1) This lab report summarizes an experiment investigating elastic and inelastic collisions between pucks. For an elastic collision, momentum and kinetic energy were conserved to within experimental uncertainty. 2) For an inelastic collision, momentum changed by 15% and kinetic energy decreased by 32%, indicating the collision was inelastic. 3) Sources of error are discussed, such as marker placement during video analysis. Overall, the goals of determining momentum and energy conservation for elastic and inelastic collisions were met, despite some unexpected results for the inelastic case likely due to experimental errors.

1. PHY 122: Collision 2D
Yousiff Alazemi; Matt, Angus; Group number 1; David Weisberger
Objective:
Determining the law of conversation of momentum for nearly elastic and perfectly
inelastic collision, and testing whether kinetic energy is conserved in these cases were the main
goals of last week lab.
Experimental Data:
(See attached)
Data Analysis/Results:
Part 1) Elastic Collision
A- Momentum in the elastic collision:
In the x direction, x: mv1=mv2+mv3.
The initial momentum in the x direction should equal the sum of the final momentum of both
pucks in the x direction. Since mass of both pucks is equal, we will notate that with the letter m.
Mass of both pucks=50 g= .050 kg with a given Δm of .00001 kg
Initial velocity of puck 1 in the x direction (v1) = .6596 +/- .006 m/s
Final velocity of puck 1 in x direction (v2) = .03013 +/-.009 m/s
Final Velocity puck 2 in x direction (v3) = .6194 +/- .01 m/s
Initial momentum = (.05)(.6596)= .03298 Kgm/s
Uncertainty = √(mΔv1)2+(v∆m)2=√(.05*.006)2+(.6596*.00001)2=.0003 kgm/s
Final momentum = (.05)(.03013)+(.05)(.6194)=.03248 kgm/s
Uncertainty=√(mΔv2)2+(v2∆m)2+(v3∆m)2+(m∆v3)2
= √(.05*.009)2+(.03013*.00001)2+(.6194*.00001)2+(.05*.01)2= .002 kgm/s
In the y direction, y: mv1=mv2+mv3.

2. Initial velocity of puck 1 in the y direction (v1) = .3281 +/- .009 m/s
Final velocity of puck 1 in y direction (v2) = .2511 +/- .01 m/s
Final Velocity puck 2 in y direction (v3) = .0904 +/- .01 .006 m/s
Initial momentum = (.05)(.3281)= .016405 Kgm/s
Uncertainty = √(mΔv1)2+(v∆m)2=√(.05*.009)2+(.3281*.00001)2=.0003 kgm/s
Final momentum = (.05)(.2511)+(.05)(.0904)=.017075 kgm/s
Uncertainty=√(mΔv2)2+(v2∆m)2+(v3∆m)2+(m∆v3)2
= √(.05*.01)2+(.2511*.00001)2+(.0904*.00001)2+(.05*.006)2= .0006 kgm/s
Total initial momentum = √px
2+py
2=√.032982+.0164052= .03689 kgm/s
Uncertainty=√(2px*∆py)2+(2py*∆px)2 = √(2*.03298*.0003)2+(2*.016405*.0003)2 = .00002
kgm/s
Total Final momentum=√px
2+py
2=√.032482+.0170752= .03669 kgm/s
Uncertainty=√(2px*∆py)2+(2py*∆px)2 = √(2*.03248*.0006)2+(2*.017075*.002)2 = .00008 kgm/s
Percent Change Momentum= (Δp/pI)*100 = (.03689-.03669)/.03689*100= .5 % change in
momentum
B- Kinetic Energy (KE)
In the x direction, KE=5mv1
2=.5mv2
2+.5mv3
2
Mass of both pucks=50 g= .050 kg with a given Δm of .00001 kg
Initial velocity of puck 1 in the x direction (v1) = .6596 +/- .006 m/s
Final velocity of puck 1 in x direction (v2) = .03013 +/-.009 m/s
Final Velocity puck 2 in x direction (v3) = .6194 +/- .01 m/s
Initial Kinetic Energy= .5mv1
2= .5(.05)(.6596)2= .01087 J
Uncertainty= √(mv1Δv)2+(.5v1
2Δm) = √(.05*.6596*.006)2+(.5*.65962*.00001)2 = .0002 J
Final Kinetic Energy= .5mv2
2+.5mv3
2= .009614 J
Uncertainty = √(mv2Δv2)2+(.5v2
2Δm)+( mv3Δv3)2+(.5v3
2Δm)= .0003 J

3. In the y direction, KE=5mv1
2=.5mv2
2+.5mv3
2
Mass of both pucks=50 g= .050 kg with a given Δm of .00001 kg
Initial velocity of puck 1 in the y direction (v1) = .3281 +/- .009 m/s
Final velocity of puck 1 in y direction (v2) = .2511 +/-.01 m/s
Final Velocity puck 2 in y direction (v3) = .0904 +/- .01 m/s
Initial Kinetic Energy= .5mv1
2= .5(.05)(.3281)2= .00269 J
Uncertainty= √(mv1Δv)2+(.5v1
2Δm) = √(.05*.3281*.009)2+(.5*.32912*.00001)2 = .0001 J
Final Kinetic Energy= .5mv2
2+.5mv3
2= .00178 J
Uncertainty = √(mv2Δv2)2+(.5v2
2Δm)+( mv3Δv3)2+(.5v3
2Δm)= .0001 J
Total initial Kinetic Energy = √Kx
2+Ky
2=√.010872+.002692= .0112 J
Uncertainty=√(2Kx*∆Ky)2+(2Ky*∆Kx)2 = √(2*.01087*.0002)2+(2*.00269*.0001)2 = .000004 J
Total Final Kinetic Energy=√Kx
2+Ky
2=√.0096142+.001782= .0098 J
Uncertainty=√(2Kx*∆Ky)2+(2Ky*∆Kx)2 = √(2*.009614*.0003)2+(2*.00178*.0001)2 = .00006 J
Percent Change Kinetic energy= (ΔK/KI)*100 = (.0098-.0112)/.0112*100= -12.5 % change in
KE
C- Rotational Energy
Initial rotational velocity puck 1 (w1) = √(vcx-vwx)2+(vcy-vwy)2/Rcw
Radius (from center to white dot) = .024 m so w= √(.6596-.6596)2+(.3281-
4554)2/.024 = 5.29 rad/s
Final Rotational velocity puck 1 (w2)= 5.244 rad/s
Final Rotational velocity puck 2 (w3)= 1.26 rad/s

4. Part 2) Inelastic Collision
Mass of pucks= 52 g or .052 kg
A- Momentum
Initial Velocity =√vx
2+vy
2= √.35812+.29052= .4611 m/s
Final Velocity= .2654 m/s
Initial momentum= .052*.4611=.02398 kgm/s
Final Momentum= .104*.2654=.0276 kgm/s
Percent Change in Momentum= (.0276-.02398)/.02398*100=15 % change in momentum
B- Kinetic Energy
Initial Velocity=.4611 Final Velocity=.2654
Initial Kinetic energy= .5mv2=.5(.052)(.4611)2=.0053 J
Final Kinetic Energy= .5(.104)(.2654)2=.00366 J
Percent Change Kinetic Energy=(.00366-.0053)/.0053= 32 % change in Kinetic Energy
C- Rotational Energy
This only applies to the rotation of the pucks after the collision had happened.
W=√(vcx-vwx)2+(vcy-vwy)2/Rcw =√(.2264-.277)2+(.1385-.2331)2/.046= 2.33 rad/s
Rotation KE= .5IW2 where I=3mR2 and the radius is .046 and mass is equal to the sum of
both the pucks masses.
1.5(.104)(.046)2(2.33)2= .0018 J

5. Conclusion
In the first collision, momentum and kinetic were conserved. With the difference percent
in momentum being a fraction of a single percent, and the kinetic energy being a loss of 12%, it
can be realized that the collision was roughly elastic but not perfectly. In the second collision, an
over a 30% loss of kinetic energy was resulted which is not as high as was expected for the
perfectly inelastic system. Moreover, about 15% change in the momentum was resulted. The
reason for this unexpected change was believed to be duo to slight mistakes in markings on the
video because collision should be conserved in all collisions. Even though the error in
momentum was rather high to be based solely on in correct markings alone, team attempted to
minimize these errors by being as consistent as possible with markings and collaborating with
other groups to confirm measurements. Nonetheless, it can be concluded that second collision
was perfectly inelastic as more than 20% of kinetic energy was lost. For future classes it would
be helpful for a higher resolution video so that data can be collected with a higher accuracy. it
was our goal to seek for a province that momentum and kinetic energy is conserved in perfectly
elastic as well as a perfectly inelastic systems. Hence, in the first collision, objective was met by
proving that kinetic energy loss was below our given 20% of acceptance and momentum is just a
fraction of a percent. However, In the second collision, a huge loss of kinetic energy and an
oddly huge loss in momentum were resulted, which means mistakes were made while
performing the experiment. But, it still can be concluded that the second collision was perfectly
inelastic as it 32% loss of kinetic energy was higher than that of the acceptance for an elastic
collision.