Collision in Two Dimensions
READ carefully before you start – ORGANIZE your work
Objectives:
 Verify conservation of momentum in two dimensions for equal and unequal colliding masses.
 Determine the average angle between the final momenta of the colliding marbles for equal and
unequal masses.
 Compare the total initial to the total final kinetic energy for each collision. Is kinetic energy
conserved in these collisions?
Important: In order to compare experimental values such as initial and final values of KE, the error
must be known. For example 2.0 +/- 0.2 J is equal to 1.8 +/- 0.3 J but 2.00 +/- 0.06 J is not equal to
1.80 +/- 0.02 J. Make sure to determine and report the experimental error in your experiment!
Background:
Conservation of Momentum means that the total vector resultant momentum before collision is equal
to the total vector resultant momentum after collision.
For a collision between two objects, conservation of momentum can be expressed as
pa + pb = pa' + pb'
where primed quantities (') mean after the collision and unprimed mean before the collision, but remember
that momentum is a vector so it must be added as a vector!! To add momentum vectors in two
dimensions, a vector diagram must be drawn.
In this lab, you will launch a moving incident ball a that will collide with a stationary target ball b, as
shown below.
If the collision is not head on, the two balls will go in different directions after the collision.
If momentum is being conserved, then when we add the two momentum vectors after the collision (pa' and
pb'), they should give the same vector as the initial momentum vector (pa).
pa = pa' + pb'
The vector diagram in this situation would look like this:
The initial velocity of the incident ball is horizontal; therefore, the momentum that we are concerned with
is also horizontal. This means that it is only the horizontal velocities that will be included in any
calculations of the momentum.
mava = mava' + mbvb'
In this lab, the actual collision will occur at a certain height above the floor. Since each ball has the same
initial vertical velocity (0 m/s) and falls the same height, the balls will be in the air for the same amount of
time. It is only during this time that they will be traveling horizontally. Since the horizontal velocity is
constant and can be expressed in terms of the displacement using v=d/t, the conservation of momentum can
be expressed as
mada = mada' + mbdb'
where da is the displacement of the incident ball if no collision had occurred, and da' and db' are the
displacements of the two balls after the collisions. Notice that the time is not a factor since it is the same for
all of the displacements.
If all of the masses are the same, it is clear that the displacement vectors can be used in place of the actual
momentum vectors to show that momentum is being conserved. If the masses are different, then some
scaling of the displacement vectors must be performed first before they can be used to represent the
momentum vectors.
Hypothesis: What do you think will happen? Why?
Materials: rolling ball two dimensional collision apparatus, carbon paper, meter sticks, balls, masking
tape, paper
Apparatus:
Procedure:
1. Clamp your apparatus to the edge of one of the rolling cabinets or the counter on either side of the
room. The ball bearing that rolls from the launcher should roll horizontally off the edge of the table
and land on the floor. Let the ball roll down a couple of times from a specific height to determine
the general area where it is landing.
2. Remember to always launch from the same height. Tape a large sheet of paper on the floor
where the ball landed. Orient the paper to give enough room for the ball bearings to land after
collision (see the figure above). Place a piece of carbon paper over the taped paper (shiny side up).
Launch the ball ten times so that it lands on the carbon paper. Lift the carbon paper and
circle these dots.
3. Procedure to adjust the screw holding the target ball bearing: It is necessary to ensure that the
screw holding the target ball is at the proper height. Place the target ball (of equal mass to the
incident ball) on the screw directly in front of the launcher so that there will be a head on collision.
Adjust the screw so that the target ball is at the same height as the incident ball at the time of
collision. Carry out several head on collisions (releasing the incident ball from the same height as
before). If the screw is set to the proper height, the target ball should land in the same spot as
the original group of dots. If it does not, adjust the height of the screw until the target ball does
land in the same area as the original incident ball by itself.
4. Equal Mass Collision:
a. Set up the target ball (of equal mass) on the screw to one side of the launcher so that when
the ball rolling down the apparatus strikes the one at rest there will be a glancing collision.
b. Hang a plumb mass (such as a screw) directly under the point where the collision will take
place. Mark the spot on the floor directly under this collision point. This is the point from
which the vectors will be drawn.
c. Run a couple of trials of the glancing collision (without carbon paper), using the same
launch height (of the incident ball) as the other trials to make sure that the balls land on the
paper. If not, tape more paper to extend your data area. Next place two pieces of carbon
paper where the balls are landing so that the balls will leave a mark when they hit the paper.
Carry out ten trials of this glancing collision. Remove the carbon paper and label the dots.
d. Draw a vector from the mark representing the point underneath the launch position to the
center of each of the three groups of dots created. Measure and record the length of these
displacement vectors and the angles (using 0o as the direction of the original displacement
of the incident ball).
5. Unequal Mass Collision:
This part of the procedure is carried out the same way as the equal mass collision, but the target ball is
changed so that it does not have the same mass as the incident ball.
Make sure to get through part 4 at least in one lab period!!! If you don’t finish this experiment you
must start with step 2 on a new sheet of paper!
Analysis:
Each group must submit the large sheet of paper with all the vector diagrams. Each student must
have free-hand sketches in his/her lab journal.
For the collision of equal masses, draw a scaled vector diagram that will show if the momentum vectors
after the collision are equal to the momentum vector before the collision. Remember that the displacement
vectors are actually proportional to the momentum vectors, so you can use the displacement vectors to
show this conservation.
Notice the angle between the two momentum vectors after the collision. Does this angle have any
significance? Is it what you expect?
For the collision of unequal masses, you need to know the ratio of the masses. If one mass is ½ the mass of
the other, its displacement vector must be cut in ½ to represent a momentum vector. Again, draw a scaled
vector diagram that will indicate if the momentum before the collision is equal to the momentum after the
collision.
What errors may have been introduced into this experiment? How might these errors be reduced?
Provide enough experimental evidence to answer all the objectives.