Phy201
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: **
David,
On my set up I am using 1 cm dia steel ball and a 2 cm steel ball. Every time I check my set up by a trial for the larger ball rolling down the ramp, the larger ball hits the straw. When I adjust the straw down for the ball to clear the straw, the small ball will not be hit in the center of the larger ball. My exit ramp is 15 cm long my acceleration ramp is 29 cm long. The height of my acceleration ramp is 8.2 cm above my exit ramp. The top on my adjustment straw is 3 cm which is the same height of the exit ramp. At this height the large ball will clear the straw but the larger ball is hitting above center of the small ball. I am going to perform this lab. I know my data will probably be inaccurate to some extent but I don't have much choice because of the time left to complete my studies for this class.
** Distances from edge of the paper to the two marks made in adjusting the 'tee'. **
3 trials-4 cm above my base of ramp
tee top is only 3 cm because the larger ball continues to hit the tee if adjusted to 4 cm
I used a scale to take these measurements and I know my error will be compounded greatly
** Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process: **
29.4, 29.5, 29.7, 29.8, 29.9
The mean = 29.66, standard deviation = .2074
The ball was place on the end of an acceleration ramp that was 8.2 cm high and 29 cm long. The exit ramp is 3 cm high on both ends and 15 cm long. All height measurements are taken from a board 61 cm long that served as the base platform. The total height from the exit ramp to the floor is 76 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. **
second small ball: 41.2cm, 41.3cm, 41.6cm, 42.3cm, 42.5cm
first large ball: 23.5cm, 23.5cm, 23.8cm, 24.1cm, 24.2cm
The mean is 41.78, the standard deviation is .5891
The mean is 23.82, the standard deviation is .3271
The first ball was place on the end of an acceleration ramp that was 8.2 cm high and 29 cm long. The exit ramp is 3 cm high on both ends and 15 cm long. All height measurements are taken from a board 61 cm long that served as the base platform. The total height from the exit ramp to the floor is 76 cm. The second ball, half the size of the first ball was placed on a tee 1.3 cm from the exit end of the ramp. The first ball was released from the top of the accel ramp then struck the second ball will both ball propelling and striking the floor. The first ball landed shorter from its original striking position on the floor due to transferring some its energy to the second ball
** Vertical distance fallen, time required to fall. **
Distance is 76 cm
Time is .375 sec
The vertical distance measured with a Stanley 5 meter tape and the time was measured with the data program taking several times and getting an average.
I assumed that I was holding the ball at the same height as the exit ramp height. Also the standing falling time would be identical to the horizontal and downward flight time would be identical.
** 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 v1 = 84 cm/s, first ball v2 = 22.75 cm/s, second ball = 79.8 cm/s
first ball vel mean + = 57.5 cm/s, vel mean - = 56.25 cm/s
first ball after collision vel mean + = 23.06 cm/.s, vel mean - = 21.87 cm/s
second ball after collision vel mean + = 40.47 cm/s, vel mean - = 39.34 cm/s
The first ball was reported to travel about 23 cm after collision, which would give it an after-collision velocity of about 60 cm/s, not 22.75 cm/s. The other velocities you report appeared to be consistent with the reported data, and with what would be expected for the two balls you used.
** 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. **
p = m * v, conv of p = m1v1 * m2v2 = m1v1' * m2v2'
p1 = 57.5 cm/s * m1
p1 = 22.75 cm/s * m1
p2 = 41.78 cm/s * m2
Above I believe you reported about 79 cm/s as the after-collision velocity of the 'target' ball.
p before = (57.5 cm/s * m1) + 0
p after = (22.75 cm/s * m1) + (41.78 cm/s * m2)
57.5 cm/s * m1 = (22.75 cm/s * m1) + (41.78 cm/s * m2)
** 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. **
57.5 cm/s * m1 = (22.75 cm/s * m1) + (41.78 cm/s * m2)
57.5 cm/s *m1 - 22.75 cm/s * m1 = (41.78 cm/s * m2)]
34.75 cm/s * m1 = 41.78 cm/s * m2, m1 = (41.78 cm/s * m2) / 34.75 cm/s
m1 = 1.2 * m2
m1 / m2 = 1.2 / 1
according to the ratio the larger ball is 1.2 times the mass of the smaller ball
** Diameters of the 2 balls; volumes of both. **
ball 1 dia = 2.6 cm, ball 2 dia = 1 cm
ball 1 vol = 8.18 cm^3, ball 2 vol - .524 cm^3
** How will magnitude and angle of the after-collision velocity of each ball differ if the first ball is higher? **
First ball collision is higher this will cause the first ball to pop up and not lose the velocity it should if the balls were to hit exactly at their centers. This would result in the first ball going farther and the second ball going less travel distance. The speeds will be at max for the smaller ball if the centers are hit dead center and at the min. for the larger ball. The smaller ball if hit on the upper side of center will not travel as far a distance as if hit in its center.
** 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 of the first ball will be greater if it hits the second ball above center because it will not transfer its momentum.
The second ball's horizontal range will be affected also because it will not receive the greatest momentum possible if hit above the center point.
** 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: **
56.25 cm/s * m1 + 0 = 23.06 cm/s * m1 + 39.34 m2
33.19 m1 = 39.34 m2
m1 / m2 = 39.35 cm/s / 33.19 cm/s
m1/m2 = 1.19 / 1
** What percent uncertainty in mass ratio is suggested by this result? **
1.19 as compared to 1.20
The percent of accuracy appears to be 1%
** 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? **
This lab took me from 10:30 am til 2:30 pm on this date for a total of 4 hours to just the General Physics portion.
** Optional additional comments and/or questions: **
I'm choosing to stop at the earlier point you informed me I could.
You have good data and most of your analysis is good, but there appear to be some errors and inconsistencies associated with the after-collision velocities. See my notes and be sure to let me know if you have any questions.
From your data it would appear that the mass ratio should be about 7 to 1, not 1.2 to 1. If the apparent 'bad numbers' in your analysis are corrected I believe you will get a result close to the ideal.
Ideally if one ball has doubled the diameter of the other and both are made of the same material, the actual ratio would be 8 to 1. As you note that the beginning that were some minor problems with the setup, but they did not appear to have had much influence.
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