Lab Report Impulse and Momentum Essay Example

Lab Report: Impulse and Momentum Paper Background: Momentum is a measure of inertia and can be defined by mass in motion ( Stan brought ,) anything with mass that is moving has momentum. Mo momentum can be calculated with p=mass x velocity, there fore momentum is dependent on both mass and velocity. Momentum is similar to energy in that both are always conserved, but energy ha s multiple forms and has to be in the right situation to be conserved. M momentum transferred is the integral of force over time, whereas kinetic c energy transferred is the integral of the force over distance. If the Orca is constant, then momentum is force multiplied by time, and energy is force multiplied by distance(physics classroom. ) Both can be applied d to a car crash, the momentum before impact is equal to after impact , and the EKE before the impact is equal to PEe+EKE+heat during impact. Because F x t=m x delta v, impulse is the change in momentum. Force and time are related to each other during in a collision through delta v/ t therefore the shorter the time the greater the force and vice versa. This is why cars have air bags and crunch on impact: slow the impact time down the reform the body feels less force equaling less injuries. We will write a custom essay sample on Lab Report: Impulse and Momentum specifically for you for only $16.38 $13.9/page Order now We will write a custom essay sample on Lab Report: Impulse and Momentum specifically for you FOR ONLY $16.38 $13.9/page Hire Writer We will write a custom essay sample on Lab Report: Impulse and Momentum specifically for you FOR ONLY $16.38 $13.9/page Hire Writer For an elastic collision n the force vs.. Time graph would look nearly to a concave down parabola, who areas an inelastic string graph would have a sharp pinnacle similar to an absolute value graph. Implementation theorem can be derived from Newtons Second Law: F=ma delta Wet mat=NV(f)NV(i) Fine=p(f)p(i) Impulse= delta p Results: String Graph Initial velocity was taken from the negative linear line on the position vs.. Time and lined up with the amplitude on the velocity vs.. Time graph because this is where t e car was accelerating due to gravity. This was taken to find the change in evolve itty later used in change in momentum. Final velocity was found in the average of the flat line just after initial velocity because this is where the car was stretching the string. The average force was calculated around the pi nacelle in the force vs.. Time graph, used later in impel SE. The time was calculated by taking If stop time Vi start time also used later for the change in momentum. Graph of elastic material showing the longer time and less force Trial If (m/s) Vi (m/s) Delta V (m/s) Avgas. F (N) Delta t (s) String 1 . 5748 . 718 1. 2928 6. 803 2 . 2854 . 4267 . 7121 5. 376 . 11 Elastic 1 . 7512 . 9856 1. 7169 2. 437 . 4 4823 . 8781 1. 3604 2. 304 . 2 Each column was calculated by the stats button o n the computer for later use to find the impulse. Trail Impulse (NSA) Delta p (m delta v) Impulse integral (NSA) . 6803 1. 272 . 680 . 5914 . 7007 . 592 . 9748 1 . 739 . 975 . 9677 1. 339 . 967 The impulse and impulse integral were nearly identical both cal elated with force times change in time. This proves there mus t be a systematic error in the change of momentum because the numbers consistently higher and not randomly lower and higher. Discussion: The experiment done did not display implementation theorem, change in momentum was much higher than either impulses. The mass is presumed t o be weighed correctly at . Keg leaving only velocity to be wrong ,therefore the e discrepancy must of been caused by miscalculation in velocity. The miscall calculation may have been caused by the averaged slopes on the graph to be mi splayed giving the wrong average, motion sensor not being placed correctly or zeroed correctly. This could of happened by placing the motion sensor too close. To attain the correct change in velocity we can set impulse equal to . 984 times velocity. The correct change in velocity should be . 68031. 984 for string 1 which is . Mm/s. The correct change in velocity for elastic 1 should be . 991 m/s. In this situation the velocity should always be . 016% greater than the impulse. The impulse integral is the best way to calculate impulse because the come utter is always more exact. Using the other impulse and the data points we c elected were averages, therefore not 100% accurate, whereas the computer wouldnt round those numbers givin g a better answer. Although it may b the best way to calculate impulse, all three if tested correctly should be roughly the same answers so any method is valid.