Experiment 7: Conservation of Energy and Momentum in a Collision
You will check for conservation of kinetic energy and
for conservation of momentum in a collision between
two ball bearings. One ball bearing is rolled down a
ramp. You let it fly off to find its speed at the end of
the ramp, indicated by how far away it lands. Where it
lands is recorded by it hitting a sheet of paper laid over
carbon paper. The ball bearing is then rolled down
again, this time with an identical ball sitting at the end
of the track. After colliding, they fall to the floor, again
landing on paper over carbon paper. The velocity of
the one ball as it reaches the end of the ramp is used to
find the energy and momentum just before the
collision; the velocities of the two balls just after they
hit are used to find final energy and momentum.
PROCEDURE: Measure the mass of both ball bearings. They should be identical. Lay the carbon
paper on the floor, carbon side up, and put a sheet of 18” by 24” paper on top, with the plumb bob
hanging over the middle of the shorter side. (“Plumb” means vertical the same way that “level”
means horizontal. The “bob” is the weight on the end of the line.) Weight the paper down.
Mark the point under the plumb bob.
Adjust the screw at the end of the track which the target ball will sit on:
a. Get your eye exactly level with the countertop. Then, adjust the screw so that its top is at the
same height as the bottom of the groove in the
ramp. This way, the incident ball will knock the
target ball cleanly off the screw.
b. Check that the screw is not leaning to one side;
it should look like a fatter extension of the plumb
line.
c. The arm should be turned about 50° or 60° to
the side, otherwise one ball won’t be going fast
enough to miss hitting the arm as it falls. In a trial
collision, check that both balls land at least 10 cm
from the plumb bob (hopefully more).
If
necessary, readjust the angle that the arm makes.
Then, recheck the screw’s height.
Once adjusted, be careful not to bump it; the arm can
be kind of floppy. Remember where you released the
ball; it must be the same throughout the experiment.
Without the target ball in place, release the other ball at least half a dozen times. Don’t move the
paper the balls land on yet.
The screw has a dimple for the target ball to sit in. Put the
target there and try a collision. d1 and d2 should be fairly
similar to each other. 10 cm is the absolute minimum for
each of them so that the falling balls clear the arm. Do at
least another half dozen trials. Have the instructor check the
results before going on.
Turn the paper over to see the dots better. Measure the distance from the plumb bob to each cluster
of impact points, as shown. The amount of scatter in the dots shows you the uncertainty: For
example, if the closest dot is 24.2 cm from the plumb bob, and the farthest is 27.6, record the
distance as 25.9 + 1.7 cm. Also, measure the angles θ1 & θ2, and the vertical height of the fall, z.
Calculations: Show how to set each calculation up in the space provided. Do not over-round: A
zero used only to show where the decimal point goes is not significant. To have three significant
digits, you need something like .00437 not .004.
Compute the time to fall to the floor: The vertical component of the ball’s motion obeys
z = vzit + ½azt2
In this case, vzi = 0 because the ball is moving horizontally at the end of the track. Also az = g,
taking down as positive. With these substitutions,
z = 0 + ½gt2 Solving, we obtain
The uncertainty in t is very small; accurately enough we can call it zero.
Next, find vi, the speed the first ball has at the end of the ramp. Once it becomes a projectile, the
horizontal component of its velocity doesn’t change, so its speed at the end of the ramp equals its
average horizontal speed while flying: vi = di/t. Assuming no uncertainty in time, v will have the
same percent uncertainty as d.
From this, compute the kinetic energy of the first ball before the collision. Since v is squared, and
m’s uncertainty is effectively zero, KE will have twice the percent uncertainty v does. Convert this
percent into joules.
Find v1 and v2 the same way you found vi. From them, find the system’s total kinetic energy after
the collision. When totaling up the final energy, remember that when adding or subtracting, you
add the uncertainties themselves, not percents. Percents go with multiplication and division.
Now, check for conservation of momentum.
components:
Momentum is a vector, so think in terms of
- Find the magnitude of the momentum before the collision, pi. Assuming no uncertainty in
mass, pi will have the same percent uncertainty as vi. Then, write what its x and y components
are. ( p
⃑ i points directly along the x axis.) Include each component’s uncertainty, in units of
momentum, calculated from the percentage.
- Find the magnitude of each ball’s momentum after the collision, p1, and p2. Again, they have
the same percent uncertainty as the velocity they came from.
- Find the components of p
⃑ 1 and p
⃑ 2 , showing how in the
space provided. Pay attention to + and - signs. As a
crude approximation for the uncertainty, let’s say that
the angle might be off by as much as p, since the dots
are scattered across the direction of motion as well as
along it. So, double the percent in p to get the percent
uncertainty in its components.
Calculate each
component’s uncertainty, in units of momentum, from
the percentage.
Find the x and y components of the total final momentum,
with their uncertainties.
Conclusions: Within the uncertainty, was kinetic energy conserved? (Don’t be surprised if a little
energy was lost. Realistically, the collision is slightly inelastic.) Was the x component of the
momentum conserved? Was the y component of the momentum conserved?
Please put the ball bearings back in their box. They tend to get lost otherwise.
PHY 121
Experiment 7: Energy and Momentum in a Collision
DATA:
m = __________ + 0 (uncertainty small enough to ignore.)
di = ___________ + ___________
(percent uncertainty = ____________%)
d1 = ___________ + ___________
(percent uncertainty = ____________%)
d2 = ___________ + ___________
(percent uncertainty = ____________%)
θ1 = ___________
θ2 = ___________
z = ___________ + 0 (also small enough to ignore.)
COMPUTATIONS:
Compute t:
Just before collision:
vi =
= ___________ + ___________ %
KEi =
= ___________ + ___________J
Just after collision:
v1 =
= ___________ + ___________%
v2 =
= ___________ + ___________ %
KE1 =
= ___________ + ___________ J
KE2 =
= ___________ + ___________ J
KEf = ___________ + ___________ J
Magnitude of pi =
pix = ___________ + ___________ kgm/s,
= ___________ + ___________ %
piy = ___________ + ___________ kgm/s
Magnitude of p1 =
= ___________ + ___________ %
Magnitude of p2 =
= ___________ + ___________ %
p1x = ___________ + ___________ kgm/s
p2x = ___________ + ___________ kgm/s
p1y = ___________ + ___________ kgm/s
p2y = ___________ + ___________ kgm/s
pfx = ___________ + ___________ kgm/s,
pfy = ___________ + ___________ kgm/s