Conservation of Momentum and Elastic Collision Lab
There are two objectives to this lab exercise. The first objective is to examine whether or not
the collision of two pool balls is elastic and conserves momentum. The second objective is to
learn a formal method for propagating errors in a physics calculation.
The equipment you will need for this lab is: two stopwatches (any device that can record time
down to tenths of a second will work), one measuring tape or ruler, and one pool (billiards)
table with balls and cue stick (there are pool tables available at the student center). You will
need at least three people to perform the lab.
Method: Place the cue (white) ball near one edge of the pool table and the eight (black) ball at
the middle of the table. (See diagram below). Measure the distance from the cue ball to the
eight ball and the distance from the eight ball to the far end of the table; call these distances
“Distance 1” and “Distance 2”.
Cue ball
Distance 1
Eight ball
Distance 2
Record the distances here: Distance 1 _________________ Distance 2 ____________________
Now take a shot: the idea is to hit the cue ball lightly with the pool cue so that it strikes the
eight ball straight on. If the cue ball is shot straight at the eight ball, the cue ball should (mostly)
stop and the eight ball should pick up all (or almost all) of the momentum. Let each person in
the lab group try a few practice shots to see if you can do this. You will need to hit the cue ball
hard enough to make the eight ball reach the other end of the table, but soft enough that you
can measure the time that each ball spends moving. Once each person in the lab group has
tried a few practice shots, pick the person who seems to have the straightest shot as the
“shooter”; the other people will be the “timers”. Now you are ready to make a timed
measurement.
For the timed measurement, when the “shooter” hits the cue ball the first “timer” records the
time between the moment that the cue ball is shot and the moment it hits the eight ball; this is
“Time 1”. When it hits the eight ball the second “timer” records the time between the moment
of the collision between the two balls and the moment the eight ball hits the back wall of the
pool table; this is “Time 2”. Don’t worry if it takes more than one trial to get this to work; the
important thing is that both timers should be able to record a reasonably accurate time.
Record the two times here: Time 1 _____________________ Time 2 ____________________
Take the masses of the two pool balls to each be exactly 170 g.
Now calculate the momentum of the system before and after the collision and record the
results below:
What is the percent difference in the momenta before and after the collision?
Now calculate the kinetic energy of the system before and after the collision and record the
results below:
What is the percent difference in the kinetic energies before and after the collision?
Do you think that momentum is conserved?
Do you think that the collision is elastic?
What do you think some sources of error might be in this experiment?
Part II – Propagation of Errors
Up to this point in the class we have been treating errors in a very informal and qualitative way.
Now we will make a first attempt at QUANTIFYING errors. For this discussion the direction of all
vectors is assumed to be towards the far end of the pool table.
For every basic measurement we make, there is an associated uncertainty. For a measurement
coming from a ruler the uncertainty is roughly equal to the smallest notch on the ruler; i.e. if
the smallest notch on the ruler is one mm = 0.1 cm, then a measurement of 40 cm should
REALLY look like:
Distance = 40.0 cm ± 0.1 cm
or Distance = 0.400 m ± 0.001 m
The 0.1 cm is the uncertainty in the measurement of 40 cm. Similarly for a time measurement,
a person may only be able to make a measurement with a stop watch down to the nearest 0.2
s, in this case a time measurement of 2.4 s should REALLY look like:
Time = 2.4 s ± 0.2 s
When combining these to form a velocity we obtain:
Velocity = Distance / Time = (0.400 m ± 0.001 m)/(2.4 s ± 0.2 s)
In order to deal with the uncertainties in the equation and find the uncertainty in the overall
velocity we use the formula:
(Velocity Uncertainty/Velocity)2 = (Distance Uncertainty/Distance)2 + (Time Uncertainty/Time)2
Think of this as a special Pythagoras-type (a2 + b2 = c2) rule for combining the uncertainties of
two measurements into one combined uncertainty. In the example above
(Velocity Uncertainty/Velocity)2 = (0.001 m / 0.400 m)2 + (0.2 s/ 2.4 s)2
= 0.000006 + 0.007 ≈ 0.007
Thus (Velocity Uncertainty/Velocity) = √(0.007) = 0.083
And if Velocity = Distance / Time = 0.400 m /2.4 s = 0.17 m/s
Then Velocity Uncertainty = 0.083 × (Velocity) = 0.083 × 0.17 m/s = 0.01 m/s
And we can express the final velocity with its uncertainty as v = 0.17 m/s ± 0.01 m/s
Now estimate the uncertainties in your time and distance measurement and calculate your
velocity with its uncertainty.
If the masses of the billiard balls including their uncertainties are given by 0.17 kg ± 0.1 kg then
use the same procedure to find the momentum before and after the collision including the
uncertainties in each momentum.
Is the difference between the two momenta larger or smaller than the sum of their
uncertainties? What does this tell you?
Now try the same procedure for kinetic energy. Keep in mind that because velocity is squared
in the kinetic energy equation, once you have the velocity and its uncertainty, the uncertainty
in the kinetic energy is given by:
(KE Uncertainty/KE)2 = (Mass Uncertainty/Mass)2 + 2 × (Velocity Uncertainty/Velocity)2
where the last term is multiplied by 2 to reflect the fact that velocity is squared in the kinetic
energy equation.
Calculate the kinetic energy before the collision including its uncertainty:
Calculate the kinetic energy after the collision including its uncertainty:
Is the difference between the two kinetic energies less than or greater than the sum of the two
uncertainties? What does this tell you?
In the end, would you say that you found momentum to be conserved in this experiment?
In the end, would you say that you found the collision to be elastic?