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'. **
3.5cm, 3.5cm
2cm
I estimate the uncertainity to be +/- 5%
The first line above gives the height that balls struck the vertical piece of paper from the edge. The second line lists the height of the tee.
** Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process: **
2.3cm, 2.4cm, 2.7cm, 2.9cm, 3.1cm
2.68cm, .33cm
The values above are the horizontal measurements of the large metal leaving the ramp, and the second line lists the mean and 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. **
11, 11.2, 11.3, 11.6, 12.5cm
1.1, 1.2, 1.3, 1.3, 1.4cm
11.52cm, .59cm
1.26cm, .11cm
The values above are the horizontal distances of the 2 balls after collision. Line 1, the little ball; line 2, the big ball. The third and fourth lines are the mean and standard deviation of the first two lines.
** Vertical distance fallen, time required to fall. **
3.5cm, 3.5cm
.845s
The time of fall was determined with the formula: dsy = Voy(dt) + .5(ay)(dt^2), and then solving for dt.
** 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. **
3.17cm/s, 1.49cm/s, 13.63cm/s
3.56cm/s, 2.78cm/s
1.62cm/s, 1.36cm/s
14.33cm/s, 12.93cm/s
The values of the first line are the horizontal velocity of the ball before collision, followed by the horizontal velocity of both balls after the collision. The lines following there after are the velocities based of the mean horiz distance(x) + and - standard deviation.
** 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. **
m1(3.17cm/s)
m1(1.49cm/s)
m2(13.63cm/s)
m1(3.17cm/s) + m2(0m/s)
m1(1.49cm/s) + m2(13.63cm/s)
m1(3.17cm/s) + m2(0m/s) = m1(1.49cm/s) + m2(13.63cm/s)
The total momentum immediately before collision is equal to the total momentum immediately after collision.
** 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(3.17cm/s) - m1(1.49cm/s) = m2(13.63cm/s) - m2(0m/s)
m1 = m2(13.63cm/s) / 1.68cm/s = m2(8.113)
m1 / m2 = 8.113
m1 / m2 = 8.113
The equation above represent the factor by which m1 is larger than m2.
** Diameters of the 2 balls; volumes of both. **
m1: 2.5cm, m2: 1.2cm
Volume(m1): 8.18cm^3, Volume(m2): 0.905cm^3
Volume was calculated with the formula, (4/3) * pi * r^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 balls are not aligned, they do not collide center to center, it will change the y distance of the equation(dsy). I belive that the speed of the balls would be less, since all the force and momentum is not focused on the center. The after collison direction could be altered, where the ball could be jumped up at an angle or shot downward at an angle.
** Predicted effect of first ball hitting 'higher' than the second, on the horizontal range of the first ball, and on the second: **
I believe either the way the first ball will travel farther, and the second ball after the collision will travel less of a distance in the x-range.
** 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 = 12.35
m1/m2 = 9.11
I used the formula used earlier: m1(min v) + m2(0cm/s) = m1(max v) + m2(max v); m1(min v) + m2(0cm/s) = m1(min v) + m2(min v)
** What percent uncertainty in mass ratio is suggested by this result? **
12.2%, 52.2%
** What combination of before-and after-collision velocities gives you the maximum, and what combination gives you the minimum result for the mass ratio? **
max = min v before = max v after
min = max v before = min v 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? **
max = m1(min v1) + m2(v2) = m1(max u1) + m2(max u2)
min = m1(max v1) + m2(v2) = m1(min u1) + m2(min 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. **
In a repeat experiment of the above with the second ball 2mm lower than the previous expirement, I have calculated the following.
Mean Horizontal Distance: Ball 1 - (before)6.76cm, (after)4cm; Ball 2 - 14.18cm
Time: Ball 1: .714s; Ball 2: .742s
Velocity: Ball 1 - (before)9.47cm/s; (after)5.60cm/s; Ball 2 - 19.11cm/s
Equation: m1(9.47cm/s) + m2(0cm/s) = m1(5.60cm/s) + m2(19.11cm/s)
Mass Ratio: m1 / m2 = 4.94
** 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? **
2.5cm, 14.18cm, -.194
Run resulted in an error when trying to calculate velocity.
19.42cm/s, 18.80cm/s
14.33cm/s, 12.93cm/s
5.48cm/s
Yes, the velocities resulting from the 2mm drop of the center of the second ball seem to be higher than the velocities when the two balls centers where aligned.
** Your report comparing first-ball velocities from the two setups: **
The velocities for the first ball in the experiment seems to travel at a higher velocity when the centers were mis-aligned by a 2mm drop in the height of the second ball than in the first experiment with the centers of the balls aligned.
** Uncertainty in relative heights, in mm: **
My would assume that the uncertainity in my experiment would be +/- .5mm.
Reason, I give this number would be the measurement on the paper after the trial. There is a possibility that I could have measured improperly; therefore, I claim +/- .5mm.
** 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. **
I think relative uncertainity could pay a significant factor in this experiment. The height of the second ball was lowered slightly to a level of 2mm less, yet the results of the experiment changed dramatically. Therefore, I will argue against the hypothesis.
** How long did it take you to complete this experiment? **
3hr 35min
** Optional additional comments and/or questions: **
Excellent results.
The time of fall doesn't affect your calculation of relative masses, since the time of fall for every projectile was presumably the same as for every other one. However the time of fall for a 3.5 cm drop is not .8 seconds; the time you got is more liklely the time for a 3.5 meter drop. If that is the source of the error, then your time of fall would be 1/10 as great and your velocities would be 10 times as great as those reported.