there is a slight overlap between the two carts, that is due to the collision time not being picked up exactly through the motion detectors facing each other. But, just like the previous collision no impulse is added or subtracted from the system and theoretically momentum should be conserved through the experiment. We are still able to find that there is a correlation of the same momentum before and after collision through the percent difference.
Final Momentum = (Mass A + Mass B) * Final Velocity A and B = 0.250 kg * -0.6666 m/s + 0.250 kg * -0.4982 m/s = 0.2912 kgm/s
Percent Difference = |0.3835 - .2912|/0.3374 = 27.36%
There is a 27.36% difference between our two momentum values, which is quite large this time. We are dealing with small numbers, so a small amount of error can result in a large percent difference, but this is most likely due to having both carts moving and different friction factoring into the forces. Additionally we can see through the position time graph that there is a slight overlap between the two carts, that is due to the collision time not being picked up exactly through the motion detectors facing each other. But, just like the previous collision no impulse is added or subtracted from the system and theoretically momentum should be conserved through the experiment. We are still able to find that there is a correlation of the same momentum before and after collision through the percent difference.
Sticky Collision - 1 Moving / 1 Stationary
A cart is moving at a constant velocity towards another cart which is stationary. Each cart has velcro attached to the end which will allow the carts to "stick" to each other displaying this type of collision.
A cart is moving at a constant velocity towards another cart which is stationary. Each cart has velcro attached to the end which will allow the carts to "stick" to each other displaying this type of collision.
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Cart A (Moving): Position (m) vs. Time (s)
Cart B (Stationary): Position (m) vs. Time (s) |
Cart A (Moving): Velocity (m/s) vs. Time (s)
Cart B (Stationary): Velocity (m/s) vs. Time (s) |
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Cart A (Moving): Momentum (kg*m/s) vs. Time (s)
Cart B (Stationary): Momentum (kg*m/s) vs. Time (s) |
Both Carts: Momentum (kg*m/s) vs. Time (s)
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Initial Momentum = Mass A * Initial Velocity A + Mass B * Initial Velocity B = 0.250 kg * 0.8013 m/s + 0.250 kg * 0 m/s = 0.2003 kgm/s
Final Momentum = (Mass A + Mass B) * Final Velocity A and B = 0.250 kg * 0.4011 m/s + 0.250 kg * 0.4011 m/s = 0.2006 kgm/s
Percent Difference = |0.2003 - .2006|/0.2005 = 0.10%
There is a 0.10% difference between our two momentum values, which is extremely small. Looking at our data we can that our initial and final momentum are almost the same, with just a difference of 0.0003 kg*m/s. In this test we can see that there was not much friction, and that significantly lowered the number and the actual results were able to show. The results being that no impulse is added to the system whatsoever and thus the momentum does not and will not change throughout the course of the experiment.
Final Momentum = (Mass A + Mass B) * Final Velocity A and B = 0.250 kg * 0.4011 m/s + 0.250 kg * 0.4011 m/s = 0.2006 kgm/s
Percent Difference = |0.2003 - .2006|/0.2005 = 0.10%
There is a 0.10% difference between our two momentum values, which is extremely small. Looking at our data we can that our initial and final momentum are almost the same, with just a difference of 0.0003 kg*m/s. In this test we can see that there was not much friction, and that significantly lowered the number and the actual results were able to show. The results being that no impulse is added to the system whatsoever and thus the momentum does not and will not change throughout the course of the experiment.
Sticky Collision - Both Moving
Both carts are moving at a constant velocity towards one another. Each cart has velcro attached to the end which will allow the carts to "stick" to each other displaying this type of collision.
Both carts are moving at a constant velocity towards one another. Each cart has velcro attached to the end which will allow the carts to "stick" to each other displaying this type of collision.
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Cart A (Moving): Position (m) vs. Time (s)
Cart B (Moving): Position (m) vs. Time (s) |
Cart A (Moving): Velocity (m/s) vs. Time (s)
Cart B (Moving): Velocity (m/s) vs. Time (s) |
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Cart A (Moving): Momentum (kg*m/s) vs. Time (s)
Cart B (Moving): Momentum (kg*m/s) vs. Time (s) |
Both Carts: Momentum (kg*m/s) vs. Time (s)
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Initial Momentum = Mass A * Initial Velocity A + Mass B * Initial Velocity B = 0.250 kg * 0.7144 m/s + 0.250 kg * -0.7285 m/s = 0.0035 kgm/s
Final Momentum = (Mass A + Mass B) * Final Velocity A and B = 0.250 kg * 0.0134 m/s + 0.250 kg * 0.0119 m/s = 0.0063 kgm/s
Percent Difference = |0.0035 - .0063|/0.0049 = 57.14%
There is a 57.14% difference between our two momentum values, which is quite large, just like the other "both moving" trial. Again, since we have two carts, there is more friction involved, which will skew the numbers a bit more. Although the number looks large, looking at the total momentum graph, we can see that there really is not much difference between the initial and final momentum. The only reason it seems that way is because the momentums are small numbers, and a small bit of error will completely throw off the numbers. However, we are still able to tell that the momentum is conserved throughout the trial and that initial and final should hypothetically be equal.
Final Momentum = (Mass A + Mass B) * Final Velocity A and B = 0.250 kg * 0.0134 m/s + 0.250 kg * 0.0119 m/s = 0.0063 kgm/s
Percent Difference = |0.0035 - .0063|/0.0049 = 57.14%
There is a 57.14% difference between our two momentum values, which is quite large, just like the other "both moving" trial. Again, since we have two carts, there is more friction involved, which will skew the numbers a bit more. Although the number looks large, looking at the total momentum graph, we can see that there really is not much difference between the initial and final momentum. The only reason it seems that way is because the momentums are small numbers, and a small bit of error will completely throw off the numbers. However, we are still able to tell that the momentum is conserved throughout the trial and that initial and final should hypothetically be equal.
Explosion
Both carts are not moving and then start moving. Each cart has a "push mechanism" that allows the carts to push away and display this type of collision.
Both carts are not moving and then start moving. Each cart has a "push mechanism" that allows the carts to push away and display this type of collision.
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Cart A: Position (m) vs. Time (s)
Cart B: Position (m) vs. Time (s) |
Cart A: Velocity (m/s) vs. Time (s)
Cart B: Velocity (m/s) vs. Time (s) |
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Cart A: Momentum (kg*m/s) vs. Time (s)
Cart B: Momentum (kg*m/s) vs. Time (s) |
Both Carts: Momentum (kg*m/s) vs. Time (s)
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Initial Momentum = Mass A * Initial Velocity A + Mass B * Initial Velocity B = 0.250 kg * 0 m/s + 0.250 kg * 0 m/s = 0 kgm/s
Final Momentum = (Mass A + Mass B) * Final Velocity A and B = 0.250 kg * 0.6512 m/s + 0.250 kg * -0.8920 m/s = -0.0602 kgm/s
Percent Difference = |0 + .0602|/0.0301 = 200%
Our results for the specific trial are very far off. The main reason is that while the two carts were pushing off of each other, we only activated one "pushing mechanism". This could've caused some more initial friction on one of the two sides, and possibly a downward force by the human pushing down on the button. Additionally, since we are comparing the number to 0, the percent difference is going to be extremely large, even if the number is close to 0. We are still able to roughly see however through the two momentum graphs that momentum should be conserved, and if we draw out the chart momentum is conserved. Although this trial did not give us the most clear results, in accumulation with the other results received we are able to tell that no impulse is added or subtracted, and that momentum is conserved.
Final Momentum = (Mass A + Mass B) * Final Velocity A and B = 0.250 kg * 0.6512 m/s + 0.250 kg * -0.8920 m/s = -0.0602 kgm/s
Percent Difference = |0 + .0602|/0.0301 = 200%
Our results for the specific trial are very far off. The main reason is that while the two carts were pushing off of each other, we only activated one "pushing mechanism". This could've caused some more initial friction on one of the two sides, and possibly a downward force by the human pushing down on the button. Additionally, since we are comparing the number to 0, the percent difference is going to be extremely large, even if the number is close to 0. We are still able to roughly see however through the two momentum graphs that momentum should be conserved, and if we draw out the chart momentum is conserved. Although this trial did not give us the most clear results, in accumulation with the other results received we are able to tell that no impulse is added or subtracted, and that momentum is conserved.