Phy 201
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: **
** #$&* Distances from edge of the paper to the two marks made in adjusting the 'tee'. **
I didn't get the carbon paper in my lab kit, so it made it a little harder to correctly calculate these numbers.
2.3, 2.1
1.1
There is a level of uncertainty here with my measurements, mainly because I don't have the carbon paper to exactly tell me the height at which the ball makes contact.
** #$&* Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process: **
20.7, 20.9, 20.9, 21.0, 21.4
20.98, 0.2588
I got the horizontal differences simply by measuring where the ball struck the paper lying on the floor. And I got the mean and st. dev by plugging the distances into the data program.
** #$&* Five horizontal ranges observed for the second ball; corresponding first-ball ranges; mean and standard deviation of the second-ball ranges; mean and standard deviation of ranges for the first ball. **
28.4, 29.1, 29.4, 29.5, 30.2
13.1, 15.0, 16.2, 19.3, 19.5
29.32, .6535
16.62, 2.769
I made my measurements just as I did before. I observed the mark the two balls left on the paper below, and measured from the origin to determine the distance. The data program helped me get the mean and standard deviation.
** #$&* Vertical distance fallen, time required to fall. **
74.4
0.3560
I know that the vertical distance is pretty precise. As for the time, I'm not too sure. I used the TIMER program and took the average of 5 trials when attempting to measure the time the ball left the ramp until it struck the floor.
** #$&* Velocity of the first ball immediately before collision, the velocity of the first ball after collision and the velocity of the second ball after collision; before-collision velocities of the first ball based on (mean + standard deviation) and (mean - standard deviation) of its uninterrupted ranges; same for the first ball after collision; same for the second ball after collision. **
58.93, 46.68, 82.36
59.66, 58.21
54.46, 38.91
84.20, 80.52
** #$&* 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. **
58.93*m1 kg*m /s
46.68*m1 kg*m /s
82.36*m2 kg*m /s
(58.93*m1 + 0) kg*m
(46.68*m1 + 82.39*m2) kg*m /s
58.93 m1 = 46.68m1' + 82.39m2'
** #$&* Equation with all terms containing m1 on the left-hand side and all terms containing m2 on the right; equation rearranged so m1 appears by itself on the left-hand side; preceding the equation divided by m2; simplified equation for m1 / m2. **
12.25*m1 = 82.39*m2
m1 = (82.39*m2) / 12.25
m1/m2 = 6.73
The ratio m1/m2 is the masses of the two balls, divided by each other.
** #$&* Diameters of the 2 balls; volumes of both. **
2.4, 1.2
7.24, 0.905
** #$&* How will magnitude and angle of the after-collision velocity of each ball differ if the first ball is higher? **
If the two balls didn't collide at exactly the center, then the force won't be transfered completely to the other ball. The speed will be less than what it could be if the balls struck off-center.
** #$&* Predicted effect of first ball hitting 'higher' than the second, on the horizontal range of the first ball, and on the second: **
I think this would cause the horizontal range of the first ball to be greater, while causing the range of the second ball to be less.
** #$&* ratio of masses using minimum before-collision velocity for the first ball, maximum after-collision velocity for the first ball, minimum after-collision velocity of the second: **
m1/m2 = 21.41
I got this ratio by solving the equation m1v1 + m2v2 = m1v1' + m2v2', and using the quantities specified in the instructions.
** #$&* What percent uncertainty in mass ratio is suggested by this result? **
a 31.43% uncertainty
** #$&* What combination of before-and after-collision velocities gives you the maximum, and what combination gives you the minimum result for the mass ratio? **
To get the highest ratio use the lowest before collision velocity and highest after collision velocity for ball 1, and the highest velocity for ball 2.
To get the lowest ration use the highest before collision velocity and the lowest after collision velocity for ball 1, and the lowest velocity for ball 2.
** #$&* 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? **
m1/m2 = u2 / (v1 - u2)
** #$&* Derivative of expression for m1/m2 with respect to v1. **
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** #$&* 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? **
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** #$&* Complete summary and comparison with previous results, with second ball 2 mm lower than before. **
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** Vertical drop of the second ball, its mean horizontal range and the slope of the line segment connecting the two centers; the velocity given by the program based on mean; velocity interval for 2-mm decrease in 2d-ball height; velocity interval from the original run at equal heights; difference in the mean-based velocities; is new velocity significantly different than original? **
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** #$&* Your report comparing first-ball velocities from the two setups: **
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** #$&* Uncertainty in relative heights, in mm: **
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** #$&* 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. **
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** #$&* How long did it take you to complete this experiment? **
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** #$&* Optional additional comments and/or questions: **
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over 3 1/2 hours!!
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&#This looks very good. Let me know if you have any questions. &#