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'. **
the large ball hit and made a mark 2.2cm from the bottom of the initial point
large ball hit the small ball and the paper making a dot 2.2cm for initial point.
+-.01
** Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process: **
24cm,22.5cm,23.2cm,23.2cm,23.5cm
23.28cm,.5449
the end of the ramp was 5cm high. the large ball left the ramp just short of the end of the table. i measured the distance from the table to the floor. i dropped a ball and found the straight line down so i could use that point to find the horizontal distances.
** 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. **
46.5cm,46.7cm,48cm,45.6cm,45cm
22.5cm,22.8cm,23.4cm,18.3cm,16.1cm
46.36,1.146
20.62,3.234.
i placed a long piece of aluminum foil on the floor and measured the horizontal distance to each mark.
** Vertical distance fallen, time required to fall. **
73.66cm,
.38sec
i measured the distance with a metric ruler and got the time using a stop watch.
Your result is close to the time we would calculate for a 74 cm fall starting with vertical velocity zero, using the known value of the acceleration of gravity. Very good timing if you used a stop watch.
** 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. **
44.1cm/sec,54.26cm/sec,122.36cm/sec
** 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. **
2.9kgcm/sec
3.63kgcm/sec
1.04kgcm/sec
2.9kg/cm/sec
4.67kgcm/sec
p1=p2+p3
** 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*44.1cm/sec)-(m1*54.26cm/sec)=(m2*122.36cm/s)
m1(-10.16=m2*122.36cm/s
12.04cm/s
** Diameters of the 2 balls; volumes of both. **
2.5cm,1cm
4.59cm^3,.293cm^3
** How will magnitude and angle of the after-collision velocity of each ball differ if the first ball is higher? **
the magnitude and the speed will be less and the velocity will remain almost constant.
magnitude will be less. the direction will almost be the same and the speed will be less if the centers are not aligned
** Predicted effect of first ball hitting 'higher' than the second, on the horizontal range of the first ball, and on the second: **
less,less
** 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: **
9.33kgcm/sec
set the momentum equation before impact equal to the sum of the max large ball momentum equation and add the small ball momentum and solve.
** What percent uncertainty in mass ratio is suggested by this result? **
24%
** 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. **
** 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: **
&#Good responses. Let me know if you have questions. &#