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course Phy 231
7/18 4:15
conservation of momentumPhy 231
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: **
** #$&* Distances from edge of the paper to the two marks made in adjusting the 'tee'. **
.96cm
.98cm
+-.03cm
** #$&* Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process: **
24.45, 25.20, 26.64, 25.51, 25.98
25.56, .8226
To get the horizontal range, I first dropped the ball from the edge of the table to determine the rest landing position. Then, I released the ball from the top of the ramp and got 5 landing positions. I then measured from the original to the landing position to get the horizontal displacement.
** #$&* Five horizontal ranges observed for the second ball; corresponding first-ball ranges; mean and standard deviation of the second-ball ranges; mean and standard deviation of ranges for the first ball. **
24.85, 24.8, 24.77, 24.82, 24.9
33.51, 33.05, 30.47, 33.62, 33.01
24.83, .04969
32.73, 1.293
&&&&&I’m not really sure what was going on here, because the big ball should be going farther when there is no obstruction because it would not lose any momentum. I probably measured the distance incorrectly or something else changed the movement of the ball. I know that it is not possible that the second ball went farther when it hit the first ball. This is because of the conservation of momentum—there has to be the same amount of momentum after the collision, so the large ball must have lost some because the small ball gained momentum from the collision. &&&&&&&&
According to this your first ball went further after colliding with the second ball than it went when the second ball wasn't in the way.
I measured the horizontal displacement with a centimeter ruler. I measured the distances from the origin, or where the ball dropped from the edge of the table.
** #$&* Vertical distance fallen, time required to fall. **
76cm
.374s
I measured the height of the drop with a centimeter ruler. I measured the time of the fall with the timer program. I assumed that as the ball fell the acceleration was constant and air resistance was negligible. The height is measured +- 1 cm. The time is measured +- .02s.
** #$&* Velocity of the first ball immediately before collision, the velocity of the first ball after collision and the velocity of the second ball after collision; before-collision velocities of the first ball based on (mean + standard deviation) and (mean - standard deviation) of its uninterrupted ranges; same for the first ball after collision; same for the second ball after collision. **
68.04, 66.03, 87.92 cm/s
70.29, 65.78
55.17, 65.89
91.54, 84.31
&&&&&&& The first ball would have to lose velocity because momentum is mass * velocity. Since the ball stayed the same mass, the velocity had to drop to allow for the drop in momentum. We know that the momentum of the large ball dropped, so the velocity had to drop as well, because the momentum was transferred to the small ball during the collision. &&&&&&
The first ball must have slowed significantly.
** #$&* 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. **
68.04 cm/s m1
66.03 cm/s m1
87.92 cm/s m2
68.04 cm/s m1 + 0 cm/s m2
66.03 cm/s m1 + 87.92 cm/s m2
68.04 cm/s m1 + 0 cm/s m2 = 66.03 cm/s m1 + 87.92 cm/s m2
** #$&* Equation with all terms containing m1 on the left-hand side and all terms containing m2 on the right; equation rearranged so m1 appears by itself on the left-hand side; preceding the equation divided by m2; simplified equation for m1 / m2. **
2.01 cm/s m1 = 87.92 cm/s m2
m1 = 43.7 m2
m1 / m2 = 43.7
m1/m2 = 43.7
The ratio is 43.7 : 1, which means that the larger ball has 43.7 times as much mass as the small ball.
** #$&* Diameters of the 2 balls; volumes of both. **
2.91 cm, 1.19 cm
103 cm^3, 7.06 cm^3
** #$&* How will magnitude and angle of the after-collision velocity of each ball differ if the first ball is higher? **
If the larger ball hits with a higher center of gravity. The velocity of the smaller ball will be steeper/at a smaller angle toward the ground. The velocity of the larger ball and the smaller ball will be less and the horizontal displacement will not be as great. With the center of mass being at the middle of the collision, the speed will be faster and the balls will be following the same angle.
** #$&* Predicted effect of first ball hitting 'higher' than the second, on the horizontal range of the first ball, and on the second: **
The horizontal ranges of the balls will be decreased is the collision of the balls is not head on.
** #$&* ratio of masses using minimum before-collision velocity for the first ball, maximum after-collision velocity for the first ball, minimum after-collision velocity of the second: **
The new ration is 216
You take the original euqation m1v1+m2v2 = m1v1' + m2v2' and substitute the proper velocities and solve as before. This gives you the new ratio of masses.
** #$&* What percent uncertainty in mass ratio is suggested by this result? **
|New value-first value| / first
216-43.7 / 43.7
394%
** #$&* What combination of before-and after-collision velocities gives you the maximum, and what combination gives you the minimum result for the mass ratio? **
Maximum: minumum large before, maximum large after, minimum small after
Minimum: max large before, min large after, max small after
** #$&* 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? **
m1v1 = m1u1 + m2u2
m1(v1+u1) = m2u2
m1 / m2 = u2 / (v1+u1)
** #$&* Derivative of expression for m1/m2 with respect to v1. **
dm1/dm2 = (-u2 dv1) / (v1+u1)^2
= -.00489 s/cm * dv1 cm/s = -.00489 dv1
&&&&&&So, this would be the change in the ratio of masses divided by the change velocity. &&&&&&
Good, but you would denote this d(m1/m2) / dv1.
** #$&* 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? **
The velocity changes by about +2cm/s
The mass ratio would change by about -.00979 times
You get the result by finding the change in the velocity and then plugging that into the derivative found above. It means that as the change in velocity increases, the mass ratio decreases. You can tell how much more mass one ball has than the other based on the change in the horizontal displacement of the balls off the table.
** #$&* Complete summary and comparison with previous results, with second ball 2 mm lower than before. **
Large ball: mean: 23.59, sd = .9326
Small ball: mean: 34.78, sd = 1.174
Mass ratio: 17.3
** Vertical drop of the second ball, its mean horizontal range and the slope of the line segment connecting the two centers; the velocity given by the program based on mean; velocity interval for 2-mm decrease in 2d-ball height; velocity interval from the original run at equal heights; difference in the mean-based velocities; is new velocity significantly different than original? **
76 cm, 34.78cm, .25
100.8 cm/s
104.5, 97.16
91.54, 84.31
12.9 cm/s
The velocity of the small ball in the second test has a higher velocity than in the first test because the ball was placed 2mm higher than before.
** #$&* Your report comparing first-ball velocities from the two setups: **
Based on the horizontal displacement of the large ball, the placement of the small ball higher did not make a significant difference on the velocity. The range of velocities was not significantly different from the previous test.
** #$&* Uncertainty in relative heights, in mm: **
.2mm
The length of the straw used to hold the ball was very small and difficult to measure, so the height of the ball was not completely exact. Also, the ball could have moved during the test, although it was re-placed after each run, and made the altitude different.
** #$&* 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. **
The uncertainty in the height of the ball should not be a major factor because the distances being dealt with are so small. Any small difference in the height of the smaller ball should not make a huge difference in the velocity of the balls.
** #$&* How long did it take you to complete this experiment? **
100.8 cm/s sin 14 degrees
24.4 cm/s
** #$&* Optional additional comments and/or questions: **
-1.41 cm/s
** **
68.04 cm/s
100.8 cm/s
66.53 cm/s
** **
26.3 cm
** **
I do not think I can give a very accurate measure of the spinning of the smaller ball; I know that it was something more than a couple of revolutions per second. Since the ball began spinning, some of the kinetic energy it had would be transferred from moving forward to the spin, so the ball might be going slower than before.
** **
.25
66.5 cm/s
76 cm for the length of the drop, 23.59 cm for the mean of the horizontal drop and 8.3 for the number of dominos becuase the slope was .25
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2 hours 30 minutes
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Excellent analysis. I do have some questions about your data. No need to redo anything or resubmit, except that I would like you to submit a brief discussion related to my comments. Just submit under the name 'momentum conservation data'.
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&#Very good responses. Let me know if you have questions. &#
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