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'. **
27.5 inches
26 inches
+/- half an inch
** Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process: **
12.5,13,13,12.5,12.5
Mean = 12.7
SD = .2739
I actually made a mistake while collecting this data I had originally measured the distance from the point of collision at the end of the ramp to the floor. However, I was able to correct this mistake by subtracting the distance from the end of the ramp to the floor and add the length of the ramp to the remaining number which results in the actual horizontal ranges. I plugged this information into the data program to obtain the mean and standard deviation.
** 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. **
13,13,13.5,13,13.5
12.5,12.5,13,13,12.5
mean = 12.7
SD = .2739
I actually made a mistake while collecting this data I had originally measured the distance from the point of collision at the end of the ramp to the floor. However, I was able to correct this mistake by subtracting the distance from the end of the ramp to the floor and add the length of the ramp to the remaining number which results in the actual horizontal ranges. I plugged this information into the data program to obtain the mean and standard deviation.
** Vertical distance fallen, time required to fall. **
26,26,26.5,26,26.5 - ball 2 (smaller ball)
25.5,25.5,26,26,25.5 - ball 1 (larger ball; coming off ramp)
I took my tape measure to record these distances as accurately as possible, and I estimate that it takes about a second or less for the balls to hit the floor.
You have not calculated the time of fall. Most of the subsequent results depend on your calculations of velocities, which depend on time of fall.
** 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. **
vAve = 'ds / 'dt = 12 inches/ 60 sec.
.2 inches/sec
Did something in this experiment accelerated uniformly through a displacement of 12 inches in 60 seconds?
Velocities are calculated by analyzing the projectile behavior of the ball, using the ball's vertical motion to determine the time of fall, and applying this time to the horizontal motion, which after collision is assumed to be at constant horizontal velocity.
Concerning the velocity of the ball immediately after collision, it should return to the original velocity following the collision because the ball will briefly stop upon collision and then gradually accelerate to the original velocity because it must regain momentum due to the force of gravity (at least) acting upon the ball.
12.9739 = mean + SD for ball 1
12.4261 = mean - SD for ball 1
12.9739 = mean + SD for ball 2
12.4261 = mean - SD for ball 2
** 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 (.2 in/sec)
p = m2 (.2 in/sec)
m1 (.2 in/sec)+ m2 (.2 in/sec) = p
** 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 (.2 in/sec) = p - m2 (.2 in/sec)
m1 = p - m2
p = -m2 / m1
** Diameters of the 2 balls; volumes of both. **
big ball has diameter of 1 inch.
small ball has diameter of .5 inch.
To find volume: D = m/V
V*D = m; V = m/D
To find volume I would need to know the density of the material the balls are made of.
** How will magnitude and angle of the after-collision velocity of each ball differ if the first ball is higher? **
Magnitude and speed will decrease after collision (concerning ball 2) if the alignment is off, also the balls will travel in a more downward direction instead of outward.
** 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 of both balls will decrease. I actually tried this to see what would happen.
** 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: **
1.04 is the ratio. I got this number by division of (m + SD) / (m - SD).
** What percent uncertainty in mass ratio is suggested by this result? **
48.1 %
I got this number by taking the original uncertainty of .05 and dividing it by the ratio of 1.04 and multiplying this answer by 100.
** What combination of before-and after-collision velocities gives you the maximum, and what combination gives you the minimum result for the mass ratio? **
Not applicable. I am a Phy 121 student.
** 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? **
Not applicable. I am a Phy 121 student.
** Derivative of expression for m1/m2 with respect to v1. **
Not applicable. I am a Phy 121 student.
** 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? **
Not applicable. I am a Phy 121 student.
** Complete summary and comparison with previous results, with second ball 2 mm lower than before. **
Not applicable. I am a Phy 121 student.
** 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? **
Not applicable. I am a Phy 121 student.
** Your report comparing first-ball velocities from the two setups: **
Not applicable. I am a Phy 121 student.
** Uncertainty in relative heights, in mm: **
Not applicable. I am a Phy 121 student.
** 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. **
Not applicable. I am a Phy 121 student.
** How long did it take you to complete this experiment? **
2 hours
** Optional additional comments and/or questions: **
The ball, when it traveled about 12 cm in the horizontal direction before striking the floor, was not moving at an average of .2 in / second in either the vertical or the horizontal direction, nor is .2 in / second a likely initial or final velocity for any part of this experiment. The 60-second interval on which this calculation was based does not appear to correspond to the time interval for any of the uniformly accelerated motion observed here.
From the point where you obtained that velocity, most of your results are based on .2 in / second. Everything from that point on will need to be revised, and based on an accurate determination of time intervals and velocities. See my notes and let me know if you have questions; then please submit a revision and/or questions.