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'. **
5.4 cm
1.2 cm
1.0 mm
** Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process: **
28.25, 28.75, 29.00, 28.75, 28.00
28.55, .4108
I set up the lab according to directions. I set the small ball on the straw at the end of the ramp. I then placed the large ball 1 cm from the very top of the ramp. Then, I released the large ball, watching (with the help of a family member) the position at which each ball landed. I measured these distances and repeated accordingly for other parts of the experiment.
** 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. **
39.25, 37.25, 39.00, 38.5, 39.00
12.5, 12.5, 12.5, 12.5, 12.5
38.6, .8023
12.5, 0
Again, with the help of a partner, we stuck pins in the carpet where the balls hit. We measured the distances of these balls (the pin locations) from the vertical drop position. These distances yielded our horizontal ranges.
** Vertical distance fallen, time required to fall. **
61 cm
.421875 seconds
I merely measured from the edge of the tabletop to the floor for the distance. For the time, I assumed the balls left the table at the same time.
** 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. **
67.7, 29.6, 91.5
28.9608, 28.1392
39.4023, 37.7977
12.5, 12.5
** 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*67.7
m1*29.6
m2*91.5
(m1*67.7)+(m2*0)
(m1*29.6)+(m2*91.5)
(m1*67.7)+(m2*0)=(m1*29.6)+(m2*91.5)
** 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*67.7)-(m1*29.6)=(m2*91.5)-(m2*0)
m1=[(m2*91.5)-(m2*0)]/38.1
(38.1*m1)/9.5=m2
m1/m2=.249
This is the number that should result when m1 (of the first ball) is divided by m2 (of ball two).
** Diameters of the 2 balls; volumes of both. **
2.5, 1.75
8.18 cm^3, 2.81 cm^3
** How will magnitude and angle of the after-collision velocity of each ball differ if the first ball is higher? **
It will push the first ball upwards and slow the velocity down to zero as it changes direction from up to down. The speed will be less than because the first ball will have to start over from zero at its peak and the second ball will be pushed down onto the tee, thus causing it to also be thrown upwards. The direction will differ in the balls going up or down, but they still should not go sideways as long as the experiment is set up correctly.
** Predicted effect of first ball hitting 'higher' than the second, on the horizontal range of the first ball, and on the second: **
Because both balls are being thrown upwards, both horizontal ranges will be shorter because the velocities of both balls are slower.
** 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: **
.41
I filled in variables and set up the equation as instructed, little by little, simplifying until on the ratio is left.
** What percent uncertainty in mass ratio is suggested by this result? **
.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? **
Combining the minimum before velocity of the first ball with the max after velocity yields the smallest ratio. The maximum of the after of the second ball and theminimum of the before yields the largest 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? **
(v1-u1)/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. **
The mean horizontal range for the first ball this time was 13.55 with a standard deviation of .84. The time of drop was .421875 seconds, so this yielded, for the first ball, an average velocity of 32.11 cm/s.
For the second ball, the mean horizontal range was 25.5, whereas the standard deviation was 1.2 and the velocity was 60.44 cm/s.
The ratio of mass one to mass two in this case is .588.
** 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? **
61, 25.5, .14
75.96
72.26, 79.67
116.2, 119.9
3.7, 7.41
The velocity appears to be significantly different.
** Your report comparing first-ball velocities from the two setups: **
First ball, first trial velocity: 36.68
First ball, second trial velocity: 39.67
The second trial appeared to be the fastest trial for the first ball.
** Uncertainty in relative heights, in mm: **
+-.3mm
** 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 heights were relatively insignificant because the results came out as predicted--there didn't appear to be large discrepancies in data that threw the final outcome or reading of the data off.
** How long did it take you to complete this experiment? **
2.5 hours
** Optional additional comments and/or questions: **
Very good work. Let me know if you have questions.
I believe I received your final exam today. Let me know if I haven't responded by tomorrow.