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: **
resubmit v3.0
could you check my work so far, especially the box labeled with $$$$$ and ****
I have put in over 4 hours on this today. I am submitting this just so I will have a hard copy for tomorrow. Do not grade this I will submit it again, finished tomorrow.
** 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. **
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. **
$$$$$$$
***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. **
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. **
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. **
** 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? **
8+ hours
** Optional additional comments and/or questions: **
Per your request I'm posting this without comment at this time.