Note that the data program is in a continual state of revision and should be downloaded with every lab.

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experiment has been revised since those videos were recorded, and any specific instructions given on the videos are superceded by the instructions given here, and by more recent practices in observing horizontal range.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 straw (or simply slide it up or down on 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 straw 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 box 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.  Make both measurements as accurate as possible, and indicate in the third line the uncertainty in your measurement:

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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, being sure to include the specifics of the ball position with respect to which you measure the range, and how you measured that position:

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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 box 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, and how you measured those positions.

The second ball will clearly travel further than the first ball, since the first ball cannot pass through the second in order to get ahead of it.

It should be clear that the collision interferes with the motion of the first ball, reducing its velocity and hence its horizontal range.  The ranges of the first ball should be less than in the previously reported trials, in which the first ball traveled from the end of the ramp to the floor without interference. 

That is, if the distances you reported for the first ball here exceed those you reported in the first question, you've done something wrong and need to fix it (or ask additional questions).

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:

If the balls are aligned as in the instructions, so that they collide immediately after the first ball leaves the ramp, and collide with their centers at the same height, they should both begin their fall with zero vertical velocity.  This makes it easy to solve the equations of uniformly accelerated motion for the vertical fall (initial velocity zero, acceleration equal to that of gravity, displacement is the drop to the floor).  The recommended solution is to use the fourth equation of motion to find the final velocity (the sign of the final velocity can be + or -, but must agree with the signs of the vertical displacement and acceleration).  Then either direct reasoning or the first (or second) equation of motion will easily yield the time of fall.

In the box 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.

The horizontal velocity doesn't change during the fall, so the horizontal velocity of each ball just after the collision is equal to its average horizontal velocity during the fall.  This is easily found by dividing the horizontal range by the time of fall.

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.

The total momentum after collision is easily found in terms of m1 and m2.  If the respective velocities are, say, 30 cm/s and 70 cm/s, the total momentum is 30 cm/s * m1 + 70 cm/s * m2.  The total momentum before collision is just the momentum of the first ball, since the second is stationary until it is hit.  If its before-collision velocity is, say, 40 cm/s, then the before-collision momentum would be 40 cm/s * m1.  If we set these two expressions equal we obtain an equation with unknowns m1 and m2.  For the present example the equation would be

30 cm/s * m1 + 70 cm/s * m2 = 40 cm/s * m1. 

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.

If the equation is

30 cm/s * m1 + 70 cm/s * m2 = 40 cm/s * m1 then we can solve for m1 / m2 as follows.  First subtract the right-hand side from both sides to get 

30 cm/s * m1 + 70 cm/s * m2 - 40 cm/s * m1 = 0 then subtract 70 cm/s * m2 from both sides to get

30 cm/s * m1 - 40 cm/s * m1 = -70 cm/s * m2.  Simplify the left-hand side to get

- 10 cm/s * m1 = -70 cm/s * m2.  Divide both sides by m2 to get

-10 cm/s * m1 / m2 = -70 cm/s, then divide both sides by -10 cm/s to get

m1 / m2 = -70 cm/s  / (-10 cm/s)  = 7. 

That is, m1 / m2 = 7.  The first ball in this example has 7 times the mass of the second.

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.

You should know that the volume of a sphere is 4/3 pi r^3, where r is its radius.  For example if a ball is 2.4 cm in diameter, its radius is 1.2 cm so its volume would be 4/3 pi ( 1.2 cm)^2 = 4/3 pi * 1.73 cm^2, which can be multiplied out (the result would be about 7 cm^3).

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 box below answer this question, and also answer the same questions for the second ball.

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How do you think this will affect the horizontal range of the first ball?  How will it affect the horizontal range of the second? 

If the first ball is higher than the second, then as a result of the collision the second will be given at least a bit of downward vertical velocity, and the first a bit of upward vertical velocity.  This will tend to cause the second ball to reach the floor in less time than before, and the second to take more time. If the horizontal speeds are the same as before, the first ball will therefore travel further (moer time at the same speed) than if the centers are at the same height, and the second won't travel as far. 

However we do expect the horizontal velocity to be affected by the fact that the collision is off-center.  We expect that the first ball will lose less of its momentum, and the second will gain less momentum, than if the centers are at the same height.  So the first ball will maintain more of its velocity, and the second will have less velocity, than before.  This will tend to further reduce the horizontal range.

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 box below:

For example if the intervals are respectively (38 cm/s, 42 cm/s) (first ball before collision), (29 cm/s, 31 cm/s) (first ball after collision) and (67 cm/s, 73 cm/s) (second ball after collision), then you would repeat the procedure for finding the mass ratio, but this time using 38 cm/s, 31 cm/s and 67 cm/s to get the equation 

31 cm/s * m1 - 38 cm/s * m1 = -73 cm/s * m2. 

The ratio m1 / m2 will be about 10.4, which significantly different than the ratio 7 obtained before.

What percent uncertainty in mass ratio is suggested by comparing this result to your original result?

The ratio 10.4 differs from the ratio 7 by 3.4, which is 3.4 / 7 = .47 of the original ratio.  This suggests a possible 47% uncertainty in the mass ratio. 

This is probably more uncertainty than you would have anticipated.  It occurs mainly because the left-hand side of the equation consists of the difference between two quantities which aren't all that different.  Slight changes in these two quantities can result is proportionally large changes in the left-hand side, which then result in large changes in the solution (e.g., 30 cm/s - 40 cm/s is only -10 cm/s, fairly small compared to the quantities themselves; when these quantities change to 31 cm/s and 38 cm/s the result is -7 cm/s, significantly different than the previous -10 cm/s).

Principles of Physics students may stop here.  General College Physics and 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?

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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:

The equation would be

m1 v1 = m1 u1 + m2 u2, which rearranges to

m1 u1 - m1 v1 = -m2 u2 or

m1 ( u1 - v1 ) = -m2 u2, with solution

m1 / m2 = -u2/ (u1 - v1).  Thus the mass ratio is the ratio of second-ball final velocity to the difference in the before- and after-collision velocities of the

Only University Physics students are required to answer the questions in this section.  Additional exercises for General College Physics and University Physics students follow this section. 

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 box 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.

The expression is

m1 / m2 = -u2/ (u1 - v1).  If we multiply numerator and denominator of the right-hand side by -1 this becomes

m1 / m2 = u2 / (v1 - u1).   The derivative of this expression with respect to v1 is expressed as follows, where the ' indicates derivative with respect to v1:

(m1 / m2) ' = - u2 / (v1 - u1) ^2.  

For example if v1 = 40 cm/s, u1 = 30 cm/s, u2= 70 cm/s, we get

(m1 / m2) ' = - 70 cm/s / (40 cm/s - 30 cm/s)^2 = -.7 cm/s / (cm/s)^2 = -.7 / (cm/s). 

(m1 / m2) ' is the rate of change of the predicted ratio (m1 / m2) with respect to v1, so that (change in m1 / m2) / (change in v1) = -.7 cm/s, approximately, with v1 in the neighborhood of 40 cm/s and v2 and u2 equal to 30 cm/s and 70 cm/s.

(m1 / m2) ' = -.7 / (cm/s) would mean that for these values of v1, u1, u2, the predicted value of (m1 / m2) decreases by about .7 for every cm/s change in v1.  Another closely related interpretation is that, if u1 and u2 were somehow known, the value of (m1 / m2) would be uncertain by about .7 cm/s for every cm/s uncertainty in v1.

[Note on taking the derivative:  You should be sufficiently familiar with the rules for taking derivatives to understand the following explanation.  If this is not the case, you should review those rules.  If you want a review in question-answer format, request a link from the instructor:

The derivative of 1/x with respect to x is -1 / x^2. 

The derivative with respect to x of 1 / g(x) is easily found by the quotient rule to be ( (1) ' * g(x) - 1 * g ' (x) ) / g^2(x) = - g ' (x) / g^2(x).  (quotient rule is (f / g) ' = (  f ' g - f g ' ) / g^2; in this example f(x) is the constant function 1, whose derivative is zero).

The derivative with respect to x of 1 / (x + b) is therefore - (x + b) ' / (x + b)^2 = -1 / (x + b)^2. 

The derivative with respect to v1 of 1 / (v1 + b) would be - 1 / (v1 + b)^2. 

The derivative of u2 / (v1 - u1) with respect to v1 is of very similar structure.  u2 and u1 are regarded as constants, and the derivative is - u2 / (v1 - u1)^2. 

If you have a hard time seeing this with v1 as the variable then take the derivative of a / (x - b), using the constant-multiple and quotient rules.  You get - a / (x - b)^2.   Then substitute into this expression as follows:  substitute a for u2, x for v1 and b for u1.  You will get - u2 / (v1 - u1)^2.

Assuming fairly typical values v1 = 50 cm/s, u1 = 35 cm/s, u2 = 60 cm/s the derivative would be 

• - u2 / (v1 - u1)^2 = -60 cm/s / (50 cm/s - 35 cm/s)^2 = -.26 (cm/s)^-1, or alternatively -.26 s / cm. 

This derivative is the rate at which the predicted mass ratio changes, with respect to the velocity v1. 

If the derivative is -.26 (cm/s)^-1, this means that the predicted mass ratio changes by .26 for every cm/s of error or uncertainty in the velocity v1.  I wasn’t sure how to do this I consider myself well versed in derivatives and integrals but I am sure my above equation
isn’t right.</p>

If you still don't see how to do the derivative:

This might look more natural to you if you use plain, unadorned letters in place of u2, v1 and u1. Let's use a, x and b, respectively. The expression becomes

a / (x - b)

and the derivative is

-a / (x - b)^2.

There are two ways to calculate the derivative:

• The function is the product of the constant a and the function (x - b)^(-1). The derivative of the latter is (x - b) ' * (-1) ( x - b)^(-2).

• The function is of the form f / g with f = a, g = (x - b), so the derivative is ( (a) ' ( x - b) - a * (x - b) ' ) / (x - b) ^ 2 = (0 - a) / (x - b)^2 = -a / (x - b)^2.

Either way it should now be clear that the derivative of u2 / (v1 - u1) = - u2 / (v1 - u1)^2.

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 are subtle but extremely important applications of calculus to the process of error analysis.

For example if your derivative in the previous was -.26 (cm/s)^-1 and your uncertainty in v1 was +-1.5 cm/s, the uncertainty in the mass ratio would be - .26 (cm/s)^-1  * (+- 1.5 cm/s) = +-.36. 

If the mass ratio determined by your data was, say, 3.8, then our conclusion (based on the uncertainty of v1) might be that the mass ratio is 3.8 +- .36. 

There are also uncertainties in v2 and u2, so the actual uncertainty in this experiment is greater than that predicted based on just the uncertainty in v1.  However we will not pursue the complete analysis of uncertainties in this course; to do so would require the use of multivariable calculus and is beyond the scope of the course.

Repeat the Experiment with Second Setup (General College Physics and 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 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

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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 box 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.

The significance would depend on the standard deviation of your landing positions. Done carefully it is very conceivable that the landing positions for a setup would cluster with a centimeter of one another, in which case the difference of about 3 cm, reported by one student, would be significant.

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In the box below give a similar report comparing the first-ball velocity obtained with the second ball 2 mm higher, with the first-ball velocity as previously obtained with the centers at the same height.

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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.

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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.

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General College Physics students may stop here.  University Physics students should continue:

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? 

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.

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?

Using these horizontal and vertical velocities, what should be the horizontal range of the first ball after collision?

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?

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.

Addendum based on student question

This student obtained a second-ball after-collision velocity that was less than the first-ball after-collision velocity.  This is clearly impossible since the first ball cannot pass through the second, and is inconsistent with the student's data.  The student otherwise has an excellent analysis, but didn't fully document the part where this result was obtained.  The instructor requested clarification.

Student work is in regular type below, instructor comments in bold.

You said:

You report that the velocity of the second ball is less than the after-collision velocity of the first. Unless the first ball
somehow passes through the second, it doesn't seem possible that the first ball would travel further. If the second ball
travels further than the first, then its after-collision velocity would be greater.

The ball velocities are determined by their horizontal ranges and their time of fall. Your .39 s time of fall looks right for the
distance of the fall. With that time of fall, a horizontal range of, say, 20 cm would imply a velocity of about 50 cm/s.

But they ask for the velocities of each ball IMMEDIATELY before and IMMEDIATELY after the collision. The
expanded commentary says that:
________________________________________________________________________
The horizontal velocity doesn't change during the fall, so the horizontal velocity of each ball just after the collision is equal
to its average horizontal velocity during the fall. This is easily found by dividing the horizontal range by the time of
fall.
_________

Is this true? There must be a horizontal acceleration since the second ball started at 0 cm/s.

First, we can separate the motion into three phases: before, during and after the collision.

• What happens before and after is very simple.
• What happens during is extremely complicated and analyzing it would take us well beyond the scope of this course.
• However nothing moves very far during the collision.

So if we simply assume that everything is in the same position immediately after collision as immediately before adds no significant uncertainty to the uncertainties already present.

Now regarding your conclusion that there must be a nonzero horizontal acceleration:

There certainly must be, but the acceleration occurs during the collision. The horizontal velocity which is present immediately after the collision is attained during the collision.

We don't make any attempt to analyze what goes on during the collision.

• During the collision the spheres compress then expand in fairly complicated ways, resulting in quite complicated force vs. time and/or force vs. position functions.  Even with the tools of multivariable calculus and differential equations, much less the first-year calculus assumed for this course, we would be unable to analyze these properly.
• The question of what happens during the collision is therefore far beyond the scope of this course.
• The acceleration during the collision is far from uniform and the motion during collision cannot be analyzed in terms of this assumption.

Immediately after the collision the ball has a horizontal velocity that doesn't change (except by a negligible amount due to air resistance) until it hits the floor, so the motion is easy to analyze.

So to repeat:

• We have three phases of motion: before the collision, during the collision, and after the collision.
• Before the collision and after the collision we have uniform-acceleration situations.
• During the collision we don't attempt to analyze the forces or the motion.

One exception:

• The spheres do move a bit during the collision, so there's a little uncertainty in just where they are immediately after the collision.
• However steel spheres colliding at these speeds compress only a tiny bit.
• Using ideas we develop late in the course, we could actually analyze this sufficiently to put upper bounds on how far they compress and how far they travel during the compression and expansion.

  This turns out to be a fairly small fraction of a millimeter, so there isn't much error in assuming they are at the same point immediately after collision as they were immediately before.

• A more obvious bound is obtained by simply considering that the balls compress less than 1 millimeter, and they compress and expand very rapidly, so they aren't likely to remain in contact for more that a couple of millimeters.

This second bound is easier to get than the first.  It isn't as rigorous and isn't as tight. However the uncertainty that results from an error of a couple of millimeters in the immediate-after-collision positions is insignificant compared to other uncertainties inherent in this experiment (e.g., the center-to-center alignment of the spheres introduces far more uncertainty).

Using the data for the mean range = 23.46 cm and the mean time = 0.38 seconds, the average velocity was 61.34 cm/s. However, IMMEDIATELY after the collision it cannot achieve this speed since it started at 0 cm/s. Therefore, using the above information, I found the acceleration to be 325 cm/s^2. To find how much it accelerated right after the collision, I
assumed a collision time of 0.1 seconds. Multiplying this by the acceleration gave me the change in velocity during this
time which is 32.5 cm/s. Adding 32.5 cm/s to an initial velocity of 0 cm/s, you get the IMMEDIATE after collision
velocity of 32.5 cm/s. </p>

Given your assumptions this is a good analysis. However the collision lasts for much less than 0.1 second (2 mm at an average velocity of 30 cm/s would require less than .01 second), and as described above you can't analyze the motion during the collision by assuming uniform acceleration.


To find the before velocity of the first ball I used the same train of thought. The before velocity was found by dividing the
sum of the ramp distances (50cm) by the mean time (1.42 s) giving an avg velocity of 35.21 cm/s. However,
IMMEDIATELY before the collision, it would not be traveling at an avg velocity of 35.21 cm/s, but a final velocity of
70.42 cm/s. Since it started at 0cm/s, it must have accelerated to that final velocity to achieve the avg velocity of 35.21
cm/s. So the first ball is also accelerating.

You can measure the horizontal range of the ball much more accurately than the time it spends on the ramp. The horizontal range, along with the slope and distance of fall, determine the final velocity on the ramp. So this is the more accurate method of finding the before-collision velocity of the first ball.

However if the ball takes 1.42 s to accelerate uniformly through a displacement of 50 cm, your analysis gives the correct velocity.


The after velocity of the first ball was found by first dividing the mean range (16.94 cm) by the mean time of fall (0.38 s).
This gave me an average velocity of 44.58 cm/s. Since the initial velocity is 70.42 cm/s (which is also the final velocity
from before the collision), I found the acceleration to be -136 cm/s^2. Multiplying this by an assumed collision time of 0.1
seconds, I get the change in velocity during collision to be equal to -13.6 cm/s. I then subtract this from the initial velocity
to get an IMMEDIATE after-collision velocity of 56.82 cm/s. </p>

The 44.58 cm/s after-collision velocity does follow correctly from your data.

The assumptions of uniform acceleration and a 0.1 second time interval during the collision are not valid assumptions, though your reasoning based on those assumptions is quite good.
 

???? I understand that the second ball must be going faster because the first ball cannot go through it. However, is it
possible that initially, right after the collision, the second ball is going slower yet accelerates much faster?

After the collision the second ball doesn't accelerate in the horizontal direction. Both balls, of course, accelerate at 980 cm/s^2 in the downward vertical direction.

Since the first ball already has a significant before-collision velocity, I find it hard to believe that in the split second of the collision, the second ball goes much faster and achieves its average velocity. Also, the Expanded Commentary says that the horizontal velocities are equal to the average velocities since they do not accelerate. Isn't this simply not true? Since the second ball starts at rest, as does the first, don't both ball's accelerate horizontally? ???????</p>

Before and during the collision there is plenty of acceleration going on, and as pointed out above some of it is very complicated. But after the collision everything is very simple. Horizontal velocities remain constant, vertical acceleration is the acceleration of gravity.


You pose a number of good questions, and have done some excellent analysis. You have some wrong assumptions about what goes on during the collision, and are probably confusing some of what happens during the collision with what happens after the collision. See my notes.