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'. **
20.5mm, 20.5mm
14.5mm
** Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process: **
322, 324, 323, 322, 323 mm
322.8, .8367
Ranges were measured from the end of the ramp
** 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. **
579, 597, 582, 577, 590
250, 254, 254, 253, 253
585, 8.337
252.8, 1.643
Ranges for the balls was determined from their respective locations at the point of impact
** Vertical distance fallen, time required to fall. **
799mm
0.404s
vf='sqrt(vo^2+2aAve*'ds) = 3.957m/s
'dt = 'dv/a = 0.404s
The vertical distance of each ball's fall was different. The larger ball necessitated the plywood cushioning on the floor to prevent damage to the floor. The smaller ball should have had the same plywood backing to have the same refernece frame but the plywood is tooo small to cover both landing sites. The vertical fall of the larger ball was 795mm and for the smaller ball 803mm so I used the average of 799mm. I understand the correct thing to do was calculate each ball separately but the average was used for the sake of expedience.
** 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. **
0.800, 0.628, 1.448
0.801, 0.797
0.630, 0.622
1.469, 1.427
All in m/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. **
0.800m1
0.628m1
1.448m2
0.800m1 + 0*m2
0.628m1 + 1.448m2
0.800m1 + 0*m2 = 0.628m1 + 1.448m2
** 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. **
0.172m1=1.448m2
m1 = 8.419m2
m1/m2 = 8.819
This is the ratio of the masses of the two bearings used in the experiment.
** Diameters of the 2 balls; volumes of both. **
2.6, 1.3 cm
9.2, 1.2 cc
** How will magnitude and angle of the after-collision velocity of each ball differ if the first ball is higher? **
The magnitude of the larger ball's velocity will be slightly greater than a head on collision and the direction will be slightly upward of its comparable direction compared to a head on collision. A full/square head on collision will transfre the most energy into the smaller (target) ball therefore the larger balls secondary speed will be less. The after collision direction of the larger ball will be at a lower angle although with smaller deviations in impact will result in logarythmically smaller results (quickly becoming immeasurable with greater accuracy of impact).
** Predicted effect of first ball hitting 'higher' than the second, on the horizontal range of the first ball, and on the second: **
The range of the second ball would be shorter and the largeer ball would be greater as less momentum in transferred.
** 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: **
8.545
I followed the eqution reduction instructions above
** What percent uncertainty in mass ratio is suggested by this result? **
3.1%
** What combination of before-and after-collision velocities gives you the maximum, and what combination gives you the minimum result for the mass ratio? **
Mean-STD for the large ball pre-collision
Mean+STD for the large ball post-collision
Mean+STD for the small ball post-collision
The three values above will result in the maximum ratio
Mean+STD for the large ball pre-collision
Mean-STD for the large ball post-collision
Mean-STD for the small ball post-collision
The three values above will result in the minimum 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*m1 + 0*m2 = u1*m1 + u2*m2
(v1-u1)*m1 = u2*m2
m1/m2 = u2/(v1-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. **
570, 567, 564, 564, 571
259, 258, 259, 261, 259
567.2, 3.271
259.2, 1.095
m1/m2 = 8.886
m1/m2 = 8.819 previously
** 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? **
79.9cm, 567.2, 0.103
147.1 cm/s
148.0, 146.3
146.9, 142.7
1.1, 3.6
The results with the target ball lowered 2mm differed from the horizontal strike by +0.7% to -2.5%.
** Your report comparing first-ball velocities from the two setups: **
The larger ball's velocity in the second trials was 2% higher. This would be due to less momentum being transferred in the off-center strike.
** Uncertainty in relative heights, in mm: **
0.3mm
I think this is the limit of visual acuity in interpreting the dots.
** 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. **
This would depend on the interpretation of the word significant. I do not think that significant uncertainty existed to adversely affect the experimantal results given the degree of accuracy desired.
** How long did it take you to complete this experiment? **
0810-1130 (minus 45 min break) = 155min
** Optional additional comments and/or questions: **
** **
Excellent work.
Also, I did receive your test today. Didn't have time to grade it, but on a quick read-through it looks very good.
Email me if you don't receive a message tomorrow.
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'. **
20.5mm, 20.5mm
14.5mm
** Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process: **
322, 324, 323, 322, 323 mm
322.8, .8367
Ranges were measured from the end of the ramp
** 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. **
579, 597, 582, 577, 590
250, 254, 254, 253, 253
585, 8.337
252.8, 1.643
Ranges for the balls was determined from their respective locations at the point of impact
** Vertical distance fallen, time required to fall. **
799mm
0.404s
vf='sqrt(vo^2+2aAve*'ds) = 3.957m/s
'dt = 'dv/a = 0.404s
The vertical distance of each ball's fall was different. The larger ball necessitated the plywood cushioning on the floor to prevent damage to the floor. The smaller ball should have had the same plywood backing to have the same refernece frame but the plywood is tooo small to cover both landing sites. The vertical fall of the larger ball was 795mm and for the smaller ball 803mm so I used the average of 799mm. I understand the correct thing to do was calculate each ball separately but the average was used for the sake of expedience.
** 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. **
0.800, 0.628, 1.448
0.801, 0.797
0.630, 0.622
1.469, 1.427
All in m/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. **
0.800m1
0.628m1
1.448m2
0.800m1 + 0*m2
0.628m1 + 1.448m2
0.800m1 + 0*m2 = 0.628m1 + 1.448m2
** 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. **
0.172m1=1.448m2
m1 = 8.419m2
m1/m2 = 8.819
This is the ratio of the masses of the two bearings used in the experiment.
** Diameters of the 2 balls; volumes of both. **
2.6, 1.3 cm
9.2, 1.2 cc
** How will magnitude and angle of the after-collision velocity of each ball differ if the first ball is higher? **
The magnitude of the larger ball's velocity will be slightly greater than a head on collision and the direction will be slightly upward of its comparable direction compared to a head on collision. A full/square head on collision will transfre the most energy into the smaller (target) ball therefore the larger balls secondary speed will be less. The after collision direction of the larger ball will be at a lower angle although with smaller deviations in impact will result in logarythmically smaller results (quickly becoming immeasurable with greater accuracy of impact).
** Predicted effect of first ball hitting 'higher' than the second, on the horizontal range of the first ball, and on the second: **
The range of the second ball would be shorter and the largeer ball would be greater as less momentum in transferred.
** 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: **
8.545
I followed the eqution reduction instructions above
** What percent uncertainty in mass ratio is suggested by this result? **
3.1%
** What combination of before-and after-collision velocities gives you the maximum, and what combination gives you the minimum result for the mass ratio? **
Mean-STD for the large ball pre-collision
Mean+STD for the large ball post-collision
Mean+STD for the small ball post-collision
The three values above will result in the maximum ratio
Mean+STD for the large ball pre-collision
Mean-STD for the large ball post-collision
Mean-STD for the small ball post-collision
The three values above will result in the minimum 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*m1 + 0*m2 = u1*m1 + u2*m2
(v1-u1)*m1 = u2*m2
m1/m2 = u2/(v1-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. **
570, 567, 564, 564, 571
259, 258, 259, 261, 259
567.2, 3.271
259.2, 1.095
m1/m2 = 8.886
m1/m2 = 8.819 previously
** 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? **
79.9cm, 567.2, 0.103
147.1 cm/s
148.0, 146.3
146.9, 142.7
1.1, 3.6
The results with the target ball lowered 2mm differed from the horizontal strike by +0.7% to -2.5%.
** Your report comparing first-ball velocities from the two setups: **
The larger ball's velocity in the second trials was 2% higher. This would be due to less momentum being transferred in the off-center strike.
** Uncertainty in relative heights, in mm: **
0.3mm
I think this is the limit of visual acuity in interpreting the dots.
** 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. **
This would depend on the interpretation of the word significant. I do not think that significant uncertainty existed to adversely affect the experimantal results given the degree of accuracy desired.
** How long did it take you to complete this experiment? **
0810-1130 (minus 45 min break) = 155min
** Optional additional comments and/or questions: **
** **
Excellent work.
Also, I did receive your test today. Didn't have time to grade it, but on a quick read-through it looks very good.
Email me if you don't receive a message tomorrow.