Phy 121
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'. **
1.9,2
1.25
+- .05cm
** Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process: **
24.7,25,28,28.3,28.3
26.86,1.842
the ball was rolled 5 times from the top of the incline through the flat ramp and onto the floor, the paper was aligned with the edge of the table and the distance of the ball was measured from the edge of the paper to the dot made by the carbon paper, the trials were put into the data program and mean and standard deviation were calculated
** 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. **
38.4,41.5,35.8,42.3,40.5
17.8,20.3,22.3,19.15,18.5
39.7,2.624
19.61,1.763
the straw and target ball were set up at the end of the ramp, the large ball was again rolled down the incline onto the flat ramp, striking the target ball and both fell to the floor, measurements were taken as before the paper was aligned with the edge of the table and the distance of each ball was measured from the edge of the paper to the dot made by the carbon paper, mean and standard deviation were calculated by putting the measurements of each ball into the data program
** Vertical distance fallen, time required to fall. **
91cm
.601s
i measured the distance from the top of the straw to the floor to get the vertical distance in cm
i used the timer program to get the time to fall the given distance, the assumption was that the ball started from rest
** 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. **
44.7,32.6,66.1
42.9,46.5
30.9,34.5
63.5,68.7
** 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. **
p1 before = m1*44.7cm/s
p1 after = m1*32.6cm/s
p2 after = m2*66.1cm/s
p1 = m1*v1 + m2*v1
p2 = m1*v2 + m2*v2
m1*v1 + m2*v1 = m1*v2 + m2*v2
** 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*v1 - m1*v2 = m2*v2 - m2*v1
m1 - m1 = m2*v2 - m2*v1 / v1-v2
m1 / m2 = v2-v1 / v1-v2
m1 / m2 = 5.5
mass 1 is 5.5 times larger than mass 2
** Diameters of the 2 balls; volumes of both. **
2.2,1.5
5.3,2.4
** How will magnitude and angle of the after-collision velocity of each ball differ if the first ball is higher? **
for ball one if the center is higher than the center of the second ball, the magnitude will be the same as will the direction of the velocity, the velocity will be the same and the after collision velocity direction will be the same, for the second ball if the first ball hits higher than the center, the magnitude will be less because the velocity will be less, this would be like a glancing blow
** Predicted effect of first ball hitting 'higher' than the second, on the horizontal range of the first ball, and on the second: **
this will not affect the horizontal range of the first ball, it will affect the horizontal range of the of the second ball, the range will be less because there will not be a direct hit and the direction of the ball will be more down and not out 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: **
m1/m2 = 7.9
v2 = 66.1 , v1 = 0 since the second ball started at rest
v1 = 42.9, v2 = 34.5, then 66.1 - 0 / 42.9 - 34.5 = 7.9
** What percent uncertainty in mass ratio is suggested by this result? **
5.5 / 7.9 = .696 * 100 = 69.6
then 100 - 69.6 = 30.4%
** 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: **
** **
** **
** **
** **
** **
2.5hr
** **
Excellent work throughout.