Read: Follow all nine text links from this page, read the content on each: http://www.physicsclassroom.com/Class/momentum/ .

This is a lot to read, but it's really good and worth every minute spent on it. Answer the questions in your notebook. Watch the animations and videos. Think about what the author is communicating all the while you read.

Problems (answer on the website, show some work):

Review of energy:

1. What is the kinetic energy possessed by a 2300 kg truck traveling at 28 m/s?

0.5•2300•28² =900000 J (about)

2. How much work is done by the brakes on the truck from 1 in each of the following instances:

a. They slow the truck down to 24 m/s.

∆E = W

kinetic energy before = 900,000 J

Kinetic energy after = .5 • 2300 • 24² = 662000 J (about)

Change in kinetic energy = 900,000 J - 662,000 J = 238,000 J of work against friction.

b. They exert 8000 N of force against the motion of the truck over a distance of 50.0 meters.

W = F • d

W = 8000 N • 50 m = 400,000 J of work means 400,000 J of lost kinetic energy

c. They exert 8000 N of force against the motion for 2.0 seconds.

∆p = m∆v = F•t = 8000 N • 2 seconds = 16,000 N•s

m∆v = 16000 N•s so ∆v = 16,000 ÷ 2300 kg = 6.96, let's say 7 m/s

v = 28 - 7 = 21 m/s

∆Kinetic energy = 0.5•2300•(28² - 24²) = 239000 J lost -- so 23,900 J of work by the brakes

Review of impulse and momentum

3. How much momentum is possessed by a 2300 kg truck traveling at 28 m/s?

p = m•v = 2300 kg • 28 m/s = 64,400 J

4. How much momentum is lost by the truck from number 3 in each of the following instances?

a. It slows down to 24 m/s.

∆p = m∆v = 2300kg(28-24)m/s = 9200 J

b. The brakes exert 8000 N of force against the motion of the truck over a distance of 50.0 meters.

See work above: truck loses 400,000 J of kinetic energy so it now has 900,000 - 400,000 = 500,000 J

Kinetic Energy = 0.5 • m•v² so v= sqrt(2E_k/m) = sqrt(2•500,000/2300) = 20.85, let's say 21 m/s

∆p = m∆v so 2300 kg •(28-21)m/s = 16,000 J

c. The brakes exert 8000 N of force against the motion of the truck over a period of 2.0 seconds.

∆p = F•t = 8000 N • 2 s = 16000 N•s = 16000 J

Review of addition of vectors at right angles to each other.

5. A right triangle has legs of length 4 meters and 3 meters. How long is the hypotenuse?

Pythagorean Theorem: hypotenuse is sqrt(4² + 3²) = 5 meters

6. When the velocity of an airplane is examined, it is found to travel 100 m north each second and 75 m east each second. What is the magnitude of its velocity?

sqrt(100² + 75²) = 125 m/s

7. A bucket of mass 12 kg and velocity 4 m/s slides freely across the ice towards the east. A 20 kg magnet with an initial velocity of 6 m/s to the north collides with and sticks to the bucket of ice. What is the new velocity (magnitude)?

p_east = 12 •4 = 48 kg•m/s

p_north = 20 • 6 = 120 kg•m/s

p =sqrt (48² + 120²) = 129 kg•m/s

v = p÷m = 129÷(20 + 12) = about 4 m/s

8.(Try this one again).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?.

Hint: Find kinetic energy of each before braking, work of each set of brakes, kinetic energy of each after braking, velocity of each after braking, momentum of each after braking, combine momentum as vectors, find velocity of pile.

see last week's answer