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'. **
2.133cm
2.106cm
.027cm
** Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process: **
18,18.3,18.2,18,18.2
18.14,.1342
I dropped the ball from the height of the table to the floor five times and found the mean. I made my measurements from this point. This point is 1.2cm from the edge of the table. All of the measurements above are in centimeters.
** 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. **
35,36,36.1,36,35.6
14.5,14.4,14.3,14.5,14.4
35.74,.4561
14.42,0.08367
I made the measurements exactly as before from the 1.2cm which marks the mean of the free fall of the large metal ball.All of the measurements above are in centimeters.
** Vertical distance fallen, time required to fall. **
66.6
3.68
The above are the distance of fall in centimeters and the time required to fall from rest in seconds. This was calculated by using `ds = v0`dt + .5a`dt^2 and solving for `dt.
It doesn't take 3.68 seconds to fall 66.6 cm. Probably .368 seconds; careful about the units.
** 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. **
49.2,39.2,97.0
49.7,48.9
39.4,38.9
98.4,95.9
These velocities are in 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. **
49.2cm/s * m1
39.2cm/s * m1
97.0cm/s * m2
49.2cm/s * m1 + 0cm/s * m2
39.2cm/s * m1 + 97.0cm/s * m2
49.2cm/s * m1 + 0cm/s * m2 = 39.2cm/s * m1 + 97.0cm/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. **
49.2cm/s * m1 - 39.2cm/s * m1 = 97.0cm/s * m2 - 0cm/s * m2
m1 = 9.7 * m2
m1/m2 = 9.7
The meaning of this is that m1 has 9.7 times more mass than m2
** Diameters of the 2 balls; volumes of both. **
2.50,1.25
7.24,0.905
** How will magnitude and angle of the after-collision velocity of each ball differ if the first ball is higher? **
If the first ball is higher than the second ball it would not give the second ball the velocity in the x-direction that it would if they collided at the center.
** 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 first ball will have a bit more horizontal range than if the balls collided at the 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: **
10.1
I used the equation 48.9cm/s * m1 + 0cm/s * m2 = 39.4cm/s * m1 + 95.9cm/s * m2 and simplified it as I did previously.
** What percent uncertainty in mass ratio is suggested by this result? **
4.12
This is a result of (10.1 - 9.7)/9.7 * 100 = 4.12%
** What combination of before-and after-collision velocities gives you the maximum, and what combination gives you the minimum result for the mass ratio? **
The largest ratio comes from the smallest difference between the first ball velocities and the greatest velocity for the second ball.
The smallest ratio comes from the maximum difference between before and after velocities for the first ball and the minimum after collision velocity for the second ball.
** 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? **
v1 * m1 + 0 cm/s * m2 = u1 * m1 + u2 * m2
(v1-u1)* m1 = u2 * m2
m1 = u2/(v1-u1) * m2
m1/m2 = u2/(v1-u1)
** Derivative of expression for m1/m2 with respect to v1. **
d/dv1 u2/(v1-u1) = -u2/(v1 - u1)^2
-.97
This is the rate of change in mass ratio 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? **
.365
-.354
The change in mass ratio was obtained by placing the v1 with a velocity that was adjusted by the standard deviation in horizontal range
This means that the mass ratio is adjusted by -.354 for every change in velocity. This is the illustration of the chain rule.
** Complete summary and comparison with previous results, with second ball 2 mm lower than before. **
x-displacement
B1: 14.5,14.3,14.2,14.3,14.4 mean: 14.34 +- .1140
B2: 35.6,35.6,35.6,34.6,35.6 mean: 35.4 +- .4472
velocity mean, max, min
B1: 39.0, 39.3, 38.7
B2: 96.9, 97.4, 95.0
** 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? **
66.6,35.4,0.108
99.79
101.1, 98.49
98.4,95.9
2.79
The velocity of the second run seems to be greater than the first run.
** Your report comparing first-ball velocities from the two setups: **
66.6,14.34,0.108,
39.66
39.98,39.34
39.4,38.9
.46
The velocity seems to be very close between the first run and second run.
** Uncertainty in relative heights, in mm: **
.27
This distance in mm between marks on the triply reduced ruler that was used to make the measurement.
** 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 was not a significant factor in the first setup because the uncertainty in the measurement was very low. It was much less than a millimeter.
** How long did it take you to complete this experiment? **
About 4 hours.
** Optional additional comments and/or questions: **
Good work. The actual mass ratio, assuming the same density for both steel balls, would be the cube of the ratio of diameters. The ratio of diameters is 2/1, so mass ratio should be 8/1, which is close to your 10/1 ratio. Uncertainties associated with the tee, and also with the different positions of the centers of mass (the second ball actually starts out half its radius in front of the point of collision, the first half its radius behind, which tends to lead to an overestimate the second-ball velocity and underestimate the first-ball velocity), make it difficult to be much more accurate.