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: **
** #$&* Distances from edge of the paper to the two marks made in adjusting the 'tee'. **
1.56, 1.56
1cm
.01mm
** #$&* Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process: **
2.8, 3.4, 3.4, 3.4, 3.3
3.26,.2608
I obtained these result by allowing the ball to run down the inclines and fall vertically from the edge of the table to a piece of carbon paper on the floor. I took measurement from the edge of the carbon paper, the same paper I used for the first step. All measurements are in cm.
** #$&* Five horizontal ranges observed for the second ball; corresponding first-ball ranges; mean and standard deviation of the second-ball ranges; mean and standard deviation of ranges for the first ball. **
.55,.54,.54, .56, .55
.53, .55, .54, .56, .55
.55, .008
.55, .011
I made the measurement above by halfing the carbon paper and placing one piece were the small ball dropped to the floor and one where the big ball dropped to the floor, I htne measured those distances from the edge of the paper for each and used the data program to find mean and standard deviation, all measurements are in cm.
** #$&* Vertical distance fallen, time required to fall. **
60cm
1.36s
It only takes about .5 seconds to fall 60 cm from rest.
The distance wa measured from the edge of the ramp, the time was taken using the timer program of the ball at rest until it hit the floor.
** #$&* Velocity of the first ball immediately before collision, the velocity of the first ball after collision and the velocity of the second ball after collision; before-collision velocities of the first ball based on (mean + standard deviation) and (mean - standard deviation) of its uninterrupted ranges; same for the first ball after collision; same for the second ball after collision. **
#$&*#$&*
55.29m/s,32.8m/s, 32.8m/s
These are not unreasonable velocities, but it's unclear how they follow from your data.
You report a horizontal displacement of a few centimeters, and a time of fall of over 1 second. This would give you a horizontal velocity of only a few centimeters / second. It's not clear how you ended up with horizontal velocities between 30 and 60 cm/s, based on your previously reported information.
55.3m/s, 55.27m/s
33.06m/s, 33.04m/s
32.81m/s, 32.7m/s
** #$&* 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. **
m155.3cm/s
m133.06cm/s
m232.81cm/s
m155.3cm/s +m2
m1 33.06cm/s + m2 32.81cm/s
m1 55.3cm/s + m2 = m1 33.06cm/s + m2 32.81cm/s
The initial velocity of the second mass is zero, so the equation would be
m1 55.3cm/s + 0 * m2 = m1 33.06cm/s + m2 32.81cm/s, or just
m1 55.3cm/s = m1 33.06cm/s + m2 32.81cm/s
** #$&* Equation with all terms containing m1 on the left-hand side and all terms containing m2 on the right; equation rearranged so m1 appears by itself on the left-hand side; preceding the equation divided by m2; simplified equation for m1 / m2. **
m155.3cm/s + m1 33.06cm/s = m2 + m2 32.81cm/s
m1= m2+m232.81cm/s/88.36cm/s
m1/m2=.004cm/s
This does not follow from the preceding line.
m1/m2=.004cm/s
The is the ratio of m1/m2, It is the ratio the masses.
If your equation is
m155.3cm/s + m1 33.06cm/s = m2 32.81cm/s
then you get
m1= m2 * 32.81cm/s/88.36cm/s.
Divide both sides by m2 to get the ratio m2 / m1.
What do you get?
** #$&* Diameters of the 2 balls; volumes of both. **
6.4cm, 4.4cm
13.4cm^3, 9.22cm^3
Your volume calculations don't follow from your reported diameters.
** #$&* How will magnitude and angle of the after-collision velocity of each ball differ if the first ball is higher? **
The magnitude and direction of the velocity of the first ball immediately after collision would be much higher than that of the second ball. The speed would be greater than if the center are at the same height. Yes the direction would differ, the second smaller ball would have a more vertical incline that if it hit at the center.
** #$&* 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 would be effected some but not as much as the smaller ball, the smaller balls horizontal range would be decreased.
** #$&* ratio of masses using minimum before-collision velocity for the first ball, maximum after-collision velocity for the first ball, minimum after-collision velocity of the second: **
m1/m2=.370
I determined this ratio using the calculation as before, but with the minimum before collision of the first ball, max after collision of the first ball, and minimum after collision velociy of the second ball.
** #$&* What percent uncertainty in mass ratio is suggested by this result? **
.370/.004 * 100 = 92.5 percent
** #$&* What combination of before-and after-collision velocities gives you the maximum, and what combination gives you the minimum result for the mass ratio? **
The before collision gives you the maximum the aftr collision gives you the minimum result for mass ratio.
** #$&* 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? **
u2/(v1+u1)= mass ratio
** #$&* Derivative of expression for m1/m2 with respect to v1. **
** #$&* 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? **
** #$&* Complete summary and comparison with previous results, with second ball 2 mm lower than before. **
Ball one had a horizontal range after collision of 3.12 +-.1483 at a max velocity of 78.19cm/s, ball two had a horizontal range of 3.14+-.5029 after collision at a max velocity of 78.45cm/s
The ratio of masses was caculated to be m1/m2=1.69
The previous mass ratio was .004, much less than the second, but I think this may be due to the extra millimeter and distance and the fact that I used a new piece of carbon paper(the old was just to used to differentiate the data).
** Vertical drop of the second ball, its mean horizontal range and the slope of the line segment connecting the two centers; the velocity given by the program based on mean; velocity interval for 2-mm decrease in 2d-ball height; velocity interval from the original run at equal heights; difference in the mean-based velocities; is new velocity significantly different than original? **
60cm, 3.64, 2
17.74cm/s
78.95cm/s, 78.44
32.81cm/s, 32.8cm/s
46.14cm/s
I would say that the velocity obtained from the 2mm increase is significantly different that the results obtained from the centers at the same level.
** #$&* Your report comparing first-ball velocities from the two setups: **
55.29cm/s- 78.95cm/s= 23.66cm/s difference.
** #$&* Uncertainty in relative heights, in mm: **
My uncertainty was about .01, the dots almost overlapped each other.
** #$&* 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. **
In the first setup the uncertainty in the relative heights of the balls I would think was a significant factor, unless the calculations in the second set-up are really off, just 2mm made a huge difference, so I argue against this hypothesis.
** #$&* How long did it take you to complete this experiment? **
** #$&* Optional additional comments and/or questions: **
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about 4 hours,
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