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'. **
.85cm, .85cm
.85cm
These are +-.1
** Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process: **
33.4, 34.5, 33.0, 32.9, 34.0
33.56, .6804
I got these results using a ramp that is 6.5cm from the table top initially. It then runs down this ramp, onto the short ramp, and then projects off the table. I recorded where it landed in relation to a point straight down from 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. **
45.5, 49.0, 50.2, 46.0, 48.5
24.5, 21.9, 22.9, 23.9, 20.3
47.84, 2.013
22.7, 1.667
I found these positions with respect to the mark I found by dropping the ball directly down from the table top.
** Vertical distance fallen, time required to fall. **
77cm
0.4sec
I did this by first finding the vf using the equation vs^2=vo^2+2(77m)(980cm/s/s). I assumed the initial vertical velocity was 0cm/s and found vf to be 388.5cm/s. Then i used the eqn vAve*dt=ds and solved for dt using a vAve of 194.24cm/s.
** 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. **
83.9cm/sec, 56.75cm/sec, 0cm/sec, 119.6cm/s
82.2, 85.6
52.6, 60.9
114.6, 124.6
** 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=m1(83.9cm/s)
p=m1(56.7cm/s)
p=m2(119.6cm/s)
p=m1(83.9cm/s)+m2(0cm/s)
p=m1(56.7cm/s)+m2(119.6cm/s)
m1(83.9cm/s)=m1(56.7cm/s)+m2(119.6cm/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(83.9cm/s)-m1(56.7cm/s)=m2(119.6cm/s)
m1=(m2(119.6cm/s))/27.2cm/s
m1/m2=119.6cm/s/(27.2cm/s)
m1/m2=4.4
m1 divided by m2 is 4.4, therefore m1 is 4.4 times m2.
** Diameters of the 2 balls; volumes of both. **
2cm, 1cm
4.18cm^3, .52cm^3
** How will magnitude and angle of the after-collision velocity of each ball differ if the first ball is higher? **
The speed of the first ball would be greater than if the balls were at the same height. This is because, not all of the momentum would be transmitted, more would stay with the first ball. The speed of the second ball will be smaller. The direction should be the same.
** 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 would be larger then when they hit directly and the horizontal range of the second will be smaller.
** 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=5.38
I determined this by using the equation m1v2+m2v2=m1vf1+m2vf2. I used the min and max velocity when stated and solved for m1/m2.
** What percent uncertainty in mass ratio is suggested by this result? **
There must be a large percent uncertainty, since the min and maxes gave an answer that is 22% larger then the mean answer.
** 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. **
average vo of m1=34.1cm/s
mean range for 1: 30.9
mean range for 2: 43.17
average vo of m2=0cm/s
average vf of m1=77.25cm/s
average vf of m2= 107.9cm/s
Based off of 5 trials with a tee 2mm higher then the original set up.
dt was .4sec.
m1/m2=13.5
** 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? **
77cm, 43.17cm, .01
108.0 cm/s
103.8 cm/s, 112.2 cm/s
114.6, 124.6
11.6cm/s
There appears to be a difference.
** Your report comparing first-ball velocities from the two setups: **
77cm, 30.9cm, .01
77.25cm/s
70.0, 84.7
52.6, 60.9
20.6cm/s
There is obviously a difference here.
** Uncertainty in relative heights, in mm: **
.5mm
I did the set up as directed, i was very precise with my cutting of the straw, but there is still room for error since the dot may not be totally accurate, and the balls may have hit oddly causing a less then optimal location for the spot.
** 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. **
This is a significant factor. As we saw above, raising the ball 2mm caused extreme changes in the velocities. This means that the results are most likely sensitive to even slight variations in the height of the tee.
** How long did it take you to complete this experiment? **
** Optional additional comments and/or questions: **
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2 hours
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Excellent work with this challenging experiment.