Phy 201
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: **
** #$&* Distances from edge of the paper to the two marks made in adjusting the 'tee'. **
2.5 cm, 2.5 cm
.9 cm
I believe these measurements are very close to accurate, in that all the marks were very close together with little variation.
** #$&* Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process: **
27.4 cm, 26.5 cm, 26.8 cm, 25.9 cm, 25.8 cm
26.48 cm, .6611 cm
My results for the horizontal range were obtained by rolling a ball down the 'long' ramp, off the edge of the desk to the floor. A measurement was taken from the edge of the ramp to where it hit the floor.
** #$&* Five horizontal ranges observed for the second ball; corresponding first-ball ranges; mean and standard deviation of the second-ball ranges; mean and standard deviation of ranges for the first ball. **
23.2 cm, 23.6 cm, 22.65 cm, 22.9 cm, 24.25 cm
15 cm, 11.3 cm, 10.2 cm, 13.2 cm, 12.8 cm
23.23 cm, .6291 cm
12.5 cm, 1.841 cm
Ball 1 was rolled down the 'long' ramp, where at the end, (edge of the desk), it collided with ball 2. Marks were made where each ball landed on the floor. Measurements were then taken from the dge o the desk to the landing point on the floor.
** #$&* Vertical distance fallen, time required to fall. **
77.5 cm
.5156 sec
The vertical distance was determined by measuring the top of the 'straw' to the floor.
The timer program was used on each trial to mark the top of the ramp start time, the collision time and the stop time.
The number expressed above is the mean of the times from collision to stop.
** #$&* Velocity of the first ball immediately before collision, the velocity of the first ball after collision and the velocity of the second ball after collision; before-collision velocities of the first ball based on (mean + standard deviation) and (mean - standard deviation) of its uninterrupted ranges; same for the first ball after collision; same for the second ball after collision. **
56.14 cm/s, 24.24 cm/s, 45.05 cm/s
49.31 cm/s, 65.16 cm/s
25.62 cm/s, 22.61 cm/s
42.62 cm/s, 47.94 cm/s
** #$&* 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. **
Momentum = m1 * 56.14 cm/s
Momentum = m1 * 24.24 cm/s
Momentum = m2 * 45.05 cm/s
Momentum = (m1 * 56.14 cm/s) + (m2 * 0 cm/s)
Momentum = (m1 * 24.24 cm/s) + (m2 * 45.05 cm/s)
(m1 * 56.14 cm/s) + (m2 * 0 cm/s) = (m1 * 24.24 cm/s) + (m2 * 45.05 cm/s)
** #$&* Equation with all terms containing m1 on the left-hand side and all terms containing m2 on the right; equation rearranged so m1 appears by itself on the left-hand side; preceding the equation divided by m2; simplified equation for m1 / m2. **
(m1 * 56.14 cm/s) - (m1 * 24.24 cm/s) = (m2 * 45.05 cm/s)
m1 = 1.412 * m2
m1 / m2 = 1.412
1.412
The mass of m1 is 1.412 times larger than m2.
** #$&* Diameters of the 2 balls; volumes of both. **
3.7 cm, 3.3 cm
26.52 cm^3, 18.81 cm^3
** #$&* How will magnitude and angle of the after-collision velocity of each ball differ if the first ball is higher? **
If the center of the first ball is higher than the second, the first ball is more likely to have a longer magnitude and greater angle of direction. The speed would be slightly less than if the centers were the same height. the second ball would have less a velocity, with a lesser magnitude and angle of direction.
** #$&* Predicted effect of first ball hitting 'higher' than the second, on the horizontal range of the first ball, and on the second: **
The horizontal range of the first ball would be larger.
The horizontal range of the second ball would be less.
** #$&* ratio of masses using minimum before-collision velocity for the first ball, maximum after-collision velocity for the first ball, minimum after-collision velocity of the second: **
m1 / m2 = 1.21
(m1 * 65.16 cm/s) - (m1 * 25.62 cm/s) = (m2 * 47.94 cm/s)
** #$&* What percent uncertainty in mass ratio is suggested by this result? **
16.7%
1.412 / 1.21 = 16.7
** #$&* What combination of before-and after-collision velocities gives you the maximum, and what combination gives you the minimum result for the mass ratio? **
Maximum ratio = 1.80, max beofre, max after, max after
Minimum ratio = 1.13, min before, min after, min after
** #$&* 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? **
m 1 = (m2 * u2) / (v1 - u1)
** #$&* Derivative of expression for m1/m2 with respect to v1. **
** #$&* 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? **
** #$&* Complete summary and comparison with previous results, with second ball 2 mm lower than before. **
The data from this trial is slightly different than the first trial with the after velocities being greater for ball one and two.
I believe the mass ratio this time to be 2.435 = m1 / m2 rather than trial 1's 1.412 = m1 / m2.
** Vertical drop of the second ball, its mean horizontal range and the slope of the line segment connecting the two centers; the velocity given by the program based on mean; velocity interval for 2-mm decrease in 2d-ball height; velocity interval from the original run at equal heights; difference in the mean-based velocities; is new velocity significantly different than original? **
79 cm, 23.32 cm, -2 mm
60.64 cm/s
62.33 cm/s, 58.96 cm/s
62.17 cm/s, 58.80 cm/s
.15 cm/s
The change in altitude has a very small difference (.15 cm/s) between the 1st trial and the 2nd, increased altitude trial.
** #$&* Your report comparing first-ball velocities from the two setups: **
78.8 cm, 12.5 cm
32.13 cm/s
36.94 cm/s, 27.34 cm/s
36.95 cm/s, 27.34 cm/s
0 cm/s
There is no change in velocity.
** #$&* Uncertainty in relative heights, in mm: **
.1 mm
My percent uncertainty found in the beginning of the lab was very small, therefore I would expect it to remain the same throughout the experiment.
** #$&* 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. **
Since both trials involving collision had no significant variation in velocity, I would say the hypothesis is correct.
** #$&* How long did it take you to complete this experiment? **
** #$&* Optional additional comments and/or questions: **
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3 hours 20 minutes
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7.3, 2.9
11.8, 7.3
They are almost proportional. I may have had some human errors in all of the calculations that prevented them from being 2:1 proportionally.
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7.3, 11.8
7.8, 8.5, 11.54
The numbers were all previously found. The magnitude of the result were the sum of the two vector lengths each squared and then the total of that amount squared.
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0.26, 2.25%
16.06, 48%
I found these values by doing what the instructions said in order to find the percent error. I also was able to find the magnitude of error by taking the difference of the two and dividing by the original.
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2 hours
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7.3, 2.9
11.8, 7.3
They are almost proportional. I may have had some human errors in all of the calculations that prevented them from being 2:1 proportionally.
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7.3, 11.8
7.8, 8.5, 11.54
The numbers were all previously found. The magnitude of the result were the sum of the two vector lengths each squared and then the total of that amount squared.
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0.26, 2.25%
16.06, 48%
I found these values by doing what the instructions said in order to find the percent error. I also was able to find the magnitude of error by taking the difference of the two and dividing by the original.
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2 hours
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7.3, 2.9
11.8, 7.3
They are almost proportional. I may have had some human errors in all of the calculations that prevented them from being 2:1 proportionally.
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7.3, 11.8
7.8, 8.5, 11.54
The numbers were all previously found. The magnitude of the result were the sum of the two vector lengths each squared and then the total of that amount squared.
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0.26, 2.25%
16.06, 48%
I found these values by doing what the instructions said in order to find the percent error. I also was able to find the magnitude of error by taking the difference of the two and dividing by the original.
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2 hours
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&#Very good responses. Let me know if you have questions. &#