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.
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Distances from edge of the paper to the two marks made in adjusting the 'tee'.
1.9cm, 1.9cm
1.0cm
0.1cm
Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process:
15.3, 15.4, 15.2, 15.2, 14.9
15.2, .1871
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.
27.7, 24.8, 26.7, 26.9, 26.5
6.6, 7.1, 7.5, 6.4, 7.4
26.52, 1.064
7.0, .4848
Vertical distance fallen, time required to fall.
75.3cm
.446s
I measured height from the floor to the collision point of the two balls before falling. I measured time as the average of 5 readings with my stopwatch. I could not orient this experiment close enough to my desktop to use the timer program.
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.
34.1m/s, 15.7m/s, 59.5m/s
33.7m/s, 34.5m/s
14.6m/s, 16.8m/s
57.1m/s, 61.8m/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.
m1*34.1m/s
m1*15.7m/s
m2*59.5m/s
m1*34.1m/s + m2*0m/s
m1*15.7m/s + m2*59.5m/s
m1*34.1m/s + m2*0m/s = m1*15.7m/s + m2*59.5m/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*34.1m/s - m1*15.7m/s = m2*59.5m/s - m2*0m/s
m1 = (m2*59.5m/s)/(18.4m/s)
m1/m2 = 3.23
This is the ratio of the mass of ball 1 to the mass of ball 2.
Diameters of the 2 balls; volumes of both.
3.0cm, 2.2cm
4.5cm^3, 1.77cm^3
How will magnitude and angle of the after-collision velocity of each ball differ if the first ball is higher?
If at collision the center of the first ball is higher than the center of the second, the magnitude of the velocity of the first ball will be increased and the direction of the velocity of the first ball will be more horizontal, while the magnitude of the velocity of the second ball will be decreased and the direction of the velocity of the second ball will be more vertical.
Predicted effect of first ball hitting 'higher' than the second, on the horizontal range of the first ball, and on the second:
The first ball would have a greater horizontal range. The second ball will have a greater vertical 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:
Using the ranges required above in the equation m1/m2 = (v2' - v2)/(v1 - v'1) = 3.38
What percent uncertainty in mass ratio is suggested by this result?
4.4%
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 3.66 given by v2'=61.8, v1=33.7, v1'=16.8
Minimum 2.87 given by v2'=57.1, v1=34.5, v1'=14.6
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?
u2/(1-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.
Ball 1 distances: 6.1, 6.1, 5.6, 6.0, 5.3
Ball 1 mean and STD: 5.82, .3564
v1=34.1m/s, v1'=13.0m/s
Ball 2 distances: 16.2, 18.1, 18.0, 19.7, 17.3
Ball 2 mean and STD: 17.86, 1.278
v2'=40.0m/s
m1/m2=40/(34.1-13)=1.90. This is 58% of 3.23
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?
75.1cm, 17.86cm, 4.20
95.79cm/s
88.65cm/s, 105.8cm/s
151.2 cm/s, 168.2cm/s
36.29cm/s
The 1st velocity is 62% of the 2nd velocity
Your report comparing first-ball velocities from the two setups:
75.3cm, 5.82cm, 12.9
27.74 cm/s
25.94 cm/s, 29.55 cm/s
31.29 cm/s, 36.16 cm/s
12.04cm/s
The 1st velocity is 57% of the 2nd velocity.
Uncertainty in relative heights, in mm:
.5mm
Using the guidelines outlined at the beginning of the experiment, I felt confident that the balls were aligned on the same milimeter tick on my ruler. I am visually able to discriminate down to about .5mm that the measurements were accurate, but I am less confident about smaller fractions of a milimeter.
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.
In the second setup, I only altered the height of the 2nd ball by 2mm and the landing distances of both balls changed considerably. This is strong evidence that even a small amount of error in aligning the balls (i.e., if the balls collide at any other position than middle-to-middle) can have a large impact in the measurements obtained after the balls collide.
How long did it take you to complete this experiment?
8hrs
Optional additional comments and/or questions:
Excellent work.