Phy 231
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'. **
18.3cm, 12.4cm
.9cm
+-.5cm
** Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process: **
24.6, 24.5, 24.1, 24.1, 24.2
24.3, .2345
I rolled the large ball down the sloped ramp and then the level ramp and saw how far from the edge of the paper the ball went. I just measured the horizontal distance. I didn't use the small ball at all. The distance from the edge of the paper and the place it landed, was on average 11.9cm and then I added 12.4cm to take into account the distance from the edge of the paper to the starting point of the small ball.
** 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. **
31.3, 31.1, 31.3, 31.5, 31.3
9.3, 9.5, 9.5, 9.1, 9.2
31.3, .1414
9.32, .1789
To find the second ball measurements (which I took to mean the small ball) I found the distance it fell from the edge of the paper and added this to the distance from the edge of the paper to the edge of the table. To find the measurements of the second ball, I observed that the ball fell short of the edge of the paper, so I just measured from the edge of the table to the contact point of the floor of the ball. I found the mean and standard deviations by using the data program.
** Vertical distance fallen, time required to fall. **
73.3cm, .4s
I found the vertical distance traveled by the two balls by measuringfFrom the top of the tee to the floor. I found the dt by using the eqn: vf^2=v0^2+2ads. ds=73.3cm, vf=0m/s, a=980m/s/s, to get a dv of 379cm/s/s. I then used the eqn, a=dv/dt to get dt=.4s. To calculate the dt, I took the assumption that the ball would accelerate at a 9.8m/s/s pace and that the ball traveling vertically would have little or no effect on the time it takes for the ball to fall to the floor 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. **
122, 47, 157
124.1, 121.9
48, 46
157.4, 156.6
** 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. **
momentum=m1*122cm/s
momentum=m1*47cm/s
momentum=m2*157cm/s
momentum=(m1*122cm/s) + (0)
momentum=(m1*47cm/s) + (m2*157cm/s)
m1*122cm/s = (m1*47cm/s) + (m2*157cm/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*122cm/s) - (m1*47cm/s) = m2*157cm/s
m1=(1.3*m2)/(1-.39)
m1=2.13m2
m1/m2 = 2.13
This means that m1 is 2.13 times greater than m2. Visually, this number looks pretty accurate.
** Diameters of the 2 balls; volumes of both. **
2.5cm, 1.3cm
8.18, 1.15
** How will magnitude and angle of the after-collision velocity of each ball differ if the first ball is higher? **
The first ball was slightly higher than the second I would have expected that the first ball would roll farther and the second ball wouldn't roll as far as it did because the force of impact wouldn't be as great if it didn't hit it right on. The second ball would also travel a shorter distance because the impact would be forcing the ball in a downward motion instead of a straight line motion.
** 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 the first ball would be greater than previously recorded. The horizontal range of the second ball would be less than previously recorded.
** 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: **
2.12
I used the numbers 121.9cm/s, 48cm/s, and 156.6cm/s to get the equation: (121.9*m1) - (48*m1) = 156.6*m2. I then solved for m1/m2.
** What percent uncertainty in mass ratio is suggested by this result? **
.47%
** What combination of before-and after-collision velocities gives you the maximum, and what combination gives you the minimum result for the mass ratio? **
to get greatest mass ratio: max ball 1 before coll, min ball 1 after coll), min ball 2 after coll.
to get least mass ratio: min ball 1 before coll, max ball 1 after coll, max ball 2 after coll.
** 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? **
m1/m2 = u2/(v1-u1)
** Derivative of expression for m1/m2 with respect to v1. **
derivative of m1/m2 = (-u2)/(v1-u1)^2
derivative of m1/m2 = (-157cm/s)/(122cm/s-47cm/s)^2
This is the rate the mass changes with respect to v1. I used the quotient rule to find the derivative.
** 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? **
If the first ball's range changed by .1414 it would change v1 very slightly because v1 is a comparatively much larger number. If v1 changed, only a slight amount, the mass ration would be effected greatly because v1 is to the exponent of 2. The mistake would big. Slight changes are very noticeable when you are squaring the answer.
** Complete summary and comparison with previous results, with second ball 2 mm lower than before. **
mass ratio = (-247m/s)/(115m/s-30m/s)^2
The small ball went further than before and the large ball didn't travel as far as before. ratio1=-.03, ratio2=-.03. The mass ratios are very similar but the second ratio is a little farther away from 0 than the first ratio.
** 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? **
I'm sorry I have no idea what triangle they are talking about. From what I understand there's a line connecting the two centers of the balls. After that, I dont see where they designate where the other two sides should be.
** 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: **
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2 hours
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Sorry I couldn't finish it.
Your work on this lab is good except that you weren't able to finish it; however even with that you passed this lab.