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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Distances from edge of the paper to the two marks made in adjusting the 'tee'.
.3,.3
1.1
.1
Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process:
17.4,18.5,18.4,18,17.6
17.98,.4817
Ball started at top of sloped ramp and rolled to edge of table where it dropped to paper. From the edge of paper I measured the mark of 5 drops in cm.
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.
20.4,19.5,19.1,16.9,16.6
15.8,14.5,14,11.5,11.2
18.5,1.669
13.4,1.986
After collison the two balls fell to the floor where I measured the impact from the edge of the paper to the marks left on the paper from the impact. The heavier ball left a darker mark.
Vertical distance fallen, time required to fall.
82.3cm
.41s
Measured top of table to the floor.
ds of 82.3 = vo + .5(980cm/s^2)dt^2 = 82.3/490 = sqrt.168 = .41s
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.
43.85,32.71.45.12
44.33,43.37
34.7,30.72
46.79,43.45
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.
P1 = m1kg * .4385m/s
P1 = m1kg * .3271
P2 = m2kg * .4512m/s
P = mkg * .4385m/s
P = m1 * .3271m/s + m2 * .4385m/s
m1*.4385 = m1 * .3271 + m2 * .45.12
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*.4385 + m1 * .3271 = m2 * .4512
2m1 = m2 *.4512 / .766
The equation
m1*.4385 + m1 * .3271 = m2 * .4512
does not follow from
m1*.4385 = m1 * .3271 + m2 * .4512.
The correct equation that follows from your original equation is
m1*.4385 - m1 * .3271 = m2 * .4512.
This will modify your result for the ratio.
To be sure you have this step correct, before making subsequent modifications, you may insert responses into a copy of this question, including your original answer and my response. Please mark my responses with &&&& and denote your additional responses with ****, and submit.
m1/m2 = .589/2
m1/m2 = .295
This is the difference in momentum between the larger ball and the smaller ball.
Diameters of the 2 balls; volumes of both.
5,2cm
65.43cm^3, 4.19cm^3
How will magnitude and angle of the after-collision velocity of each ball differ if the first ball is higher?
The first ball will have greater velocity after collision because it will not be head on. Depending on where the first ball hits the second ball will tell the dirction after collison. The second ball will not have as great a velocity because it will not have the full impact of the first ball.
Predicted effect of first ball hitting 'higher' than the second, on the horizontal range of the first ball, and on the second:
The horizontal range may be greater, depending on where the first ball hits the second.
The second ball will not have as great a horizontal range because it will not have as great a velocity.
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:
.35
I did a ration of 43.37 over 34.7 and equal to 43.45 over X. I crossed multiplied and divided to get x equals .35.
What percent uncertainty in mass ratio is suggested by this result?
They are very close. .05 difference
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
Optional additional comments and/or questions:
Your data look good, but there is an incorrect step in your analysis of the m1/m2 ratio. Please see my note and respond as requested.