course PHY 201
Prof. Smith, This is a Conservation of momentum lab, review.
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conservation of momentum
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.
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Hello Prof. Smith,
Here is my Conservation of Momentum lab.
Tanya
Distances from edge of the paper to the two marks made in adjusting the 'tee'.
2, 2, 0
Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process:
10.3, 10.3, 10.2, 10.26, 10.4
10.29, 0.07294
I measured the fall position of the first ball from the point of the free fall of a 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.
4.0, 4.1, 4.2, 4.2, 4.15
10.3, 10.3, 10.2, 10.26, 10.4
4.13, 0.08367
10.29, 0.07294
I measured the positions from the free ball position to each center of the ball prints.
This appears to indicate that the first ball travels further than the second, which would be impossible unless the first ball passes through the second. Please clarify this.
Vertical distance fallen, time required to fall.
71 cm
2.140625
The distance I measured from the place where the ball falls to the place on the floor where it lands.
For timing the falling ball, I used the Timer program. I pushed the Timer button at the same time when I released the ball. I stopped the Timer when the ball landed.
This would indicate the time for the ball to roll down the ramp then fall to the floor. The time of fall is the time spend in free fall, after leaving the end of the ramp and before reaching the floor. This time is determined using the acceleration of gravity, the distance to the floor, etc.
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.
26.3, 10.29, 4.13
23.876, 26.893
10.363, 10.217
4.214, 4.046
These velocities don't follow from the time of fall you reported and the horizontal ranges reported earlier. Can you clarify how you obtained these results?
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.
p1 = m1 * 26.3 m/s
p2 = m1 * 10.29 m/s
p3 = m2 * 4.13 m/s
p4 = m1 * 26.3 m/s
p5 = (m1 + m2) * 14.42 m/s
p5 would imply the two masses 'stuck together', which is not the case in this experiment. I do not believe a velocity of 14.42 m/s was observed.
m1 * 26.3 m/s = (m1 + m2) * 14.42 m/s : m1 * 1.82 m/s = m1 + m2
This is based in part on the p5 momentum you used above, which as I said does not appear to be related to your observations. The p1, p2, p3 and p4 are related to your observed quantities (but see my previous notes and make sure these quantities are all correct). If your observed quantities are correct, you need only these four momenta to determine the mass ratio.
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 * 1.82 m/s - m1 = m2
m1 = m2 / (1.82 m/s - 1)
m1 / m2 = 1/ 0.82 = 1.22
m1 / m2 = 1.22
It means that the masses of the balls are in the ratio of 122 over 100. Or the mass of the first ball is 122 (or 61) and the mass of the second ball is 100 (or 50).
Diameters of the 2 balls; volumes of both.
2 cm, 2.4 cm
4.1888 cm^3, 2.424 cm^3
How will magnitude and angle of the after-collision velocity of each ball differ if the first ball is higher?
1. In my opinion, the target ball will jump and the distance of its landing would be shorter; however, the velocity probably would be higher because the ball will fall from the higher position. The speed would be greater than if the centers would be at the same height. The direction of the ball would be different because the ball will make a loop.
2. If it would happen, the second ball will hit the first ball from the top, so the first ball will change the direction and will have a deep drop. The magnitude will be shorter and the direction of the velocity would be decreased. The speed would be less because the second ball would reach an obstacle. The direction of the velocity would differ.
Predicted effect of first ball hitting 'higher' than the second, on the horizontal range of the first ball, and on the second:
In both cases the horizontal range of the balls would change, since all measurements are changed.
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:
0.6105
0.53
I used the same way to calculate these numbers as I used previously, but for the first number I used maximum values and for the second one minimum values.
What percent uncertainty in mass ratio is suggested by this result?
50 %
43%
What combination of before-and after-collision velocities gives you the maximum, and what combination gives you the minimum result for the mass ratio?
Combination of the maximum of the before-collision velocities of the first ball, and the maximum of the first ball after collision will give me maximum.
Combination of the minimum of the first ball after collision, and the minimum of the second ball after collision will give me the minimum.
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?
p1 = m1 * v1
p2 = m1 * u1
p3 = m2 * u2
p4 = m1 * v1
p5 = (m1 + m2) * (u1+u2)
m1 * v1 = (m1 + m2) * (u1+u2)
m1 * v1/(u1+u2) - m1 = m2
m1 = m2 / (v1/(u1+u2))
m1 / m2 = 1/ (v1/(u1+u2))
Derivative of expression for m1/m2 with respect to v1.
Dv1 (m1 / m2) = -(u1+u2) / v1^2
Dv1 (m1 / m2) = -(10.29 m/s+ 4.13m/s) / 26.3m/s^2 = -0.0208
It is probably rate of masses
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?
by 2.424
by - 2.4749
By this number the mean v1 different from the standard deviation, and the second number I obtained by plugging in the equation of the derivative with respect to v1.
Complete summary and comparison with previous results, with second ball 2 mm lower than before.
m1 / m2 = 0.55
The new ratio is smaller by 0.67 than the first one.
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?
71, 0.4783,
10.90
10.34, 10.45
10.363, 10.217
0.023, 0.233
According to my obtained results, I can conclude that the velocity resulting from the 2 mm decrease in 2d-ball altitude appears to be significantly different than that I obtained when the centers were at the same level.
Your report comparing first-ball velocities from the two setups:
According to my obtained results, I can conclude that the velocity resulting from the 2 mm decrease in first ball altitude appears to be significantly different than that I obtained when the centers were at the same level.
Uncertainty in relative heights, in mm:
0.023, 0.233
The uncertainties could appeared when I measured the distances of the dots made by the carbon paper when the balls collided
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.
The hypothesis that the uncertainty in the relative heights of the balls was a significant factor in the first setup. As a result, I can determine now how accurate I made my measurements.
How long did it take you to complete this experiment?
3 hours
Optional additional comments and/or questions:
See my notes and please send me a copy of this document with your clarifications indicated by *&*&
*&*&*&*&*&*&*&*&*&*&*&*&*&*&*&*&*&*&*&*&*&*&*&*&*&*&*&*&
Distances from edge of the paper to the two marks made in adjusting the 'tee'.
2, 2, 0
Five horizontal ranges of uninterrupted large ball, mean and standard deviation and explanation of measuring process:
10.3, 10.3, 10.2, 10.26, 10.4
10.29, 0.07294
I measured the fall position of the first ball from the point of the free fall of a 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.
10.3, 10.3, 10.2, 10.26, 10.4
4.0, 4.1, 4.2, 4.2, 4.15
10.29, 0.07294
4.13, 0.08367
I measured the positions from the free ball position to each center of the ball prints.
Vertical distance fallen, time required to fall.
71 cm
.203125
The distance I measured from the place where the ball falls to the place on the floor where it lands.
For timing the falling ball, I used the Timer program. I pushed the Timer button at the same time when I released the ball. I stopped the Timer when the ball landed.
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.
27.0423 cm/sec (1.109375 sec, 30 cm)
22.027 cm/sec (.1875 sec, 4.13 cm)
29.9345 cm/sec (.34375 sec, 10.29 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.
p1 = m1 * 27.0423 cm/s
p2 = m1 * 22.027 cm/s
p3 = m2 * 29.9345 cm/s
p4 = m1 * 27.0423 cm/s
p5 = m2 * 29.9345 + m1 * 22.027 cm/s
m1 * 27.0423 cm/s = m2 * 29.9345 cm/s + m1 * 22.027 cm/sec
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 * 5.0153 cm/s = m2 * 29.9345 cm/sec
m1 = m2 * 5.9686 cm/sec
m1 / m2 = 5.9686 cm/sec
It means that the masses of the balls are in the ratio of 59686 over 10000. Or the mass of the first ball is about 6 units and the mass of the second ball is 1 unit.
Diameters of the 2 balls; volumes of both.
2 cm, 2.4 cm
4.1888 cm^3, 2.424 cm^3
How will magnitude and angle of the after-collision velocity of each ball differ if the first ball is higher?
1. In my opinion, the target ball will jump and the distance of its landing would be shorter; however, the velocity probably would be higher because the ball will fall from the higher position. The speed would be greater than if the centers would be at the same height. The direction of the ball would be different because the ball will make a loop.
2. If it would happen, the second ball will hit the first ball from the top, so the first ball will change the direction and will have a deep drop. The magnitude will be shorter and the direction of the velocity would be decreased. The speed would be less because the second ball would reach an obstacle. The direction of the velocity would differ.
Predicted effect of first ball hitting 'higher' than the second, on the horizontal range of the first ball, and on the second:
In both cases the horizontal range of the balls would change, since all measurements are changed.
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:
0.6105
0.53
I used the same way to calculate these numbers as I used previously, but for the first number I used maximum values and for the second one minimum values.
What percent uncertainty in mass ratio is suggested by this result?
50 %
43%
What combination of before-and after-collision velocities gives you the maximum, and what combination gives you the minimum result for the mass ratio?
Combination of the maximum of the before-collision velocities of the first ball, and the maximum of the first ball after collision will give me maximum.
Combination of the minimum of the first ball after collision, and the minimum of the second ball after collision will give me the minimum.
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?
p1 = m1 * v1
p2 = m1 * u1
p3 = m2 * u2
p4 = m1 * v1
p5 = m1 * u1 + m2 * u2
m1 * v1 = m1 * u1 + m2 * u2
m1 * (v1- u1) = m2 * u2
m1 = m2 (u2 / v1- u1)
m1 / m2 = u2 / v1- u1
Derivative of expression for m1/m2 with respect to v1.
Dv1 (m1 / m2) = Dv1(u2 / v1- u1) = - u2 / (v1- u1)^2
Dv1 (m1 / m2) = - 29.9345 / (27.0423 - 22.027)^2 = - 1.19
It is probably rate of masses
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?
by -1.10633
by – 7.1586
By this number I obtained by adding together the number which correspond the standard deviation (0.08367) of the uninterrupted ball and the corresponding number v1, and the second number I obtained by plugging in the equation of the derivative with respect to v1.
Complete summary and comparison with previous results, with second ball 2 mm lower than before.
m1 / m2 = 0.55
The new ratio is smaller by 0.67 than the first one.
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?
71, 0.4783,
10.90
10.34, 10.45
10.363, 10.217
0.023, 0.233
According to my obtained results, I can conclude that the velocity resulting from the 2 mm decrease in 2d-ball altitude appears to be significantly different than that I obtained when the centers were at the same level.
Your report comparing first-ball velocities from the two setups:
According to my obtained results, I can conclude that the velocity resulting from the 2 mm decrease in first ball altitude appears to be significantly different than that I obtained when the centers were at the same level.
Uncertainty in relative heights, in mm:
0.023, 0.233
The uncertainties could appeared when I measured the distances of the dots made by the carbon paper when the balls collided
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.
The hypothesis that the uncertainty in the relative heights of the balls was a significant factor in the first setup. As a result, I can determine now how accurate I made my measurements.
Please submit this once more and mark each modified response with *&*& so that I can easily identify what has been changed and what has not.
Note also that the time required to travel down the ramp is not used to calculate the horizontal velocity of the projectile, which is based only on the horizontal range and the dsitance of the drop.