Physics Assignment 3

Read over the Momentum Basics page linked on the homepage of physicsclub.

Work through the following physlets (linked at http://www.physicsclub.net/physletIndex/momentumImpulse.html and within the question). Just follow the directions on each and keep notes on paper. The answers will be with the answer to the web question, but you need not include it in your web response.

1. Ball and Flower Pot

The Titan Ball experiences twice as much change in momentum since it "bounced". p = m Δv and for the bounce, Δv was from -v to + v where for the flower pot it was only from -v to 0.

2. Hammer and Nail

First find the momentum of the hammer before colliding: p = mv = 2 kg * 8 m/s = 16 kgm/s (I get the velocity by finding a displacement and a time and dividing).
Then find the time required to stop the hammer and divide (16 kgm/s ÷ .34 s = 47 N). You might find the time as much as .5 s for a force as small as 32 N.

3. Explosion CM (CM stands for center of mass)

The system was at rest before so after the CM must stay put. V = 0 m/s.

4. Cars and Walls

The car that bounces sees a much larger change in momentum. This makes that collision all the more dangerous for occupants.

Reading:

1. Section §4.5 of your text. Be sure you can define mass and weight. Name the m-k-s units for each. Be able to convert from one to the other on earth and on the moon.

2. §5.3 of Hewitt and §6.2 of Hewitt. Be able to compare the force that a baseball exerts on a pitcher to the force that the pitcher exerts on the baseball.

3. Chapter 7 in your text.

4. "The Physics Classroom" http://www.physicsclassroom.com/Class/momentum/ : a nice alternative to a traditional text with animations etc.

Answer the following text questions in your notebook (not on the website, but before working on the website).

Review questions: 2, 4, 5, 6, 8, 10, 11 page 100

Simple Problems about Rosebud over a Period of Coasting on a Horizontal Field of Snow (or Were There References to Classical Mechanics in Classic Movies of the Day?)

1. How much momentum is possessed by a 5 kg sled moving horizontally at 8 m/s?

5kg ×8 m/s = 40 kg m/s

2. How much acceleration is experienced by this sled if it slows down to 3 m/s over the next 10 seconds?

a = Δv/t = (8-3)m/s ÷10 s = 0.5 m/s/s

3. By how much does the sled's momentum change in that time?

before = 40 after = 15 so change is 40 - 15 = 25 kg m/s

4. The "rate of anything" is usually just "anything" ÷ by how much time it took. At what rate did the sled's momentum change? **Important note: the rate of change of momentum is usually called "Net Force".

F = Δp/t = 25 kg m/s ÷ 10 s = 2.5 N

5. So what is the net force on the sled during the 10 seconds here?

the rate of change of momentum is net force, so the net force is 2.5 N

6. Essay: Describe what happened to the sled in a sentence or two, using the words "momentum", "net force", "velocity" and "impulse".

As the problem began the sled had 40 kg m/s of momentum. It continued to coast against friction and over the next ten seconds the snow produced an impulse opposite the motion of the sled equal to 25 N s which reduced the sled's momentum. This is the result of a net force of 2.5 N owed to that friction.

Simple 1-D Inelastic Collisions

1. A bug and a bus, each moving 20 m/s but in opposite direction collide head on.

1a. compare the force experienced by the bug to the force experienced by the bus. The same.

1b. compare the change in momentum experienced by the bug to the change in momentum experienced by the bus. The same.

1c. compare the acceleration experienced by the bug to the acceleration experienced by the bus. The bug had a MUCH LARGER ACCELERATION.

2. A 5000 kg rail car moves west at 12 m/s and collides with (and "couples" with) a 6000 kg rail car moving east at 8 m/s. Assuming that momentum is conserved, at what rate does the resulting two car train move?

first car before: 5000 × -12 = -60000 kg m/s (the negative sign indicates west) second car before:6000 × 8 = 48000. -60000 + 48000 = - 12000 kg m/s of momentum after the collision. Now we have 11000 kg of train, moving west at 12/11 (1.1) m/s.

2a (is kinetic energy conserved? yes or no and how do you know?)

No. Check with the kinetic energy formula. Before: 360000 + 192000 = 552 000 J After: 5600 J. Most of the kinetic energy is gone. Note that energy is NOT a vector and note that the negative sign for the first velocity is "squared out". Kinetic energy is not conserved in inelastic collisions though strict energy still is -- the missing energy was converted into heat and sound and likely the deformation of the metal cars (this would be quite a collision).