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'. **
1.8cm, 1.0cm
.03cm
** Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process: **
12cm, 12.4cm, 12.6cm, 12.3cm, 12.7cm
12.4cm, .2739
I measured the distance with a ruler from where the ball hit and then using hte measured distance, I used the data program to find 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. **
19.0cm, 18.5cm, 18.9cm, 19.3cm, 19.2cm
4.0cm, 3.9cm, 4.6cm, 4.9cm, 3.6cm
18.98cm, .3114
4.2cm, .5339
I measured the distance with a ruler from where the ball hit and then using the measured distance, I used the data program to find the mean and standard deviation.
** Vertical distance fallen, time required to fall. **
75cm
1.04sec
The ball will fall 75 cm in less than .4 seconds. Actual velocities will be more than 2.5 times the velocities you get based on a 1.04 sec time of fall.
I used my ruler and meausred the distance from the floor to the table top and then I used the timer program to determine the time from the release to the collision to the fall to the floor.
The fall to the floor will be a free fall, with 0 initial vertical velocity, under the acceleration of gravity. Using this you can calculate the 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. **
7cm/s, 9.6cm/s, 43.4cm/s
4.73cm, 3.66cm
4.73cm, 3.66cm
19.29cm, 18.66cm
You need to explain how these results are connected to your data. They do not seem to follow from the time of fall you reported and the horizontal ranges; note also that the time of fall should be corrected.
** 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. **
m1= m1(7cm/s)/7cm/s
m1=m1(9.6cm/s)/9.6cm/s
m2=m2 (43.4cm/s)/43.4cm/s
m1(7cm/sec) +m2(31.6cm/s)= m1(7cm/sec) +m2(31.6cm/s)
m1(9.6cm/sec) +m2(43.4cm/s)= m1(9.6cm/sec) +m2(43.4cm/s)
m1(7cm/sec) +m2(31.6cm/s)= m1(9.6cm/sec) +m2(43.4cm/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(7cm/s)/m1(7cm/s) / m2(31.6cm/s/m2(31.6cm/s)
m1= 7cm/s / m2(31.6cm/s)
m1/m2= 7cm/s/ 31.6cm/s
This isn't a correct solution for m2 / m1.
Note that if we divide through the velocity units, which are common to all terms of the equation, the equation gives us simple number multiples of m1 and m2. As a simple example of solving for m2 / m1, suppose we have the following equation:
5 m1 + 9 m2 = 3 m1 + 14 m2. We could get all the m1 terms on one side and all the m2 terms on the other if we subtract 3 m1 + 9 m2 from both sides:
5 m1 + 9 m2 - (3 m1 + 9 m2) = 3 m1 + 14 m2 - (3 m1 + 9 m2) gives us
5 m1 + 9 m2 - 3 m1 - 9 m2 = 3 m1 + 14 m2 - 3 m1 - 9 m2 so that
2 m1 = 5 m2. Then to ge m2 / m2 divide both sides by 5 to get
2 / 5 * m1 = m2 and divide both sides by m1 to get
2/5 = m2 / m1.
.2
The meaning of the ratio m1/m2 is the m1 velocity divided by the m2velocity
** Diameters of the 2 balls; volumes of both. **
2.5cm, 1cm
8.18, .52
** How will magnitude and angle of the after-collision velocity of each ball differ if the first ball is higher? **
The first ball will drop stargith down making the magnitude and direction less tahn the second. The speed will be less and the velocity will be less as well. And for the second ball, the magnitude, direction, speed and velcoity will all be less as well.
** Predicted effect of first ball hitting 'higher' than the second, on the horizontal range of the first ball, and on the second: **
It think that the horizontal range would be less. Second would be less as well because it wouldn't have enought speed behind it as it does on the same height
** 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: **
.36
.19
I used m1/m2 and I just used the numbers of the the minumum before on the first ball, which I previously recorded and I used the maximum after collision velcoity of the first ball and the minimum after collision velocity of the second and divided m1/m2 and received my results.
** What percent uncertainty in mass ratio is suggested by this result? **
.01%
** 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 before-and after-collision velocities that would give you the maximum would be the minimum before the collision on the first ball divided by the minimum after the collision of the second ball and the combination that gives you the minimum result for the mass ratio is the maximum after the collision of the first ball divided by the minimum after the collision of the second.
** 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
** Derivative of expression for m1/m2 with respect to v1. **
I am a general physics 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? **
I am a general physics student!
** Complete summary and comparison with previous results, with second ball 2 mm lower than before. **
The 5 horizontal ranges for ball 1 were 5.0cm, 5.6cm, 5.7cm, 5.1cm, 5.0cm with a mean and standard deviation of 5.28cm, and .3421. The 5 horizontal ranges for ball 2 were 20.9cm, 20.3cm, 20.1cm, 20.5cm, 20.7cm with a mean and standard deviation of 20.5cm, and .3162. Ball 1 before collision was 9.9cm/s, and after the collision was 7.6cm/s, Ball 2 before collision was 38.1cm/s and after the collision was 29.8cm/s the total mometum before was 1.302 and the total momentum after was 1.300. The previous total mometum was .2 and therefore now having a difference of 1.1
** 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? **
74.8cm, 20.5cm/s, .006cm/s
52.25cm/s
20.18cm/s, 20.82cm/s
19.29cm/s, 18.66cm/s
1.52cm/s
Yes because the velocities were greater for the 2mm ball, with a difference of 1.52cm/s.
** Your report comparing first-ball velocities from the two setups: **
The 2mm lower first ball had a velocity of 5.28cm/s and the first ball velocity with the centers at the same height was 4.2cm/s with a difference of 1.08cm/s
** Uncertainty in relative heights, in mm: **
Mine was .1mm uncertainity because my original result was .19 and the ending results was .2, resulting in a .1mm uncertainity.
** 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. **
It is a significant factore based on the velocity changes. The first was .2 and the second time it was 1.30, a difference of 1.10.
** How long did it take you to complete this experiment? **
2hours
** Optional additional comments and/or questions: **
Most of your steps are good. However the step where you calculated your velocities doesn't seem to follow from your data, and this will modify some of your subsequent responses.
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