Collisions in Two Dimensions
AP Physics - Becke
Name ______________________________
Question:
What is the evidence that momentum is conserved in a two-dimensional collision? In what manner does the
evidence support the law of momentum conservation?
Purpose:
To conduct an analysis of a two dimensional collision in an effort to gather convincing evidence that the total
system momentum is conserved AND to describe how the evidence supports the law of momentum
conservation.
Background:
In this lab, we will launch a moving incident sphere A that will collide with a
stationary target sphere B, as shown.
If the collision is not head on, the two
spheres will go in different directions
after the collision.
If momentum is being conserved, then when we combine the two momentum vectors after the collision (pa' and
pb'), they should be equal to the initial momentum vector (pa).
The velocity of the incident sphere when it leaves the ramp is horizontal; therefore, the momentum that we are
concerned with is also horizontal. This means that it is only the horizontal components of the velocities that will
be included in any calculations of the momentum.
In this lab, the actual collision will occur at a certain height above the floor. Since each sphere has the same
initial vertical velocity (0 m/s) and falls the same height, the spheres will be in the air for the same amount of
time. During this time they will be traveling horizontally as they accelerate towards the ground.
Materials: two dimensional collision apparatus, carbon paper, meter sticks, spheres, masking tape, paper,
protractor
Procedure:
 Set up your apparatus so that the sphere that rolls from the launcher launches horizontally beyond the
edge of the table and lands on the floor.
 Measure the vertical distance from the floor to the point where the sphere will leave the launcher.
 Hold the sphere at the top of the launcher against the guide screw. Allow the sphere to roll down a
couple of times from this height to determine the general area that it is landing.
 Note that the hanging mass is located directly under the point where the collision will later take place.
 Tape a large piece of paper on the floor that will include both where the sphere lands and the location
of the collision.
 Mark the spot on the floor directly under the hanging mass. This is the point from which you will
measure the horizontal distance travelled.
 Place a piece of carbon paper over the taped paper (shiny side up).
 Launch the sphere 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 sphere is at
the proper height.
o Place the target sphere (of equal mass to the incident sphere) on the screw directly in front of
the launcher so that there will be a head on collision.
o Adjust the screw so that the target sphere is at the same height as the incident sphere at the
time of collision.
o Carry out several head on collisions (releasing the incident sphere from the same height as
before). If the screw is set to the proper height, the target sphere 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
sphere does land in the same area as the original incident sphere by itself.
Equal Mass Collision:
Set-up:
 Adjust the location of the target sphere so that when the sphere rolling down the apparatus strikes the
one at rest there will be a glancing collision (the more glancing, the better!). Be careful not to disturb
the location of the ramp!
Data collection:
 Run a couple of trials of the glancing collision (without carbon paper), using the same launch height (of
the incident sphere) as the other trials. Tape an additional piece of paper to the floor where the two
spheres are landing if they do not land on your big piece of paper.
 Next place two pieces of carbon paper at these
Incident ball landing
locations so that the spheres will leave a mark
spot after collision
when they hit the paper.
 Carry out ten trials of this glancing collision.
Remove the carbon paper and label the dots.
Incident ball landing
 Draw lines from the mark representing the
θ₁
spot without collision
point underneath the launch position to the
x
center of each of the three groups of dots
θ₂
created. Measure and record the length of
Location of
collision
these lines and the angles (using 0° as the
(hanging mass)
Target ball landing
direction of the original displacement of the
spot after collision
incident sphere).
Unequal Mass Collision:
 Replace the metal target sphere with the glass marble and repeat the “Data collection” procedure
above, using a different color pencil or pen to circle the landing positions after the collision and to draw
the lines between launch position and the landing positions.
Data:
Mass of steel sphere: 7.60 g = 0.00760 kg Height of Launcher: _______ m
Incident ball
with no collision
Length (m)
θ (°)
0°
Incident ball
after collision
Length (m)
θ₁ (°)
Equal mass collision
Unequal mass collision
Target ball
after collision
Length (m)
θ₂ (°)
Analysis:
1. Determine the time it takes for the spheres to reach the ground:
2. Calculate the following (note that the x and y subscripts do not indicate vertical motion, but instead
represent the two different axes of motion within the horizontal plane)
velocity = x/t
vx = v cos (0°)
px = mvx
*Incident Sphere
before collision
Total momentum
before
vy = v sin (0°)
py = mvy
0
0
0
(*The reason you marked the landing point of the incident sphere without a collision was to be able to
fill in the first row in the above table)
Equal Mass Collision
3. Calculate the final (post-collision) velocity and momentum of the incident sphere and the target sphere.
velocity = x/t
Incident Sphere
after collision
Target Sphere
after collision
Total momentum
after
Error
before vs after
Percent Error
before vs after
vx = v cos θ
px = mvx
vy = v sin θ
py = mvy
Unequal Mass Collision
4. Assuming momentum is conserved; determine the mass of the target sphere.
Initial Momentum (from question 2)
velocity = x/t
vx = v cos θ
velocity = x/t
vx = v cos θ
px = mvx
vy = v sin θ
Incident Sphere
after collision
vy = v sin θ
Target Sphere after
collision
Expected momentum of
Target Sphere after collision
Total (Pythagorized) momentum of
Target Sphere after collision
Calculated Mass of
Target Sphere
Actual Mass of
Target Sphere
Percent Error
2.42 × 10–³ kg
py = mvy