#$&*
course Phy 231
3/29 6pm
Experiment 19. When two objects moving along a common straight line collide and maintain motion along the same line as before collision, the total of their momenta immediately after collision is
equal to the total immediately before collision.
See CD EPS01 for a general overview of Lab Kit Experiment 19. Note, however, that this experiment
has been revised since those videos were recorded, and any specific instructions given on the
videos are superceded by the instructions given here, and by more recent practices in observing
horizontal range.
The basic setup for this experiment is pictured below. The small metal ball (referred to here as
the 'target ball') from the kit is supported on a short piece of drinking straw, and the large ball
rolls down a ramp (the end of this first ramp appears at far right), then along a horizontal second
ramp until it reaches the end of the ramp and collides with the target ball.
The small ball is positioned near the edge of the table according to the following criteria:
The large ball collides with the small ball just a very short distance after it has lost
contact with the ramp.
The collision occurs when the centers of mass of the two balls are at the same height above the
table.
After collision both balls fall uninterrupted to the floor. Their velocities after collision are
easily determined by the distance to the floor and their horizontal ranges. The velocity of the
first ball prior to collision is determined by removing the second ball and allowing it to run down
the inclined track and across the horizontal track before falling uninterrupted to the floor.
In the picture above, the 'tee' holding the smaller ball consists of a piece of plastic tubing
inserted into a glob of hot glue.
A piece of the drinking straw that came with your lab kit slips over the tubing, allowing you
to easily adjust the vertical position of the ball.
The base of the 'tee' is flat and keeps the system stable, while providing no significant
interference with the motion of either ball.
This 'tee' was not included in the kit for Spring 2006; if you are a Spring 2006 student you
may submit your email address and the instructor will be glad to send you a few.
The alternative to using the 'tee' is to balance the ball just on the drinking straw section,
which is feasible (students have been doing it for years) but which can be unnecessarily time
consuming.
If you use the tee, you should trim about 1 cm, maybe just a little more (if you trim it a little
too long it's easy to shorten it), from the end of the straw. The flat end of the straw should be
the one that supports the ball; the end you trim will tend to be less straight and the ball might
tend to roll off.
The straw might or might not fit tightly enough that the weight of the ball does not cause it to
slide up or down. If this is not the case, you can insert a thin strip of paper into the straw,
before slipping it over the plastic tubing in order to 'shim' the inside and ensure a tighter fit.
Set up the system:
The motion of the balls before and after collision should be horizontal:
To analyze the projectile motion of the balls it is very helpful if their initial velocities
are both in the horizontal direction. Due to slight irregularities in the shapes of the balls and
to the fact that the first ball is spinning, this cannot be completely assured, but if the centers
of the two balls are at the same vertical height when they collide, the two will come out of the
collision with their initial velocities very close to horizontal.
To adjust the height so this will be the case, proceed as follows:
Place a piece of carbon paper in contact with a piece of white paper just past the end of the
ramp, as shown below. The ball should strike the paper just after it loses contact with the ramp.
The smooth edge of the white paper should be contact with the tabletop. The mark made on the
paper when the balls collide will lie at a distance from this edge which is equal to the height of
the point of collision above the tabletop.
If you place a hard, flat solid object (which should also be either cheap or unbreakable and
not subject to denting) behind the paper, oriented so that its flat side is vertical, then when the
ball strikes it will leave a mark indicating the height of its center above the table.
You should do this three times, positioning the system so that the ball will collide just after
leaving the ramp. The centers of your three marks should all line along or very close to a single
straight horizontal line, all at very nearly the same distance from the edge of the paper. If
necessary repeat your trials until you are sure the system is properly set up to give you
consistent results.
If you place the smaller ball on the tee behind the paper, then the collision will produce a
mark at the point of contact of the two balls.
If the center of the mark made by colliding the balls is at the same height as the centers of
the marks made by the ball against the flat object, then the centers of the balls will be at equal
heights. If not, adjust the length of the 'tee'.
Use the long 'track' as the incline, and the short piece of track as the horizontal section. The
high end of the long incline should be about 5 cm higher than the short end. This vertical
distance should be measured and should be kept the same throughout your trials, but it doesn't have
to be exactly 5 cm. For example 4 cm, or 6 cm, or 5.37 cm would be fine, as long as it is measured
accurately and checked repeatedly to be sure it doesn't change after being set up.
The height of the top of the 'tee' supporting the target ball should also be checked throughout
your trials to ensure that it doesn't change.
You should therefore check the height of the end of the ramp, and the height of the top of the
straw, periodically throughout the experiment. You should note these 'maintenance checks' in your
lab notebook, noting when they were done and verifying that the slope of the ramp and the height of
the 'tee' has not changed.
The large ball must be just out of contact with the ramp at the instant of collision, being no more
that a couple of millimeters from the point where it leaves the ramp, and after collision both
balls should 'clear' the edge of the table as they fall. If they don't, then the system can be
moved closer to the table's edge, and/or the slope of the inclined ramp could be increased to give
the ball greater velocity.
Proceed to adjust the height of the 'tee' until the balls collide with their centers at the same
vertical altitude. In the space below, give in the first line the measurement from the edge of the
paper to the mark made by the ball as it strikes the vertical object, and from the edge of the
paper to the mark made by the collision of the two balls. In the second line give the height of
the top of the 'tee' above the tabletop. In the third line give the distance from tabletop to
floor. Make both all measurements as accurate as possible, and indicate in the fourth line the
uncertainty in each of your measurements and how these were estimated:
-------->>>>>>>> collision pt 1 ball against vert and coll pt 2 balls, ht of top of tee above
tabletop, tabletop to floor, uncertainties
Your answer (start in the next line):
1.7, 1.72
1.08
74.5
Collision point of large ball with book, in cm from edge of paper, and collision of two balls, in
cm from edge of paper. Height in cm of tee from tabletop. Height in cm of tabletop from floor.
Uncertainty +-1mm for collision marks (as marks are faint and difficult to distinguish the center
of the mark), and +-2mm for measurements involving tabletop and floor, as both are older wood and
somewhat scratched.
#$&*
Run your first set of trials:
Now you will remove the 'tee' and release the ball from the rest at the high end of the sloped
track. You will use the same procedures as in previous experiments for observing the horizontal
range of the ball as it falls to the floor.
Be sure the ramps remain well aligned, and if necessary 'shim' the end of the inclined ramp to
ensure that there is no 'bump' when the ball moves from one ramp to the next.
Conduct 5 trials, and in the first line give 5 horizontal ranges; in the second line give the mean
and standard deviation of the range of the ball. Starting in the third line explain in detail how
you got your results.
-------->>>>>>>> 5 ranges uninterrupted, mean & sdev, explanation
Your answer (start in the next line):
25.80, 25.79, 25.25, 24.70, 24.52
25.21, .59
Horizontal range in cm from tee, larger ball alone; mean and std dev of horizontal range. Observed
by allowing larger ball to roll from about .5cm from top of angled ramp (there is a bend in one of
the rails of the track at the very top, which was making the results inconsistent until I lowered
my starting point). Marked center of landing point as marked on paper, measured to edge of paper +
12.5cm (distance from edge of paper to point on floor directly below the tee).
#$&*
Now place the target ball at the edge of the table, as described earlier. Measure the distance in
cm from the edge of the ramp to the closest point on the straw.
Align the target ball so that after the collision, the 'forward' paths of both balls are in the
same direction as that of the uninterrupted first ball. That is, make sure the collision is
'head-on' so that one ball doesn't go to one side and the other to the opposite side of the
original path.
Divide the carbon paper into two pieces, and position the two in such a way that after collision
the two balls will leave clear marks when they land. Do this until you get marks for five trials.
Be sure to note which second-ball position corresponds to which first-ball position (e.g., number
the marks).
Using your marks, determine the horizontal ranges of the two balls after collision.
In the first line of the space below, give the five horizontal ranges observed for the second ball,
using comma-delimited format. In the second line give the corresponding first-ball ranges. In the
third line give the mean and standard deviation of the second-ball ranges, and in the fourth line
give the same information for the first ball. Starting in the fifth line specify how you made your
measurements, and as before specify the positions with respect to which you found your ranges, as
well as how you measured those positions.
-------->>>>>>>> five ranges target ball, five ranges first ball, mean and std second, mean and std
dev first ball, details
Your answer (start in the next line):
39.22, 38.81, 39.25, 40.30, 40.55
21.91, 21.78, 21.05, 20.80, 20.32
39.63, .75
21.17, .67
Horizontal ranges of small ball, post-collision; horizontal ranges of large ball, post-collision;
mean and std dev of small ball; mean and std dev of large ball.
Leaving the paper exactly 12.5cm from the point on the floor corresponding with the tee, marked
center of each pair of landing positions, measuring to edge of paper and adding 12.5cm.
#$&*
Do not disassemble the system until you are sure you are done with it. General College Physics and
University Physics students will use the system again in subsequent activities, and should leave it
as it is.
Analysis of Results from First Setup:
Give in the first line below the vertical distance through which the two balls fell after
collision, and in the second line the time required to fall this distance from rest. Starting in
the third line, explain precisely how you determined these distances, how you determined the time
of fall and what assumptions you made in determining the time of fall:
-------->>>>>>>> vertical fall, time to fall, explanation
Your answer (start in the next line):
75.58
.39
Vertical displacement of both balls, in cm; time required to free-fall that distance from rest, in
s.
Distance is sum of floor-to-tabletop and tabletop-to-tee measurement. Time calculated as follows:
vf^2=v0^2+2a`ds
vf=sqrt(2*980cm/s^2*75.58cm)=384.88cm/s
vAve=192.44cm/s
192.44cm/s=75.58cm/`dt
`dt=.39s
This assumes uniform acceleration and negligible air resistance.
#$&*
In the space below give in the first line the velocity of the first ball immediately before
collision, the velocity of the first ball immediately after collision and the velocity of the
second ball immediately after collision, basing your calculations on the time of fall and the mean
observed horizontal ranges. In the second line give the before-collision velocities of the first
ball based on (mean + standard deviation) and (mean - standard deviation) of its uninterrupted
ranges. In the third line do the same for the first ball after collision, and in the fourth line
for the second ball after collision.
-------->>>>>>>> velocity first ball before, first ball after, second ball after collision; mean +-
std dev first ball before, after, 2d ball after
Your answer (start in the next line):
64.6, 54.3, 101.6
64.6+-1.5 (63.1-66.1)
54.3+-1.7 (52.6-56.0)
101.6+-1.9 (99.7-103.5)
Velocity of first ball before collision in cm/s calculated using uninterrupted horizontal range and
fall time; velocity of each ball after collision in cm/s calculated using post-collision
horiztontal range and fall time.
Each velocity then given +- standard deviation.
#$&*
The masses of both balls are unknown. Using momentum conservation, you will determine the ratio of
their masses:
Let m1 stand for the mass of the large ball and m2 the mass of the small ball. In terms of m1
and m2 write expressions for each of the following:
The momentum of the first ball immediately before collision, using the velocity you
reported above (the velocity based on the mean range and distance of fall). Be sure to use both
the numerical value of the velocity and its units. This will be reported in the first line below.
The momentum of the first ball immediately after collision, using the velocity you reported
above. This will be reported in the second line below.
The momentum of the second ball immediately after collision, using the velocity you
reported above. This will be reported in the third line below.
The total momentum of the two balls immediately before collision. This will be reported in
the fourth line below.
The total momentum of the two balls immediately after collision. This will be reported in
the fifth line below.
The total momentum immediately before collision is equal to the total momentum immediately
after collision. Set the two expressions equal to obtain an equation. Report this equation in the
sixth line below.
-------->>>>>>>> equation for momentum conservation
Your answer (start in the next line):
64.6cm/s*m1
54.3cm/s*m1
101.6cm/s*m2
(64.6cm/s*m1 + 0)
(54.3cm/s*m1 + 101.6cm/s*m2)
64.6cm/s*m1 = 54.3cm/s*m1 + 101.6cm/s*m2
#$&*
Rearrange your equation so that all terms containing m1 are on the left-hand side, and all terms
containing m2 are on the right-hand side. Report this equation in line 1 below.
Divide both sides by the appropriate quantity so that m1 appears by itself on the left-hand side.
Report the resulting equation in line 2.
Divide both sides of the equation by m2, and report the resulting equation in line 3.
Simplify the right-hand side, if you have not already done so, to obtain a single number. If you
have done your calculation correctly, the units will cancel out. Report the resulting equation in
line 4. The left-hand side will be m1 / m2 and the right-hand side will be a single decimal number
or, if you prefer, a reduced fraction.
Starting in the fifth line discuss the meaning of the ratio m1 / m2.
-------->>>>>>>> equation solution in steps, meaning of ratio m1 / m2
Your answer (start in the next line):
64.6cm/s*m1-54.3cm/s*m1=101.6cm/s*m2
10.3cm/s*m1=101.6cm/s*m2
m1/m2=101.6cm/s/ 10.3cm/s
m1/m2=9.9
#$&*
Measure and report the diameter of ball 1 and the diameter of ball 2, in comma-delimited format in
the first line below.
Calculate the volumes of the two balls and report them in the second line.
-------->>>>>>>> diameters, volumes
Your answer (start in the next line):
2.42cm, 1.35cm
7.43cm^3, 1.23cm^3
#$&*
Physics 121 students are not required to continue, but may do so
Error Analysis for First Setup:
If at collision the center of the first ball is higher than the center of the second, how will this
affect the magnitude and direction of the velocity of the first ball immediately after collision?
Will the speed be greater or less than if the centers are at the same height? Will the direction
of the after-collision velocity differ, and if so how?
In the space below answer this question, and also answer the same questions for the second ball.
-------->>>>>>>> if first ball higher what is effect on its motion, same question for second ball
Your answer (start in the next line):
I don't believe that the velocity of the first ball would change very much, particularly in my case
here where m1 is so much greater than m2 (I used two glass marbles and didn't even realize they
differed that much). The velocity will still be slower than it was when uninterrupted, because
some of the KE will transfer to m2, but its direction, I would expect to be pretty much the same.
The second ball would likely not travel as far, because the force of the first ball would be
directed at a downward angle onto the second (smaller) ball, so its net force will include LESS
horizontal force and MORe downward vertical force-- meaning it will cover less horizontal distance
during the time it takes to fall. In other words, its horizontal velocity will be less from the
outset, and its vertical velocity will be more.
#$&*
How do you think this will affect the horizontal range of the first ball? How will it affect the
horizontal range of the second?
-------->>>>>>>> effect on horizontal ranges
Your answer (start in the next line):
The first ball will travel about the same horizontal distance-- perhaps very slightly farther.
The second ball will travel less horizontal distance.
#$&*
For the first ball before collision you reported an interval of velocities based on mean + std dev
and mean - std dev of observed horizontal ranges. You did the same for the first ball after
collision, and the second ball after collision. Each of these intervals includes a minimum and a
maximum possible velocity.
What do you get for the ratio of masses if you use the minimum before-collision velocity in the
interval reported for the first ball, the maximum after-collision velocity for the first ball, and
the minimum after-collision velocity of the second? Report how you determined this ratio in the
space below:
-------->>>>>>>> mass ratio using min before, max after 1st ball, min after 2d
Your answer (start in the next line):
m1/m2=14.04
63.1*m1=56.0*m1+99.7*m2
7.1*m1=99.7*m2
m1/m2=14.04
#$&*
What percent uncertainty in mass ratio is suggested by comparing this result to your original
result?
-------->>>>>>>> % uncertainty suggested by previous
Your answer (start in the next line):
70%
#$&*
General College Physics students may stop here. University Physics students will continue with
additional error analysis, then with an investigation of how errors in the relative heights of the
two balls at collision might affect results.
Suppose you can choose either the maximum or minimum of the velocities in each of the reported
velocity intervals. What combination of before-and after-collision velocities gives you the
maximum, and what combination gives you the minimum result for the mass ratio?
-------->>>>>>>> combination of three obs velocities to get max, min mass ratio
Your answer (start in the next line):
Max pre-collision v for ball 1, min post-collision v for ball 1, max post-collision v for ball 2:
m1/m2=lowest, at 7.7
Min pre-collision v for ball 1, max post-collision v for ball 1, min post-collision v for ball 2:
m1/m2=highest, at 14.0
#$&*
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? You will
solve the same equation in the same manner as before, but use the symbols v1, u1 and u2 instead of
the numerical results you used earlier:
-------->>>>>>>> mass ratio using symbols
Your answer (start in the next line):
m1/m2=u2/(v1-u1)
#$&*
You just obtained an expression for m1 / m2 in terms of v1, u1 and u2. Treating v1 as a
variable and u1 and u2 as constants, take the derivative of this expression with respect to v1.
Give your expression in the first line of the space below. In the second line give the value of
this expression for your velocities v1, u1 and u2, as you reported them earlier based on the mean
value of each. Include units in your result. Starting in the third line explain what you think
this quantity might mean and how it might be related to error analysis.
-------->>>>>>>> derivative wrt v1;value for observed v1, u1, u2 incl units; interpret
Your answer (start in the next line):
f'(v1)= -u2/(v1-u1)^2
f'(v1)=-101.6cm/s /(64.6cm/s-54.3cm/s)^2
f'v1=-.95cm/s
This figure is the rate at which v1 changes with respect to changes in the predicted mass ratio.
This tells us that errors in calculating either quantity can affect the other quantity by as much
as .95 units.
#$&*
The derivative you just reported is the rate at which the predicted mass ratio changes with
respect to v1; that is, your derivative is the approximate value of
approximate derivative = (change in mass ratio) / (change in v1)
Again, this is an approximation, for small values of the change in v1.
It follows that the change in the mass ratio as predicted by this experiment is equal to
the value of the derivative, multiplied by the change in v1:
approximate change in predicted mass ratio = approximate derivative * change in v1
v1 is the before-collision velocity of the first ball.
Answer the following:
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? Answer in the first line below.
If v1 changes by this amount, then based on the numerical value you reported for above
the derivative with respect to v1, by how much would the predicted mass ratio change? Answer in
the second line below.
Starting in the third line explain how you got your results and what they mean in terms of
this experiment. You might also address how this chain of reasoning illustrates the chain rule for
derivatives. You might also consider how this analysis illustrates the meaning of the differential
of a function. These applications of calculus to the process of error analysis are subtle but very
important.
-------->>>>>>>> change in v1 if range changes by std dev ; resulting change in mass
ratio; discussion
Your answer (start in the next line):
If the first ball's range changes by .59cm, the predicted value of v1 changes by 1.5cm/s.
If v1 changes by 1.5cm/s, then:
-.95cm/s=change in mass ratio / 1.5cm/s
@&
I believe the units of the -.95 quantity are reciprocal (cm/s), so that the units would divide out, leaving the change in the mass ratio unitless. Since the mass ratio is a unitless quantity, this is as expected.
*@
For a change in v1 equaling -.95cm/s, our predicted mass ratio will change by -1.43 (unitless
ratio). This is another indication of how an error can be magnified through each time the
derivative is taken (in other words, an error in one calculation shows up as a more significant
error when looking at a rate involving that calculation).
#$&*
Repeat the Experiment with Second Setup (University Physics Students Only):
The second setup will change the system so that the target ball is 2 mm higher than before. The
same observations will be made. Analysis will take account of the non-horizontal direction
velocities of the balls immediately after collision.
Repeat the preceding experiment with the target ball 2 mm higher than before:
Set up the system with the second ball 2 millimeters higher than before.
Repeat the preceding experiment with this new setup.
Analyze as before, including the previous assumption that the two balls both have negligible
vertical velocity after collision. This assumption is no longer completely valid, and we will soon
correct for this.
Give a summary of your data and your prediction of mass ratio, and compare to your previous
prediction of mass ratio
-------->>>>>>>> summary expt repeated 2 mm higher
Your answer (start in the next line):
25.21, .59
20.10, 19.88, 19.24, 19.05, 18.70
19.39, .58
42.85, 42.68, 42.60, 41.35, 41.12
42.12, .82
Mean horizontal distance in cm, standard dev of horizontal distance for uninterrupted ball 1
(calculated earlier), for ball 1 post-collision (with 5 trials) and ball 2 post-collision (with 5
trials).
64.6, 49.72, 108.0
Horizontal velocity, in cm/s, for uninterrupted ball 1, ball 1 post-collision and ball 2 post-
collision
m1/m2=7.26
Predicted mass ratio usuing m1/m2=u2/(v1-u1)
This is relatively close to the lowest possible mass ratio predicted using the earlier method.
#$&*
Make a detailed sketch of the two balls at collision:
Sketch a horizontal line on a sheet of paper.
Place a dot on that horizontal line, representing the position of the center of the first ball
at the instant of collision.
Sketch a circle representing the first ball, making the circle about the same size as the ball.
Sketch a dot which represents the center of the second ball at collision, and a circle
representing that ball, again more or less actual size. The sketch should be more or less to
scale, with the second dot 2 mm higher than the horizontal line.
The velocity of the second ball immediately after collision will be in the direction of the line
connecting the centers of the two balls. Find the slope of this line:
Sketch a right triangle whose hypotenuse runs from the center of the first ball to the center
of the second, with one vertical and one horizontal leg. The vertical leg of the triangle will
consist of the 2 mm vertical segment from the line to the center of the second ball.
You know the length of the hypotenuse and the vertical leg of the triangle. What therefore is
the length of the horizontal leg? Is this length significantly different than the hypotenuse?
The vertical and horizontal legs represent the rise and the run between the centers of the two
balls. What therefore is the slope of the line segment connecting the centers?
Recall that the data analysis program will give you the velocity of a projectile if you give it the
horizontal range and initial slope of its path. This feature was applied in a previous experiment
to a ball rolling off an incline with given slope. It can also be applied to this situation.
Run the data analysis program.
Click on 'Experiment-Specific Calculations'
Enter the number for the Ball and Ramp Projectile Experiment.
When prompted, enter the vertical drop of the second ball.
When prompted, enter the mean horizontal range observed for the second ball.
Rather than the slope, you will be asked for the 'number of dominoes'. The program uses the
number of dominoes to calculate the slope, based on a 30 cm ramp; for small slopes each domino
changes the slope of a 30 cm ramp by approximately .03 so if you divide the slope by .03 you get
the equivalent number of dominoes (e.g., if the slope is .16 the equivalent number of dominoes is
.16 / .03 = 5.33). However note the following qualification:
If the slope is not small then the number of dominoes can be calculated by dividing the slope
by .03, as before, then dividing the result by sqrt( 1 + m^2), where m is the slope.
You obtained a standard deviation for the observed ranges of the second ball.
Repeat the calculations, this time using the horizontal velocity predicted by mean + std
deviation of the horizontal ranges of the second ball.
Repeat once more using the velocity predicted by mean - std deviation of horizontal ranges.
In the space below:
Report the vertical drop of the second ball, its mean horizontal range and the slope of the
line segment connecting the two centers.
In the second line enter the velocity given by the program, based on the mean horizontal range.
In the third line enter the velocity interval obtained this run of the experiment, giving the
second-ball velocities based on mean horizontal range + standard deviation and on mean horiz range
- std dev. Give these quantities in cm/sec as two numbers in comma-delimited format.
In the fourth line enter the velocity interval obtained in the original run of the experiment,
giving the second-ball velocities based on mean horizontal range + standard deviation and on mean
horiz range - std dev. Give as two numbers in comma-delimited format.
In the fifth line give the difference in the mean-based 2d-ball velocity for the original run,
and the mean-based velocity for the current run.
Starting in the sixth line state, based on the information given above, whether the velocity
resulting from the 2 mm increase in 2d-ball altitude appears to be significantly different than
that obtained when the centers were at the same level.
-------->>>>>>>> target 2 mm higher 2d ball vert drop mean range slope of segment connecting
centers; velocity based on mean horiz range; vel interval, vel interval 1st run, difference in vel
between runs, is difference significant
Your answer (start in the next line):
75.8cm, 42.12cm, .01
106.3cm/s
104.2cm/s, 108.3cm/s
99.7 cm/s, 103.5cm/s
The difference is significant-- when the second ball started .02cm higher, it experienced a vAve
5cm/s greater-- but I'm not sure how the .2mm difference plays into this. The velocity vector for
this ball is positive, whereas in previous experiments it would have been negative (the dominos, or
slope, would result in the ball starting out ""downhill,"" whereas this experiment has it starting
out ""uphill"" after being struck lower on its surface.)
#$&*
In the space below give a similar report comparing the first-ball velocity obtained with the second
ball 2 mm higher, with the first-ball velocity obtained with the centers at the same height.
-------->>>>>>>> same info 1st ball
Your answer (start in the next line):
49.1 cm/s
47.61 cm/s, 50.53 cm/s
52.6 cm/s, 56.0 cm/s
Again, the difference is significant, but in the opposite ""direction."" When it strikes below the
center of ball 2, ball 1 acutally experiences a lower velocity, by about 5cm/s.
@&
It isn't certain that in the first setup the balls actually struck in a perfectly 'head-on' manner. This can be difficult to ascertain, especially since the velocity of the first ball was in fact a bit below horizontal.
We could have investigated this, but this experiment is already long enough.
*@
#$&*
The first run was set up so the two centers would be at the same altitude. You took a certain
amount of care to ensure that this was so, within the limits of precision possible with available
apparatus. What do you think were the limits of your precision?
Report in the following manner.
Since they were ideally at the same altitude, the relative height of the second ball relative
to the first was ideally 0; its actual altitude is therefore 0 +- uncertainty. If you were careful
at the beginning of the experiment, measuring as precisely as possible the distances of the dots
made by the carbon paper when the balls collided, your uncertainty will probably be less than 1
millimeter.
In the first line, give the uncertainty in millimeters (e.g., .5 mm, .8 mm, .03 mm), and state
beginning in the second line the reasons for the number you give.
-------->>>>>>>> uncertainty in relative ball hts, reasoning
Your answer (start in the next line):
.75mm
The dot itself has a size, a radius. Trying to estimate the ""center"" of the dot, when they are
slightly different shapes (I suppose based on slight imperfections in the surface of the marbles),
is not entirely exact. I wouldn't expect to be off by a whole millimeter, but it is possible that I
was off by nearly that.
#$&*
Based on the results you have obtained to this point, argue for or against the hypothesis that the
uncertainty in the relative heights of the balls was not a significant factor in the first setup.
-------->>>>>>>> argue for and against hypothesis
Your answer (start in the next line):
Since we see that 2mm deviation from a center-to-center collision can result in very significantly
different results in the velocity calculation, then yes, a .75mm error for the first section of the
lab could be quite significant. If, for instance, the center of the 2nd ball actually started out
more than a half millimeter below the center of the first ball, it would have a downward vertical
component to its velocity as the larger ball would be striking it from slightly above.
#$&*
Based on the slope of the initial after-collision path of the second ball and its velocity, as
determined in the second setup, what is the vertical component of the second ball's immediate
after-collision velocity?
-------->>>>>>>> init vert comp of 2d ball vel
Your answer (start in the next line):
1.1cm/s
#$&*
Based on this result and on the mass ratio determined in the first setup, what is the vertical
velocity of the first ball immediately after collision? Note that vertical momentum is conserved,
and that since immediately before collision nothing was moving in the vertical direction, the total
vertical momentum is zero immediately before collision.
-------->>>>>>>> infer init vert vel of 1st ball
Your answer (start in the next line):
.11cm/s
#$&*
What is the horizontal velocity of the first ball before collision and the horizontal velocity of
the second ball after collision? Based on the mass ratio obtained in the first setup, what
therefore is the horizontal velocity of the first ball after collision?
-------->>>>>>>> horiz vel of 1st before and of 2d after; iner horiz of 1st after
Your answer (start in the next line):
64.6cm/s, 106.3cm/s
53.87cm/s
#$&*
Using these horizontal and vertical velocities, what should be the horizontal range of the first
ball after collision?
-------->>>>>>>> predicted horz range of 1st
Your answer (start in the next line):
Need to find `dt when there is an initial upward vertical velocity:
vf^2=v0^2+2a`ds
vf^2=(.11cm/s)^2+2(-980cm/s^2)(-75.8cm)
vf=-385.4cm/s
vAve=-192.7cm/s
-192.7cm/s=-75.8cm/`dt
`dt=-75.8cm / -192.7cm/s
`dt=.4s
Applied to horizontal range of first ball:
`ds=.4s*53.87cm/s
`ds=21.2cm
This is just under 2cm farther than the mean distance I actually observed.
#$&*
Before collision, the first ball was actually spinning fairly rapidly. When the balls collide, the
coefficients of static and kinetic friction are those between two smooth steel surfaces. Can you
make a reasonable estimate of the relative spinning rates of the two balls after collision? Would
the fact that the first ball is spinning affect the path of the second ball, or just its spin?
-------->>>>>>>> estimate rel spin rates after collision, effect on path of 2d ball
Your answer (start in the next line):
I would imagine that since the first ball is spinning forward and is larger than the second, it strikes the second ball at a ""downward"" angle even if the centers are even. The surface of the larger ball is spinning toward the srufece of the smaller ball from above, so it's possible that it causes a slight downward vertical component to the smaller ball's velocity even if the collision is head-on.
The circumfrence of the larger ball is about 7.6cm.
The circumfrence of the smaller ball is about 4.3cm.
I'm not sure how to estimate the spinning rate, except that based on the two circumfrences, the larger ball would have to be spinning nearly twice as fast if the two were travelling at the same rate.
#$&*
What is the slope of the line connecting the centers of the two balls? Hint: The rise is 2 mm.
The run is very nearly equal to the length of the line segment between the centers. You will
report this slope in the first line below.
Using this slope and the mean of the observed second-ball ranges, what does the program give you
for the after-collision speed of the second ball? Report in the second line.
Starting in the third line report exactly what numbers you gave the program, and how you determined
each of these numbers.
-------->>>>>>>> slope of segment connecting centers, after coll speed of 2d based on this slope
and on ball range, numbers into program and how determined
Your answer (start in the next line):
Slope is .01.
Calculated speed: 106.3cm
Vertical drop: 75.8
Horiztonal range: 42.12
Number of dominoes: .33 (based on your instruction to convert to number of dominoes by dividing slope by .03)
Again, I'm confused because the slope seems to be facing the opposite direction from the other times we've used this program. The rise here is positive, correct?
@&
You would use a negative slope, as you previously used a positive number for the negative slope. However the .01 slope won't make a lot of difference in any case.
*@
Also, I'm thrown by this question appearing now because this is the method I already used to determine the velocity of the second ball earlier in the analysis of the second set-up. So I may need to go back and do that section a different way.
@&
You're doing very well on this experiment. I won't be asking you to redo anything.
*@
#$&*
Your instructor is trying to gauge the typical time spent by students on these experiments. Please
answer the following question as accurately as you can, understanding that your answer will be used
only for the stated purpose and has no bearing on your grades:
Approximately how long did it take you to complete this experiment?
-------->>>>>>>>
Your answer (start in the next line):
3.5 hours
#$&*
@&
Excellent work, but do check out my notes.
*@