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: **
I am sorry this lab is 25 minutes late. I did the labs early this morning and just hadn't had time to put them into format. I've had to work nearly full time this week and take tests, do the normal physics assignments and go to the doctor for appointments. Every spare moment I have had have been put into this class. I am getting ready to fill in the answers to the next lab that I have already completed as well. I'm sorry.
** Distances from edge of the paper to the two marks made in adjusting the 'tee'. **
2.4 cm, 2.5 cm
2.4 cm
+- 0.10 cm
** Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process: **
29.50, 29.50, 29.25, 30.00, 29.25
29.50, .3062
I set up the apparatus as depicted and marked my release from rest point so that I had a place to start for each trial. I then released the ball and had the carbon paper set up to where the ball would put a mark on the white paper so I had a point to measure from.
** 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. **
29.50, 29.20, 29.25, 30, 29.25
12.25, 12.50, 13.40, 14.35, 15.20
29.5, .3062
13.54, 1.243
The first ball had the same measurements as the textbox above because there was not a deviation in how the first ball was set up and the trial ran. The second ball that was place on the tee was considered the projectile and it had less of a distance to travel therefore it did not travel as far. The mass of the second ball was a lot less than the first ball.
After colliding with the second ball, the moving ball will have slowed down. Its after-collision speed will be less than the before-collision speed indicated by its uniterrupted path.
** Vertical distance fallen, time required to fall. **
69.00 cm
.20 s
The vertical distance through which the two balls fell after collision was 69.00 cm when measured with a ruler. The time was .20 s measured by a second hand on a watch. I knew that the time would be really quick with just that vertical height of the table.
You can't measure a time interval that short using a watch.
If the collision was set up as instructed, so that the velocities immediately before and immediately after collision are both horizontal, the time of fall will be identical to that of a freely falling object released from rest. To fall 69 cm would require a little under .4 second. You need to calculate this time of fall and use it when figuring your velocities.
** 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. **
29.5 cm/s, 147.5 cm/s, 67.7 cm/s
The first ball is slowed by the collision and will not have a velocity greater than if no collision occurred.
29.8062 cm/s, 29.1938 cm/s
147.8062 cm/s, 147.1938 cm/s
68.993 cm/s, 66.407 cm/s
The first ball is the one on the tee and the second ball the one on the ramp.
This reverses your 'first' and 'second' ball references from before, where you stated that the ssecond ball was the one on the tee. This clarifies my previous question, but it is also not possible for the ball on the ramp to have less velocity after collision than the ball that hit it. The moving ball cannot pass through the target ball.
** 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. **
29.5 cm/s
147.5 cm/s
67.7 cm/s
29.5 cm/s
215.2 cm/s
m1 / 29.5 cm/s = m2 / 215.2 cm/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 / 29.5 cm/s = m2 / 215.2 cm/s
m1 = 29.5 cm/s *(m2 / 215.2 cm/s)
m1/m2 = .13708
m1/m2 = 295/2152
The momentum of the first ball is a lot less than that of the second because it is being used as a projectile starting from rest and v0 = 0 cm/s while as the second ball already had an initial velocity other than 0 cm/s.
** Diameters of the 2 balls; volumes of both. **
1 cm, 2.5 cm
1.05 cm^3, 5.24 cm^3
Volume = 4/3*pi*r^2
** How will magnitude and angle of the after-collision velocity of each ball differ if the first ball is higher? **
The magnitude and direction of the velocity of the first ball immediately after collision if the first ball was higher than the center of the second ball would less than if the centers of the ball were matched equally. The speed will be less than if the centers are at the same height. Since it would not be hitting the second ball in the center then after collision velocity would not be as forceful therefore it would be less. All of the answers apply to the second ball because when the first ball hits the second ball there will not be enough momentum behind the velocity to send the ball in a positive direction as it would if it hit the center
** 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 would be shorter and off track they would not line up nor be at great lengths.
** 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: **
292/2152
59/432
.1372
Ex:
m1/m2 = 29.1938 / 215.2
m1/m2 = 292/2152
** What percent uncertainty in mass ratio is suggested by this result? **
+-.2%
When you divide out each number and multiply it by a hundred each number changes by approximately +-.2%.
** 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? **
1.5 hours
** Optional additional comments and/or questions: **
See my notes and see if you can clarify the points I raise by explaining how the system was set up and exactly what was measured.