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.
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This is a RESUBMIT of Conservation of Momentum. The ranges I previously reported were right, but I placed the incorrect measurement out from them. I had the ranges in cm/s instead of cm.
Distances from edge of the paper to the two marks made in adjusting the 'tee'.
Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process:
16.5cm, 17.9cm, 17.5cm, 19.7cm, 18.8cm/s
18.1cm, 1.23
I took the speed of the object horizonatlly anf multiplied it by the time of the fall.
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, 43.0cm, 43.4cm, 41.6cm, 42.0cm
17.5cm, 20.7cm, 19.3cm, 18.5cm, 18.4cm
41.96cm, 1.409
18.88cm, 1.201
I took the speed of the object horizonatlly anf multiplied it by the time of the fall.
Vertical distance fallen, time required to fall.
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.
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 * m1
55.5cm/s * m1
123cm/s * m2
46.6cm/s * m1
178.5cm/s
One momentum is a multiple of m1, the other a multiple of m2. The two momenta are therefore not like terms and you can't just add the coefficients. Also m1 and m2 don't disappear from the expression.
The total momentum after collision would be 55.5 cm/s * m1 + 123 cm/s * m2.
I also note that it doesn't seem possible that the first ball should be moving faster when it doesn't collide with the second than when it does. Did the first ball really travel further after colliding with the second than it did when its motion was uninterrupted? If so, were the centers of the two balls at the same height?
46.6cm/s = 178.5 cm/s
The equation will involve m1 and m2. See my previous note.
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*m1 + 55.5cm/s*m1 = 123cm/s
102cm/s*m1 /123cm/s*m2 = 123cm/s / 123cm/s
102cm/s*m1 / 123cm/s*m2 = .83
Diameters of the 2 balls; volumes of both.
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?
Optional additional comments and/or questions:
This is a RESUBMIT.
See my notes and please submit a revised solution for this problem; you are of course also welcome to ask questions. denote any question, revisions and your answers to questions with multiple asterisks (e.g., ******).