phy 121
Your 'conservation of momentum' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Your optional message or comment: **
I only completed the portion of the lab required for physics 121 students.
** #$&* Distances from edge of the paper to the two marks made in adjusting the 'tee'. **
12.1cm, 19.2
.8cm
I think that the uncertainty of my measurement is high. Perhaps within 2 or 3cm. I tried to use my piece of carbon paper to get specific marks of where the balls fell, but as my paper was about 3cm by 8cm, it was difficult to find a perfect spot to put it where it could capture both marks.
** #$&* Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process: **
14.4, 14.9, 16.2, 16.7, 17.8
16, 1.373
I obtained the above results in the first line by being able to actually position my carbon paper in a spot where it would make impressions on my paper so that I could accurately measure where the ball hit. From here I measured the horizontal distance from the end of my paper. I then used the data program to calculate the mean and the standard deviation.
** #$&* Five horizontal ranges observed for the second ball; corresponding first-ball ranges; mean and standard deviation of the second-ball ranges; mean and standard deviation of ranges for the first ball. **
15.2, 18.5, 13.8, 14.2, 17.3
17.0, 21.4, 16.1, 16.3, 19.3
15.8, 2.029
18.02, 2.277
I obtained the above results in the first two lines by being able to actually position my carbon paper in a spot where it would make impressions on my paper so that I could accurately measure where the ball hit. I had to run the trail more than 5 times so I could get a total of 5 hits for each ball during each trial. The carbon paper was cut in half as specified in the instructions. From here I measured the horizontal distance from the end of my paper. I then used the data program to calculate the mean and the standard deviation.
** #$&* Vertical distance fallen, time required to fall. **
75 cm
.46875 s
I used the timer program to calculate the time of the fall. I timed how long it took for the ball to roll to the stationary ball and then time how long it took for the stationary ball to fall when hit. I think that the speed of the stationary ball falling was much faster falling after it was hit by the rolling ball then if it had just been dropped by me and timed
accelerating downward at 9.8 m/s^2 the ball would fall in about .4 seconds; difficult to time that accurately and you did a good job. Still calculating the time of fall using the vertical displacement and the acceleration of gravity would have been easier and a bit more accurate.
** #$&* Velocity of the first ball immediately before collision, the velocity of the first ball after collision and the velocity of the second ball after collision; before-collision velocities of the first ball based on (mean + standard deviation) and (mean - standard deviation) of its uninterrupted ranges; same for the first ball after collision; same for the second ball after collision. **
66.2 cm/s, 188.235 cm/s, 165.51724
64.171 to 68.229
186.206 to 190.254
163.24024 to 167.79424
you don't say how you got these results; they don't appear to follow from your data and previous results
however you've used these results appropriately below
** #$&* First ball momentum before collision; after collision; second ball after collision; total momentum before; total momentum after; momentum conservation equation. All in terms of m1 and m2. **
P=m1 * 15.8cm/s
P=m1 * 94.117cm/s
P=m2 * 165.517cm/s
P= m1 * 15.8cm/s + m2 * 165.517cm/s
P= m1 * 94.117cm/s + m2 * 165.517cm/s
m1 * 15.8cm/s + m2 * 165.517cm/s = m1 * 94.117cm/s + m2 * 165.517cm/s
** #$&* Equation with all terms containing m1 on the left-hand side and all terms containing m2 on the right; equation rearranged so m1 appears by itself on the left-hand side; preceding the equation divided by m2; simplified equation for m1 / m2. **
m1 * 15.8cm/s +m1* 94.117cm/s = m2 * 165.517cm/s+ m2 * 165.517cm/s
m1 = m2+ 18.33
m1/m2= 18.33
m1/m2= 18.33
If we were to take the mass of our first object and divide it by the mass of the second object our result would be 18.33.
** #$&* Diameters of the 2 balls; volumes of both. **
1.7cm, 1.3 cm
2.57cm^3, 1.15 cm^3
** #$&* How will magnitude and angle of the after-collision velocity of each ball differ if the first ball is higher? **
I think that if the center of the first ball is higher than that of the center for the second ball that it will cause the magnitude of the first ball to be greater and also cause the velocity to be greater. It would then cause the velocity and magnitude of the second ball to be smaller.
If the centers are at the same height, I think that the speeds will be closer together but not the same. The after collision velocity will differ if the balls are lined up so the centers are off.
** #$&* Predicted effect of first ball hitting 'higher' than the second, on the horizontal range of the first ball, and on the second: **
If the centers of each ball are not lined up, I think it will also alter the horizontal range of the first ball as well as the second ball. The first ball will travel further than the second ball because it is not losing as much momentum or transferring as much of its energy. The second ball will fall almost straight down because it is not being hit as hard or fast as it would have been if hit dead center.
** #$&* ratio of masses using minimum before-collision velocity for the first ball, maximum after-collision velocity for the first ball, minimum after-collision velocity of the second: **
m1 * 64.171 cm/s +m1 * 190.254cm/s = m2 * 163.24024cm/s+ m2 * 163.24024cm /s
m1= m2 + 26647.37596/12208.78943
m1/m2=2.18
** #$&* What percent uncertainty in mass ratio is suggested by this result? **
18.33-2.18= 16.15
** #$&* What combination of before-and after-collision velocities gives you the maximum, and what combination gives you the minimum result for the mass ratio? **
** #$&* In symbols, what mass ratio is indicated if the before-collision velocity of ball 1 is v1, its after-collision velocity u1 and the after-collision velocity of the 'target' ball is u2? **
** #$&* Derivative of expression for m1/m2 with respect to v1. **
** #$&* If the range of the uninterrupted first ball changes by an amount equal to the standard deviation, then how much does the predicted value of v1 change? If v1 changes by this amount, then by how much would the predicted mass ratio change? **
** #$&* Complete summary and comparison with previous results, with second ball 2 mm lower than before. **
** Vertical drop of the second ball, its mean horizontal range and the slope of the line segment connecting the two centers; the velocity given by the program based on mean; velocity interval for 2-mm decrease in 2d-ball height; velocity interval from the original run at equal heights; difference in the mean-based velocities; is new velocity significantly different than original? **
** #$&* Your report comparing first-ball velocities from the two setups: **
** #$&* Uncertainty in relative heights, in mm: **
** #$&* Based on the results you have obtained to this point, argue for or against the hypothesis that the uncertainty in the relative heights of the balls was a significant factor in the first setup. **
** #$&* How long did it take you to complete this experiment? **
** #$&* Optional additional comments and/or questions: **
** **
** **
** **
** **
** **
2-3 hours (i took breaks in bewtween)
** **
&#Good work. Let me know if you have questions. &#