CHAPTER-9 Momentum & Its Conservation 9.1 Impulses And Momentum Relating Impulse in addition to Momentum Ft =
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9.1 Impulses And Momentum How is Velocity Affected by Force The as long as ce changes over time. What is the physics, instant by instant, behind serving a tennis ball The as long as ce acting on the tennis ball quickly increases, maxes out, in addition to then quickly decreases. Just after contact the ball de as long as ms or is squeezed, the strings are stretched in addition to the as long as ce acting on the ball is increased. The as long as ce reaches its maximum, the ball begins to snap away from the strings in addition to the ball will begin to regain its shape.
The as long as ce F acting on the tennis ball begins at 0N, then rapidly increases to a as long as ce >1000 times the weight of the ball, then the F will quickly drop back to 0N. The entire event last only thous in addition to ths of a second. What would the as long as ce vs time graph look like GRAPH Relating Impulse in addition to Momentum Newtons Second Law equation = F = ma Where a = v/t (t is time of contact) F = m v/t Multiplying each side by t yields Ft = mv Impulse = Ft (units Ns)
Impulse = Impulse is the product of the average as long as ce (Fnet) exerted on an object in addition to the time interval over which it acts on the object. Impulse = Ft Back to: mv m(v2 v1) mv2 mv1 Ft = mv2 mv1 Linear momentum = = mv Impulse = Impulse-Momentum Theorem = Ft = = 1 2 = mv = mv2 mv1 Demo: I need one volunteer to jump from a lab table, then discuss jump in detail. Q: Why did — bend his/her knees when l in addition to ing on the floor A: To lesson the as long as ce of the impact on the knees, back, feet, so he/she would not be hurt. Bending of the knees absorbs the shock of the jump. Q: How/why does bending the knees absorb the shock A: By extending the time it takes to stop. Q: What does time have to do with it
Draw sketch on board in addition to explain why lengthening time decreases the amount of as long as ce exerted on the knees, back, feet, Ft = = 2 1 = mv = mv2 mv1 Ft = mv2 mv1 mv1 = just be as long as e hits ground, max speed, maximum momentum mv2 = stopped, no motion, no momentum Ft = (mv2-mv1) is a fixed value in addition to cannot change in addition to if t increases, then the as long as ce (F) exerted on the knees, back, feet must decrease. Using the Impulse-Momentum Theorem What is the of the tennis ball if the area under the graph is 1.8 What units would apply to area The units = Ns The area = Impulse = = 1.8Ns(kgm/s)
The ball began at rest , = 0, in addition to then gained impulse/speed while in contact with the racket to a final = 1.8Ns (kgm/s) Q: If the mass of the tennis ball is 0.055kg, what is the final velocity of the ball = 2 1 = mv2 mv1 1.8Ns = 0.055kg(v2) 0 1.8Ns = 0.055kg(v2) V2 = 1.8(kgm/s)/0.055kg V2 = 32.7m/s Because velocity is a vector quantity, so is Impulse/momentum the signs (+/-) are important Using the Impulse-Momentum Theorem to Save Lives A large occurs when there is a large Impulse. A large impulse occurs when: 1. large as long as ce acting over a short period of time 2. small as long as ce acting over a long period of time
EGG EXAMPLE If you remember nothing else about this class, remember this example, it could save your life some day. Car Accident seat belt/air bag vs windshield/tree/pole An 80kg person is traveling in a car at 60m/s, the car hits a tree in addition to stops in 0.04 seconds. The driver A is wearing a seat belt increasing the stopping time to 0.95 seconds. What as long as ce will the driver experience during the stop Ft = mv v is the same with/without the air bag in addition to /or seat belt Ft = mv2 mv1 Ft = 0 mv1 Ft = mv1 F = mv1/t F = (80kg)(60m/s)/0.95s F = 5052.6kgm/s2 F = 5052.6N
An 80kg person is traveling in a car at 60m/s, the car hits a tree in addition to stops in 0.04 seconds. The driver B is not wearing a seat belt. What as long as ce will the driver experience during the stop F = (80kg)(60m/s)/0.04s F = 120,000kgm/s2 F = 120,000N vs 5052.6N (24X greater F) REMEMBER THIS EXAMPLE FOREVER! Example-Stopping an SUV A 2200kg SUV traveling at 94km/hr (-m/s) can be stopped in a)21s by gently applying the brakes, b)5.5s by a panic stop in addition to c)0.22s if it hits a wall. What is the average as long as ce exerted on the SUV as long as each of the stops SOLUTION Draw a sketch, table known/unknown, equations, steps
Ft = Ft = 2 1 2 = 0 Ft = – 1 Fa = – 1/t Fa = -mv1/t Fa = -(2200kg)(26m/s)/21s Fa = -2724kgm/s2 Fa = -2724N Fb = -mv1/t Fb = -(2200kg)(26m/s)/5.5s Fb = -10,400kgm/s2 Fb = -10,400N Fc = -mv1/t Fc = -(2200kg)(26m/s)/0.22s Fc = -260,000kgm/s2 Fc = -260,000N 9.2 THE CONSERVATION OF MOMENTUM Law of Conservation of Momentum = The total momentum of a system be as long as e a collision/explosion is equal to the total momentum of the system after the collision/explosion as long as there is no net external as long as ce acting on the system. p1 = p2 pA1 + pB1 = pA2 + pB2
Momentum in a Closed System Closed System = A system (collection of objects) that does not loose or gain mass. Internal Forces = All the as long as ces within a closed system. External Forces = All the as long as ces outside the system. Isolated System = When the net external as long as ce on a closed system is zero. No system on Earth is truly closed in addition to isolated. There is always some interaction between a system in addition to its environment. Usually these interactions are small in addition to can be neglected by physics problems.
Example Problem 2 An astronaut in space fires a pistol, the thrust expells 35g of gas at -875m/s. the combined mass of the astronaut/pistol is 84kg. What is the velocity in addition to direction of the astronaut after firing the pistol Two-Dimensional Collisions A 2kg ball (A) moving at a speed of 5m/s collides with a stationary ball (B) that also has a mass of 2kg. After the collision, ball-A moves off 30o to the left, ball-B moves off 90o to the right of ball-A. How fast are they moving after the collision Possible Bonus/white board/
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