Snow Day Momentum
For this blog, we were tasked with demonstrating the law of conservation of momentum. This law simply says that the momentum of any two objects before a collision will equal the momentum after the collision (in an isolated system). In mathematical terms it can be expressed as:
m1v1 + m2v2 = m1v1f + m2v2f
With m1 and m2 equaling the respective masses, v1 and v1 being their velocities before the collision, and v1f v2f as their velocities after the collision.
To prove this, I decided that a collision between two basketballs would make for an accurate experiment. Not only would their spherical shape make a collision easy to produce, but their size and distinctive color makes them easier to track on LoggerPro, which would give me more accurate data.
After filming this experiment, I put the video into LoggerPro to analyze the data.
Firstly, I found the velocity of m1 (the ball that is initially moving) by tracking its position during each frame. Below is the position vs time graph for the first ball (m1).
According to the best fit line, the initial velocity of the ball of -2.875 m/s. It is negative because the video was filmed from an angle where the ball was traveling left instead of right.
Since the second ball was not moving, it has an initial velocity of 0. Now we can move on to finding the velocities of the balls after the collisions. Below is the position vs time graph for the first ball (m1) after the collision.
It wasn't a very clean collision, but the slope gave us a velocity of -0.0832 m/s.
Lastly, we need to find the velocity of the second ball after the collision (m2). Below is the position vs time graph for that ball.
The slope of this graph gives us a velocity of -0.8378 m/s.
Now that we have all of the velocities, we can just weigh the balls and plug our values into the equations. Using a scale, I found the mass of m1 to be about 620 g, which comes out to about 0.62 kg. I found m2 (which was severely under inflated) to be around 350 g, which is 0.35 kg.
Using these values, we can test to see if the experiment gave us an accurate demonstration of the conservation of momentum. When plugging into the conservation of momentum equation, we get:
m1v1 + m2v2 = m1v1f + m2v2f
(0.35 kg)(-2.553 m/s) + (0.62 kg)(0) = (0.35 kg)(-0.0832 m/s) + (0.62)(-0.8738)
-0.89 kg * m/s = -0.57 kg * m/s
So there was a difference of about 3.2 kg * m/s between our two momentum. This may have been due to several factors. Firstly, the surface that the collision took place on was very rough and not frictionless. It was also icy and wet, contributing to the error in my results. Additionally, it was rather windy when I filmed the demonstration. This may have altered the velocities of the balls both before and after the collision. Air resistance may have also played a factor in this. Also, it was hard to produce a 'perfect' collision. I struggled to get the balls to hit each other dead on, resulting in them drifting off at different angles. And lastly, the surface that I used was at a bit of an incline, which could have impacted the velocities of the masses. Given all of these potential errors, my results seem to be reasonable. So it looks like the law of conservation of momentum is true after all. Who would have known?!
Finally, since this is an inelastic collision, kinetic energy was lost during the collision. This can be proven by finding the kinetic energy of the masses both before and after the collision using the formula
KE = 1/2(mv^2)
Since the velocity of m2 before the collision was 0, its KE was also 0. Therefore we just need to find the KE of m1 before the collision and the KE of m1 and m2 after the collision.
KEi =1/2(m1v1^2)
KEf = 1/2(m1v1f^2) + 1/2(m2v2f^2)
KEi = 1/2(0.35 * -2.553^2) = 1.141 N
KEf = 1/2(0.35 * -0.0832^2) + 1/2(0.62 * -0.8738^2) = 0.237 N
1.141 - 0.237 = 0.9, so about 0.9 N of energy was lost during this collision.
m1v1 + m2v2 = m1v1f + m2v2f
With m1 and m2 equaling the respective masses, v1 and v1 being their velocities before the collision, and v1f v2f as their velocities after the collision.
To prove this, I decided that a collision between two basketballs would make for an accurate experiment. Not only would their spherical shape make a collision easy to produce, but their size and distinctive color makes them easier to track on LoggerPro, which would give me more accurate data.
The attempt
After filming this experiment, I put the video into LoggerPro to analyze the data.
Firstly, I found the velocity of m1 (the ball that is initially moving) by tracking its position during each frame. Below is the position vs time graph for the first ball (m1).
Since the second ball was not moving, it has an initial velocity of 0. Now we can move on to finding the velocities of the balls after the collisions. Below is the position vs time graph for the first ball (m1) after the collision.
Lastly, we need to find the velocity of the second ball after the collision (m2). Below is the position vs time graph for that ball.
Now that we have all of the velocities, we can just weigh the balls and plug our values into the equations. Using a scale, I found the mass of m1 to be about 620 g, which comes out to about 0.62 kg. I found m2 (which was severely under inflated) to be around 350 g, which is 0.35 kg.
Using these values, we can test to see if the experiment gave us an accurate demonstration of the conservation of momentum. When plugging into the conservation of momentum equation, we get:
m1v1 + m2v2 = m1v1f + m2v2f
(0.35 kg)(-2.553 m/s) + (0.62 kg)(0) = (0.35 kg)(-0.0832 m/s) + (0.62)(-0.8738)
-0.89 kg * m/s = -0.57 kg * m/s
So there was a difference of about 3.2 kg * m/s between our two momentum. This may have been due to several factors. Firstly, the surface that the collision took place on was very rough and not frictionless. It was also icy and wet, contributing to the error in my results. Additionally, it was rather windy when I filmed the demonstration. This may have altered the velocities of the balls both before and after the collision. Air resistance may have also played a factor in this. Also, it was hard to produce a 'perfect' collision. I struggled to get the balls to hit each other dead on, resulting in them drifting off at different angles. And lastly, the surface that I used was at a bit of an incline, which could have impacted the velocities of the masses. Given all of these potential errors, my results seem to be reasonable. So it looks like the law of conservation of momentum is true after all. Who would have known?!
Finally, since this is an inelastic collision, kinetic energy was lost during the collision. This can be proven by finding the kinetic energy of the masses both before and after the collision using the formula
KE = 1/2(mv^2)
Since the velocity of m2 before the collision was 0, its KE was also 0. Therefore we just need to find the KE of m1 before the collision and the KE of m1 and m2 after the collision.
KEi =1/2(m1v1^2)
KEf = 1/2(m1v1f^2) + 1/2(m2v2f^2)
KEi = 1/2(0.35 * -2.553^2) = 1.141 N
KEf = 1/2(0.35 * -0.0832^2) + 1/2(0.62 * -0.8738^2) = 0.237 N
1.141 - 0.237 = 0.9, so about 0.9 N of energy was lost during this collision.
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