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: **
submitted
** Distances from edge of the paper to the two marks made in adjusting the 'tee'. **
.35 cm, .4cm .4 cm +-.05
** Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process: **
12.9, 13.2, 13.4, 13.6, 13.62, 13.34, 0.3011 to obtain my results i rolled the big ball down the incline and allowed it to hit the floor. I marked where it hit the floor each time.
** 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. **
9.9, 10.3, 10.5, 11.1, 11.5 ( big ball ) 12.2, 12.3, 12.6, 12.8, 13.1 ( small ball ) 12.60, 0.3674 ( small ball ) 10.66, 0.6387 ( big ball )
** Vertical distance fallen, time required to fall. **
9.9, 10.3, 10.5, 11.1, 11.5 ( big ball ) 12.2, 12.3, 12.6, 12.8, 13.1 ( small ball ) 12.60, 0.3674 ( small ball ) 10.66, 0.6387 ( big ball ) ** Vertical distance fallen, time required to fall. ** 9.9, 10.3, 10.5, 11.1, 11.5 ( big ball ) 12.2, 12.3, 12.6, 12.8, 13.1 ( small ball ) 12.60, 0.3674 ( small ball ) 10.66, 0.6387 ( big ball ) ** Vertical distance fallen, time required to fall. ** 62.42 cm - ball 1, 62.78 cm - ball 2 .357 s, .358 s To determine just how far the balls feel I used Pyth Theory. The distance from the table to the floor was 61.5 cm (A^2). I then took the avg distance (B^2). I added A^2 + B ^2 and received C^2. I took the squareroot of C^2 and got the total distance. 61.5 ^ 2 + 10.66 ^ 2 = square root of ( 3782.25 + 113.64 ) = 62.42 cm 61.5 ^ 2 + 12.60 ^ 2 = square root of ( 3782.25 + 158.76 ) = 62.78 cm To determine the time I used an equation for uniform acceleration: x = x 0 + v 0 t + 1 / 2 a t ^ 2 62.42 = 0 + 0 + 1/2 * 980 cm / s ^2 * t ^ 2 .357 = t x = x 0 + v 0 t + 1 / 2 a t ^ 2 62.78 = 0 + 0 + 1/2 * 980 cm / s ^2 * t ^ 2 .358 = t Note: you told me that Pyth Theory would not work here, but I could keep this stat just because it is close enough to the exact value.
** 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. **
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***LINE ONE*** (length of ramp / time) 45 cm / 1.5 s = 29.68 cm / s 10.66 cm +-.6387 (length 1st ball travels from ramp to edge of table = cm) / .357 s = 31.65 cm/s and 28.07 cm/s 12.60 +-.3674 (length 2nd ball travels from straw to edge of table = cm / .358 s = 36.22 cm/s and 34.17 cm/s
***LINE TWO*** 13.34, +- 0.3011 13.34 + 0.3011 = 13.64 cm / .357 = 37.89 cm/s 13.34 - 0.3011= 13.04 cm / .357 = 36.22 cm/s
***LINE THREE*** 10.66 cm +-.6387 (length 1st ball travels from ramp to edge of table = cm) / .357 s = 31.65 cm/s and 28.07 cm/s Length big ball travels from ramp to edge of table velocity should remain consistent w/ velocity going down ramp = 28.68 cm/s
***LINE FOUR*** 12.60 +-.3674 (length 2nd ball travels from straw to edge of table = cm / .358 s = 36.22 cm/s and 34.17 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. **
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Momentum of 1st ball before collision U1 = 37.1 cm/s V1=29.6 cm/s V2=35 cm/s m1u1+m2u2=m1v1+m2v2 m1(37.1 cm/s) + 0 = m1(29.6 cm/s) + m2 (35 cm/s) m1*37.1 cm/s - 29.6 cm/s * m1 = m2 * 35 m1* 7.5 = m2 * 35 ( ratio ) m1 = 4.7 : m2 = 1 *****Line 1 M1*37.1 cm/s = p *****Line 2 M1* 29.6 cm/s = p : *****Line 3 M2* 12.60 cm/s *****Line 4 M1*37.1 cm/s + m2 (0) = p *****Line 5 M1*29.6 cm/s + m2 * 12.60 cm/s = 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. **
*****LINE 1 37.1 cm/s * m1 – 29.6 cm/s M1 = m2 35 cm/s *****LINE 2 M1 = 4.7 * M2 *****LINE 3 M1=1” m2 = 1/4.7 *****LINE 4 M1/m2 = 4.7 *****LINE 5 Mass of m1 = 4.7 * mass of m2
** Diameters of the 2 balls; volumes of both. **
Ball 1 = 2cm = 4.19 ml Ball 2 = 1 cm = .52 ml Diameter=measured w/ruler, volume = 4/3 pie * r ^ 3. To get r I divided diameter by 2. Ball 1 = 2cm = 4.19 ml Ball 2 = 1 cm = .52 ml
** How will magnitude and angle of the after-collision velocity of each ball differ if the first ball is higher? **
If the first ball is higher then the magnitude will be effect each ball accordingly: The big ball rolling down the ramp will hit the smaller ball and push it down, the distance it travels will be greatly reduced. The big ball will hit the smaller ball and quite possibily use the smaller ball as a ramp, which will make the bigger ball go even further then before. If the first ball is higher then the angle will be effect each ball accordingly: As stated earlier it will push the smaller ball down and thus the angle it falls will be closer to a straight drop. The bigger ball’s distance will increase thus be closer to keeping its horizontal velocity.
** Predicted effect of first ball hitting 'higher' than the second, on the horizontal range of the first ball, and on the second: **
as stated earlier the first ball rolling down will have a longer range than its inital value. the smaller ball will cover a much smaller distance then before.
** 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: **
first ball before collison = 13.34 cm +- .3011 first ball after collision = 10.66 cm +- .6387 second ball after collision = 12.60 cm +- .3674 minimum before collision velocity of first ball = 36.22 cm/s maximum after collision velocity of first ball = 37.89 cm/s minimum after collision velocity of the second = 33.98 cm/s 36.22 : 37.89 : 33.98 36:38:34 18:19:17
** What percent uncertainty in mass ratio is suggested by this result? **
the very most I rounded was when I took .22 and rounded it to 0. Less than 1%.
** What combination of before-and after-collision velocities gives you the maximum, and what combination gives you the minimum result for the mass ratio? **
maximum = maximum before collision velocity of first ball, minimum after collison velocity of 2nd ball minimum = minimum before collision velocity of first ball, maximum after collison velocity of 2nd ball.
** 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? **
m1u1+m2u2 = m1v1+m2v2 m1v1+m2u2 = m1u1+m2u2 m1v1-m1u1 =0
** 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. **
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Velocity of 1st ball after collision 12.70 +-.2185 Velocity of 2nd ball after collision 12.69 +-.7270 The mass of the balls remain constant in this experiment. The only real difference is the horizontal location of the balls. Which land basically in the exact same location almost every trial, regardless of height of ball. I changed the height to make sure my calculations were correct (I took horizontal measurements at 61.5 cm, 108.5 cms, and 115 cm repectively. I then took the five values I had obtained from the 61.5 cm drop, and averaged out the velocities for each in the same fashion as I did the initial experiment.
** 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? **
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*****LINE 1 61.5, 12.69, .15 *****LINE 2 35.34 cm/s *****LINE 3 (12.69+.7720) / .36 = 37.39 cm/s (12.69-.7720) / .36 = 33.11 cm/s *****LINE 4 (12.60+.3677)/.36 = 36.02 cm/s (12.60-.3677)/.36 = 33.98 cm/s *****LINE 5 12.65 cm/s is the mean of the two runs. A difference of .10 cm/s *****LINE 6 yes the velocity will decrease slightly when dropped 2 mm.
** Your report comparing first-ball velocities from the two setups: **
$$$$$$$$$$*****LINE 1 61.5, 12.70, .14 *****LINE 2 36.98 cm/s *****LINE 3 (12.70+.2185) / .36 = 35.88 cm/s (12.70-.2185) / .36 = 34.67 cm/s *****LINE 4 (10.66+.6387)/.36 = 31.39 cm/s (10.66-.6387)/.36 = 27.84 cm/s *****LINE 5 11.68 cm/s a difference of 1.02 cm/s *****LINE 6 When the balls were even the small ball went further, now that the small ball is .2 mm lower, the balls land in almost exactly the same spot (regardless of table height) for each trial.
** Uncertainty in relative heights, in mm: **
Less than 1 mm. I spent several hours during this lab. Whether it was calculating horizontal ranges, talking with my tutor about making a plan about answering certain questions, or even going on line looking for ideas on how to solve this problem. When I made my measurements I used the at most percision from a straight edge obtained from the college library. I felt this was one aspect of the lab that I had complete control over, and there were no excuses for not being as accurate as possible. I really feel as if there is very limited human error on this portion of the lab.
** 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. **
There has to be certainty in the heights of the balls when trying to determine horizontal velocity while the balls eventually fall to the ground. After doing this lab and adjusting the height of the 2nd ball I came to this conclusion. An epiphany, more like an analogy, I came to was imagining a baseball sitting on a tee. If you swing completely level then the ball will go on a line drive. If you swing on top half of the ball, the ball will go towards the ground quicker then if you swung level. This in a nut shell is basically what I did when doing this experiment. Finding that if you adjust the smaller ball lower then the bigger ball, the bigger ball will strike it and push it down to the ground closer to the tee as compared to if the balls connected on an ideally level plane.
** How long did it take you to complete this experiment? **
8.5 hours
** Optional additional comments and/or questions: **
The second ball should land considerably further from the straight-drop position than the first. However you say that they landed at nearly the same point, and if this is the case then your analysis is correct.