Momentum Read Your Textbook: Introduction to Physical Science Chapter 3.5 Chapter 4 Practice Homework Exercises

Motion Review Velocity = change in displacement = Dx Speed change in time Dt Acceleration = change in velocity = Dv How fast are you change in time Dt getting faster. Force = mass x acceleration = m Dv a = F/m Dt A look at the two definitions of acceleration….

Force and Acceleration F = m Dv = m a Dt If a Force acts occurs over a short time, a small acceleration results. If a Force acts over a long time, a large acceleration results.

Cannon Ball! F Dt = m Dv F = m Dv Dt For the same Force (amount of powder), why is the speed of a cannon ball greater when fired from a longer cannon barrel? F = m Dv Dt

Interaction Time F Dt = m Dv The longer cannon barrel gives the cannon ball a larger impulse and therefore more momentum. The Force (F) is allowed to act for a longer time Dt to build up velocity (Dv).

Impulse and Momentum acceleration = acceleration a = a F = Dv F Dt = m Dv m Dt Impulse Momentum

Impulse and Momentum F = Dv F Dt = m Dv m Dt Impulse Momentum If a change in velocity (momentum) occurs over a short time, a large force must act. If the change in velocity (momentum) occurs over an extended time, a small force is acting. Recall the Egg Toss Game A Boxer Bobs and Weaves His Head Bending Legs Upon a Parachute Landing

Conservation of Momentum Momentum is a conserved quantity, that is, for any isolated system, the total momentum remains unchanged. Momentum = mass x velocity P = m v

Conservation of Momentum Momentum is a conserved quantity, that is, for any isolated system, the total momentum remains unchanged. Momentum = mass x velocity P = m v Consider the following collision: Before After v M m V

Conservation of Momentum Momentum is a conserved quantity, that is, for any isolated system, the total momentum remains unchanged. Momentum = mass x velocity P = m v Consider the following collision: Before After v V’ M m m M v’ V

Conservation of Momentum Momentum is a conserved quantity, that is, for any isolated system, the total momentum remains unchanged. Momentum = mass x velocity P = m v Consider the following collision: Before After Total Momentum MV + mv = total momentum = MV’ + mv’ v V’ M m m M v’ V

Ice Ball Toss v M m Total Momentum Before: M V + m v 60 kg ( 0 km/hr) + 20 kg (10 km/hr) = 200

Ice Toss Momentum After (must be identical to momentum before) = 200 = (M+m) v’ 200 = (M+m) v’ 200 = (60+20) v’ v’ = 200/80 = 2.5 km/hr

Conservation of Momentum What is the total momentum of the debris from a firecracker? Before After M V = 0 = total momentum before Total Momentum After = m1v1 + m2v2 + m3v3 + … m2 m1 m3 m4

Conservation of Momentum What is the total momentum of the debris from a firecracker? Before After M V = 0 = total momentum before Total Momentum After = m1v1 + m2v2 + m3v3 + … = 0 m2 m1 m3 m4

Rifle Shot Let mbullet = 0.3 kg, Mrifle = 5 kg, and vbullet = 370 m/s mbulletvbullet + MrifleVrifle = 0.3 kg (370 m/s) + 5kgVrifle 0 = 0.3 kg (370 m/s) + 5kgVrifle

Rifle Shot If momentum is conserved, why doesn’t a rifle kill you upon recoil after firing a bullet? Before: mbulletvbullet + MrifleVrifle = 0

Rifle Shot If momentum is conserved, why doesn’t a rifle kill you upon recoil after firing a bullet? Before: mbulletvbullet + MrifleVrifle = 0 After: = 0 Let mbullet = 0.3 kg, Mrifle = 5 kg, and vbullet = 370 m/s

Rifle Shot Let mbullet = 0.3 kg, Mrifle = 5 kg, and vbullet = 370 m/s mbulletvbullet + MrifleVrifle = 0.3 kg (370 m/s) + 5kgVrifle 0 = 0.3 kg (370 m/s) + 5kgVrifle -0.3(370) = 5 kg Vrifle Vrifle = - 0.3(370)/5 = - 2.2 m/s

Mriflevrecoil = mbulletVbullet Rifle Shot Let mbullet = 0.3 kg, Mrifle = 5 kg, and vbullet = 370 m/s mbulletvbullet + MrifleVrifle = 0.3 kg (370 m/s) + 5kgVrifle 0 = 0.3 kg (370 m/s) + 5kgVrifle -0.3(370) = 5 kg Vrifle Vrifle = - 0.3(370)/5 = - 2.2 m/s Shoulder aches, BUT your alive! Mriflevrecoil = mbulletVbullet

Now you try: What is the velocity of a bullet (m = 0.15 kg) after being fired from a 10 kg rifle (NOTE: rifle recoils with a velocity of 3 m/s). A. 30 m/s B. 1.5 m/s C. 200 m/s D. 310 m/s E. none of these mbulletvbullet + MrifleVrifle = 0.15 kg (vbullet) + 10kg (3 m/s) 0 = 0.15 kg (vbullet) + 30 kg m/s -30 kg m/s = 0.15 kg vbullet vbullet = - 30 kg m/s /(0.15 kg) = - 200 m/s

Train Link An train engine runs into a stationary box car weighing 4x more than itself to link up. If the engine was traveling 10 mph before link up, how fast does the train move after?

Train Link An train engine runs into a stationary box car weighing 4x more than itself to link up. If the engine was traveling 10 mph before link up, how fast does the train move after? MOMENTUM BEFORE = MOMENTUM AFTER MVBC + 0 = (M + 4M) VAC

Train Link An train engine runs into a stationary box car weighing 4x more than itself to link up. If the engine was traveling 10 mph before link up, how fast does the train move after? MOMENTUM BEFORE = MOMENTUM AFTER MVBC + 0 = (M + 4M) VAC M(10) = (5M) VAC 10 = 5 VAC 2 = VAC

Angular Momentum L Angular Momentum: A combination of... m Mass v Speed of Rotation r Mass Position (with respect to rotational axis) L = m v r Conservation Examples: Spins of Dancers or Ice Skaters Those Funky Coin Vortexes in Stores Tops and Gyroscopes Riding a Bicycle

Precession and the Earth 1 complete cycle takes 26,000 years

Orbit Application Angular Momentum L, is the product of a planet's mass (m), orbital velocity (v) and distance from the Sun (R). The formula is simple: L = m v R, where R = a function of e the eccentricity.

Faster, Closer Conservation of Angular Momentum: L = L m V r m v R

Summary Impulse and Momentum Conservation of Momentum Angular Momentum F Dt = m Dv Conservation of Momentum Total Momentum (P = M V) Angular Momentum L = m v r