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'. **
2.1 cm, 2.5 cm
1.8 cm
I think my measurements are pretty accurate because the mark made by the carbon paper was just a small dot so I could measure that pretty accurately. There may be a small amount of error when measuring the length of the straw becuase I couldn't get right next to the straw but had to look at the ruler from about a centimeter away.
** Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process: **
25.5, 26, 26, 25.8, 25.5
25.76, .2509
I got this measurement by putting the horizontal part of the ramp to the edge of the table. Then, I placed two pieces of paper on the floor in line with the edge of the table and did a trial run to see approximately where the ball would drop. Then I placed the piece of carbon paper where the ball landed and did five trials. I measured from the end of the paper that corresponded to where the edge of the table was to where the ball landed for each of the five trials.
** 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. **
28.2, 27.8, 24.6, 25.3, 27.3
14.9, 9.9, 10.6, 11.8, 12.2
26.64, 1.595
11.88, 1.923
I got these positions in the same way as above. I placed the two carbon pieces at approximately the same positions as where the trial balls dropped. Also, I listed the small ball first because I thought that was what you were talking about being the second ball.
** Vertical distance fallen, time required to fall. **
64 cm
small ball - .37 sec, big ball - 1.5 sec
I got the vertical distance by measuring up from the floor to the top of the straw from where the balls fell. I got the time for the big ball by timing from the time of release all the way down the ramp to the time the big ball hit the floor. I got the time for the small ball by timing from the time of collision to the time the small ball hit the floor.
** 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. **
69.62 cm/s, 31.68 cm/s, 72 cm/s
26.0109, 25.5091
37.305, 26.897
28.235, 25.045
** 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 * 69.622 cm/s
momentum = m1 * 31.68 cm/s
momentum = m2 * 72 cm/s
momentum = (m1 * 62.622 cm/s) + (m2 * 0)
momentum = (m1 * 31.68 cm/s) + (m2 * 72)
(m1 * 62.622 cm/s) + (m2 * 0) = (m1 * 31.68 cm/s) + (m2 * 72)
** 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 * 69.622 cm/s) - (m1 * 31.68 cm/s) = (m2 *72 cm/s) - (m2 * 0)
m1 = (m2*72) / 30.942
m1/m2 = 72 cm/s / 30.942 cm/s
m1/m2 = 2.327
The ratio of m1/m2 is the ratio of mass according to the velocities of the balls before and afeter collision.
Very good. Note that you could have divided through by the units in the first step and you would have been done with them early, but it's good that you did include the units.
** Diameters of the 2 balls; volumes of both. **
2.4 cm, 1.2 cm
7.238, .90477
** How will magnitude and angle of the after-collision velocity of each ball differ if the first ball is higher? **
Both balls will not experience the full force of the other ball since only a fraction of the balls collided. I think the magnitude would be less and there's no really telling the direction of the ball for the first ball. Now for the second, I think the direction will be affected because the first ball is kind of pushing it down instead of going straight off at a horizontal. I think the speed of the first ball would be increased if the balls didn't hit right in the center and the speed of the second ball would be decreased if they didn't hit right in the middle. The second ball isn't getting the full force of the first ball if it's not hitting directly in the center. This will affect the velocity in the same way.
By Newton's Third Law the forces exerted by the two balls on one another are equal and opposite.
** Predicted effect of first ball hitting 'higher' than the second, on the horizontal range of the first ball, and on the second: **
I think that the horizontal range of the first ball would be longer if the first ball hit higher than the second. I think the horizontal range of the second ball would be shorter because the larger ball seems to be pushing down on the second ball and wouldn't allow it to shoot out horizontally.
** 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: **
2.1395
My equation to obtain this number was based on the characteristic above:
(m1 * 25.5091)+(m2 * 0) = (m1 * 13.803) + (m2 * 25.045)
** What percent uncertainty in mass ratio is suggested by this result? **
When I compared this to my original, I got 2.1395/2.327 = .91 or 91% uncertainty. This seems pretty large even though I think I did a pretty good job with accuracty in this experiment.
** 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? **
2 hours
** Optional additional comments and/or questions: **
You did good work and obtained good results. I see nothing wrong with your analysis. See my brief note(s).