Analysis of collisions
Randy Draeger
Preassessment: A perfect elastic collision is where kinetic energy is not lost. The kinetic energy
is the same pre/post collision. This is usually not the case on the larger scale due to
deformation and energy transformation. In an inelastic collision momentum is conserved,
allowing us to use the momentum conservation equation. Also for an inelastic collision kinetic
energy is transferred. Knowing the initial and final momentum allows us to find the energy
pre/post collision. A perfectly inelastic collision is where the two involved objects “stick
together”. This scenario the final velocity is the same for both objects. For both types the
conservation of momentum equation can be used. Also the energy can be found, however
using slightly different methods, due to the differences with kinetic energy. Inelastic collisions
are considered isolated due to the fact that kinetic energy is transformed into potential energy
or internal energy.
The coefficient of restitution is the scientific value for bounciness regarding collisions. It is a
velocity ratio, shows the effect of collision on the speed and direction. A coefficient of one
means the collision was elastic. Less than one means an inelastic collision. A zero means no
motion post-collision. The displacement is proportional to momentum. Displacement is
proportional to momentum. A massive target head-on collision results in approximate reversal
of shooter’s velocity and the target’s velocity is around zero. A head-on with equal mass results
in the transfer of the shooter’s velocity.
1
1
1
1
𝑚𝑣 2 + 𝑚𝑣 2 = 𝑚𝑣 2 + 𝑚𝑣 2
2
2
2
2
𝑥
𝑣=𝑚
Objective: Analyze elastic collisions with different masses and (head-on and nonhead-on)
1. A
The experiment measured angle and displacement. Using the conservation of momentum
equation allows finding the unknowns.
1. F
The teacher instructed us that the displacement was proportional to momentum. Then the
experiment used the equation to isolate velocity using this new information.
2. A
The experiment forces the users to use the new information to complete the experiment. Then
experiment shows the relationships in one and two-dimensional collisions. The protractor used
to measure angles.
2. B
Use the coefficient of restitution and conservation of momentum and energy to analyze elastic
collisions.
2. D
Use given equations and data to find the relation, and then work backwards.
3. A
Using the data and vector drawings can make conclusions about the collisions and its
momentum.
3. D
Analyze the data from head-on and nonhead-on collisions using the drawing and equations.
4.
This lab report demonstrates this.
Safety: When moving objects at high speeds, the group must take care to the objects on their
predicted path. Fling the washers away from people, and only fling the washers to complete the
experiment.
Material:





Meter Stick
Washers (4)
Protractor
Paper
Pencil
Procedure:
1. Measure the masses of washers
2. Place paper on table
3.
4.
5.
6.
Place two washers on the paper
Flick the washer at the other
Draw the resulting collision, and measure the angles and distances
Do this to get three head-on and nonhead-on collisions for equal and non-equal masses.
Washer
Data
Masses: 18.7g, 16.8g, 6.16g, 6.32g
Equal mass
Head-on
nonHead-on
Starting distance
x1
16.2 cm
32.3 cm
4.1 cm (75) 42.1 cm (10)
14.1 cm
32.3 cm
28.2 cm (26) 33.9 cm (43)
14.8 cm
26 cm
15.2 cm
15.7 cm
17.4 cm
13.1 cm (27)
x2
17.6 cm(57)
Unequal Mass
nonHead-on
Head-on
4.8 cm
7.2 cm
Starting distance
x1
5.1 cm
8 cm
21.9 cm
4.1 cm (29) 16.3cm (70)
5.1 cm
4.8 cm
21.9 cm
6.2 cm (19) 9.2 cm (108)
Starting Distances: 115.7 cm, 19 cm, 16.8
16.9 cm
6.5 cm (19)
x2
12.4 cm (104)
*note (for all data): all measurement are in centimeters and numbers in the parentheses are
the angles in degrees.
The coefficient of restitution
Equal mass head-on
1.02
1.11
.97
Equal mass non
head-on
1.18
.17
.97
Non Equal mass
head-on
.76
.69
.59
nonEqual mass
nonhead-on
.53
.14
.34
Calcs
𝑐=
𝑣−𝑣
14.8 − 0
=
= 1.02
𝑢−𝑢
14.5 − 0
𝑣=
𝑥
. 145
=
= 23.5𝑚/𝑠
𝑚 . 00616
𝑚𝑣 + 𝑚𝑣 = 𝑚𝑣 + 𝑚𝑣 = 14.5 + 14.8 = 29.3
𝑘𝑔𝑚
𝑠
1
1
1
1
𝑚𝑣 2 + 𝑚𝑣 2 = 𝑚𝑣 2 + 𝑚𝑣 2 = .5 × 14.52 − 14.82 = 4.395 𝐽
2
2
2
2
Analysis
The momentum is proportional (equal) to the displacement. Then set momentum equal to its
definition and isolate velocity. Ending up with x/m. However, when plugging this substitute
back in the original definition; mass cancels out. This leaves p=x. The total momentum and
kinetic energy is conserved pre/post collision, looking at displacement for equal head-ons. The
starting distance 14.5cm and average resulting displacement is 14.7 cm. almost the same
showing they are equal. Mass cancels out when calculating kinetic energy for these problems.
So since the displacement is now the only factor, if it is conserved in momentum; then it will be
the same for energy. This is for equal mass head on collision. For unequal head-on, the unequal
starting distance is 17.2 cm, and the ending displacement is 6.6 cm. But, the “shooter” washer
moves back about 5 cm from the collision site. So the original washer reverses in a third of its
original speed, and the secondary washer moves forward in a 1/3 of the original velocity. The
third of lost velocity is lost due internal energy due to the mass difference. Which makes sense
because there is a 3:1 mass ratio. Therefore moving an object with more inertia requires more
energy. Vector addition with equal mass non head-on collision proves the conservation of mass.
Also looking at average angles for the emnho collision, this concludes collision angles are 90
degrees. The angle seems to be related to the final velocity and displacement. The higher
displacement (velocity), the larger the angle. This makes sense because a high velocity means it
takes more of the initial kinetic energy. Looking at the nemnho collisions, the data shows the
role of mass. The smaller shooter keeps most of its velocity, allowing it to shoot off at an obtuse
angle. The inertia of more massive object “absorbs” some of its kinetic energy into internal
energy, making its displacement smaller. The emho collisions are proved elastic based on the
coefficient. Also the emnho collisions have a coefficient of one. However, unequal mass
collisions seem to be a different story. Both head-on collisions seem to closet to being elastic.
Having a more massive target, interval time of contact making more inelastic.
𝑚𝑣1 = 𝑚𝑣2 cos 𝜑 − 𝑚𝑣/ cos 𝜃
𝑚𝑣1 = 𝑚𝑣2 sin 𝜑 − 𝑚𝑣 / sin 𝜃
𝑚𝑣2 cos 𝜑 −
𝑚𝑣
𝑚𝑣
/ cos 𝜃 = 𝑚𝑣2 sin 𝜑 −
/ sin 𝜃
1
1
cos 𝜑 sin 𝜑
=
, 𝜑 + 𝜃 = 90
cos 𝜃 sin 𝜃
Conclusion
The point of this experiment was to analyze elastic head-on/nonhead-on collisions with
differing masses. This was successfully accomplished, the vector diagrams and coefficients
allowed data to show trends and relationships. With emho collisions, the displacement shoed
the shooter transferred its velocity to the target, confirming preassesment knowledge. Also it
confirmed conservation of momentum and energy. The initial energy and momentum of the
shooter are equal to the final momentum and energy of the target. Using v=x/m , and then
using that in the equations that have equal mass, and the mass doesn’t matter because it’s the
same on the both sides. Averaging the results shows the collision for equal masses should be 90
degrees, confirming classroom knowledge. Regarding unequal mass non head-on collisions, the
mass of target compared to the shooter is important. The mass ratio predicts the displacement
and angles. Also, unequal mass head on collision show the importance of mass. A high mass
target wills a lighter shooter because of newton’s laws (inertia and equal and opposite forces),
so the shooter will opposite direction but equal magnitude velocity. The coefficient shows out
velocity compared to in velocity. A ratio of velocity is constant, and nothing overall has
changed. However, the unequal mass collisions have grater time interval, causing the
coefficient to slightly inelastic.
Not all the data confirms what we have learned. Ripples in paper added potential
energy to the equation, skewing the results. Also, there is the possibility of measurement
errors. Each washer wasn’t flicked with the same force. Also the masses were not exactly
equal. The rounded edges of washers allowed them to spin, entering rotational momentum. In
class, the lessons have been centered on linear momentum. The focus was on isolated elastic
collisions. This experiment allows for direct applications of lectures. Tools learned in the class
can determine the validity and purpose of data. Also the lab expands what wasn’t taught in
class: The purpose and effect of the coefficient of restitution. How the change of velocity is
related to elasticity and the coefficient cancels out in the equation. Also the analysis of angles
regarding two-dimensional elastic collisions and it effects. The analysis forced the students to
discover the relationship between the two angles. The lab is a real-life example of homework
problems. (Nothing to change)
1. F when told that momentum is proportional to displacement it was necessary to apply this
new information. Using manipulation and substitution, found the unknown needed to complete
the lab.2.an Using previously stated information, the lab had velocity without actually finding it,
making the lab the easier. Also the protractor and meter makes the needed measurements. 2.
B The design is built to compare differing masses and different collision types. Also to look at
aspects not covered in class. 2. D Using knowledge and equations the data produced allows the
lab to work backwards to find the original conditions. 3. D analyzing the coefficients of
restitution. 4. This lab report shows how science method and previous knowledge is applied to
science to gain new knowledge by confirming it with data analysis and conclusion.
NOTE: emho-equal mass head on,nemho-nonequal mass head on emnhoequal mass non head on, nemnho-nonequal mass non head on