Phy 231
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, 2.1
1.5
The values in the first line are accurate to within a millimeter due to difficulty reading faint marks left by the carbon paper.
** Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process: **
10.85, 11.10, 10.95, 10.90, 11.00
10.96, 0.09618
In order to best determine the range I first determined the point directly below the end of the horizontal section of the table edge. Directly below this point I made a line on the blank sheet of paper used to detect the impact points. The table edge is 1.6 cm from the edge of the ramp; therefore, the values above are the raw data points plus 1.6 cm to give me the horizontal distance traveled by the ball from the edge of the ramp until it reached the surface beneath.
** 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. **
20.60, 21.25, 21.35, 22.10, 22.35
8.70, 8.80, 8.85, 8.85, 9.10
21.53, 0.7023
8.86, 0.1475
** Vertical distance fallen, time required to fall. **
11.6
0.154
I determined by measuring the line I used to determine the point directly below the edge of the table, and by measuring this line was able to determine the height of the edge of the table, and I then added the distance from the edge of the table to the bottom of the steel ball. I determined the time of fall with the assumption of zero for initial velocity giving me the general equation 'dt = 'sqrt( 2 * 'ds / a), which is this case 'ds = 11.6 cm and a = 980 cm/s^2.
** 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. **
79.45 , 63.00, 174.2
80.22, 78.68
64.13, 61.87
181.4, 167.2
** 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 * 79.45
m1 * 63.00
m2 * 174.2
m1 * 79.45 + m2 * 0
m1 * 63.00 + m2 * 174.2
m1 * 79.45 + m2 * 0 = m1 * 63.00 + m2 * 174.2
** 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 * 79.45 - m1 * 63.00 = m2 * 174.2
m1 = (m2 * 174.2)/ (16.45)
m1/ m2 = 174.2 / 16.45
m1/ m2 = 10.59
This ratio indicates that the mass of the first ball is 10.59 times larger than the mass of the second ball.
** Diameters of the 2 balls; volumes of both. **
2.6, .8
9.20, 0.268
** How will magnitude and angle of the after-collision velocity of each ball differ if the first ball is higher? **
It would cause the speed of the first ball to be no more or less compared to if it hit off center. Yes the direction after collision would change. If the second ball was higher than the first then the first would have a lower direction than a centered hit; however, if the second ball was lower than the first then the first ball would have a higher direction than a centered hit.
** Predicted effect of first ball hitting 'higher' than the second, on the horizontal range of the first ball, and on the second: **
If second ball is two high as made by the assumption above the first ball would have a the same magnitude in velocity; however, the there would be a direction change as discussed above that would shorten the horizontal range of the first ball. The second ball of the other hand would have a higher direction than a centered hit, which would extend its horizontal 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: **
11.49
I determined this ratio by using 78.68 as the before-collision velocity of the first ball, 64.13 as the after-collision velocity of the first ball, and 167.2 as the after-collision velocity of the second ball. This gave me an initial expression of m1 * 78.68 = m1 * 64.13 + m2 * 167.2, which when simplified for the expression m1/m2 gives me 167.2 / (78.68 - 64.13) and this equal m1/m2 = 11.49 as given above.
** What percent uncertainty in mass ratio is suggested by this result? **
8.50
** 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 maximum mass ratio would be found by minimum before collision velocity of the first ball, the maximum after-collision velocity of the first ball, and the maximum after-collision velocity of the second ball. The minimum mass ratio would be found by the maximum before-collision velocity of the first ball, the minimum after-collision velocity of the first ball, and the minimum after-collision velocity of 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? **
m1/ m2 = u2/ (v1 - u1)
** Derivative of expression for m1/m2 with respect to v1. **
(u2 * u1 *(v1-u1)^-2)
.644
This quantity most likely represents the amount of error that could be added or subtracted to my previously found mass ratio.
I would like to comment that you ask for units for the value in the second line; however, by my calculations we have the units of cm/s * cm/s * (cm/s)^-2 in the expression above, which cancels out all units when solved, which is why I do not have a unit assigned above.
** 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? **
.77
.496
The value in the first line is the difference in v1 between the average and the maximum values. The value in the second line is the .77 times .644 value found by the approximate derivative above. This values shows us how much we can expect the mass predictions to fluctuate due to experimental error; however this is less than the actual fluctuations found earlier by about half. This goes to show that experimental error can exceed those predicted by calculations.
** Complete summary and comparison with previous results, with second ball 2 mm lower than before. **
The after-collision velocity I found this time was 61.01 for ball 1, and 173.9 for ball 2; therefore, my prediction for mass ratio is now 9.43
** 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? **
11.6, 21.5, -.156
170.1
174.1, 166.2
178.1, 169.8
3.8
By analyzing the percent 3.8 cm/s is of the original 173.9 cm/s velocity we find that it is 2.19%, which means that by lowering the height of the second ball by only .2 cm or 2 mm we lowered the velocity of the ball on average by 2.19%. I find this to not be a significant difference, and also when considering the large overlap in the range of velocities it is possible this small percentage was caused by experimental error only.
** Your report comparing first-ball velocities from the two setups: **
By using the same methods we find the velocity of the first ball to be 61.01, which means compared to the original runs velocity of 64.13 we have a difference of 3.12. This gives us a percentage of 4.87% difference. This difference is roughly twice that of the difference found by the smaller second ball; however, again this is not a significant difference and the range of velocity possible due to experimental error allow for the possibility that the velocities are the same.
** Uncertainty in relative heights, in mm: **
.75
The reason I give .75 mm as the uncertainty in my measurements is due to the spread of the 5 data points 4 of which were within a circle of 1.5 mm diameter. The 5th data point could be seen as an outlier or abnormality, and if we discard it we find the rest to be less than .75 mm away from the center point of this circle.
** 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. **
Just by looking at how the impact points spread on the paper as indicated by the marks made by the carbon paper I see a similar spread in range for both runs, and comparable consistency; therefore, I do not believe the relative height was a significant factor in this experiment.
** How long did it take you to complete this experiment? **
3 hours and 50 minutes
** Optional additional comments and/or questions: **
You asked further questions for University student above however, there is not a box for me to place any of the answers requested as indicated by the questions.
&#This looks very good. Let me know if you have any questions. &#