ISM Physics 2021
Welcome!
Welcome!
Super Green
Problem-Solving Steps
While there is no simple step-by-step method that works for every problem, the following general procedures facilitate problem solving and make it more meaningful. A certain amount of creativity and insight is required as well.
Step 1
Examine the situation to determine which physical principles are involved. It often helps to draw a simple sketch at the outset. You will also need to decide which direction is positive and note that on your sketch. Once you have identified the physical principles, it is much easier to find and apply the equations representing those principles. Although finding the correct equation is essential, keep in mind that equations represent physical principles, laws of nature, and relationships among physical quantities. Without a conceptual understanding of a problem, a numerical solution is meaningless.
Step 2
Make a list of what is given or can be inferred from the problem as stated (identify the knowns). Many problems are stated very succinctly and require some inspection to determine what is known. A sketch can also be very useful at this point. Formally identifying the knowns is of particular importance in applying physics to real-world situations. Remember, “stopped” means velocity is zero, and we often can take initial time and position as zero.
Step 3
Identify exactly what needs to be determined in the problem (identify the unknowns). In complex problems, especially, it is not always obvious what needs to be found or in what sequence. Making a list can help.
Step 4
Find an equation or set of equations that can help you solve the problem. Your list of knowns and unknowns can help here. It is easiest if you can find equations that contain only one unknown—that is, all of the other variables are known, so you can easily solve for the unknown. If the equation contains more than one unknown, then an additional equation is needed to solve the problem. In some problems, several unknowns must be determined to get at the one needed most. In such problems it is especially important to keep physical principles in mind to avoid going astray in a sea of equations. You may have to use two (or more) different equations to get the final answer.
Step 5
Substitute the knowns along with their units into the appropriate equation, and obtain numerical solutions complete with units. This step produces the numerical answer; it also provides a check on units that can help you find errors. If the units of the answer are incorrect, then an error has been made. However, be warned that correct units do not guarantee that the numerical part of the answer is also correct.
Step 6
Check the answer to see if it is reasonable: Does it make sense? This final step is extremely important—the goal of physics is to accurately describe nature. To see if the answer is reasonable, check both its magnitude and its sign, in addition to its units. Your judgment will improve as you solve more and more physics problems, and it will become possible for you to make finer and finer judgments regarding whether nature is adequately described by the answer to a problem. This step brings the problem back to its conceptual meaning. If you can judge whether the answer is reasonable, you have a deeper understanding of physics than just being able to mechanically solve a problem.
When solving problems, we often perform these steps in different order, and we also tend to do several steps simultaneously. There is no rigid procedure that will work every time. Creativity and insight grow with experience, and the basics of problem solving become almost automatic. One way to get practice is to work out the text's examples for yourself as you read. Another is to work as many end-of-section problems as possible, starting with the easiest to build confidence and progressing to the more difficult. Once you become involved in physics, you will see it all around you, and you can begin to apply it to situations you encounter outside the classroom, just as is done in many of the applications in this text.
While there is no simple step-by-step method that works for every problem, the following general procedures facilitate problem solving and make it more meaningful. A certain amount of creativity and insight is required as well.
Step 1
Examine the situation to determine which physical principles are involved. It often helps to draw a simple sketch at the outset. You will also need to decide which direction is positive and note that on your sketch. Once you have identified the physical principles, it is much easier to find and apply the equations representing those principles. Although finding the correct equation is essential, keep in mind that equations represent physical principles, laws of nature, and relationships among physical quantities. Without a conceptual understanding of a problem, a numerical solution is meaningless.
Step 2
Make a list of what is given or can be inferred from the problem as stated (identify the knowns). Many problems are stated very succinctly and require some inspection to determine what is known. A sketch can also be very useful at this point. Formally identifying the knowns is of particular importance in applying physics to real-world situations. Remember, “stopped” means velocity is zero, and we often can take initial time and position as zero.
Step 3
Identify exactly what needs to be determined in the problem (identify the unknowns). In complex problems, especially, it is not always obvious what needs to be found or in what sequence. Making a list can help.
Step 4
Find an equation or set of equations that can help you solve the problem. Your list of knowns and unknowns can help here. It is easiest if you can find equations that contain only one unknown—that is, all of the other variables are known, so you can easily solve for the unknown. If the equation contains more than one unknown, then an additional equation is needed to solve the problem. In some problems, several unknowns must be determined to get at the one needed most. In such problems it is especially important to keep physical principles in mind to avoid going astray in a sea of equations. You may have to use two (or more) different equations to get the final answer.
Step 5
Substitute the knowns along with their units into the appropriate equation, and obtain numerical solutions complete with units. This step produces the numerical answer; it also provides a check on units that can help you find errors. If the units of the answer are incorrect, then an error has been made. However, be warned that correct units do not guarantee that the numerical part of the answer is also correct.
Step 6
Check the answer to see if it is reasonable: Does it make sense? This final step is extremely important—the goal of physics is to accurately describe nature. To see if the answer is reasonable, check both its magnitude and its sign, in addition to its units. Your judgment will improve as you solve more and more physics problems, and it will become possible for you to make finer and finer judgments regarding whether nature is adequately described by the answer to a problem. This step brings the problem back to its conceptual meaning. If you can judge whether the answer is reasonable, you have a deeper understanding of physics than just being able to mechanically solve a problem.
When solving problems, we often perform these steps in different order, and we also tend to do several steps simultaneously. There is no rigid procedure that will work every time. Creativity and insight grow with experience, and the basics of problem solving become almost automatic. One way to get practice is to work out the text's examples for yourself as you read. Another is to work as many end-of-section problems as possible, starting with the easiest to build confidence and progressing to the more difficult. Once you become involved in physics, you will see it all around you, and you can begin to apply it to situations you encounter outside the classroom, just as is done in many of the applications in this text.
There may be a time when time becomes meaningless.
08/24/2021
First Day Stuff
1. Learning to think. Confusion is good.
2. We are all Scientists. Scientists love not knowing.
We Are All Connected
First Day Stuff
1. Learning to think. Confusion is good.
2. We are all Scientists. Scientists love not knowing.
We Are All Connected
7. How To Learn Physics
| ap_lab_report_overview_and_rubric.pdf |
your brain
your brain on physics
Click the files below to download your Revision Guides
| 02_lectureoutline.pdf |
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projectile motion
Classic Monkey-Hunter simulation.
2D projectile motion.
Original Music.
2D projectile motion.
Original Music.
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Play the video below to see HW 1 problems being worked out.
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Everything you will learn in Physics I
can be classified as Newtonian Mechanics.
Newtonian Mechanics reminds me of
Non-Newtonian Mechanics
Which reminds me of
Non-Newtonian Fluids
Which reminds me of
The only good thing about
DUB STEP
Making
Stuff
Dance
can be classified as Newtonian Mechanics.
Newtonian Mechanics reminds me of
Non-Newtonian Mechanics
Which reminds me of
Non-Newtonian Fluids
Which reminds me of
The only good thing about
DUB STEP
Making
Stuff
Dance
This is real.
Really good headphones
or Sonos Sub required
Really good headphones
or Sonos Sub required
THIS IS NOT OPTIONAL. YOU MUST CREATE AN ACCOUNT !
HOMEWORK SERVER. CHECK IT OUT!
Once students are provided with the access code, they should go to www.PhysicsLE.com and click on "Physics LE for Students" for complete instructions on setting up their account.
WELCOME TO THE FUTURE!
Once students are provided with the access code, they should go to www.PhysicsLE.com and click on "Physics LE for Students" for complete instructions on setting up their account.
WELCOME TO THE FUTURE!
How to enter an answer in scientific notation.
| 02_lectureoutline.pdf |
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Tuesday: Derive one more equation. Work problems. Talk about free fall. READ about dropping things and throwing things in the air. It's all in CH 2.
For example. I throw a ball straight up and it returns to its initial position in 4 seconds.
a. What was its initial velocity?
b. How high did it go?
c. What was its final velocity?
Wednesday: Free fall problems.
Friday: A Lab! On free fall!
For example. I throw a ball straight up and it returns to its initial position in 4 seconds.
a. What was its initial velocity?
b. How high did it go?
c. What was its final velocity?
Wednesday: Free fall problems.
Friday: A Lab! On free fall!
| physics1sep4.pdf |
| phy1sep5.pdf |
| phys1sep11.pdf |
| phys1sep14.pdf |
PRACTICE TEST BASIC EQUATIONS WILL BE PROVIDED
1. A car starting from rest, accelerates at 2m/s/s for 10 seconds.
a. What is the car's final velocity?
b. How far did the car travel?
a. Vf = Vo +at, Vo = 0m/s so Vf = at = 2m/s/s X 10 s = 20 m/s
b. deltaX = Vot + 1/2 a t ^2, again Vo = 0 m/s so delta X = 1/2 a t^2 = 1/2 (2m/s/s)(10s)^2 = 100m
2. A duck walking with a constant velocity of 1 m/s , walks 20m.
a. How long was the duck walking?
b. What was the duck's acceleration?
c. How does a duck know what direction south is? And how to tell his wife from all the other ducks?
a. V = X/t so t = X/V = 20m/1m/s = 20 s
b. since V is constant, acceleration = 0 m/s/s
3. A car with an initial velocity of 30m/s slows at a uniform rate to a stop in a distance of 100m.
a. What was the car's acceleration?
b. How long did the car take to stop?
a. Vo = 30m/s
Vf = 0m/s
delta X = 100 m
t= ?
so use Vf^2 - Vo^2 = 2ax, solve for a:
a = (Vf^2 - Vo^2)/2x = 0m/s - (30m/s)^2 /( 2 x 100m) = -4.5m/s/s
b. t = (Vf - Vo)/a = (0m/s - 30m/s)/-4.5m/s/s = 6.7 s
4. A puppy dog starting at rest, reaches a velocity of 10 m /s in 5 seconds.
a. What was the puppy dog's acceleration?
b. How far did the puppy dog travel?
c. What is the puppy dog's name?
a. a = (Vf - Vo)/t = (10m/s - 0m/s)/5s = 2m/s/s
b. x = Vot + 1/2 at^2 , Vo = 0m/s so x = 1/2(2m/s/s)x5^2 = 25 m
c. Sugar Bear
5. A BMW traveling with a constant velocity of 20m/s passes by a police car at rest.
At that moment, the police car accelerates at a constant rate of 2m/s/s.
a. How long will it take the police car to catch up to the BMW?
b. How FAR away do the two cars meet?
c. Who was driving the BMW?
a. For the BMW velocity is constant so x = vt so x = 20m/s x t
For the police Vo = 0m/s and a = 2m/s/s so x = 1/2at^2 so x = 1/2(2) x t^2 or x = t^2
x for BMW and x for police are equal sooooo
20t = t^2
20 = t or t = 20s
b. Then use the value for t in either equation for motion: BMW x = vt = 20m/s x 20 s = 400m
c. gonzo
6. You drop your physics book from the top of a 45m tall building.
a. How long does it take the book to reach the ground?
b. What is the velocity of the book as it strikes the ground?
a. Choose an appropriate coordinate system. Down = positive would be good. Initial position of book = 0m is good.
So Vo = 0m/s. a = 10 m/s/s. and X = 45m
Use x=Vot + 1/2 at^2 and solve for t
Since Vo = 0m/s equation is X = 1/2 at^2
solving for t gives us
t = square root (2x/a) = sqrt( 2*45/10) = 3 s
b. Vf = Vo + at: Vf = 0 + 10m/s/s X 3s = 30 m/s
7. You throw your physics book straight up with a velocity of 10 m/s.
a. How high will the book go?
b. How long will it take the book to return to you?
c. What is the acceleration of the book at it's highest point?
For this problem, up is the positive direction and down in negative. So a = -g = about -10m/s/s.
Vo = 10m/s.
Since the book momentarily stops at the highest position we can say Vf = 0m/s.
Then use Vf^2 - Vo^2 = 2ax and solve for x
x = (Vf^2 - Vo^2)/2a = (0 - 10^2)/(2 x -10) = 5 m
b. To find time, realize that for the whole problem delta x = 0m (the book returns to its initial position)
use deltaX = Vot + 1/2at^2 and solve for t
0 = 10t +1/2(-10)t^2
5t^2 = 10t
5t = 10
t = 2 s
c. acceleration = about 10m/s/s the entire time.
8. Standing 30m above the ground, you throw your physics book straight down with a velocity of 5m/s.
a. How long will it take the poor book to reach the ground?
b. What is the velocity of the book as it strikes the ground?
Since everything in this problem is DOWNWARD I would let down be the + direction.
So Delta X = 30m, Vo = 5m/s and a = g = about 10 m/s/s.
You could solve for t using a quadratic equation, but I would solve for Vf first.
Vf^2 = Vo^2 + 2ax. Vo = 0m/s, sooooo
Vf = sqrt(2ax) = sqrt(2*10m/s/s*30m) = 25 m/s (about)
Now solve for t.
t = (Vf - Vo)/a = (25m/s - 5m/s)/10m/s/s = 2s
9. From a position 20m above the ground, you toss a penguin straight up with a velocity of 10m/s.
a. How long till the penguin reaches the ground?
b. What is the velocity of the penguin as it reaches the ground?
c. How high above its initial position will the penguin travel?
For this problem I would let the initial position of the penguin =0m. Since the problem involves motion up and down, I will let up be positive and down be negative.
So Vo = 10m/s
delta x = -20m
a = g = -10 m/s/s
The easiest way to do this problem is to find Vf first.
Vf = +/- sqrt(Vo^2 + 2ax)
Vf = +/-sqrt(10m/s^2 + 2(-10m/s/s)(-20m)
since the penguin is moving down, Vf is negative
Vf = -sqrt(100 + 400) = -sqrt(500) = -22 m/s
Now find t using t =( Vf -Vo)/a = (-22m/s - 10m/s)/-10 = 3.2 s
To find how high the penguin goes use 2ax = Vf^2 - Vo^2 and solve for x. At the highest point Vf = 0m/s
x = (- (10m/s^2)/2(-10m/s/s) = 100/20 = 5m
There will be some questions about motion graphs.
1. A car starting from rest, accelerates at 2m/s/s for 10 seconds.
a. What is the car's final velocity?
b. How far did the car travel?
a. Vf = Vo +at, Vo = 0m/s so Vf = at = 2m/s/s X 10 s = 20 m/s
b. deltaX = Vot + 1/2 a t ^2, again Vo = 0 m/s so delta X = 1/2 a t^2 = 1/2 (2m/s/s)(10s)^2 = 100m
2. A duck walking with a constant velocity of 1 m/s , walks 20m.
a. How long was the duck walking?
b. What was the duck's acceleration?
c. How does a duck know what direction south is? And how to tell his wife from all the other ducks?
a. V = X/t so t = X/V = 20m/1m/s = 20 s
b. since V is constant, acceleration = 0 m/s/s
3. A car with an initial velocity of 30m/s slows at a uniform rate to a stop in a distance of 100m.
a. What was the car's acceleration?
b. How long did the car take to stop?
a. Vo = 30m/s
Vf = 0m/s
delta X = 100 m
t= ?
so use Vf^2 - Vo^2 = 2ax, solve for a:
a = (Vf^2 - Vo^2)/2x = 0m/s - (30m/s)^2 /( 2 x 100m) = -4.5m/s/s
b. t = (Vf - Vo)/a = (0m/s - 30m/s)/-4.5m/s/s = 6.7 s
4. A puppy dog starting at rest, reaches a velocity of 10 m /s in 5 seconds.
a. What was the puppy dog's acceleration?
b. How far did the puppy dog travel?
c. What is the puppy dog's name?
a. a = (Vf - Vo)/t = (10m/s - 0m/s)/5s = 2m/s/s
b. x = Vot + 1/2 at^2 , Vo = 0m/s so x = 1/2(2m/s/s)x5^2 = 25 m
c. Sugar Bear
5. A BMW traveling with a constant velocity of 20m/s passes by a police car at rest.
At that moment, the police car accelerates at a constant rate of 2m/s/s.
a. How long will it take the police car to catch up to the BMW?
b. How FAR away do the two cars meet?
c. Who was driving the BMW?
a. For the BMW velocity is constant so x = vt so x = 20m/s x t
For the police Vo = 0m/s and a = 2m/s/s so x = 1/2at^2 so x = 1/2(2) x t^2 or x = t^2
x for BMW and x for police are equal sooooo
20t = t^2
20 = t or t = 20s
b. Then use the value for t in either equation for motion: BMW x = vt = 20m/s x 20 s = 400m
c. gonzo
6. You drop your physics book from the top of a 45m tall building.
a. How long does it take the book to reach the ground?
b. What is the velocity of the book as it strikes the ground?
a. Choose an appropriate coordinate system. Down = positive would be good. Initial position of book = 0m is good.
So Vo = 0m/s. a = 10 m/s/s. and X = 45m
Use x=Vot + 1/2 at^2 and solve for t
Since Vo = 0m/s equation is X = 1/2 at^2
solving for t gives us
t = square root (2x/a) = sqrt( 2*45/10) = 3 s
b. Vf = Vo + at: Vf = 0 + 10m/s/s X 3s = 30 m/s
7. You throw your physics book straight up with a velocity of 10 m/s.
a. How high will the book go?
b. How long will it take the book to return to you?
c. What is the acceleration of the book at it's highest point?
For this problem, up is the positive direction and down in negative. So a = -g = about -10m/s/s.
Vo = 10m/s.
Since the book momentarily stops at the highest position we can say Vf = 0m/s.
Then use Vf^2 - Vo^2 = 2ax and solve for x
x = (Vf^2 - Vo^2)/2a = (0 - 10^2)/(2 x -10) = 5 m
b. To find time, realize that for the whole problem delta x = 0m (the book returns to its initial position)
use deltaX = Vot + 1/2at^2 and solve for t
0 = 10t +1/2(-10)t^2
5t^2 = 10t
5t = 10
t = 2 s
c. acceleration = about 10m/s/s the entire time.
8. Standing 30m above the ground, you throw your physics book straight down with a velocity of 5m/s.
a. How long will it take the poor book to reach the ground?
b. What is the velocity of the book as it strikes the ground?
Since everything in this problem is DOWNWARD I would let down be the + direction.
So Delta X = 30m, Vo = 5m/s and a = g = about 10 m/s/s.
You could solve for t using a quadratic equation, but I would solve for Vf first.
Vf^2 = Vo^2 + 2ax. Vo = 0m/s, sooooo
Vf = sqrt(2ax) = sqrt(2*10m/s/s*30m) = 25 m/s (about)
Now solve for t.
t = (Vf - Vo)/a = (25m/s - 5m/s)/10m/s/s = 2s
9. From a position 20m above the ground, you toss a penguin straight up with a velocity of 10m/s.
a. How long till the penguin reaches the ground?
b. What is the velocity of the penguin as it reaches the ground?
c. How high above its initial position will the penguin travel?
For this problem I would let the initial position of the penguin =0m. Since the problem involves motion up and down, I will let up be positive and down be negative.
So Vo = 10m/s
delta x = -20m
a = g = -10 m/s/s
The easiest way to do this problem is to find Vf first.
Vf = +/- sqrt(Vo^2 + 2ax)
Vf = +/-sqrt(10m/s^2 + 2(-10m/s/s)(-20m)
since the penguin is moving down, Vf is negative
Vf = -sqrt(100 + 400) = -sqrt(500) = -22 m/s
Now find t using t =( Vf -Vo)/a = (-22m/s - 10m/s)/-10 = 3.2 s
To find how high the penguin goes use 2ax = Vf^2 - Vo^2 and solve for x. At the highest point Vf = 0m/s
x = (- (10m/s^2)/2(-10m/s/s) = 100/20 = 5m
There will be some questions about motion graphs.
| 03_lectureoutline.pdf |
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