Hello, can someone please help me interpret my lab results? I have attached pictures of my results and the the actual lab instructions. I have no idea if I verified the law of conservation of momentum because I am not sure what my data even means. I need to analyze this for my lab report, but physics is such a confusing subject for me.

Picture 1 X JX B D E F G H Part A: Perfectly inelastic collision K L width of cart 1 flag L1= (0.012)m width of cart 2 flag L2= (0.012)m Collision type 1: m1= (0.26305) | m2= (0.26423) kg Please record the coresponding photogate readings that will be used to calculate each velocity above the velocity cell( if the cart was at rest or stopped after the collision then write at rest or stopped and fill the corresponding velocities with 0 ) Photogates readings: m1*v1i Time vli m2v2i V2 (m1+m2)vf Time Final Vf pli+p2 P1i pf Trial 1 P2i 0.0208 Pf Ptotal_ o 0.58 0.026 0.462 Ptotal_f %Change (in P) Trial 2 0.0153 0.15 0.244 0.78 0.035 o Trial 3 .343 ol 0.15 0.244 52.609 0.0172 0.7 0.21 0.031 0.181 0.387 ol 0.21 0.181 13% 0.18 0.204 0.18 0.204 13.30% Average % Change (p): 29.63 29.63 Kinetic energy table 0.5*m1*V1i^2 0.5*m2*V2i^2 0.5*(m1+m2)*Vf^2 K1i+K2 Kf - must below [(Ktotalf-Ktotali)/Ktotali ]*100% Kli K2i K2f Ktotal_i Ktotal_ Trial 1 %Change(K) o 0.044 0.056 0.044 D.056 Trial 2 0.103 27.27% 0.031 0.103 Trial 3 0.031 0.092 69.90% 0.039 0.092 0.039 57.61% Average % Change(K)= 51.59 Collision type 2: m1= (.26305) k m2= (.51423) kg Photogates readings: m1*V1i m2*V2i (m1+m2)*V V1 VF P1i+P2i Pf V2 P1 [(Ptotalf-Ptotali)/Ptotali ]*100% P2 Ptotal_i Trial 1 Ptotal_f 0.71 0.39 %Change (in P) Trial 2 0.37 0.66 0.3 0.38 0.37 0.3 0.34 18.90% Trial 0.3 0.78 0.34 0.43 0.3 0.4 0.33 11.769 Time 0.4 0.33 0.0169 0.030 17.50% 0.0183 0.0317 Average % Change (p): 6.05% 0.0158 0.0281 Kinetic energy table 0.5*m1*V1i^2 0.5*m2*V2i^2 0.5*(m1+m2)*Vf^2 Kli+K2 Kf Kli [(Ktotalf-Ktotali)/Ktotali ]*100% K2i K2f Ktotal_ Ktotal_f Trial 1 %Change(K) 0.13 0.059 Trial 2 0.13 0.059 0.11 54.61% 0.056 0.11 0.056 Trial 0.15 19.09% 0.071 0.15 0.071 52.66% Collision type 3: m1= (0.51305) | m2= (0.26423) kg Average % Change(K)= 52.01 Photogates readings: .012 / time m1*v1i m2*V2i (m1+m2)*Vf V1i Pli+P2 V2i Vf pf P1 [(Ptotalf-Ptotali)/Ptotali ]*100% P2 PF Trial 1 Ptotal_i Ptotal_f 0.81 0.25 |%Change (in P) Trial of 0.21 0.19 0.85 0.26 0.21 0.19 Trial 0.22 9.52% 0.2 0.77 0.24 0.22 0.2 0.2 Time 0.19 9.09% 0.0148 0.0476 0.2 0.19 5.00% Average % Change (p): 0.0142 0.0451 7.87 0.0155 0.0506-41 X V fx B C D E F G H M N 0 P Q Part B: Elastic collision width of cart 1 flag L1= (0.012)m width of cart 2 flag L 0.012 Collision type 1: 1 with initial velocity = 0 m1= (0.26305) kg m2= (0.26423) kg Please rrecord the coresponding photogate readings that will be used to calculate each velocity above the velocity cell( if the cart was at rest or stopped after the collision then write at rest or stopped and fill the corresponding velocities with 0 ) Photogates readings: m1*vli m2*V2i m1*v1f m2*V2f Pli+P2i P1f+P2f [(Ptotalf-Ptotali)/Ptotali ]*100% Vli V21 V1f V2f P1i P2i P1f P2f Ptotal i Ptotal f %Change (in P) Trial 1 0.714 0.642 0.188 0.168 0 0.188 0.168 10.63% Trial 2 0.75 0.625 0.198 0.164 o 0.198 0.164 Trial 3 o 17.17% 0.714 0.678 0.188 0.178 o 0.188 0.178 5.32% Average %Change (P)= Time 11.04% 0.0168 0.0187 0.016 0.0192 0.5*m1*V1i^2 0.5*m2*V2i^2 0.5*m1*V1f^2 0.5*m2*V2f^2 Kli+K2i K1f+K2f [(Ktotalf-Ktotali)/Ktotali ]*100% 0.0168 0.0177 K1i K21 K1f K2 Ktotal i Ktotal f %Change 0.067 0.054 0 0.067 0.054 19.40% po 0.074 0.051 o 0.074 0.051 31.08% 0.067 0.06 o 0.067 0.06 10.44% Collision type 2: Average %Change(K)= 20.31% m1= (0.26305) m2= (0.51423) kg m2 heavier mass Photogates readings: m1*vli m2*V2i m1*vif m2*V2f P1i+P2i P1f+P2f [(Ptotalf-Ptotali)/Ptotali ]*100% Vli Vzi Vif V2f Pli P2i PIf P2f Ptotal i Ptotal f %Change (in P) Trial 1 0.49 0.502 0.059 0.252 0.132 0.03 0.252 0.162 35.71% Trial 2 0.52 0.556 0.098 0.267 0.146 0.05 0.267 0.196 Trial 3 o 26.59% 0.58 0.66 0.13 0.298 0.173 0.067 0.298 0.24 19.46 Time 0.0243 0.0239 0.201 Average %Change (P)= 27.25% 0.023 0.0216 0.1225 0.0207 0.0181 0.0906 0.5*m1*V1i^2 0.5*m2*V2i^2 0.5*m1*V1f^2 0.5*m2*V2f^2 K1i+K2i KIf+K2f [(Ktotalf-Ktotali)/Ktotali ]*100% K1i K21 K21 Ktotal_i Ktotal f %Change (K 0.0617 0.033 8.95E-04 0.0617 0.0338 45.22% D.06 0.041 0.0024 0.069 0.0434 7.109 0.086 0.057 0.0339 0.086 0.090 5.69% Average %Change(K)= 29.33% Collision type 3: m2= (0.26423) kg m1 heavier mass m1= (0.51305) kg Photogates readings: m1*Vli m2*V2i m1*vif m2*V2f Pli+P2i P1f+P2f [(Ptotalf-Ptotali)/Ptotali ]*100% Vli V21 V1f V2f P1 Pzi P1f P2 Ptotal i Ptotal f %Change (in P Trial 1 o 0.74 0.43 0.212 0.195 0.22 0.056 0.195 0.276 41.53% Trial 2 o 0.75 0.449 0.218 0.198 0.23 0.057 0.198 0.287 Trial 3 o 0.603 0.364 0.165 44.94% 0.159 0.19 0.043 0.159 0.233 Time 46.54% 0.0162 0.0276 0.0566 Average %Change (P)= 14.33% 0.0159 0.0267 0.0548 0.0199 0.033 0.0728 0.5*m1*V1i^2 0.5*m2*V2i^2 0.5*m1*V1f^2 0.5*m2*V2f^2 Kli+K2i K1f+K2f [(Ktotalf-Ktotali)/Ktotali ]*100% K1i K21 Kif K2f Ktotal i Ktotal f %Change (K) 0.072 0.047 0.005 0.072 0.052 27.77% 0.074 0.051 0.006 0.074 0.057 22.97% 0.048 0.033 0.004 0.048 0.037 22.92% Average %Change(K)= 24.55Experiment # 5 Conservation of linear momentum 1. Synopsis You will verify the law of conservation of momentum for both elastic and inelastic collisions. 2. Objectives * Setup a nearly-elastic collision between carts using a track. * Measure the momentum of objects. * Measure the momenta of a system of colliding carts before and after the collision using varied combinations of cart masses and initial speeds. * Verify the law of conservation of momentum. * Check the conservation of kinetic energy in elastic and inelastic collisions. 3. Preparation e Read sections 4-7 of this write-up. 4. Introductory Remarks One of the most important physical values describing motion of a point-like object is its velocity. The importance of the velocity has been understood for more than 2300 years (Aristotle). However, it was only Newton who discovered (circa 1684) what causes velocity to change. The central idea in Newton's study of motion was inertia, which is the reluctance to change the state of motion. This included something more than the velocity of the moving object, it included a dynamical characteristic of the object called mass. The mass of an object is a measure of its inertia ~ its capacity to resist changes of its velocity. Naturally, Newton chose the simplest (mathematical) combination of mass (a scalar quantity) and velocity (a vector) to produce a value describing the motion with inertia. This combination is now referred to as linear momentum (p) or just momentum. Newton then stated three laws that included momentum as a key player. The first law states that if the net external force acting on a point-like object is zero, the object must be moving at a constant velocity, 'HY 2171 EXPERIMENT #4 i.e., the object is in an inertial state. The second law says that to change the state of inertia of an object in an inertial frame, an external agent, a net external force is required. The rate of change of momentum is equal to that force. The third law stipulates that if two parts of a closed system interact with each other, they act on each other with equal and opposite forces, so that the total change of momentum of the system s zero, in other words, in a closed system, linear momenturn s conserved. The principle of linear momentum conservation is fundamental. It is also quite often the only tool available for the analysis of the interaction between two objects. The first reason for this is the universality of momentum conservation it is conserved regardless of the type of interaction. The second reason is that it is the most tractable property if the mass and the velocity of the participant of interaction are known, so are their momenta. There is nothing hidden about it, there are no latent forms of momentum! For instance, imagine two particles collide and produce some (they may be quite different) particles as a result of the collision. The detected particles with their momenta serve as the smoking gun of the interaction event. It is due to this smoking gun, the momenta, that the interaction scenario can be studied and discovered. In this laboratory, you will study several types of collisions between two objects and analyze the momentum exchange taking place during the collisions. One type, called elastic collision, encompasses only collisions in which the kinetic energy is also conserved. All other collisions, where some kinetic energy is transformed into other forms of energy, are called inelastic. Among the latter, a perfectly inelastic collision is such that the participants of an interaction \"stick\" together after the collision. Photogate A Photogate B_ IEn M, Track Figure 1. A diagram of the positions of the photogates and glider cars. In all of these experiments, the sum of the initial momenta of the system of cars is equal to the sum of the final momenta of the system of cars: 2= ER BT In the case of two objects: D1+ P, = Puy + Pz, > muby + maby, = mydy, +maby, Again in the case of two objects, we have for the initial (before the collision) and final (after the collision) Kinetic energies: 1 2 1 2 K= 5muvl +5mavd, 'HY 2171 EXPERIMENT #4 If K; = Ky then the collision is elastic; if K; > K then the collision is inelastic. (Note that K;