Your work on conservation of momentum 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'.
1.7cm, 1.3cm
Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process:
16.5cm/s, 17.9cm/s, 17.5cm/s, 19.7cm/s, 18.8cm/s
18.1cm/s, 1.23
I took the speed of thae ball down the ramp and multiplied it by the time it took to reach the floor from the end of the ramp to the floor.
This question asked for your horizontal ranges, not your speeds. Horizontal ranges are in cm, not in cm/s.
The ranges are quantities you actually measured; the speeds are inferred. Basic principle: Report what was measured, then move to what is inferred.
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.
39.8cm/s, 43.0cm/s, 43.4cm/s, 41.6cm/s, 42.0cm/s
17.5cm/s, 20.7cm/s, 19.3cm/s, 18.5cm/s, 18.4cm/s
41.96cm/s, 1.409
18.88cm/s, 1.201
Using the distance of the markings and the timings of roll and fall.
Vertical distance fallen, time required to fall.
73cm.
.34sec.
Measured the distance and took the time in seconds to travel from rest.
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.
46.6cm/s, 55.5cm/s, 123cm/s
46.6cm/s +- .8944
55.5cm/s +- 1.20
123cm/s +- 1.41
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.
46.6cm/s
55.5cm/s
123cm/s
46.6cm/s
178.5cm/s
46.6cm/s = 178.5 cm/s
The masses (m1 or m2, depending on which ball) are not included in your expressions for momentum.
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.
46.6cm/s + 55.5cm/s = 123cm/s
102cm/s /123cm/s = 123cm/s / 123cm/s
102cm/s / 123cm/s = .83
Again masses are not included in your expressions or your equations.
The ratio m1/m2 means the difference in the mass of one object as opposed to the other being tested.
Diameters of the 2 balls; volumes of both.
2.5cm, 1.9cm
8.2g, 3.6g
How will magnitude and angle of the after-collision velocity of each ball differ if the first ball is higher?
This will cause the second ball to not go as far as it would with the centers being equal. The speed will be greater. The velcoity will go up and fly out farther. If the centers are off the second ball will shoot down or up.
Predicted effect of first ball hitting 'higher' than the second, on the horizontal range of the first ball, and on the second:
This will cause the horizontal range to be shorter, because the ball's centers are not the same.
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:
m1/m2
45.7cm/s + 56.7cm/s / 121.6cm/s = .84
What percent uncertainty in mass ratio is suggested by this result?
1% uncetainty
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 max before collision and the minimum after collision gives you the max result and the minimum before collision and the max after collision gives you the minimum result.
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.
My data seems to correlate almost identical to the preceeding experiment, with only a couple of ranges being different. I believe that by moving the tubing 2mm., this will cause the mass ratios to be closer together.
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?
.33cm, 42.3cm/s
122.9cm/s
43.972cm/s, 40.628cm/s
43.369cm/s, 40.551cm/s
2.818cm/s, 3.344cm/s
The velocity doesnt seem to be signifacatly different at all. The 2mm drop didnt seem to effect the 2hd ball velocity that much at all.
Your report comparing first-ball velocities from the two setups:
.33cm, 19.8cm/s
54.26cm/s
21.28cm/s, 18.33cm/s
20.08cm/s, 17.68cm/s
2.4cm/s, 2.95cm/s
The velocity doesnt seem to be signifacatly different at all. The 2mm drop didnt seem to effect the 2hd ball velocity that much at all.
Uncertainty in relative heights, in mm:
.5mm, when using the tubing it was hard to get a completely actual reading of its measurement, because the wieght of the ball could effect the tubing hieght and also the super glue compress when set on the table, especially with a metal ball resting on it.
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 in heights did not affect the the balls in the first setup. If you increased the tube more or less than 2 mm it would effect the velcoity and horizontal range slightly, but that is all. You would have to move the target ball up or down by cm instead of mm to obtain an answere that gave a significant difference in horizontal range than in the first trial.
How long did it take you to complete this experiment?
4 HOURS
Optional additional comments and/or questions:
There was an error in your reporting of ranges and in your explanation for how ranges were determined. See that note and please clarify.
Your expressions for momentum did not include mass.
To be sure you have this step correct, before making subsequent modifications, you may insert responses into cipies of these questions, including your original answers and my responses. Please mark my responses with &&&& and denote your additional responses with ****, and submit.
Your work on conservation of momentum 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'.
1.7cm, 1.3cm
Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process:
16.5cm/s, 17.9cm/s, 17.5cm/s, 19.7cm/s, 18.8cm/s
18.1cm/s, 1.23
I took the speed of thae ball down the ramp and multiplied it by the time it took to reach the floor from the end of the ramp to the floor.
This question asked for your horizontal ranges, not your speeds. Horizontal ranges are in cm, not in cm/s.
The ranges are quantities you actually measured; the speeds are inferred. Basic principle: Report what was measured, then move to what is inferred.
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.
39.8cm/s, 43.0cm/s, 43.4cm/s, 41.6cm/s, 42.0cm/s
17.5cm/s, 20.7cm/s, 19.3cm/s, 18.5cm/s, 18.4cm/s
41.96cm/s, 1.409
18.88cm/s, 1.201
Using the distance of the markings and the timings of roll and fall.
Vertical distance fallen, time required to fall.
73cm.
.34sec.
Measured the distance and took the time in seconds to travel from rest.
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.
46.6cm/s, 55.5cm/s, 123cm/s
46.6cm/s +- .8944
55.5cm/s +- 1.20
123cm/s +- 1.41
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.
46.6cm/s
55.5cm/s
123cm/s
46.6cm/s
178.5cm/s
46.6cm/s = 178.5 cm/s
The masses (m1 or m2, depending on which ball) are not included in your expressions for momentum.
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.
46.6cm/s + 55.5cm/s = 123cm/s
102cm/s /123cm/s = 123cm/s / 123cm/s
102cm/s / 123cm/s = .83
Again masses are not included in your expressions or your equations.
The ratio m1/m2 means the difference in the mass of one object as opposed to the other being tested.
Diameters of the 2 balls; volumes of both.
2.5cm, 1.9cm
8.2g, 3.6g
How will magnitude and angle of the after-collision velocity of each ball differ if the first ball is higher?
This will cause the second ball to not go as far as it would with the centers being equal. The speed will be greater. The velcoity will go up and fly out farther. If the centers are off the second ball will shoot down or up.
Predicted effect of first ball hitting 'higher' than the second, on the horizontal range of the first ball, and on the second:
This will cause the horizontal range to be shorter, because the ball's centers are not the same.
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:
m1/m2
45.7cm/s + 56.7cm/s / 121.6cm/s = .84
What percent uncertainty in mass ratio is suggested by this result?
1% uncetainty
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 max before collision and the minimum after collision gives you the max result and the minimum before collision and the max after collision gives you the minimum result.
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.
My data seems to correlate almost identical to the preceeding experiment, with only a couple of ranges being different. I believe that by moving the tubing 2mm., this will cause the mass ratios to be closer together.
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?
.33cm, 42.3cm/s
122.9cm/s
43.972cm/s, 40.628cm/s
43.369cm/s, 40.551cm/s
2.818cm/s, 3.344cm/s
The velocity doesnt seem to be signifacatly different at all. The 2mm drop didnt seem to effect the 2hd ball velocity that much at all.
Your report comparing first-ball velocities from the two setups:
.33cm, 19.8cm/s
54.26cm/s
21.28cm/s, 18.33cm/s
20.08cm/s, 17.68cm/s
2.4cm/s, 2.95cm/s
The velocity doesnt seem to be signifacatly different at all. The 2mm drop didnt seem to effect the 2hd ball velocity that much at all.
Uncertainty in relative heights, in mm:
.5mm, when using the tubing it was hard to get a completely actual reading of its measurement, because the wieght of the ball could effect the tubing hieght and also the super glue compress when set on the table, especially with a metal ball resting on it.
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 in heights did not affect the the balls in the first setup. If you increased the tube more or less than 2 mm it would effect the velcoity and horizontal range slightly, but that is all. You would have to move the target ball up or down by cm instead of mm to obtain an answere that gave a significant difference in horizontal range than in the first trial.
How long did it take you to complete this experiment?
4 HOURS
Optional additional comments and/or questions:
There was an error in your reporting of ranges and in your explanation for how ranges were determined. See that note and please clarify.
Your expressions for momentum did not include mass.
To be sure you have this step correct, before making subsequent modifications, you may insert responses into cipies of these questions, including your original answers and my responses. Please mark my responses with &&&& and denote your additional responses with ****, and submit.