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'. **
6cm
4cm
2cm
+/- .5cm
** Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process: **
10.0,10.1,10.2,10.25,10.40
10.19
.1517
** 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. **
10.1,10.1,10.0,10.2,10.3
9.95,10.0,9.9,10.0,10.1
10.14,.1140
9.99,.07416
After colliding with the ball on the 'tee', the traveling ball would not travel nearly as far as it did in the previous trials when it did not collide with the that ball. However according to the data you report here, that is what happened.
** Vertical distance fallen, time required to fall. **
48cm
3.195s
An object falls 48 cm in well under half a second. It doesn't require over 3 seconds. You didn't explain this calculuation so I can't be sure, but I suspect you used the distance in centimeters but used the acceleration of gravity in meters/second. The units don't work out when you do that, and the result is therefore not correct; however if this is the error, you did otherwise use the correct equation and procedure.
** 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. **
4.06cm/s,4.05cm/s
4.01cm/s
4.055cm/s,4.062cm/s
4.517cm/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. **
4.06cm/s
4.05cm/s
4.00cm/s
4.06cm/s
8.05cm/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=1.006
m2=.9938
** Diameters of the 2 balls; volumes of both. **
2.5cm,2.45cm
approx.15.625cm^3,14.706cm^3
** How will magnitude and angle of the after-collision velocity of each ball differ if the first ball is higher? **
There would be less transfer of energy. The speed would be less if the centers where not equal.The ball would not travel as far horizontally.
** 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 greater for the first ball. The second ball would not travel as far horizontally.
** 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: **
1.006,.9938
** What percent uncertainty in mass ratio is suggested by this result? **
+/- .005
** 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 highest two and the lowest two.
** 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? **
u1 = (v2+u2)/v1
u2 = (v1+u1)/v2
** Derivative of expression for m1/m2 with respect to v1. **
not required beyond this point
** 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? **
** Optional additional comments and/or questions: **
A good bit of your analysis is either correct or nearly correct, but there are some errors and omissions, and some aspects of your data are hard to reconcile. See my notes and please clarify.