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'. **
3.1 cm, 2.8 cm
2.5 cm
about .2 cm
** Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process: **
76 cm, 74 cm, 79 cm, 75 cm, 80 cm
76.8 cm, 2.588 cm
i measure the horizontal range by measuring the distance from free fall from the edge of the ramp, to the floor. You measure it by measuring straight-drop landing positions.
What you report is the vertical drop, not the horizontal range.
The horizontal range is how far the ball traveled in the horizontal direction after leaving the edge of the ramp.
** 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. **
70 cm, 73 cm, 69 cm, 74 cm, 76 cm
72 cm, 74 cm, 73 cm, 78 cm, 71 cm
72.4. cm, 2.881.
73.6 cm, 2.702.
i measure the horizontal range for both the second and first ball by measuring the distance from free fall from the edge of the ramp, to the floor. You measure it by measuring straight-drop landing positions.
** Vertical distance fallen, time required to fall. **
146 cm
.37 s
i found the this by using the equation `ds/v= `dt
** 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. **
387.9 cm/s , 376 cm/s, 379.8 cm/s
394.46 cm/s
384.12 cm/s
386.71 cm/s
** 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= m2 * 394.46 cm/s
p= m2 * 384.12 cm/s
p= m1 * 386.71 cm/s
p= m1 + m2 * 770.83 cm/s
p= m1 + m2 * 770.83 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. **
p-m1=m2 * 770.83 cm/s
m1= m2 * 770.83 cm/s + p
m1/m2= 770.83 cm/s + p
** Diameters of the 2 balls; volumes of both. **
2.1 cm. 1.2 cm
6.23 cm^3, 1.35 cm^3
** How will magnitude and angle of the after-collision velocity of each ball differ if the first ball is higher? **
The speed will probably be greater than if the centers are at the same height. The velocity will probably differ and the velocity will become greater.
The speed for the second ball would probably be less and so would the velocity.
** Predicted effect of first ball hitting 'higher' than the second, on the horizontal range of the first ball, and on the second: **
It will make the horizontal range of the first ball less and it will make it greater for the second ball.
** 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: **
394.46 cm/s, 384.12 cm/s 386.71 cm/s
** What percent uncertainty in mass ratio is suggested by this result? **
probably by about 10 percent
** What combination of before-and after-collision velocities gives you the maximum, and what combination gives you the minimum result for the mass ratio? **
the combination of the velocity before for the first ball and after for the second would give the maximum and the combination for the velocity for the first ball after and for the second ball after would give the minimum.
** 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? **
v1/u1=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? **
76 cm, 73.2 cm, .16
210.1 cm/s
215.6 cm/s
** 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? **
about 2 hours
** Optional additional comments and/or questions: **
Most of your work is based on the vertical drop rather than the horizontal ranges. You will need to redo most of this. The vertical drop does determine the time of fall.