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'. **
2.2cm, 2.13cm
1.4cm
Measurments should be accurate to +-.07cm
** Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process: **
25.7cm, 25.8cm, 25.9cm, 26cm, 26.1cm
25.9cm, .1581
First I secured the paper to the floor then droped the ball from the edge of the table five times. Then I let the ball go down the ramp five times. I drew a straight line through the points on the paper and picked an origin before the five drop marks from the edge of the table. I measured the drop marks from the edge of the table to be 1cm, 1.2cm, 1.5cm, 1.5cm, 3.2cm which had a mean of 1.7cm. I then measured the five trials off the ramp and got 25cm, 25.1cm, 25.2cm, 25.3cm, 25.4cm and subtracted the first drop point from them and added 2.4cm for the length from the end of the ramp to the end of the table to get the ranges then took the mean and standard deviations of the ranges.
** 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. **
48.6cm, 49.5cm, 49.9cm, 50.3cm, 50.8cm
25.1cm, 25.2cm, 25.3cm, 25.4cm, 25.5cm
49.82cm, .8349
25.3cm, .1581
I measured the distances using the same method as the last box dropping the ball from the edge of the table and taking all my measurments from that spot.
You should have obtained the ranges of the first ball after collision. The first-ball ranges you report do not appear to differ significantly from the ranges of the uninterrupted first ball.
** Vertical distance fallen, time required to fall. **
75.6cm
.4s
`dt=sqrt(`ds/(.5a))= sqrt(.756m/(.5*9.8m/s^2))= sqrt(.756m/4.9m/s^2)= sqrt(.154s^2)= .4s
In determining the time the objects fell I made the assumption that air resistance was small and did not factor it into the equation.
** 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. **
.1036m/s, .101m/s, .199m/s
.104m/s, .103m/s
.102m/s, .1005m/s
.203m/s, .196m/s
A horizontal range of 40 cm would imply a horizontal velocity of 40 cm / (.4 sec) = 100 cm/s or 1 m/s. None of the results you report above are consistent with horizontal velocities in the neighborhood of .1 or .2 m/s.
Can you explain how you got these results, and/or correct your results?
** 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*.104m/s
p=m1*.101m/s
p=m2*.199m/s
pTotal=m1*.104m/s+m2*0m/s
pTotla=m1*.101m/s+m2*.199m/s
m1*.104m/s+m2*0m/s=m1*.101m/s+m2*.199m/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*.003m/s= m2*.199m/s
m1=(m2*.199m/s)/.003m/s
m1/m2=.199m/s /.003m/s
m1/m2=66.3
This means that the mass of m1 is 66.3 times greater than the mass of m2
** Diameters of the 2 balls; volumes of both. **
2.6cm, 1.3cm
7.07 cubic cm, 1.68 cubic cm
** How will magnitude and angle of the after-collision velocity of each ball differ if the first ball is higher? **
It will force the first ball to take a slightly higher direction and won't transfer as much of its energy to the second ball.
The speed will be less because more energy will be transfered to the second ball.
Yes it will have a slightly higher trajectory.
Magnitude will be less than if it had taken a direct hit and its angle of direction will be lower.
The speed will be greater because more energy could be transfered from the first to the second.
After collision trajectory will be lower.
** Predicted effect of first ball hitting 'higher' than the second, on the horizontal range of the first ball, and on the second: **
The first ball will have a slightly higher trajectory and transfered less energy and go further.
The second ball will have a slightly lower trajectory and recieved less energy and won't travel as far.
** 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: **
196/1
I subtracted the after velocity for the first mass from the before velocity to get .001m/s then divided the after velocity of the second mass by this .196m/s /.001m/s to get 196 or a ratio of 196 to 1
** What percent uncertainty in mass ratio is suggested by this result? **
66.3/196= .338
.338*100%= 34%
** 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? **
4hrs
** Optional additional comments and/or questions: **
Most of your analysis is good, but you don't seem to have accurate after-collision ranges for the first ball, and your velocity calculations don't appear to follow from your data and time of fall.
Can your correct and/or explain these discrepancies? If you have questions they are of course welcome.