Collisions in Two Dimensions
Purpose:
To determine if momentum is conserved in two-dimensional collisions.
Theory:
When you analyzed one dimensional collisions, you could show that in an isolated system the momentum of each object
before the collision added up to equal the total momentum after the collision. This still applies in two-dimensional
collisions, but remember that momentum is a vector so it must be added as a vector!! For a collision between two objects,
conservation of momentum can be expressed as
where primed quantities (') mean after the collision and unprimed mean before the collision. In a one-dimensional
situation, the vector nature of the momentum could be addressed using positive or negative values for the velocities.
In two dimensions, the vector nature of momentum does not allow simple algebraic operations. Although you can still
express the conservation of momentum, special attention must be paid to the vector nature of momentum. To add
momentum vectors in two dimensions, a vector diagram must be drawn.
In this lab, we 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).
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.
mva = mva' + mvb'
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
mda = mda' + mdb'
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.
Materials: rolling ball two dimensional collision apparatus carbon paper
meter sticks
balls
masking tape
paper
Procedure: Set up your apparatus so that the ball bearing that rolls from the launcher is rolling horizontally off the edge of the table
and landing on the floor. Let the ball roll down a couple of times from a specific height to determine the general area that it
is landing. Remember to always launch from the same height. Tape a piece of paper on the floor where the ball
landed. 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.
Before collecting any more data, 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.
Equal Mass Collision:
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. 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.
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. Tape two pieces of paper to the floor where the two balls are landing. Next place two pieces of carbon
paper at these locations so that the balls will leave a mark when they hit the paper. Carry out ten trials of this glancing
collision. Remove the paper and label the dots.
Draw a vector (chalk lines can be used) 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 0 as the direction of the original displacement of the incident ball).
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.
Results
Experiment
incident ball before collision
length, da / m
angle, a
incident ball after collision
length, da’ / m
angle, a’
target ball after collision
length, db’ / m
angle, b’
equal mass
collision
unequal mass
collision
Analysis: 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 1⁄2 the mass of the other,
its displacement vector must be cut in 1⁄2 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.
Conclusion: Does your evidence indicate that momentum was conserved in these collisions? What errors may have been introduced
into this experiment? How might these errors be reduced?