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'.
2.1, 2.1
1.7
1.9
Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process:
27.5, 28.0, 28.5, 28.4, 27.8
28.04, .4159
I started the ball from the top of the incline, it rolled onto the horizontal ramp, projected off the table(height of 78cm) and landed in these positions that were measured horizontally from the edge of the table.
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.
36.7, 37.0, 37.5, 37.0, 36.3
14.1, 14.0, 13.4, 14.8, 14.1
36.9, .4416
14.08, .4969
I consider the first ball to be the larger one-the first one in motion, and the second to be the smaller one, the second one to move.I started the large ball from the top of the incline, it collided with the smaller ball at the end of the horizontal ramp, both were projected off the table(height of 78cm) and landed in these positions that were measured horizontally from the edge of the table.
Vertical distance fallen, time required to fall.
78cm
.52 s
I converted te 78 cm to .78m. I then established the vf using the fourth equation of motion. From that I calculated the vAve, divided it into the
`ds to obtain the time interval.
The time of fall for a 78 cm drop will be closer to .4 seconds than .52 sec. The velocities you report are therefore a little low; however the resulting error is uniform over all calculations and won't affect your final conclusions regarding relative masses.
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.
.70,.27, .71
your reported horizontal ranges are significantly different; they won't give you velocities as close as .70 m/s and .71 m/s.
.71, .69
.28, .26
.71, .70
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 * .7m/s
M1 * .27m/s
M2 * .71m/s
M1 * .7m/s * m2
No term in these equations would contain the product of the two masses.
Total momentum before is m1 * .7m/s, not M1 * .7m/s * m2
(M1 * .27m/s) + (M2 * .71m/s)
(M1 * .7m/s )* m2 = (m1 * .27m/s )* (m2 * .71m/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 * .7m/s)/ (m1 * .27m/s) = (m2 * .71m/d) / m2
m1 = ??? I'm lost now.
The correct equation for your reported velocities would be
(m1 * .7m/s ) = (m1 * .27m/s ) + (m2 * .71m/s) rather than
(M1 * .7m/s )* m2 = (m1 * .27m/s )* (m2 * .71m/s).
Diameters of the 2 balls; volumes of both.
Did you measure the diameters of the two spheres?
How will magnitude and angle of the after-collision velocity of each ball differ if the first ball is higher?
Predicted effect of first ball hitting 'higher' than the second, on the horizontal range of the first ball, and on the second:
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:
What percent uncertainty in mass ratio is suggested by this result?
What combination of before-and after-collision velocities gives you the maximum, and what combination gives you the minimum result for the mass 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?
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.
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?
Your report comparing first-ball velocities from the two setups:
Uncertainty in relative heights, in mm:
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.
How long did it take you to complete this experiment?
3 hours 31 minutes
Optional additional comments and/or questions:
See my notes and see if you can modify accordingy the equation for momentum conservation, and solve it for the mass ratio.
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'.
2.1, 2.1
1.7
1.9
Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process:
27.5, 28.0, 28.5, 28.4, 27.8
28.04, .4159
I started the ball from the top of the incline, it rolled onto the horizontal ramp, projected off the table(height of 78cm) and landed in these positions that were measured horizontally from the edge of the table.
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.
36.7, 37.0, 37.5, 37.0, 36.3
14.1, 14.0, 13.4, 14.8, 14.1
36.9, .4416
14.08, .4969
I consider the first ball to be the larger one-the first one in motion, and the second to be the smaller one, the second one to move.I started the large ball from the top of the incline, it collided with the smaller ball at the end of the horizontal ramp, both were projected off the table(height of 78cm) and landed in these positions that were measured horizontally from the edge of the table.
Vertical distance fallen, time required to fall.
78cm
.52 s
I converted te 78 cm to .78m. I then established the vf using the fourth equation of motion. From that I calculated the vAve, divided it into the
`ds to obtain the time interval.
The time of fall for a 78 cm drop will be closer to .4 seconds than .52 sec. The velocities you report are therefore a little low; however the resulting error is uniform over all calculations and won't affect your final conclusions regarding relative masses.
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.
.70,.27, .71
your reported horizontal ranges are significantly different; they won't give you velocities as close as .70 m/s and .71 m/s.
.71, .69
.28, .26
.71, .70
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 * .7m/s
M1 * .27m/s
M2 * .71m/s
M1 * .7m/s * m2
No term in these equations would contain the product of the two masses.
Total momentum before is m1 * .7m/s, not M1 * .7m/s * m2
(M1 * .27m/s) + (M2 * .71m/s)
(M1 * .7m/s )* m2 = (m1 * .27m/s )* (m2 * .71m/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 * .7m/s)/ (m1 * .27m/s) = (m2 * .71m/d) / m2
m1 = ??? I'm lost now.
The correct equation for your reported velocities would be
(m1 * .7m/s ) = (m1 * .27m/s ) + (m2 * .71m/s) rather than
(M1 * .7m/s )* m2 = (m1 * .27m/s )* (m2 * .71m/s).
Diameters of the 2 balls; volumes of both.
Did you measure the diameters of the two spheres?
How will magnitude and angle of the after-collision velocity of each ball differ if the first ball is higher?
Predicted effect of first ball hitting 'higher' than the second, on the horizontal range of the first ball, and on the second:
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:
What percent uncertainty in mass ratio is suggested by this result?
What combination of before-and after-collision velocities gives you the maximum, and what combination gives you the minimum result for the mass 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?
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.
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?
Your report comparing first-ball velocities from the two setups:
Uncertainty in relative heights, in mm:
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.
How long did it take you to complete this experiment?
3 hours 31 minutes
Optional additional comments and/or questions:
See my notes and see if you can modify accordingy the equation for momentum conservation, and solve it for the mass ratio.