Otterbein University Department of Physics
Physics Laboratory 1500-8
EXPERIMENT 1500-8
MOMENTUM & COLLISIONS
NAME:
INTRODUCTION
Momentum is a property of moving objects. In one dimension, the momentum p of an
object is equal to its mass times its velocity:
p = mv
Note that while kinetic energy is always a positive number, momentum can be negative.
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(In general, momentum is actually a vector quantity: p  mv .) Like energy, the total
momentum of an isolated system, i.e. a system on which no external forces act, is a
conserved quantity. Namely, if you add up all the momenta of the constituents of the
isolated system at a time tinitial, and compare it to the sum of the momenta at a later time
tfinal, you will find that the sums are exactly equal, although the individual momenta will
have changed.
Consider what happens if two carts (1 and 2) with equal mass and speed hit head-on.
(Feel free to try this with your carts!) What is the total momentum of this system?
p = p1+p2 is _____ (Positive, negative or zero?)
What is the total kinetic energy?
K = K1 + K2 is _____ (Positive, negative or zero?)
Momentum conservation implies p1,initial + p2,initial = p1,final + p2,final.
After the carts collide, what has to be true about their velocities, according to momentum
conservation?
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Otterbein University Department of Physics
Physics Laboratory 1500-8
Interestingly, momentum is always conserved in a system – unless there is an external
force acting on the system. However, as we saw last week, a system can lose kinetic
energy, e.g. to heat.
In the experiment, we will be colliding two carts. The carts are fitted with magnets so
that they don’t actually touch. This is to ensure a gentle collision where there is no energy
lost to heat or sound; this kind of collision is called elastic.
Collisions are classified according to what happens to the kinetic energy in the collision.
1. Completely elastic collision – Kinetic energy is conserved.
2. Partially inelastic collision – Some kinetic energy is lost.
3. Completely inelastic collision – Maximum amount of kinetic energy is lost and the
objects stick together.
Newton’s third law is related to the law of conservation of momentum. It states that
whenever one body exerts a force on another, the second body exerts a force on the first
that is equal in magnitude and opposite in direction to the other force.
How does this relate to our lab? It simply means that the force of cart 1 on cart 2 and its
Newton-III-partner, the force of cart 2 on cart 1, add up to zero. These forces occur
within the system (made up of cart 1 and cart 2), and are called internal forces. It is
therefore clear from Newton III that the net internal force of any system is zero by
construction. Without a net force, there is no change of velocity (Newton I). Keep in
mind, however, that this velocity is the velocity of the system, not that of an individual
cart. Therefore the velocity of the system or center-of-mass velocity does not change. The
forces add up to zero, but they are not zero themselves. There is, then, a force acting on
each cart, so the velocities of the individual carts do change. It is these changing
velocities, the velocities after a collision, which we wish to calculate. The fact that the
sum of the momenta of the carts are unchanged by a collision helps us, because it is one
equation (actually, n equations in n dimensions, as it is a vector equation) we can use that
involves the velocities we want to calculate.
It would be great to have another equation, since we have two unknown quantities, v1,final
& v2,final (sometimes called v1,before & v2,after or v’1 & v’2 ). If the collision is completely
elastic, we can use the conservation of total kinetic energy. The sum of kinetic energies
of the carts before the collision is the same as the sum of kinetic energies after:
1
2
m1v12,initial  12 m2 v22,initial  12 m1v12, final  12 m2 v22, final
On the other hand, if the collision is completely inelastic, we can use the fact that the
carts stick together, and thus have the same final velocities v1,final = v2,final ≡ vfinal, where
we defined a new name for the unambiguous final velocity. In any case, we have two
equations for two unknown quantities, and can solve the collision problem.
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Otterbein University Department of Physics
Physics Laboratory 1500-8
TESTING THE CONSERVATION LAWS
1. Secure the motion detectors at opposite ends of the track. Make sure that the motion
detectors are aimed down the length of the track. To keep the motion detector from
receiving echoes from the track, the head should be tipped up a little bit, making an
angle of 82 -83 with the body.
2. Make sure the motion detectors are connected to ports Dig/Sonic 1 and 2 of the
LabPro. Launch Logger Pro. Set up the detectors by typing the room temperature
(lab thermometer) in the appropriate field.
3. Place a cart midway between the motion detectors. Click the Zero button, and select
zero all sensors. This defines the origin of coordinates at the center of the track.
4. Click Collect. Move the cart back and forth, and make sure the motion detectors are
giving sensible readings. Note which cart is labeled as ‘1’ and ‘2’ on the screen (blue
and red).
5.
6. Measure the mass of each cart:
m1 = ________________
m2 = ________________
A. ELASTIC COLLISIONS, SIMILAR MASSES, ONE CART AT REST
Runs 1-3. Place two “collision carts” on the track. Note that they have magnets in them
that make them repel. In a collision between these two carts, the forces between the carts
are magnetic, but Newton’s laws should apply to magnetic forces as well as any other.
Because the collision is gentle, there is no opportunity for kinetic energy to be lost, so the
collision will be approximately elastic.
Start with one cart 1 at rest (“target”). Start the data collection, and gently roll cart 2
(“projectile”) into cart 1. Sketch the velocities of the two carts below, indicating the
time of the collision:
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Otterbein University Department of Physics
Physics Laboratory 1500-8
Now, highlight a region of the graph before the collision and use the statistics button to
find the average velocity of each cart, and the uncertainty (standard deviation) on the
uncertainty of each cart. Enter these values into the table below. Then do the same for the
velocities of the carts after the collision.
Repeat this experiment 3 times, filling out the data for runs 1,2, and 3. Remember to add
minus signs where appropriate!
Run
Initial Velocity
Cart 1
v1 (m/s)
Δv1 (m/s)
Final Velocity
Cart 2
v2 (m/s)
Δv2 (m/s)
1
2
3
Describe what the collisions looked like.
4
Cart 1
v1 (m/s)
Δv1 (m/s)
Cart 2
v2 (m/s)
Δv2 (m/s)
Otterbein University Department of Physics
Physics Laboratory 1500-8
Those are a lot of numbers. Let’s simplify by estimating the average percent uncertainty
on velocity. (Estimate or take the average of the above results. Don’t use velocities that
are tiny.)
Δv
/v = _____________%
Now let’s do some calculations. Find the momentum for each cart, and the total
momentum before and after the collision. Use the percent uncertainty on velocity to find
the percent uncertainty on momentum. (That is: Δp/p = Δv/v).
Run
Initial
p1
p2
Final
ptotal
Δptotal
p1
p2
ptotal
Δptotal
1
2
3
Comment on your results: do the initial momenta agree with the final momenta, to within
the uncertainty?
We can do the same with energies. Remember that KE = 1/2 mv2
To find uncertainty on energy, use this rule of thumb:
ΔE/E = 2 Δv/v
The ‘2’ appears in this equation because the velocity is squared.
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Otterbein University Department of Physics
Run
Physics Laboratory 1500-8
Initial
E2
E1
Etotal
Final
ΔEtotal
E1
E2
Etotal
ΔEtotal
1
2
3
Comment on your results: do the initial energies agree with the final energies, to within
the uncertainty?
B. ELASTIC COLLISIONS, DIFFERENT MASSES, ONE CART AT REST
Run 4: Let’s do the exercise again, this time adding two mass bars to the stationary cart.
This time, ignore uncertainties and do it once. Complete the tables below.
m1 = ____________
m2 = ____________
Initial
Cart 2
Cart 1
Final
Total
Cart 1
Cart 2
v1 (m/s)
v2 (m/s)
v1 (m/s)
v2 (m/s)
p1
p2
p1
p2
(kg
(kg
(kg
(kg
m/s)
m/s)
m/s)
m/s)
E1 (J)
E2 (J)
E1 (J)
E2 (J)
6
Total
Otterbein University Department of Physics
Physics Laboratory 1500-8
Which quantities are conserved: velocity, momentum, or energy?
Run 5: Repeat the experiment this time adding the mass bars to the moving cart.
m1 = ____________
m2 = ____________
Initial
Cart 2
Cart 1
Final
Total
Cart 1
Cart 2
v1 (m/s)
v2 (m/s)
v1 (m/s)
v2 (m/s)
p1
p2
p1
p2
(kg
(kg
(kg
Total
(kg
m/s)
m/s)
m/s)
m/s)
E1 (J)
E2 (J)
E1 (J)
E2 (J)
Which quantities are conserved: velocity, momentum, or energy?
Run 6: Repeat the experiment, this time removing all the bars, but rolling both carts at
each other.
m1 = ____________
m2 = ____________
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Otterbein University Department of Physics
Physics Laboratory 1500-8
Initial
Cart 2
Cart 1
Final
Total
Cart 1
Cart 2
v1 (m/s)
v2 (m/s)
v1 (m/s)
v2 (m/s)
p1
p2
p1
p2
(kg
(kg
(kg
Total
(kg
m/s)
m/s)
m/s)
m/s)
E1 (J)
E2 (J)
E1 (J)
E2 (J)
Which quantities are conserved: velocity, momentum, or energy?
C. INELASTIC COLLISIONS
This time, use the cart labeled ‘dynamics cart’ as one of the two carts. Notice that it’s got
no magnets, but it does have velcro, so the two carts will stick together instead of bounce.
Repeat the experiment. (No bars, but remember to weigh the new cart).
m1 = ____________
m2 = ____________
Initial
Cart 2
Cart 1
Final
Total
Cart 1
Cart 2
v1 (m/s)
v2 (m/s)
v1 (m/s)
v2 (m/s)
p1
p2
p1
p2
(kg
(kg
(kg
(kg
m/s)
m/s)
m/s)
m/s)
E1 (J)
E2 (J)
E1 (J)
E2 (J)
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Total
Otterbein University Department of Physics
Physics Laboratory 1500-8
Which quantities are conserved: velocity, momentum, or energy?
Describe what is different this time in terms of momentum and energy. What did we lose,
and where did it go?
How does our experiment today bear on Newton’s third law?
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