Name: Nellie Newton Period: 9 Date: November 1, 2012

Name: Nellie Newton Period: 9 Date: November 1, 2012 Slow Car Analysis Formulas you may need: average v = d/t vf = average v x2 a = (vf – vi)/t p = mv GPE = mgh KE = ½ mv2

Measurements: (original) (modified) Distance (length of ramp): ____2.42 m______2.42 m_____ Height of ramp: ____0.785 m______0.785 m____ Average time: ____1.24 s______2.12 s_____ Average velocity: ____1.95 m/s______1.14 m/s__ Initial velocity: _____0 m/s______0 m/s_____ Final velocity: _____3.9 m/s______2.28 m/s___ Acceleration: ___ 3.14 m/s 2______1.07 m/s 2__ Mass of wooden car: __0.0313 kg______0.0475 kg__

Name: Nellie Newton Period: 9 Date: November 1, 2012 Slow Car Analysis Formulas you may need: average v = d/t vf = average v x2 a = (vf – vi)/t p = mv GPE = mgh KE = ½ mv2

Measurements: (original) (modified) Distance (length of ramp): ____2.42 m______2.42 m_____ Height of ramp: ____0.785 m______0.785 m____ Average time: ____1.24 s______2.12 s_____ Average velocity: ____1.95 m/s______1.14 m/s__ Initial velocity: _____0 m/s______0 m/s_____ Final velocity: _____3.9 m/s______2.28 m/s___ Acceleration: ___ 3.14 m/s 2______1.07 m/s 2__ Mass of wooden car: __0.0313 kg______0.0475 kg__ Name: ______Period: _____ Date: ______Slow Car Motion practice analysis You should be able to do all of the following for the analysis day (test). You will not have to do them all, but you don’t know which ones you will have to do, so be prepared. Since this is just a practice, you can use your notes or book. The only thing you get during the real analysis day is a sheet with the formulas and both sets of your data (wooden car & slow car), like on the yellow ½ sheet.

Ch 10 (Motion) Calculate average velocity for Slow Car. Show your work. G U E S S d = 2.42m v=? v = d/t 2.42/2.12 1.141509434 t = 2.12s 1.14 m/s

Calculate final velocity for Slow Car. Show your work. G U E S S

av v = 1.14 vf =? vf = average v x 2 1.14 x 2 = 2.28 m/s

Calculate acceleration for Slow Car. Show your work. G U E S S vf = 2.28 m/s a = ? a = (vf – vi)/t (2.28 – 0)/2.12 1.075471698 2 vi = 0 m/s 2.28/2.12 1.08 m/s t = 2.1.2 s

Sketch a graph of data for distance and time for both sets of data (is it going up or down? At a constant rate or not?) Sketch a graph of data for velocity and time for both sets of data (is it going up or down? At a constant rate or not?)

Ch 11: Forces Describe a time in the experiment when the forces are balanced and how you can tell. At the top before you let go and when it stops It’s not changing motion (staying still)

Describe a time in the experiment when the forces are unbalanced and how you can tell. When it’s rolling It’s accelerating

Describe a time in the experiment where we see the following about Newton’s Laws: 1st: it’s an object at rest acted upon by balanced forces At the top before you let go and when it stops

1st: it’s an object in motion acted upon by balanced forces After it’s off the ramp and still moving (constant v)

1st: it’s an object in motion acted upon by unbalanced forces When it’s rolling (accelerating)

2nd: How have the force making it move, its mass and acceleration changed (or not changed) due to your modifications? F: increased, if mass increased

m: increased

a: decreased due to increase in friction/air resistance 3rd: Name the action or reaction force: Gravity pulls down on the car: _car pulls up on the Earth_

The car pushes through the air: _air pushes back (air res.)_

_The car pushes down on the board_ : the board pushes up on the car (FN)

Here are three of the forces that affect motion (there is no applied force). How did they change (if they did) from the original car to your slow car?

Fg: increased, due to increase in mass

FN: increased due to increase in mass

Ff: increased due to your modifications

Draw a free body diagram when the car was…. At the top of the ramp (not moving) (could also have Ff backwards (static friction) as well as Fa going forward (your hand holding it in place)

Moving down the middle of the ramp

Moving on the floor after it left the ramp.

Does your car have more momentum before or after the changes? Explain conceptually (without numbers). Could be either…you have to calculate. It has more mass, which would give it more momentum, but less velocity, which would give it less momentum, so it depends which has changed more (m or v)

Calculate momentum for the Slow Car at the bottom of the ramp. Show your work. G U E S S m = .0475 kg p = ? p = mv.0475 x 2.28 .1083 v = 2.28 m/s .108 kg x m/s

Energy (ch. 12) Calculate GPE at the top of the ramp for the Slow Car. Show your work. G U E S S m = .0475 kg GPE = ? GPE = mgh .0475 x 10.0 x .785 .372875 g = 10.0 m/s2 .373 J h = .785 m

Calculate KE at the bottom of the ramp for the Slow Car. Show your work. G U E S S m = .0475 kg KE = ? KE = ½mv2 .5 x .0475 x 2.282 .123462 v = 2.28 m/s .123 J

What is the efficiency of this set up? For the Slow Car data, take the KE/GPE then multiply by 100 to get a percent.

KE/GPE .123/.373 .329758713

State the Law of Conservation of Energy. Energy can not be created or destroyed; it just changes form.

Explain why your data above disproves (or does not disprove) the Law of Conservation of Energy (the two numbers are different, but the KE came from the GPE, didn’t it?). Because of friction and air resistance, some of the energy is converted to nonmechanical forms such as sound and heat.

Discuss energy transformations (from what form to what form and what force/work is being done):

At the top of the ramp as soon as you let go GPE is being converted into KE by the force of gravity (gravity is doing the work on the car) (some energy is also being converted to heat/sound because of the force of friction/air resistance)

As the car rolls down the ramp from top to bottom GPE is being converted into KE by the force of gravity (gravity is doing the work on the car) (some energy is also being converted to heat/sound because of the force of friction/air resistance)