Unformatted text preview: Laboratory 7
In Laboratory 5, we showed that linear momentum is conserved in both inelastic and elastic collisions, whereas kinetic energy is only conserved in elastic collisions. We will be repeating these
analyses again this week in the context of two-dimensional collisions. The major difference in the
analysis is that in last week’s laboratory, we used a single calculation of total momentum before
and after the collision to establish whether momentum was conserved. In this week’s laboratory,
we will use two calculations: the total momentum in the x- and y- directions. This illustrates an
important point: every extra dimension in a problem leads to one more equation for vector quantities like momentum. However, for scalar quantities like energy, the conservation law can be
represented by a single equation no matter how many dimensions a problem involves. This is why
conservation of energy is such a powerful tool in multidimensional problems.
We will also look at one unique feature of inelastic collisions in two dimensions. When two
extended objects collide inelastically in two dimensions, they usually begin to rotate about their
combined center of mass. This makes their motion after the collision much more challenging
to analyze, which is why the collision problems you see in lecture usually deal only with point
masses. This lab is an introduction to the complex nature of collisions as they occur in the real
world. Theoretical Background: Collisions Revisited
The most important facts to remember when solving collision problems are:
1. Linear momentum is always conserved, in every dimension, for every type of collision, as
long as no external forces act on the colliding objects. When a collision involves only two
objects, we write this as
1 + 2 = constant.
Because momentum is a vector, we can break this equation into components and say that the
total linear momentum in the x-direction is constant and that the total linear momentum in
the y-direction is constant.
2. Total energy is conserved in the absence of external forces. The Law of Energy Conservation,
like the Law of Conservation of Linear Momentum, cannot be violated.
57 3. In elastic collisions, kinetic energy is conserved, so that
Ki = K f .
4. In inelastic collisions, kinetic energy is not conserved because some of it is converted into the
internal energies of the colliding objects. But since the total energy must still be conserved
(see #2) you can calculate exactly how much internal energy was gained during the collision
by looking at how much of the total kinetic energy was lost during the collision. Theoretical Background: The Center of Mass
In discussing momentum conservation, it is important to deﬁne one further quantity, which is the
center of mass of a system of particles - in our case, pucks. The center of mass of a system is a
speciﬁc point at which we can consider all of the system’s mass to be concentrated when solving
certain types of problems. For example, in physics we routinely consider extended objects such as
cars and elephants to be “point particles” when we are describing their motion using kinematics
equations or Newton’s Laws. In reality, what we are really doing is treating these objects as though
all of their mass is located at the center of mass.
The center of mass of a system of particles is deﬁned as an average of their vector positions i ,
weighted by their masses mi , or
cm = i
mtot = ∑ mi .
i This means that the center of mass of a system of particles will be closer to heavier particles and
further from lighter ones. This deﬁnition works for any number of point particles, and we can also
use it with extended objects (such as elephants) by considering them to be inﬁnite collections of
Problem 7.1 If a hockey puck of mass m1 is situated on the x-axis at x1 m,
and another hockey puck of mass m2 is situated on the x-axis at x2 m, where
is the center of mass of the system located? Where is the center of mass for
two pucks of equal mass?
We can also deﬁne a quantity called the center of mass velocity, which is deﬁned as
∑i mi i ∆ cm
∆t 58 (2) Note that the numerator in the middle equation is just the total linear momentum of the system.
Therefore, if momentum is conserved in a particular collision, the center of mass velocity should
be the same before and after the collision. Theoretical Background: Moment of Inertia Revisited
We ﬁrst encountered the concept of moment of inertia in Laboratory 6, in the context of rotational
motion. It is also important for any analysis of two-dimensional collisions between extended
objects. In particular, in an inelastic collision between two extended objects, some of their initial
translational kinetic energy from before the collision gets turned into rotational kinetic energy of
the combined object after the collision.
As we already know, one cannot obtain an object’s rotational kinetic energy from its rotational
speed, ω , without knowing its moment of inertia. In a situation where multiple extended objects
are colliding inelastically and sticking together, the moment of inertia of the combined object can
become quite difﬁcult to calculate. We can exploit two properties of the moment of inertia to
simplify our calculations.
First, in 2-D the moment of inertia is a scalar. That means that we can ﬁnd the moments of inertia
of the different parts of the combined object separately and then add them up algebraically to get
the total moment of inertia. Therefore, we should start by looking for ways to break the object into
pieces of simple geometric shapes.
Second, we can use the parallel axis theorem. The parallel axis theorem is a very powerful tool
in the study of rotational motion. If you know the moment of inertia of a body about an axis that
passes through its center of mass, you can ﬁnd its moment of inertia about any other axis parallel
to that axis with the parallel axis theorem, given by
I = Icm + Mh2 , (3) where M is the mass of the object and h is the perpendicular distance between the two parallel
axes. In other words, the moment of inertia of an object about any axis is equal to the moment of
inertia it would have about that axis if all of its mass were concentrated at its center of mass (Mh2 )
plus its moment of inertia about its own center of mass (Icm ).
Problem 7.2 Say you have two circular pucks of mass m and radius r sitting next to each other such that their edges are just barely touching. What is
their combined moment of inertia about the point at which they are touching? What if you had three pucks all sitting in a row like this, with their
edges just barely touching? What would be their combined moment of inertia about the center of the middle puck? 59 b V Figure 1: The image shows an impending inelastic collision. If the impact parameter is greater
than zero, rotation will occur. The length of the red line is equal to b, the impact parameter. The
impact parameter is the perpendicular distance between the projected velocity vector of the moving
particle and the center of mass of the stationary particle. Theoretical Background: Rotational Motion in Inelastic Collisions
When the collision of two extended objects is perfectly inelastic, they will rotate about their combined center of mass. The combined center of mass may reside within one of the objects, or it may
not, depending on the objects’ relative masses and geometries. (For example, the center of mass
of the two-puck system, with m1 = m2 , in Problem 7.1 lies not within either puck, but along each
puck’s edge at the sticking point.)
To see why this occurs, consider the two puck system shown in Figure 1. The blue puck is at rest,
while the green puck has velocity V . The anlge between the velocity vector and the line joining
each pucks’ center of mass is θ . If the velocity is collinear with the center of mass line (i.e. θ = 0),
the two pucks will not rotate about their combined center of mass after the collision; this is the
situation we encountered in Laboratory 5. However, if θ = 0, the two pucks will rotate about their
combined center of mass. The rate of rotation resulting from this sort of 2 dimensional encounter
is related to the impact parameter, b, which is the shortest distance between the velocity vector of
the moving object and the center of mass of the other. The larger the impact parameter, the higher
the rate of rotation after the collision. To keep the situation as simple as possible, we will only
analyze situations where one puck is initially at rest.
In Laboratories 1-5, we explored situations where all of the objects were treated as point particles:
their motion was purely translational. In Laboratory 6, we began to explore rotational motion, in
which extended objects rotate about their own centers of mass. In general, the motion of an object
or system of objects can always be broken into two parts:
1. The motion of the object/system’s center of mass.
2. The motion of each particle of the system relative to the system’s center of mass. Each
particle may rotate around the system’s center of mass, get closer or further from the center
of mass, or some combination thereof.
60 Theoretical Background: Reference Frames
For both of the collisions we will study in this laboratory, we will subtract off the motion of the
center of mass and only look at the motion of the components of the system relative to that center
of mass. Because momentum is conserved in the absence of external forces, the center of mass
velocity should be the same before and after the collision; therefore, removing that motion from our
analysis should not affect the conservation of energy or any other important aspect of the collision.
Analyzing data in this way is called “working in the center of mass reference frame”.
In general, whenever you take a measurement of an object’s motion, you are doing it relative to
some reference frame, whose motion you take to be uniform (stationary or moving at a constant
velocity). For example, in Laboratories 1 through 3, we took data in the reference frame of the
ultrasonic sensor, which we took to be stationary. In Laboratory 4, our reference frame was the air
table. In Laboratory 6, it was the inclined plane. In this laboratory, our reference frame will start
out as the air table. By subtracting the motion of the pucks’ center of mass relative to the air table,
we end up in the reference frame of the pucks’ own center of mass. You will learn more about
reference frames in your future studies of special relativity.
Problem 7.3 Say one puck moves toward another [stationary] puck along
a straight line on an air table. In the reference frame of the air table, only
the ﬁrst puck is moving. In the reference frame of the pucks’ center of
mass, how many pucks are moving? Why?
Problem 7.4 A [straight] river ﬂows along at r m/s, and a crocodile in the
river moves at c m/s upstream, where c is measured from the reference frame
of the river water. What are the speed and direction of the crocodile from
the reference frame of the river bank? What are the speed and direction of
the river bank from the reference frame of the crocodile? What are the speed
and direction of the river water from the reference frame of the crocodile? Experiments
Experiment 1: Two-Dimensional Elastic Collisions
Here we verify the conservation laws for energy and linear momentum in the context of elastic
collisions. We will use the same air table apparatus used in Laboratory 4, but we will use two
pucks and track the motion of each in the x and y directions.
1. Open 2dcollision.cmbl from the Logger Pro template folder. The video capture and video
analysis for the elastic collision is nearly completed for you in this ﬁle. Use the following
procedures to complete the analysis.
61 2. Use linear ﬁts in LoggerPro to ﬁnd the x and y-velocities of each puck before and after the
collision (see Appendix A Section 1.2).
3. Paste the graphs into your lab report. State what is moving and what the reference frame is.
4. Fill out the data table marked “FOR LAB FRAME DATA” in which you calculate the kinetic
energies of the pucks before and after the collision.
5. Leave your data and your video in the background of Logger Pro! You will need them
in Experiment 3! Also note that you will leave some graph boxes in Experiment 1 empty
on your lab template before coming back to them.
Experiment 2: Two-Dimensional Inelastic Collisions
Here we investigate a more complicated situation: an inelastic collision in two dimensions. The
procedure is much the same, except that after the collision the situation will be reversed from that
of the elastic collision; the pucks will orbit their center of mass but will not translate toward or
away from it.
1. Set up the air table similarly to how it was in Laboratory 4, except make sure it is level. (A
puck placed in the center should not move in any preferred direction.) If necessary, make
use of the leveling screws on the underside of the table for this purpose.
2. Position two equal-mass pucks (with Velcro bands) on the air table, with one puck stationary
at the center of the table and the other placed in the puck launcher. You will want to launch
this puck with substantial speed, so use a stiff rubber band or one of the front sets of launcher
3. Using the same video capture procedure that you used in Laboratory 4, take video of the
pucks as the one in the launcher is released, moves toward the other puck, and bounces off.
See Appendix A Sections 1.8 and 1.9 for how to set up your camera and capture video.
4. Analyze the video to obtain graphs of the x and y positions of each puck as functions of time.
Use the Add Point, Set Origin, and Set Scale buttons described in Appendix A Section 1.10 to analyze your video. After tracking the motion of one puck, click the
Set Active Point button, rewind the video, and track the motion of the other puck. Logger
Pro will automatically create a graph containing four separate data sets (x and y positions for
each puck). You will separate the graph into two graphs: one for the x-dimension and one
for the y-dimension. Refer to the procedure from Appendix A Section 1.3 for instructions
on how to make extra graphs. 62 5. Calculate the moment of inertia of the combined puck about its center of mass (the point
at which the two pucks meet, since they are of equal mass and radius). Note that you will
need to use Equation 3 in your calculation; see your solution to Problem 7.2 for a similar
6. Once again, save your data and video, and don’t worry about the empty graph boxes on
your template just yet.
Experiment 3: Combined Analysis
Here you will make some new calculated columns for your data that “ﬁlter out” the motion of the
center of mass. If you do this correctly (and if you saved both of your original data sets. . . ), the
columns should show up automatically for both your elastic and inelastic data sets, which will save
you a lot of time.
1. Create two new calculated columns in LoggerPro for the x and y positions of the center of
mass, using the formula for the center of mass location given in the theoretical section of this
lab. These columns should appear for both the elastic and inelastic data sets. See Appendix
A Section 1.5 for how to do this.
2. Make graphs displaying the x and y positions of the center of mass as functions of time (for
both the elastic and inelastic collisions). Paste these graphs into your lab report and state
what is moving and what the reference frame is.
3. Create four new calculated columns that contain the following functions:
y1 − ycm y2 − ycm x1 − xcm x2 − xcm for both data sets. Make sure you understand what these calculated columns are doing.
4. Make two graphs per data set of these four columns as functions of time. One graph should
be for the x-positions, the other for the y-positions. Since you’re making graphs for both
collision types, you will end up with a total of four graphs (see Appendix A Section 1.3).
5. For the elastic data set: Fit straight lines to the graphs to obtain the x and y velocities of
both pucks before and after the collision. (Note: You already did this for the graph of the
original data set in Experiment 1. Now you need to do it for the data set in the center of
mass reference frame.) Fill out the data table in Experiment 1 marked “FOR COM FRAME
DATA” by calculating the pucks’ kinetic energies in the new reference frame before and after
the collision. 63 6. Paste the two elastic graphs into your lab report, and state what is moving and what the
reference frame is.
Problem 7.5 Compare the total kinetic energy before and after the
elastic collision in the air table reference frame. Are the numbers similar?
Compare the total kinetic energy before and after this collision in the
center of mass reference frame. Are these numbers similar? Give your
best explanation of why the before and after numbers should be identical
within each respective reference frame, given what we know about energy
7. For the inelastic data set: In order to get the ﬁnal rotational kinetic energy, you will need to
calculate the angular velocity (ω ) of the combined puck system. You can do this by ﬁtting a
sine curve to one of the four position vs. time graphs after the collision. The ﬁt function will
be of the form
y = A sin(Bx + C) + D,
and the parameter B is equal to ω (see Appendix A Section 1.4 for how to ﬁt a sine curve).
Note: You will understand this functional form better after you study simple harmonic motion in Laboratory 10. Fill out the data table for Experiment 2 by calculating the pucks’
initial translational kinetic energies and ﬁnal rotational kinetic energy.
8. Paste the two inelastic graphs (with the ﬁt lines included) into your lab report and state what
is moving and what the reference frame is.
Problem 7.6 From the perspective of the center of mass reference frame,
what is the major difference between elastic and inelastic collisions? Is
more of the original translational kinetic energy of the pucks converted into
rotational kinetic energy when the collision is elastic or inelastic? After
which collision type are the pucks moving away from their center of mass?
After which collision type do the pucks stay a constant distance from their
center of mass? Explain.
Problem 7.7 Is it possible to have a two-dimensional inelastic collision in
which there is no rotation about the center of mass after the collision? If so,
what would such a collision look like?
Problem 7.8 Hopefully you observed that the x- and y-positions of the
center of mass were straight lines with respect to time. What would it mean
if one of them (the y-position graph, for instance) were parabolic instead of
straight? Would energy be conserved in this system? Why or why not? 64 ...
View Full Document