For ERHS AP Physics C

Textbook: Physics for Scientists and Engineers by Serway, Jewett. 9th edition. 
(If you do a Google search for the textbook, you will be able to find a pdf file of the textbook.) 


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PGCPS Letter about AP Exams 12_9-18-2017.pdf : Please click on link to read the letter. 

AP Physics C — Schedule for Mechanics Unit 1: Kinematics (Ch. 2, 4)

Date

In class

Homework

9/6/2017 Wed

Introduction.

Unit conversion: km/h to m/s. Average speed.

1-dim kinematics definition, unit, vector vs. scalar.

1-dim kinematics graphs, incl. slope and area under.

Serway: The textbook has Objective Questions, Conceptual Questions, and Problems at the end of each chapter. Please be sure to find the correct items to do for homework assignments.

Review: Ch. 2: ObjQ(Objective Questions): p.49: 11. Problems: 7, 11, 14, 39

9/7

1-d kinematics: equations vs. graphs.

Review: Ch. 2: Problems: 79

Graph & equation: Ch. 2: Problems: 17, 19, 24, 29, 33

9/8

1-d kinematics: equations vs. graphs.

Special kinds of motion: including const. v motion, const. a motion and free fall.

Graph & equation: Ch. 2: Problems: 21, 38, 59a) acceleration only, 59b), and c).

(Serway 21, 38 has x(t) as 2nd degree equation. 59: v(t) is 2nd degree. All need to do d/dt!!!)

Const. a and falling obj.: Ch. 2: ObjQ: 14, 15, 16, 17.

9/11

Mon

1-d kinematics: equations vs. graphs.

Special kinds of motion: including const. v motion, const. a motion and free fall. Review for quiz.

Const. a and falling obj.: Ch. 2: Problems: 50, 52, 54, 55, 56

Ch. 2 Problems: 85

9/12

Review.

2-dim or 3-dim kinematics: including projectile motion.

Using vectors: r: directions of v, ac or aR & aT.

Review.

9/13

Quiz on 1-d motion.

2-dim. kinematics: including projectile motion.

Ch. 4 Obj.Q. 1, 4, 7, 11, 14. Concept Q. 1, 2, 3, 4. 

9/14

2-d or 3-d motion.

Ch.4 Prob. 2, 3 (for prob. 3 add part c: What is the shape of the particle’s trajectory?)Prob. (22, 29: shot horizontally), (15: symm)  

9/15

Projectile motion.

Ch. 4 Prob. (asymm: 21, 23)

9/18
Mon

Relative motion.

Ch. 4 Prob. 49, 53, 55

9/19

Frame of reference.AP problem.

9/20

Review. 

Study. 

9/21

Review.

Study.

9/22

Test: mechanics unit 1.

APC Kinematics Videos on integration: Videos 1 (5:07 to end), 5 and 6.

  AP Physics C – For Mechanics Unit 1 Homework

Answers and hints for Ch. 2 problems: Try to do the problems without the hints first. It’s more fun that way!!

OQ11. b                 To obtain a(t), we need to first obtain v(t) from the x vs. t graph.

7. a) -2.4 m/s, b) about -3.8 m/s, c) about 4s                  a) ave. v = ∆x/∆t, c) v = slope of x vs. t graph, so v = 0 when the slope of the graph is zero.

11. a) 5m, b) 4875s                             Hint: Both hare and tortoise run at constant speed if you ignore the waiting time of the hare.
Solution: Total running time for hare = 1000m/(8m/s) = 125s, Total running time for tortoise = 1000m/(0.2m/s) = 5000s. So the hare has to wait 5000-125= 4875 s. At first the hare runs for 0.8km/(8m//s) = 100 s, so the hare must begin to run again 125-100 = 25s before the race ends. That means the tortoise must be (0.2m/s)(25s) = 5 m from finish line.

14. 13400m/s2                        ave. a = ∆v/∆t. Pay attention to the direction of the velocities.

39. a) against (i.e. not equal), b) for (i.e. equal)            Hint: The glider goes through constant acceleration motion, so its ave. v = (v0 + v) /2.
Solution: Since the glider goes through constant acceleration motion, its velocity changes at a constant rate, it reaches the (v0 + v) /2 half way through time. I.e. it reaches the ave. v half way through time. This is not the same as the v half way through space, because the glider speeds up, so it takes less time to travel the second half of the space than the time it takes to travel the first half of the space.

79. 1.6m/s2                            From problem 39, we know that for constant acceleration motion, the average velocity is the velocity half way through time, and we have enough information to find the average v when each car goes by. So we can find the velocities of the cars when they pass Liz half way in time.

17. a) 1.3 m/s2, b) at t = 3 s, 2 m/s2, c) at t = 6s and t = about 10.3 s to 12 s, d) 8s, -1.5 m/s2               a) Use definition of ave. a. b) steepest part of the graph, c) where slope =0, d) steepest negative slope.

19. a) at 10s: 20m/s, at 20s: 5m/s, b) 263m                    a) use ∆v = area of graph, b) Can plot a v vs. t graph and then find area of graph: Since we are looking for distance, we have to add +∆x and the absolute value of -∆x together. Another way to do this is to look at the motion in 3 separate segments (0 to 10s, 10 to 15s, and 15 to 20s), and use kinematics equations to find the displacement traveled in each segment. In this case the object does not turn around, otherwise we have to be careful to pay attention to when the object turns around in order to find the correct distance traveled

21. a) 2m, b) -3m/s, c) -2 m/s2          a) Plug in t. b) Take time derivative and then plug in t. c) Take one more time derivative and then plug in t.

24. 160ft               Use the v2 equation for constant acceleration motion. v02 is proportional to ∆x.

29. -16 cm/s2       

33. a) 35s, b) 15.7m/s                         Either plot a v vs. t graph or use kinematics equations and work through the 3 segments of the motion one at a time.

38. a) 2.56m, b) -3m/s                        a) What’s special about turning point?  b) When x = 0 again.

59. a) –(1×108)t + 3×105 m/s2 , b) 3×10-3s, c) 450 m/s                 a) a = dv/dt, b) make a = 0 and solve for t, c) plug the previous answer into the v(t) equation.

OQ14. b)               The 2nd ball’s motion is like the reverse of the 1st ball’s. And they meet at the same time, so it must be half way in time.

OQ15. b)               Can use kinematics or energy conservation.

OQ16. e)              

OQ17. a) A=C=E>B=D, b) D>A=B=C=E                     

50. 7.96s (if use g = 9.8m/s2)                            Find h(t=2) to find initial height. Take time derivative of h to find v(t) and plug in t = 2 to find initial velocity of the mailbag.

52. 0.6s                  When the two balls meet, they are at the same position at the same time. So if we use the same coordinate system for both balls and write one y(t) for each ball. The balls would meet when y1(t) = y2(t). Solve for t for time.  (Of course, you can also write the upward displacement of 1st ball as a function of t. Write distance fallen by 2nd ball as a function of time. Then add these 2 together to equal to 15m. Solve for t.)

54. a) (h/t) + (gt/2), b) (h/t) –(gt/2)

55. (if use g = 9.8 m/s2) b) 0.782s, a) 7.82 m                 b) Look at vertical motion.a) Look at horizontal motion.

56. a) vi + gt, b) gt2/2, c): a) The absolute value of (vi – gt), b) gt2/2                        In this case the speed is the absolute value of the velocity. To find the distance between the helicopter and the package, we can find the difference in position of the two. So if we use the same coordinate system for both the helicopter and the package to write one y(t) for each. The absolute value of the difference in their y(t) would be the distance between the two.

85.  (if use g = 9.8 m/s2) a) 26.4 m, b) 28.2 m: 6.82% error                        a) Let t be the time it takes for rock to hit water, then the time it takes for sound to reach man is (2.4 –t). Use t to express the distance traveled by rock and the distance traveled by the sound. These 2 distances should be equal which gives us a quadratic equation. Solve the quadratic equation to find t. Then use t to find depth of the well. b) Treat 2.4 s as the time it takes for the rock to fall.

Ch. 4

OQ1. e)                  What is the definition of ave. a?

OQ4. d)                 Use the v2 equation.

OQ7. d)                 First find out by what factor the time changes. Then use ∆x = vt.

OQ11. a)               T = 2πr/v

OQ14. d)              

2. 2.5 m/s              Which component of the velocity represents the velocity of the shadow?

3. a) (1i + 0.75j) m/s, b) v = (1i + 0.5j) m/s, speed = 1.12m/s, c) parabola                             a) b) Do the x- and y- components separately. c) because y(x) is a second degree equation.

CQ1. Parabola                     The spacecraft has no acceleration in one direction and has a constant acceleration in another direction.

CQ2. For the first loop, the velocity is always tangent to the circle and with increasing magnitude. The acceleration slants forward and towards the center of the circle. For the second loop, the velocity is always tangent to the circle and has a constant magnitude (no smaller than the largest velocity vector in the first loop). The acceleration always points toward the center of the circle and has constant magnitude.                                 For acceleration, first look at tangential and radial accelerations separately.

CQ3. You can determine the average velocity, but not the instantaneous velocity.

CQ4. a) Follow a straight path. b) Follow a straight path (a = constant = 0), or follow a circular path (a = non-zero constant magnitude)

22. a) ∆x = 1.18 m, so the space is wide enough for a walkway. But a very tall person will have to walk carefully close to the wall to avoid getting his/her head wet. b) 0.491 m/s                        a) Projectile shot horizontally. b) ∆x = vx t. When the height fallen changes by a factor of 1/12, the ∆y = 0 + ½ at2 changes by a factor of 1/12, so t2 changes by a factor of 1/12. Since we also want ∆x to change by a factor of 1/12, we can find the factor by which vx has to change.

29. a) (0, 50m), b) (18m/s, 0), c) Constant acceleration motion with downward a = - g = - 9.8 m/s2. d) Constant velocity motion with v = 18 m/s, e) vx = vi ,  vy = -gt, f)  x = vit ,  y = y0 - (1/2)gt2, g) 3.19s, h) 36.1m/s, 60.1º below the horizontal.

15. 53.1º                Start with an initial velocity v0 and initial angle θ. (Since v0 does not matter, we can also say the initial velocity is 1m/s or 100m/s. It won’t matter.)  Find max. height and ∆x in terms of v0, θ, and g. Then make ∆x = (3 times the max. height). v0 should cancel. Solve for the angle.

21. d(tanθi ) – (gd2 )/(2vi2cos2 θi)                      Just like a normal projectile motion problem. Separate the horizontal and vertical sides. Use the horizontal side to find t and then use that t on the vertical side to find h.

23. a) The ball clears the crossbar by 0.89m. b) While falling.                                  a) It’s just like problem 21. b) Using the initial velocity and the initial angle, we can find the time it takes the ball to reach maximum height. Compare this time to the time found in part a).

49. a) The acceleration has a downward component of 9.8 m/s2 and a southward component of 2.5 m/s2. I.e. the acceleration is 10.1 m/s2, 14.3º to the south from the vertical. b) a downward acceleration of 9.8 m/s2. c) The bolt starts from rest and has a constant acceleration of 10.1 m/s2, 14.3º to the south from the vertical. So its trajectory is a straight line that goes 14.3º to the south from the vertical. d) The bolt has an initial horizontal velocity that equals to the train’s velocity at the moment the bolt leaves the ceiling. So the bolt is like a projectile shot horizontally. The trajectory is parabolic.

53. 15.3m                              Since the observer standing on the ground observes the ball shot straight up, the ball must have been thrown with a horizontal velocity that equals to 10m/s backwards. Using trig, we can find the vertical component of the ball’s velocity which can be used to calculate for the ball’s max. height.

55. 54.4 m/s2                        Radial acceleration is the same as the centripetal acceleration.                  


For 2017-2018 AP Physics C Students: 

According to the feedback provided by the class of 2017 AP Physics C students, I should still give you the hefty summer assignment. (Sorry, but they think it's important for you to do all the summer assignment.) The class of 2017 students also said that I should still cover all the topics and not to remove the most challenging problems. However, they recommended that we spend less time during the school year on Mechanics, so we can spend more time on Electricity & Magnetism. 

Below are the summer assignments: ALL students have summer assignments, but those who took AP Physics 2 have a little less to do. 

2017-2018: AP Physics C – Summer Assignment – Notes and work are due on the first day of school: Sept. 6, 2017

Suggested Day#

Watch these AP Physics 1 and AP Physics 2 lesson videos and:

·         Please be sure to review for a few minutes between lessons.

·         For static electricity: Take notes on them

·         For kinematics, forces, work and energy, and momentum: Do all problems in the videos

·         For rotation: Those who did not take AP Physics 1: Take notes on them

·         For rotation: Those who took AP Physics 1: Do all problems in the videos

 

For those who have not taken AP Physics 2

For those who have taken AP Physics 2

1

Static electricity videos: 5, 6, 7, 8, 9

Kinematics videos: 3 (3:00 to 6:38: review terms), 5, 6, 7, 14

2

Static electricity videos: 10, 11, 12, 13, 14

Kinematics videos: 16 (2:23 to 12:20), 18, 19, 20, 21, 25

3

Static electricity videos: 15, 16, 17, 19, 20, 21

Forces videos: 5 (to 8:15), 8, 10, 14, 16, 17.

4

Static electricity videos: 22, 23, 24, 25

Forces videos: 18, 20, 25, 26, 27.

5

Static electricity videos: 26, 27, 28, 29, 30

Work and energy videos: 3, 4, 12, 14, 17, 18

6

Static electricity videos: 31, 32, 33, 34, 35, 36

Work and energy videos: 21 (0:37 to 1:17), 22, 23

7

Static electricity videos: 37, 38, 39, 40, 41

Momentum videos: 4, 9, 10, 11, 12, 18 (to 2:47)

8

Kinematics videos: 3 (3:00 to 6:38: review terms), 5, 6, 7, 14

Momentum videos: 19, 20, 21, 22, 23

9

Kinematics videos: 16 (2:23 to 12:20), 18, 19, 20, 21, 25

Rotation videos: 1 to 8

10

Forces videos: 5 (to 8:15), 8, 10, 14, 16, 17.

Rotation videos: 9 to 13

11

Forces videos: 18, 20, 25, 26, 27.

Rotation videos: 14 to 22

12

Work and energy videos: 3, 4, 12, 14, 17, 18

Rotation videos: 23 to 28

13

Work and energy videos: 21 (0:37 to 1:17), 22, 23

Rotation videos: 29 to 35

14

Momentum videos: 4, 9, 10, 11, 12, 18 (to 2:47)

Rotation videos: 38 to 45

15

Momentum videos: 19, 20, 21, 22, 23

 

16

Rotation videos: 1 to 8

 

17

Rotation videos: 9 to 13

 

18

Rotation videos: 14 to 22

 

19

Rotation videos: 23 to 28

 

20

Rotation videos: 29 to 35

 

21

Rotation videos: 38 to 45

 

 



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