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'. **
2.3, 2.3
1.7
+-.025
** Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process: **
24.0, 24.0, 24.0, 24.0, 24.0,
24.0, 0
I first established the edge of the table to be point zero by dropping the ball several times vertically from the table’s edge, each of the points lined up so I drew an axis beginning here. To measure horizontal ranges, I released the stationary ball from the top of the ramp, after ensuring it would run smoothly, and marked with carbon paper the drop positions. The all matched perfectly, and so I was able to easily measure distance from the origin 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. **
32.5, 33.1, 33.7, 34.1, 34.7
17.0, 17.8, 19.3, 19.7, 19.7
33.62, .8556
18.7, 1.231
I used the cleanest of five trials (the tee often seems to interfere as the large ball hits it on the way off the ramp) and measured from the specified origin the ranges of the small target ball and the large ball after collision
** Vertical distance fallen, time required to fall. **
44cm
.300s
I measured distance from the floor to the top of the table to be .44m, and from this and g=9.8m/s, I set PE (top) = mgh = E and KE (floor) = .5mv^2 = E and solved for final velocity to be 2.936m/s. I solved for average velocity assuming initial vertical velocity was zero by halving final velocity. To get time I solved for t=h/Vave
** 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. **
.8m/s, .623m/s, 1.121m/s
.8 +- 0m/s
.623 +- .041m/s
1.121+- .029m/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. **
p=m1(.8 +- 0m/s)
p=m1(.623 +- .041m/s)
p=m2(1.121+- .029m/s)
p(tot)=m1(.8 +- 0m/s) + m2(0m/s)
p(tot)= m1(.623 +- .041m/s) + m2(1.121+- .029m/s)
m1(.8 +- 0m/s) = m1(.623 +- .041m/s) + m2(1.121+- .029m/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(.8m/s) - m1(.623m/s) = m2(1.121m/s)
m1[(.8m/s)- (.623m/s)] m2(1.121m/s)
(m1/m2) [(.8m/s)- (.623m/s)]= (1.121m/s)
(m1/m2)=6.333
The ratio m1/m2 is the ratio of masses of ball 1 and ball2
** Diameters of the 2 balls; volumes of both. **
2.5cm, 1.3cm
8.181cm^3, 1.150cm^3
** How will magnitude and angle of the after-collision velocity of each ball differ if the first ball is higher? **
If the center of the first ball is higher at collision, upon collision it will direct the second ball at an angle below horizontal and initial downward velocity will not be zero. The horizontal component will also be less than if both balls were even because the newly introduced vertical component of velocity will detract from it, given that net velocity should still be the same. If the second ball is higher at collision, then it will be directed slightly upwards and initial vertical velocity will again not be zero. This will actually increase the horizontal range of the second ball.
** Predicted effect of first ball hitting 'higher' than the second, on the horizontal range of the first ball, and on the second: **
When the first ball is higher, its horizontal range will be longer and the second ball’s will be shorter. If the second ball is higher, its horizontal range will be longer and the first ball’s 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=8.029
ratio was found just as before, by equating before and after total momentums and solving for m1/m2
** What percent uncertainty in mass ratio is suggested by this result? **
26.78%
** What combination of before-and after-collision velocities gives you the maximum, and what combination gives you the minimum result for the mass ratio? **
Low pre-collision m1 and high post-collision m1 and m2 give maximum mass ratio
High pre-collision m1 and low post-collision m1 and m2 give minimum 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? **
m1/m2 = u2 / (v1- u1)
** Derivative of expression for m1/m2 with respect to v1. **
d(m1/m2)/dv1 = (-u2) / (v1-u1)^2
d(m1/m2)/dv1 = -35.782(m/s)^-1
this is the rate of change of the mass ratio in regards to the change in v1. If we know the error in a v1 measurement, we can use this equation to estimate the mass ratio
** 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? **
0
0
I did not have a standard of deviation to worry with, as all v1 trials were identical. There is therefore no range of mass ratio values in relation to a range of v1 values
** Complete summary and comparison with previous results, with second ball 2 mm lower than before. **
For the large ball, there was a mean range of 10.68 cm with standard deviation of .1304, and out of this a mean velocity of u1=.356m/s. The small ball had a mean range of 29.94cm and standard deviation of .6025, and it had a mean velocity of .998m/s. these values produced a mass ratio of m1/m2=2.248, as compared to the previous mass ratio m1/m2=8.029
** 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? **
44cm, 29.94cm, .106
104.7cm/s
106.9cm/s, 102.5cm/s
114.9cm/s, 109.2cm/s
7.4cm/s
The 2mm decrease run appears significantly different, as there is no overlap between the two velocity intervals. There would be overlap at two standard deviations though.
** Your report comparing first-ball velocities from the two setups: **
44cm, 10.68cm, .106
36.38cm/s
36.83cm/s, 35.93cm/s
58.2cm/s, 66.4cm/s
25.92cm/s
The 2mm decrease is significantly different, there is a suspicious amount of distance between the velocity intervals. Something may have gone wrong
** Uncertainty in relative heights, in mm: **
Uncertainty of .5mm, I could not judge confidently below this interval on the ruler being used, but was confident to a level below 1mm regardless.
** 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. **
Although the differences between velocities, especially for the larger ball were significant, I can imagine the system set up bearing most responcibility for error. The tee as mentioned before had a tendency of impeding the large ball on its way to the floor. Given the discrepancy in results, I can not rule out the possibility that uncertainty in relative heights was a significant factor.
** How long did it take you to complete this experiment? **
4 hours
** Optional additional comments and/or questions: **
Excellent results and as far as I can see your analysis is flawless. Good insight throughout.