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Impulse and Momentum Lab Report Mr. Johnson How are force and impulse related to linear momentum and conservation of momentum? Part 1: Qualitative Introduction Math [5.3] βπ = πΉβπ‘ ππ£ β ππ£π = πΉβπ‘ βππ£π = πΉβπ‘ Procedure Independent: speed of cart Dependent: amount of force needed to stop cart Controls: mass the same, same launching mechanism, same level table 1. Carefully measure the mass of both carts to that they are the same. 2. Using two rulers, launch both carts simultaneously into a personβs hands. Pull one ruler farther back to give one of the carts a larger velocity. 3. Discuss with catcher which cart took more force to stop it. Independent: mass of cart Dependent: amount of force needed to stop cart Controls: speed of both carts, same launching mechanism, same level table 1. Carefully measure the mass of both carts to that one is more massive than the other. 2. Using two rulers, launch both carts simultaneously into a personβs hands. Pull both rulers back the same distance so the velocity stays the same. 3. Discuss with catcher which cart took more force to stop it. Narrative of Results After completing both parts of this section, it was noticed that when mass is the same, the cart with the greater velocity took more force to stop. When velocities were the same, the more massive cart required more force to stop. I did assume the speeds of the carts were the same when I varied the masses of the carts. This assumption could lead to large uncertainties if it were quantitative. [4.4] Discussion The results of this experiment were what I expected. Looking at the relationship βππ£π = πΉβπ‘ I can see that mass and initial velocities are indirectly proportional. [5.3] Assuming the same time is needed to stop something, doubling the mass would require twice the stopping force and doubling the velocity would require twice the stopping force. A practical application for this section would be air bags in cars. The air bag increases the time for the collision of the personβs head with the steering wheel. By increasing the time, less force is needed to stop the personβs head which leads to fewer injuries. [7.2] Error Since there were no quantitative measurements there were no meaningful errors associated with this section. Part 2: Colliding Carts Math [4.1] Prediction: The change in cart 1βs momentum will equal the change in cart 2βs momentum in all three collisions. [6.4] βπ1 = ββπ2 Collision 1: Cart1βs change in momentum βπ1 = π1 π£ β π1 π£π = 0.0552ππ β 0.19 π π β 0.0552ππ β 0.85 π π = β0.036 ππ π π Collision 1: Cart 2βs change in momentum βπ2 = π2 π£ β π2 π£π = 0.0558ππ β 0.6 π π β 0.0558ππ β 0 π π = 0.033 ππ π π -0.036 kg*m/s versus 0.033 kg*m/s Collision 2: Cart 1βs change in momentum βπ1 = π1 π£ β π1 π£π = 0.2558ππ β 0.67 π π β 0.2558ππ β 0.95 π π = β0.072 ππ Collision 2: Cart 2βs change in momentum βπ1 = π1 π£ β π1 π£π = 0.0558ππ β 0.7 -0.072 kg*m/s versus 0.039 kg*m/s π π β 0.0558ππ β 0 π π = 0.039 ππ π π π π Collision 3: Cart 1βs change in momentum βπ1 = π1 π£ β π1 π£π = 0.0558ππ β 0 π π β 0.0558ππ β 0.65 π π = β0.036 ππ π π Collision 3: Cart 2βs change in momentum βπ1 = π1 π£ β π1 π£π = 0.2558ππ β 0.2 π π β 0.2558ππ β 0 π π = 0.051 ππ π π -0.036 kg m/s versus 0.051 kg*m/s Procedure [4.2] Independent: type of collision Dependent: velocity of carts before and after colliding Controls: Same surface, same planet, same person filming, same person colliding carts 1. Set up track 2. Measure mass of both carts [4.3] 3. Keep one cart stationary and film it as the second cart collides with it [4.3] 4. Study the velocity vs time graph in Tracker 5. Add mass to the colliding cart while no mass is added to the stationary cart a. Record new mass of the more massive cart [4.3] 6. Film the more massive cart colliding with the stationary less massive cart [4.3] 7. Study the velocity vs time graph in Tracker 8. Switch the carts, have the less massive cart collide with the more massive cart [4.3] 9. Study the velocity vs time graph in Tracker Narrative of Results In the first collision I had a cart hitting a stationary second cart of equal mass. In that collision the first cartβs change in momentum was -0.036 kg*m/s and the second stationary cartβs change in momentum was 0.033 kg*m/s. These two numbers are very close to each other which is what I expected to happen. In the second collision I had a more massive cart colliding with a less massive stationary cart. The massive cart had a change in momentum of -0.072 kg*m/s and the second less massive cart saw its change of momentum to be 0.039 kg*m/s. I would have liked to have seen these numbers closer to each other. In my third and final collision, I had a less massive cart strike a stationary more massive cart. In this collision the less massive cart had a change in momentum of -0.036 kg*m/s while the more massive cart had a change in momentum of 0.051 kg*m/s. Again I would have liked to have seen these numbers closer together. Discussion The law of conservation of momentum states that the total momentum of a system needs to remain unchanged. Since the first cart was moving, its momentum is the momentum of the system and this relationship needs to be true: βπ1 = ββπ2 . In these three collisions, I only feel comfortable saying momentum was conserved in the first collision where I had a cart collide with a second cart of equal mass. In that collision, the change in momentum of the two carts was only three thousandths apart; this could be explained by the error in the setup. My second two collisions were farther apart. According to conservation of momentum, βπ1 = ββπ2, this means the two numbers should equal each other but have opposite directions. In my calculations, I did see the opposite directions with the first cart having negative change in momentum and the second cart having positive change in momentum. [5.3] When I study the velocity vs time graphs for the collisions, I do see the correct relationships present. A practical application for this experiment would be CSI for police officers studying car crashes. Using conservation of momentum, police officers should be able to see if any of the carβs involved were speeding or driving left of the center line. [7.2] Tables of Data [4.3] Collision 1, Cart 1 colliding with stationary Cart 2 of equal mass. Cart 1 Data Cart 2 Data m1 = 0.0552 kg m2 = 0.0558 kg time in seconds time in seconds velocity in m/s velocity in m/s Collision 2, Cart 1 colliding with stationary Cart 2 of less mass Cart 1 Data Cart 2 Data m1 = 0.2558 kg m2 = 0.0558 kg time in seconds time in sec velocity in m/s velocity in m/s Collision 3, Cart 1 colliding with stationary Cart 2 of more mass Cart 1 Data Cart 2 Data m1 = 0.0558 kg m2 = 0.2558 kg time in sec time in seconds velocity in m/s velocity in m/s Graphs Collision 1: Cart colliding with a second cart of equal mass Collision 2: More massive cart colliding with a stationary less massive cart Collision 3: Less massive cart colliding with stationary more massive cart Error There were multiple sources of error in this experiment. Systematic errors include calibrating the videos, choosing to only do one trial per collision and the fact I choose the points in the Tracker software. Random errors include choosing to only do one trial per collision. Since I only did one trial per collision, I cannot calculate deviations to study any fluctuations in my experiment. I have no way to find the precision of this experiment. [4.4] Part 3: Explosions Math [4.1] Prediction: I predict that the total momentum of each system will remain zero [6.4] Ξ£π = 0 Explosion 1: Carts have near equal mass π π π π Ξ£π = π1 + π2 = π1 π£1 + π2 π£2 = 0.2ππ β (β0.9 ) + 0.15ππ β (0.95 ) = β0.038 ππ π π Explosion 2: Carts have unequal mass π π Ξ£π = π1 + π2 = π1 π£1 + π2 π£2 = 0.9ππ β (β0.4 π ) + 0.15ππ β (1.0 π ) = β0.21 ππ π π Procedure [4.2] Independent: amount of mass added to carts Dependent: velocities before and after explosion Controls: Same elastic force, same surface, same person filming, same planet 1. Measure mass of both carts using luggage spring scale [4.3] 2. Arm the spring by pushing it in to the first setting [4.3] 3. Rest both carts next to each other so that they will push apart when the pin is pushed 4. Push pin with a smooth surface so friction does not interfere with results [4.3] 5. Film the carts pushing away from each other. [4.3] 6. Study the velocity vs time graph in Tracker 7. Add mass to the spring cart and measure its new mass [4.3] 8. Arm the spring by pushing it in to the first setting [4.3] 9. Film carts pushing apart [4.3] 10. Study the velocity vs time graph in Tracker Narrative of Results In explosion 1, I calculated the net momentum to be -0.038 kg*m/s. I expected it to be zero. In explosion 2, I calculated the net momentum to be -0.21 kg*m/s. Again, I expected it to be zero. Discussion The law of conservation of momentum states that the total momentum of a system should remain unchanged. In an explosion, the initial momentum of the system is zero; so the total final momentum of the system needs to be zero as well. I was pleased with the result from the first explosion, 0.038 kg*m/s is very close to zero. However, the result from the second explosion is farther from zero. Itβs still less than one so I can argue that momentum is still conserved. [5.3] A practical application for this is the study of shell fragments after a bomb was detonated. By studying the pattern of the shell fragments, one would be able to gain information about the bomb itself. [7.2] Tables of Data [4.3] Explosion 1: Carts of near equal mass Cart with spring Cart w/out spring m1 = 0.2 kg m2 = 0.15 kg time in sec time in sec velocity in m/s velocity in m/s Explosion 2: Carts with unequal mass Cart with spring Cart w/out m1 = 0.9 kg m2 = 0.15 kg time in sec time in sec velocity in m/s velocity in m/s Graphs Explosion 1: Carts have near equal mass Explosion 2: Carts have unequal mass Error There were multiple sources of error in this experiment. Systematic errors include calibrating the videos, choosing to only do one trial per collision and the fact I choose the points in the Tracker software. Random errors include choosing to only do one trial per collision. Since I only did one trial per explosion, I cannot calculate deviations to study any fluctuations in my experiment. I have no way to find the precision of this experiment. [4.4]