okay, okay, okay already -- I'll take a breather on the physlets this week (maybe look at a few in class, but I won't ask you to look at any on your own). I still expect you to do some work online though and once again due on Friday, September 23.

Let us talk about inelastic collisions.

Imagine a 5 kg cart A traveling 4 m/s to the east. Imagine a second 5 kg cart moving 3 m/s to the west. Imagine that the carts collide and couple so that a train of two carts continues in one connected motion.

Answer the following questions (don't forget that momentum is a vector and that it therefore gets a sign according to its direction):

1. What is the momentum of cart A before the collision? 5 • 4 = 20 kg•m/s
2. What is the momentum of cart B before the collision? 5 • -3 = -15 kg•m/s
3. What is the momentum of cart A and B together before the collision?20 + -15 = 5 kg•m/s
4. Assuming that momentum is conserved, what is the momentum of carts A and B together after the collison?5 kg•m/s
5. Assuming that momentum is conserved, what is the velocity of carts A and B together after the collision? 5 kg•m/s ÷ 10 kg = .5 m/s

Consider a fully loaded boxcar of 130,000 kg traveling east at 2.6 m/s that overtakes a "train" of 4 coupled 30,000 kg empty boxcars already traveling east at 1.8 m/s. Assuming that momentum is conserved, what is the velocity of the resulting 5 car train?

130,000 kg • 2.6 m/s + 4 • 30,000 kg • 1.8 m/s = 250,000 • V
338,000 kg•m/s + 216,00 kg m/s = 250,000 • V
554,000 kg•m/s = 250,000 •V

V = 2.216 m/s Note that the faster car slowed down while the slower group sped up.

Consider a 2500 kg sportscar traveling south on Cicero at a rate of 40 m/s. A3000 kg pickup truck traveling east on 167th is moving at 25 m/s. At about the same instant both drivers realize that the signal has failed and that a collision is imminent. The brakes on the sportscar provide 18000 Newtons of braking force vs. the street. The pickup is able to produce18000 Newtons as well. After the collision the state trooper is able to determine from skid marks that the car had braked for 60 meters and the truck for 30 meters.

Upon collision the two vehicles entangled and the pile-up proceeded towards the (approximate) south east. How fast was the pile moving? Bonus: in exactly what direction?

Kinetic Energy of Car = .5 • 2500 • 40 • 40 = 2,000,000 J

Work done by brakes to slow car: 18000 N • 60 m = 1,080,000 J

Kinetic Energy of car at collision: 2,000,000 - 1,080,000 = 920,000 J

Velocity of car at collision: sqrt(2•kinetic energy/mass) = sqrt (2•920,000/2500) = 27 m/s

Momentum of car at collision: 2500 kg•27 m/s = 67,500 kg•m/s South

Kinetic energy of truck = .5 • 3000 • 25 •25 = 937,500 J

*Note original solution followed from 40 m of truck skidmarks where problem said 30 m.

Work done by brakes to slow truck: 18000 N • 40 m 30 m= 720,000 J, 540,000 J

Kinetic energy of truck at collision: 937,500 - 720,000 J 540,000 J= 217,500 J 397,500 J

Velocity of truck at collision: sqrt(2•217,500397500/3000) = 12 16.3 m/s

Momentum of truck at collision: 3000 kg • 12 16.3 m/s = 36,000 48,800 kg•m/s East

Momentum of pile. Momentum is a vector so momenta at right angles add by Pythagorean Theorem: sqrt(67,500² + 36,000 48,800²) =76,500 83300 kg•m/s of momentum

Velocity of pile: 76,500 83300 divided by (2500 +3000) kg = 13.9 15.1m/s

Direction: ArcTan (67,500/36,000 48,800) = 62 54˚ south of east