Circular Motion

Hi everybody!!! This is Hao : ) HAHA, Spring Break is coming. So I wish everybody will have fun and don't forgot check this blog. I try to post some useful information on this blog everyday. Today, i am so lazy so just post the answer link of " Free -Body Exercises: Circular Motion" sheet.

http://www.slideshare.net/ekozoriz/circularfbd

WHAT EVER YOU WISH TO CALL THIS...

hey everyone sorry this is a bit late.

okay we started a new unit on monday where we were introduced to PERIODIC MOTION.

you should have finished the first paragraph of chapter 7, section 2 study guide but just in case here are the fill ins for the blanks:

constant

changing, changing

perpendicular, tangent

instantaneous velocity

length

radius, centre

centripetal

directly, inversely

second

force

centripetal force

straight, tangent

ALSO please finish the two worksheets the were given on tuesday or monday...i don't remember when but it was surly this week?!??

AND HAO BECAUSE YOU ASKED SO NICELY YOU CAN SCRIBE NEXT

PHYSICS!!!! YAAARGH!!!! (:

Ok so now that the Fashion Show is finally over (I hope you guys came to watch!), I can finally start and/or finish this scribe post! Ok so it's pretty late and I don't really want to surprise anyone by telling them they are scribe so I might as well just finish the week right? So I'll just gather my information for tomorrows class and scribe for then too. However there is a piece of information that I must share with you.


The next unit test (Momentum and Projectile Motion) is on MONDAY!!!!


Yup, Monday. So study study study!


For now, it's jayp signing out. (:

Ok so now it's time to review what happened this week.

Tuesday, March 16th
Ok so Ms.Kozoriz wasn't here this day (and unfortunately neither was I) but from my understanding, we just got a few sheets to work on. I will go over the answers to the sheets in just a moment.

Wednesday, March 17th
Once again, Ms. Kozoriz was away. Again, I believe we just got some more sheets to work on as well as getting time to work on the previous sheets. Please, correct me if I am wrong.

Thursday, March 18th
Ms. Kozoriz came back. I believe the answers to Projectile Motion 1 and the Chapter 7 Review were gone over. Projectile Motion 2 (back of Projectile Motion 1) and textbook questions were assigned. If you happen to have a textbook with you, the questions were at the end of Chapter 7 (page 152) Problems 6-10. The answers will be gone over later on in the post.

Friday, March 19th
All went back to normal this day. The class was filled and physics continued to roll onwards. The answers to Projectile Motion 2 and the textbook questions. We also listened to another physics song entitled "Harry the Human Cannonball". It was a song about a man who won a woman's heart by applying the concepts of projectile motion. We also got a wordsearch and one of those fill in the blanks sheets for Periodic Motion which we are of course to complete. Also, don't forget we have a TEST on MONDAY on MOMENTUM and PROJECTILE MOTION!

OK so now I will go over the answers for the various sheets we have recieved. Let's start with Projectile Motion 2 (since Projectile Motion 1 is on the site already with explanations and correct answers).

*links to explanations are currently not working. will be up soon*

1)141.2m
solution

2)10.6 m
solution

3)a) 4.42s
solution

3)b)55m
solution

4)70.0m/s
solution

5)a)16.51m/s
solution

5)b)10.02m
solution

Next we move on to the textbook questions:

6)6.5m/s
solution

7)3.17m
solution

8)a)30.6m
solution

8)b)212m
solution

9)30.99m/s
solution

10)a)0.2s
solution

10)b)8.7m (win)
solution

Lastly the review (only the front page because we haven't dealt with circular motion yet).

1)0.25m
solution

2)7.8m
solution

3)a)6.5s
solution

3)b)210m
solution

4)a)0.82s
solution

4)b)N/A
solution

Ok so now just a couple pointers to remember for the test.

Ft=impulse
mv=momentum

impulse=momentum
Ft=mv

momentum = kg*m/s

momentum before collision = momentum after collision (in an isolated, closed system)
in a non-sticky collision (elastic)  m₁v₁+m₂v₂=m'₁v'₁+m'₂v'₂
in a sticky collision m₁v₁+m₂v₂=(m'₁+m'₂)v'

When net force is equal to 0, change in momentum is equal to 0 (in a closed system).

Adding momentum vectors is like adding normal vectors.

Horizontal velocity (for projectile motion) is constant.
Vertical velocity (for projectile motion) changes accroding to g*t (V_v=gt)
V_vf=V_v+gt
d_v=(1/2)gt²
d_h=v_h*t

OK! Well good luck everyone! Study study study! And get lots of sleep and/or rest! (:

Good luck and see you in class my fellow future physicists!

jayp, signing off!

P.S. You get the privilage of being the next scribe...........(drum roll)

Muhanad!

Congrats! You name was pulled out of the hat! Have fun! (:

Projectile Motion 1 Answers

Projectile motion 1 answers

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PROJECTILE MOTION

Hey everyone, this is Karen.

Anyway, it slipped my mind that I had to scribe, so I'm only doing to now. On Friday, we learned about Projectile Motion.




We learned that an object that is thrown or released has no horizontal force acting on it, therefore, there is no horizontal acceleration. Also, both a thrown object and a released object are accelerated downward by the force of gravity.


In class, Ms. K gave us a sheet that shows two spheres falling, one with a horizontal force and one without one.. (I'll add the picture later when I find the cord for my printer)
Also, she gave us the Chapter 7 Study Guide. And a worksheet called Projectile Motion I and on the back side is Projectile Motion II.



We haven't finished the Projectile Motion worksheet, and I think she expects us to have it done for tomorrow.
The answers for the Chapter 7 Study Guide is in the slide share uploaded by Ms. K.





That's pretty much sums up friday's class. The next scribe is jayp

Projectile Motion

Proj Motion

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Momentum & Impulse

Hi guys! In today's class, we just corrected the sheet Chapter 9 and Chapter 9 questions 16-19 on the green book.

Chapter 9 Review:

1. 1600 N
2. 1.3x10ˉ³ s.
3. 6 700 N [W]
4. 0.05 m/s
5. 8.8 m/s [W]
6. 81.25 m/s
7. 0.058 m/s
8. 10 kg·m/s, 36.9° [E of N]
9. OMIT


Text Book Chapter 9 (#'s 16-19):

16. a) 779 kg·m/s
b) -779 kg·m/s
c) -779 kg·m/s
d) -6.0x10ˉ⁴ kg·m/s
e) -6.1 m/s

17. a) 0.001 kg
b) -6.0x10ˉ⁴ kg·m/s
c) sorry, i didn't get the answer for this one.
d) 1.6x10ˉ³ kg·m/s

18. a) -100 kg·m/s
b) -500 N [S]

19. 10.83 m/s


Reminder: Hand in Impulse and Momentum/Conservation of Momentum sheet by tomorrow. First 3 questions on each pages are to be marked. Show all your work.


Next scribe is RENSTER :)

Momentum Notes

Momentum

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Wednesday Mar. 10

Today in class...

People who did not complete their lab from the previous class had the opportunity to continue their lab in class today. From the lab, u basically learn that the sum of the momentum of the target sphere and the incident sphere should be equal to the original momentum as such.

We also reviewed some things in class that we already learned. This includes:

P=mv
P=Ft
Ft=mv

For elastic collisions before and after collision:
m1v1 +m2v2 = m1v1 + m2v2

For non-elastic ("sticky") collisions in which the two objects travel together after the collision:
m1v1 +m2v2 = (m1 + m2)v2


Finally, we also received another worksheet in class today. It was a double-sided worksheet, one side entitled "Impulse and Momentum" and the other side entitled "Conservation of Momentum." You should complete the sheets in order to get practice and better understand the unit. I do not believe it is worth any marks, but i could be wrong.

Our test for the units on momentum and projectile motion will be held on Mar. 22. So don't forget to study. I believe that basically covers everything we did in class today so I guess I'm done. The next scribe will be Jeamille!

Super Mega Awesome Scribe Post!!

So I was supposed to scribe for friday annnnnd.... that didn't happen. So I said I'd scribe for both monday AND friday to make up for that, seemed like a good plan to me. And then..... that didn't happen either. Today is Tuesday. And it's time for me to do it. And... it's not gunna happen.. jk here it goes:

FRIDAY:
We had a substitute, Mr. Tiede, brother of a champion. Me and Jonno had an epic duel, and then it was down to some serious business.

Assigned work of this day:

#1-6 for... Reviewing Concepts and #1-5 for applying concepts. Or something like that.

MONDAY:
We corrected the aformentioned questions with Mrs. K.

The jist of it is, momentum is always conserved, and that even a small force can apply the same impulse as a large force if it acts for a longer amount of time. Also, smaller mass can have a higher momentum if their speed is greater. Just sorta stuff like that.

TUESDAY (TODAY):
Today we did a lab. It has something to do with momentum, a string, some shiny metal balls, and carbon paper. Quite frankly I don't really understand how it works, but using the two vectors of the 2 different balls, we can somehow calculate our initial momentum with angles and magic. We get a chance tomorrow to finish it up, so don't fret if you didn't manage to complete it.

Also Jonno, I'm sorry about your ruler. And to prove how sorry I am, you get to be the scribe tomorrow =D

Isn't Physics fun you guys???

Thursday, March 3, 2010

Today, most of us spent the class doing a lab that can be found in the textbook on page 182. While others spent the class finishing up page 193(1-6) of the textbook, along with the new study guide we picked up today. I think thats about it but anyways here are the answers to Page 193(1-6)

1. 351 kg.m/s

2. 4.8 N.s

3. 42 m/s

4. a) 60 N.s
b) 20 m/s

5. a) 2.04 x 10⁴
b) 300 N

6. a) 2.3 x 10⁴ kg.m/s
b) 2.6 x 10⁴ N

anyway the next scribe can be the "Pokemon Champion"

Dynamics Test Answers

Dynamics Test

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MARCH 03, 2010

Great ... thanks Dave.

Okay, so today we went over the front page of Chapter 9 Study Guide. Here are the answers if you missed any ...

9.1 IMPULSE AND CHANGE IN MOMENTUM
Momentum and Impulse
The amount of force needed to change the motion of a moving object depends on the velocity and mass of the object. The momentum of a body is the product of the body’s velocity and mass. Momentum is a(n) vector quantity. Its direction is the same as the direction of the velocity. The equation used to calculate momentum is p = mv. In this equation, p stands for momentum, m stands for mass, and v stands for velocity. The unit for momentum is the kilograms meters/second. The product of the force applied to an object and the time interval over which it acts is called the impulse. The direction of the impulse is the same as the direction of the force that is applied. The unit for impulse is the Newton • second. The impulse given to an object is equal to the change in the object’s momentum. This equality is the impulse-momentum theorem. It is also another statement of Newton’s second law of motion.

We were also suppose to finish the back of Chapter 9 Study Guide today, here are the answers …

9.2 THE CONSERVATION OF MOMENTUM
Newton’s Third Law and Momentum
If one object collides with another, the momentum of each object changes. The first object exerts a(n) force on the second object, and the second object exerts a(n) force or equal magnitude and opposite direction on the first object. If objects neither leave nor enter a system, the system is described as closed. If no external forces act on a system, the system is described as isolated. When there is a collision within such a system, the net change in momentum is zero. The total momentum before the collision is the same as the total momentum after the collision.
Law of Conservation of Momentum
In a closed, isolated system, the momentum does not change. This statement is the law of Conservation of Momentum. When two objects within the system collide, the magnitude of the momentum lost by one of the objects is equal the momentum gained by the other object. Momentum can be transferred from one object in the system to another.
Internal and External Forces
Internal forces act between objects in a closed system. External forces are exerted by objects outside the system. The total momentum of a system is conserved only when there are no external forces acting on the system.
Conservation of Momentum in Two Dimensions
The law of conservation of momentum does not depend on the direction in which objects move before and after colliding. The momentum of two objects in a system can be represented by two vectors, which can be resolved into vertical and horizontal components. After all vectors are added, the final sum must equal the original momentum of the system.

We also did questions from the textbook on Page 173, # 1 - 4. Here are the answers.

Okay next scribber is Kasey La :)

March 1, 2010

Today, we watched a video about momentum. We were told to write down 5 to 7 points about momentum and hand it in at the end of the period. If you weren't paying any attention to the video, "momentum is inertia in motion". Momentum is also the product of mass and the velocity (momentum = mv). Onward, "impulse" occurs as force is applied to an object over a period of time (impulse = Ft). As said in the video, impulse is also equivalent to momentum. Which, would be expressed as: Ft = mv.

Further into the video, the guy showed a few examples to give a visual image to better understand the concept. He used this train thing that shoots air to allow the pieces of metal to lie on a cushion of air to simulate a near frictionless surface.

Anyway, on the train thing, he showed object A collide with object B; object A ceased while object B continued forward. He continued to explain that momentum is conserved in collisions. The guy went on about object B having an impulse at the opposite direction when the two objects collided. Which canceled out the momentum of object A (putting it to a stop) while the conserved momentum allowed object B to continue moving on. Another example of this would be Newton's Cradle (the thing with the balls bouncing).

Another example on the train thing was when he showed that object A would stick to object B when colliding with each other. Object A travelled towards object B along the track, and when colliding (and attaching to object B), the same inertia is maintained from the initial momentum of object A travelling by itself. As a result, the attached objects A and B travelled at a slower speed than the velocity of what object A initially had. The guy explained that the momentum is the same, because the mass doubled and the velocity changed in proportion (Δmv = same momentum) .

I apologize if this is a bit confusing. Anyway, Ms. K didn't give our tests back yet... although, feel free to ask her what your score is.

Next scribe is Sayana! Good luck.