Phy 201
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'. **
19mm,19mm
13mm
+-1mm
** Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process: **
28.2,28.5,28.2,28.4,28.5
28.4,.1517
I first did a vertical drop with the middle of the big ball centerd under the edge of the table and dropped it. I marked that position. I then measured the distance from the edge of the ramp to the edge of the table. Then after I got my horizontal marks I measured from the vertical drop mark to the horizontal mark then added the distance from the edge of the ramp to the edge of the table to get my total horzontal distance
** 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. **
32.8,33,32.9,33,32.9
22.5,22.5,22.5,22.6,22.4
32.92,.08367
22.5,.0707
I got my 1st ball after impact horizontal measurement by measuring from the vertical ball drop mark to the actual landing spot, and then adding the distance from the edge of the table to thecenter of the tee + half the diameter of the 2nd ball+ half the diameter of the 1st ball.This would give me a total distance from the landing position to the center of the 1st ball at impact. The second ball distance was using the disatnce from the vertical drop mark to the landing position + the distance from the edge of the table to the center of the tee. This gives me the distance from the center of the small ball to the landing position.
** Vertical distance fallen, time required to fall. **
45.8
.30
The measurement are in cm and secs. I got the distance by measuring from the top of the tee that is setting on the table to the floor where the paper is laying on the floor.The time was gotton the same way. I held the ball at the top of the tee and dropped it. I clicked the timer at release and then as it hit the floor. I recorded the time for each ball acouple of times and then took the mean of the trials to get a better time. For both balls it was pretty much the same so we will use the same time for both balls.
** 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. **
94.7,75,109.7
95.2,94.2
75.2,74.8
110,109.3
These measurements are in cm
** 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*94.7cm/s
m1*`75cm/s
m2*`109.7cm/s
m1*94.7cm/s+m2*`0cm/s
m1*`75cm/s+m2*`109.7cm/s
m1*94.7cm/s+m2*`0cm/s=m1*`75cm/s+m2*`109.7cm/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*19.7cm/s=m2*109.7cm/s
m1=m2*5.57cm/s
m1/m2=5.57cm/s
m1/m2=5.57
Does it mean that the mass of ball number 1 is 5.57 times heavier than that of ball number 2.I dont even thik I did this right
Right. Your measurements indicate a mass ratio of 5.57.
Note that 'heavier' (which relates to weight) is not a relevant term here. Weight is the force exerted by gravity on an object, and the gravitational force is irrelevant to the collision, immediately before and after which everything moves (almost entirely) in the horizontal direction.
Weight is of course the force that (through the action of its component parallel to the incline) causes the ball to accelerated before the collision, and which causes the vertical acceleration after the collision.
** Diameters of the 2 balls; volumes of both. **
3.4,1.2
20.6,.90
** How will magnitude and angle of the after-collision velocity of each ball differ if the first ball is higher? **
I believe the 1st ball will be directed upwards just a little and the velocity will not be affected as much as it does when it hits the center. I believe the speed will be greater after impact then that of hitting the center. No left or right direction just upwards a little
** Predicted effect of first ball hitting 'higher' than the second, on the horizontal range of the first ball, and on the second: **
I believe the first ball will go a little further because it was not slowed as much and the ball 2 will be directed downward instead of being knocked horizontally and should shorten the distance it travels
Good. It will also get a bit of vertical velocity, which will cause it to take longer to get to the floor and hence tend to increase horizontal range.
** 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: **
m1/m2=5.75
I did this the same way as before when we found the ratio but this time I used my mean and standard deviation velocities as requested. As before I dont think I have done this right
** What percent uncertainty in mass ratio is suggested by this result? **
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? **
The maximum would be the +deviation of the ball 2 after velocity, the - deviation of the ball 1 before velocity and the, + deviation of the after ball 1 velocity. The minimum would be the - deviation of the ball2 after velcoity,the - deviation of the ball 1 before velocity and the, + deviation of ball 1 after velocity
** 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? **
u2/v1-u1
** 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. **
Ball 1 before impact data,H.distance mean 28.4cm,deviation .1517. Velocity ranges 94.2,94.7,95.2
Ball 1 after impact data,H.distance mean 21.22,deviation .08367. Velocity ranges 70.5,70.7,71
Ball 2 after impact data,H.distance mean 33.22,deviation .1304. Velocity 110.3,110.7,111.2
The mean ratio is m1/m2=4.6
The ratio with ball 1 minimum before velocity,ball 1 maximum after velocity,and ball 2 minimum after velocity. m1/m2=4.75
** 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? **
45.6,33.22,.017
107.3
107.7,106.8
110,109.3
2.4
It was not significantly different. 2.4cm is not that much
** Your report comparing first-ball velocities from the two setups: **
The mean and standard deviation of the center impact of ball 1 after was 75.2,75,74.8 and for the velocity after impact for the 2mm drop was 71,70.7,70.5. The difference between the two means are 4.3cm. Little bit more of a difference with the big ball
** Uncertainty in relative heights, in mm: **
+-1mm, because I did not have any carbon paper. It was only a little piece the size of a fifty cent piece and I did not have anymore
** 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 did not seem to make taht much of a difference in my results. The mean distances for the big ball and the small ball was 1 to 2 cm and the velcoity was off 2.4 to 5cm/s. I guess it depends on how precise your data needs to be to say whether or not it was a significant difference
** How long did it take you to complete this experiment? **
3hrs
** Optional additional comments and/or questions: **
&#Good work. See my notes and let me know if you have questions. &#