Phy 231
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: **
** #$&* Distances from edge of the paper to the two marks made in adjusting the 'tee'. **
4.5 cm, 4.5 cm
1cm
There is always room for error since i had to use my own eyes and a ruler to make measurements. For the first set of numbers I was not entirely sure what i was exactly supposed to measure.
** #$&* Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process: **
19.5, 19, 19.6, 19.2, 19.8
19.42,.3194
These values are measurements of the distance between where the impact was and the edge of the table.
** #$&* Five horizontal ranges observed for the second ball; corresponding first-ball ranges; mean and standard deviation of the second-ball ranges; mean and standard deviation of ranges for the first ball. **
15.4, 16.7, 17.4, 17.4, 17.3
26.4, 23.5, 24, 22.4, 21
The first line would probably be the after-collision ranges of the first ball; the second line the ranges of the second ball.
23.46, 2.007
16.94, .8562
** #$&* Vertical distance fallen, time required to fall. **
71 cm
0.37 s
ds is the height of the table and i used the equation ds = v0 * dt + 1/2 * a * dt ^2 to find dt.
** #$&* Velocity of the first ball immediately before collision, the velocity of the first ball after collision and the velocity of the second ball after collision; before-collision velocities of the first ball based on (mean + standard deviation) and (mean - standard deviation) of its uninterrupted ranges; same for the first ball after collision; same for the second ball after collision. **
70.423, 56.821, 32.52
These velocities seem a little off; I need more detail on the specific quantities you used to calculate each.
The last one seems to be off more than the others. The second ball travels furthest, so would have the greatest velocity.
69.563, 71.284
55.964, 57.678
30.54, 34.53
I had to use the acceleration from before and then used the change in velocity to get the initial velocities of the balls.
** #$&* 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. **
m1* 70.423
m1* 56.821
m2*32.52
m1* 70.423+m2*0
m1* 56.821+m2*32.52
m1* 70.423=m1* 56.821+m2*32.52
Right idea, but I don't think you used correct velocities.
** #$&* Equation with all terms containing m1 on the left-hand side and all terms containing m2 on the right; equation rearranged so m1 appears by itself on the left-hand side; preceding the equation divided by m2; simplified equation for m1 / m2. **
m1* 70.423 - m1* 56.821 = m2*32.52
m1 = (m2*32.52)/(70.423 - 56.821)
m1/m2 = 32.52 / 13.9
= 2.34
Since it is a ratio, it is showing that m1 is larger than m2 by 2.34 If you look at it as m1/n2 = 2.34/1 then yous ee that m2 = 1 and m1 = 2.34,
** #$&* Diameters of the 2 balls; volumes of both. **
2.5, 1
8.18, 0.52
** #$&* How will magnitude and angle of the after-collision velocity of each ball differ if the first ball is higher? **
The first ball will lose momentum and the second one will gain some but not as much. The second ball will have a smaller velocity and the first ball will have a larger velocity. The range will be shorter.
** #$&* 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 and the horizontal range of the second ball will be shorter.
** #$&* ratio of masses using minimum before-collision velocity for the first ball, maximum after-collision velocity for the first ball, minimum after-collision velocity of the second: **
m1/m2 = 2.57
I just had to put the new (max/min) values in.
** #$&* What percent uncertainty in mass ratio is suggested by this result? **
7.5%
** #$&* What combination of before-and after-collision velocities gives you the maximum, and what combination gives you the minimum result for the mass ratio? **
The maximum can be found by min/(max-min) and the minimum can be found by max /(min-max)
** #$&* 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*v1 = m1*u1 + m2*u2
which can later be made as:
m1/m2 = u2/(u1-v1)
** #$&* Derivative of expression for m1/m2 with respect to v1. **
-u2 / (v1-u1)^2
-0.1757
** #$&* 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? **
The predicted value of v1 changes by multiplying the uncertainty in v1 by the derivative found in the problem before.
+- 0.45
** #$&* Complete summary and comparison with previous results, with second ball 2 mm lower than before. **
For the second ball, the average distance is 38 cm and for the first ball it is 24 cm.
The first balls velocity before collision comes out to be 91 cm/sec and after collision it is 76.35 cm/sec. For the second ball, the veloicty is 80.42 cm/sec.
The velocity of the second ball should be greater than the after-collision of the first ball.
My prediction for mass ratio is 3.59.
** Vertical drop of the second ball, its mean horizontal range and the slope of the line segment connecting the two centers; the velocity given by the program based on mean; velocity interval for 2-mm decrease in 2d-ball height; velocity interval from the original run at equal heights; difference in the mean-based velocities; is new velocity significantly different than original? **
71, 38, .038
109.1
106.5, 111.7
61.15, 65.15
36
It looks different
** #$&* Your report comparing first-ball velocities from the two setups: **
It increase from 45 cm/sec to 63 cm/sec.
** #$&* Uncertainty in relative heights, in mm: **
0.1 mm
I found this value by measuring with a ruler but i had to be extremely precise since the value was so small.
** #$&* 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. **
Both uncertainties were about the same, so it probably didnt change many values in the experiment overall.
** #$&* How long did it take you to complete this experiment? **
1.84 cm/sec.
** #$&* Optional additional comments and/or questions: **
It would be -1.84 cm/sec so that the momentum would end up being 0 cm/sec.
** **
70.42 cm/sec
32.5 cm/sec
52 cm/sec
** **
19.8 cm
** **
It would not necessarily affect the ball because the spin is not long enough and strong enough.
** **
0.093
98.63
the slope was 0.093, the height was 71, the range was 38.
** **
3 1/2 hours
** **
Your data and most of your work look good, but those velocities don't seem to follow from your data.
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