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1. Impulse and Momentum and Collisions and Stuff We will now quantify exactly how it hits the fan.

2. Momentum • Momentum = product of mass X velocity • Symbol for momentum is “p” (don’t ask me why) • So, p = mv • Momentum is also a vector and points in the direction of velocity • Units for mass = kg, units for velocity = m/s • So, units for momentum = kg•m/s • Sadly, there’s no other, more convenient unit for it • Plural is ‘momenta’

3. A closer look at momentum • Think of momentum as being how easy/hard it is to get something to stop. • Remember, p = mv • So if mass or velocity is small, it’s likely that the object has a small momentum, unless the other of the two is very large • What would be the momentum of an object at rest?

4. 5 Scenarios • Let’s look at the momenta of…. • A car at rest • Mass = 1000 kg, velocity = 0 m/s • P = mv = (1000kg)(0m/s) = 0 kg m/s • A bumblebee in flight (low mass, low velocity) • Mass = 0.05 kg, velocity = 2 m/s • P = mv = (0.05kg)(2m/s) = 0.1 kg m/s • Projectile from BB gun (low mass, high velocity) • Mass = 0.01 kg, velocity = 150 m/s • P = (0.01 kg)(150 m/s) = 1.5 kg m/s

5. 5 Scenarios continued • A giant land tortoise (large mass, low velocity) • Mass = 300 kg (I looked it up!), velocity = 0.5 m/s • P = (300kg)(0.5m/s) = 150 kg m/s • Tony Stewart at the Brickyard (large mass, high velocity) • Mass = 1000 kg, velocity = 90 m/s • P = (1000kg)(90 m/s) = 90,000 kg m/s

6. Understanding momentum • Which is easier to stop… • A slow moving baseball or a fast moving baseball? • A chihuahua racing down the hall towards you or Mr. Barbini racing down the hall towards you? • A speeding Volkswagen beetle or a speeding dump truck?

7. I need a volunteer • Rank these momentafrom lowest to highest • A fast flying bee • The earth in orbit around the sun • A parked garbage truck • A slowly flying bee • Your grandmother driving down the street in her 1959 Edsel • An oil tanker sailing the seven seas. • Tony Hawk grinding on a rail

8. Impulse • Impulse = Product of Force and how long that force acts. • Impulse = F∆t • Impulse is a vector. It acts in the same direction as the force • Units of force = Newtons, units of time = seconds. So, units for impulse = Newton•Seconds

9. Okay, but what does that really mean? • Let’s say you have a force of 10 N. You apply this force to a hockey puck on (frictionless) ice for 2 seconds. • Impulse = (10N)(2 sec) = 20 Newton seconds • Now, let’s say you apply this same force for 4 seconds • So the impulse now = 40 Newton seconds

10. Impulse-momentum theorem • Remember the work-energy theorem? This is kind of like that. • Impulse = change in momentum • In general, the mass of the object(s) in questions stays constant • ∆p = pF – pI = mvF – mvI • So F∆t = mvF – mvI

11. Well, Duh… • Let’s think about 2 situations: • Situation 1: pushing someone in a rolling chair • If you push on them with a certain force for two different times, the longer time will result in a greater change in momentum • So if the chair was at rest to begin with, it will end up going faster at the end for the longer time the force was applied

12. Less Duh • Situation 2: crashing your 1,000 kg car • Let’s say you are driving along at 45 m/s (about 96 mph) and you crash into one of two objects: a solid wall or a series of water-filled plastic bins (like they have on the highway) • If your car goes from 45 m/s to 0 m/s, you have a change in momentum of: • P = mv, so momentum = (1000kg)(45m/s) = 45,000 kg•m/s

13. The car crash continued • Recall that impulse = change in momentum, so Ft = mv = 45,000 kg•m/s • So, Force X time = 45,000 kg•m/s • Now, would you rather have that force spread out over a long time, or over a short time?

14. Compare stopping times • Change in momentum is 45,000 kg•m/s • For a stopping time of 5 sec (slamming on your brakes): (F)(5 sec) = 45,000 kg•m/s • So F = 9,000 N • For a stopping time of 1.5 sec (smashing into the plastic water bins): (F)(1.5 sec) = 45,000 kg•m/s • So F = 30,000 N • For a stopping time of 0.2 sec (smashing into the concrete wall): (F)(0.2 sec) = 45,000 kg•m/s • So F = 225,000 N

15. Let’s look at those numbers more closely… • Remember we have a 1,000 kg car and F = ma • If F = 9000N, then accel = 9 m/s (just under acceleration due to gravity) • If F = 30,000N, then accel = 30 m/s (just over 3X acceleration due to gravity) • If F = 225,000N, then accel = 225 m/s (about 23X the acceleration due to gravity) • So, which is the most survivable?

16. Let’s look at some examples • Hitting a baseball (mass = 0.14 kg) • Let’s say that a baseball is travelling to the left (negative velocity) at -38m/s. After being hit, it moves to the right at 58 m/s. What is the impulse applied to the ball? If the contact time was 1.6X10-3 sec, what was the force? • Impulse = ∆p = pF – pI = mvF – mvI • Impulse = (0.14kg)(58m/s) – (0.14kg)(-38m/s) • Impulse = 13.4 kg m/s • Impulse = F∆t, so F = impulse/∆t • Force = (13.4 kg m/s)/(1.6X10-3 sec) = 8400 N

17. Conservation of Momentum • In general, momentum is conserved • This means that the momentum at the beginning is the same as the momentum at the end • I.e. momentum is not created or destroyed • Remember, symbol for momentum is ‘p’ • So PI = ‘initial momentum’ and PF = ‘final momentum’ • Let’s look at some examples

18. Conservation of Momentum: Example 1 • Cannon and cannonball • What is momentum before cannon is fired (PI)? • Zero • So what does final momentum (PF) have to be if momentum is conserved? • Zero http://www.sparknotes.com/testprep/books/sat2/physics/chapter9section3.rhtml

19. Conservation of Momentum: Example 1 • Let’s say that the cannon has a mass of 1000 kg and Cannonball has a mass of 10 kg • So total PI = PI of ball + PI of cannon • PI of ball = (10kg)(0m/s) = 0 kg m/s • PI of cannon = (1000kg)(0m/s) = 0 kg m/s

20. Conservation of Momentum: Example 1 • Now let’s say that the cannonball moves to the right at 75 m/s. How fast does the cannon move to the left to conserve momentum? • PI = PF • So PF must equal zero

21. Conservation of Momentum: Example 1 • PF must equal zero • Total PF = PF of ball + PF of cannon • Remember, moving to the right = positive velocity • Moving to the left = negative velocity

22. Conservation of Momentum: Example 1 • PF = 0, so • (10kg)(75 m/s) + (1000kg)(Vcannon) = 0 • So -Vcannon= (10kg)(75m/s) / (1000kg) • Vcannon = -0.75 m/s • If ball moves to the right, cannon moves to the left, so Vcannonshould be negative, which it is

23. Conservation of Momentum: Example 2 • Newton’s cradle (or “Executive ball clicker” or, more crassly, “Newton’s Balls”) • How does it work? • If one ball is used, how many come up on the other side? • If two are used…?

24. Now, Dr. Mason, remind us again when conservation of momentum holds… • Momentum is conserved when there are NO EXTERNAL FORCES ACTING ON THE SYSTEM. • If any net external force acts, momentum will NOT be conserved. • Let’s look at a contrived example involving pool balls.

25. Billiards • Imagine the scenario where one pool ball is stationary on a pool table. A second pool ball smacks into it. • The ‘system’ is the two pool balls. • Are there any net external forces? • Now, imagine that just before they hit, a hole opens up underneath them. • The ‘system’ is the two pool balls. • Now are there any net external forces? • Okay, this time, no sudden trapdoor. Same two pool balls collide. • The ‘system’ this time is just one pool ball. • Now are there any net external forces?

26. Freight trains example • A freight train is being assembled out of two boxcars. • Car 1 has a mass of 65,000 kg and is moving to the right at V01 = 0.8m/s. • Car 2 has a mass of 92,000 kg and is also moving to the right at V02 = 1.3 m/s. • Car 2 collides with car 1 and couples with it. What is the speed of the train after the coupling? • Momentum is conserved. • (m1 + m2)Vf = m1V01 + m2V02 • Vf = 1.1 m/s

27. Ballistic Pendulum • A ballistic pendulum is a device that is used to determine the muzzle velocity of a gun. http://session.masteringphysics.com/problemAsset/1010989/26/1010989A.jpg

28. Collisions in 2-D • Conservation of momentum holds in 2 dimensions as well • As one may expect, you can treat the X- and Y-components independently • So, Pf = Pi • And PfX = PiX • And PfY = PiY