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Dive into the concepts of momentum, impulse, and friction through real-world scenarios like dropping an object on different surfaces, explaining Newton's third law, and understanding momentum transfer. Learn about key equations and practical applications of these physics principles.
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A quick thought • A glass plate is dropped, would the impulse be greater if it fell on a concrete floor or a carpeted floor? • In both cases the impulse would be the same (assuming it stopped upon impact) because the change in momentum is the same. • The carpeted floor would have a higher time of impact (therefore a smaller force), but an equal impulse to the concrete floor.
What if it doesn’t stop upon impact • Two balls of equal mass are dropped from the same height. Upon hitting the ground, one bounces the other does not. Which ball will feel a greater impulse? • The ball that bounces • Impulse causes a change in momentum, that ball that bounces has a larger change in velocity, than the one that stops.
Another thought on Newton’s third law, momentum, impulse and friction • In the winter, several years ago I was stuck behind a Camaro spinning its wheels on the ice. • Why couldn’t it go forward? • Very little friction between its tires and the ground. Cars use friction to move, the tire pushes on the ground, and it pushes on the car. • The coefficient of friction was very low and the weight of the car was very small in the rear end. • Fr =Fnμk
Quick tidbit about most sports cars • Most of the weight of any car is in the front end (engine, transmission etc.) the only thing in the back is a trunk (low weight if its empty). • Sports cars are normally rear wheel drive meaning all power comes from the back tires, the front tires do not apply any force. • Sandbags in the trunk increase the weight in the rear end of the car and increase the friction on those tires.
Back to the Camaro on ice • I got out to push it. I pushed and went backwards the car went nowhere (it did move but it was too small to notice) Why? • Mass, friction and impulse • Friction between my feet and the ground was also small, and when Newton’s 3rd law kicked in (instantly) I went backwards. The car had a much greater mass than me so it went (almost) nowhere.
Where impulse fits in • Any force will cause an acceleration, but unless the force is huge it has to be for a sustained period of time to cause a (observable) change in motion (momentum). • I could only push for a fraction of a second before Newton’s 3rd law pushed me off of the car. I had to find a way to push the car over a period of time (impulse) to give it a forward velocity (momentum) • Ft = mv
How I moved it • No, she did not want me to crack her new Camaro’s bumper with mine, so pushing her car with mine was out. • I pulled my car right behind hers and pushed her car with my foot on my bumper. What did this do? • Gave me more friction and mass (inertia). • Mass is obvious, friction increased because the weight of the car was much greater which increased the Fn (and Fr) which allowed me to push for several seconds.
Momentum transfer • Momentum can be transferred from one object to another. • My forward momentum went from me to the Camaro. • After I pushed I had no momentum, and the car had momentum. • The net momentum of the system (myself my car and the Camaro) did NOT change.
Law of Conservation of momentum • The net momentum of a system cannot change. • My initial forward momentum (and the momentum of the Camaro 0) equaled the final momentum of the car (and my final momentum 0) • If both objects were at rest to start, the forward momentum of what was pushed will be equal to the backward momentum of what pushed it. • Momentum can be transferred to the Earth.
Representing this mathematically • The sum of the momenta of all objects to start will equal the sum of all final momenta. • mvi = mvf • The Greek letter (sigma) means “the sum of” • so the sum of initial moment equals the sum of final momentum
Bigger problems • If an 85 kg running back is running at 5.1 m/s and is hit (directly backwards) with 820 N force for 0.15 s, what will his velocity be? • -820N (.15s) = 85(v)- 85(5.1) • v = 3.7 m/s (forward)
Homework • If a 92 kg running back is running at 5.1 m/s and is hit (directly backwards) by a linebacker with 410 N force for 2.1 s, what will his velocity be?