equations that I should use? As you see the examples, you will end up with a system of equations that you have to solve

Preface The common question is Which is the equation that I should use?. I think I will rephrase the question as Which are the equations that I should use? As you see the examples, you will end up with a system of equations that you have to solve simultaneously. See the following URL: http://people.physics.tamu.edu/kamon/teaching/phys201/math. htm. You will need to be familiar Level 1, 2 an 3 before Exam 1. As to Trigonometric, see Chap. 1.8. This will be good enough for review and for spotting a blind spot on this subject. 1 Step 1 Review Card Which equation should be used? - It seems to be a very common question. One can re-phrase this as which subject is asked. front If a problem is testing momentum conservation but you try to use the energy conservation, you will not be able to solve. Use 4x6 index cards. On each card, Free-body diagram Newton s 2 nd law Friction, coefficient of friction Front write a textbook problem Back write a key subject and/or equation(s). Identifying the subject correctly and quickly is Step 1 of success in Exam. back

4 & 5 3 Structure of Newtonian Mechanics Inertial Reference Frame (Newton s 1 st Law) F = m a (Newton s 2 nd Law) Action-Reaction (Newton s 3 rd Law) Mass (m) Kinematics (r, v, a) The Nature of Force The Nature of Object The Nature of Motion 4

Normal force (n, N, F N ) Frictional force (f, F f ) Tension (T, F T ) Weight (w, mg, F G ) spring force (F spr ) Types of Force 5 Forces and Free Body Diagrams An example on page 108. The vertical components of forces are in equilibrium so there is no motion. This is a good example of forces in statics. Free Body Diagram is to identify all forces acted on a given body and to understand the motion. [Translation] Move forward, but speeding up, no motion in vertical [Translation] Move forward, but slowing down, no motion in vertical 6

Example 1: A warehouse worker pushes a 10-kg crate along the frictionless floor, as shown in the figure, by a force of 10 N that points downward at ban angle of 40 o below the horizontal. (a) (10 pts) Draw a free-body diagram for the crate. (b) (10 pts) Find the acceleration of the crate. 40 o [Example 1 - What If] An 85-N box of oranges is being pushed across a horizontal floor, as it moves, it is slowing at a constant rate of 0.900 m/s each second. The push force has a horizontal component of 20 N and a vertical component of 25 N downward. Calculate the coefficient of kinetic friction between the box and floor. Newton s Laws of Motion 9 [What If] Example 1 - Workbook Formula sheet Free-body diagrams Solving a system of equations slowing at a constant rate of 0.900 m/s each second. negative acceleration of 0.900 m/s 2 Newton s Laws of Motion 10

40 Newton s Laws of Motion 11 P5-40 Workbook Part (c) is asking this case. f k (= k n) < f s max (= s n) f s < f s max (= s n) Newton s Laws of Motion 12

P5-40 Workbook [II] A person pushes on a stationary 125 N box with 75N at 30 o below the horizontal, as shown in the figure. The coefficient of static friction between the box and the horizontal floor is 0.800. Question: What is the friction force on the box? The normal force was calculated to be 162.5N, so I assumed that I just multiply the normal force by the coefficient which is 0.800 which would make the answer: 130N, but I got the answer wrong. 95 * cos(30 o ) f s ( ) = 0 f s = 95 * cos(30 o ) = 82.27 = 82.3 N (smaller than 130 N). 130 N ( ) is the maximum static friction (f s max ) in which the box in just before moving. f s < f s max (= s n) 13 Intentionally left with blank page 14

Example 2: A block has mass m = 7.00 kg and lies on a smooth frictionless plane tilted at an angle = 22.0 o to the horizontal. (a) Determine the acceleration of the block as it slides down the plane. (b) If the block starts from rest 12.0 m up the plane from the base, what will be the block s speed when it reaches the bottom of the incline? Newton s Laws of Motion 15 Example 2 Workbook 16

[What If] Example 2: A block has mass m = 7.00 kg and lies on a plane tilted at an angle = 22.0 o to the horizontal. The coefficient of static friction between mass and the table is 0.400, whereas the coefficient of kinetic friction is 0.300. (a) Determine the acceleration of the block as it slides down the plane. (b) If the block starts from rest 12.0 m up the plane from the base, what will be the block s speed when it reaches the bottom of the incline? [What If] P5-39: A 25.0-kg box of textbooks rest on a loading ramp that makes an angle with the horizontal. The coefficient of kinetic friction is 0.250, and the coefficient of static friction is 0.350. (a) As the angle is increased, find the minimum angle at which the box starts to slip. (b) At this angle, find the acceleration once the box has begun to move. (c) At this angle, how fast will the box be moving after it has slid 5.00 m along the loading ramp? Newton s Laws of Motion 19 [What If] P5-39 Workbook 20

[What If] 34 Newton s Laws of Motion 21 P5-34 Workbook 22

Two boxes, one with mass m 1 = 10.0 kg and the other with mass m 2 = 12.0 kg, sit on the frictionless surface, connected by a light rope (see the figure below). A motion is set by a pulling force F P = 40.0 N directed = 30.0 o above the horizontal. (a) (5 pts) draw free-body diagram for the 10.0-kg box; (b) (4 pts) draw the freebody diagram for the 12.0-kg box; (c) (16 pts) using the free-body diagrams and newton s 2 nd law to find tension on the rope, acceleration, and normal force on each of two boxes. 23 Workbook 24

Problem 3: The coefficient of static friction between mass m 1 and the table is 0.400, whereas the coefficient of kinetic friction is 0.300. (a) (5 pts) Draw the free-body diagram for each of m 1 and m 2. (b) (10 ps) Express the acceleration of the block in terms of m 1, m 2, and g. (c) (5 pts) What minimum value of m 1 will keep the system starting to move for m 2 = 2.00 kg? (d) (5 pts) What value(s) of m 1 will keep the system moving at constant speed for m 2 = 2.00 kg? Newton s Laws of Motion 25 Newton s Laws of Motion 26

23 Problem 3: The coefficient of static friction between mass m 1 and the table is 0.400, whereas the coefficient cient of kinetic friction is 0.300. (a) (5 pts) Draw the free-body diagram for each of m 1 and m 2. (b) (10 ps) Express the acceleration of the block in terms of m 1, m 2, and g. (c) (5 pts) What minimum value of m 1 will keep the system starting to move for m 2 = 2.00 kg? (d) (5 pts) What value(s) of m 1 will keep the system moving at constant speed for m 2 = 2.00 kg? 24 Newton s Laws of Motion 27 P5-24 Workbook 1 2 3 28

Exam Problem is held in plane stay at rest 29 Formula sheet Half-Time Free-body diagrams Solving a system of equations Newton s Laws of Motion 30

Example 5.2: Express the tensions in each of three chains when the weight of a car engine is W. Newton s Laws of Motion 31 Example 5.2 Workbook 32

59 Newton s Laws of Motion 33 P5-59 Workbook T 2 T 3 O T 1 34

Prob. 5-59 Draw a Diagram FBD Newton s Laws Free-body diagram Newton s 2 nd law Friction, coefficient of friction Newton s Laws of Motion 35 Draw a Diagram FBD Newton s Laws Prob. 5-59 Free-body diagram Newton s 2 nd law Friction, coefficient of friction Newton s Laws of Motion 36

Example 5 You are weighing 600 N on a bathroom scale containing a stiff spring. In equilibrium the spring is 1.0 cm under your weight. Find the spring constant k. Solution: Springs or other elastic material will exert force when stretched or compressed. Spring Force (Restoring Force): The spring exerts its force in the direction opposite the displacement. F S (x) = k x 37 Example 5 You are weighing 600 N on a bathroom scale containing a stiff spring. In equilibrium the spring is 1.0 cm under your weight. Find the spring constant k. Solution: Find the spring constant k k = F max / x max = (600 N) / (0.010 m) = 6.0 x 10 4 N/m 38

Example 6 A light spring having a force constant of 125 N/m is used to pull a 9.50 kg sled on a horizontal frictionless ice rink. If the sled has an acceleration of 2.00 m/s 2, by how much does the spring stretch if it pulls on the sled horizontally. http://www.clker.com/cliparts/d/r/m/x/x/u/sled-md.png What are they asking?. x Where do I get x?. F = k x Now I see k is given. So I need to know F to get x. Where can I get F?... F = m a 39 Intentionally left with blank page 40

3 41 Understanding Projectile Motions I II III Kinematics (2D) 42

Review Card 3-1 R =? 2 x = (v 0 cos 0 ) t y = (v 0x ) t ½ g t 2 1 v x = v 0x v y = v 0y g t = tan -1 (v y /v x ) v 0x = v 0 cos 0 v 0y = v 0 sin 0 3 Solve a system of equations 43 Review Card 3-1 (Back) 2 R = (v 0x ) t 3.05 = (v 0x ) t ½ g t 2 1 v x = v 0x v y = v 0y g t = tan -1 (v y /v x ) v 0x = 20 cos v 0y = 20 sin 3 Solve a system of equations 44

For x: For y: 45 46

Your Review Card x = (37cos 53.1) t = tan -1 (v y /v x ) v 0x = 37 cos 53.1 v 0y = 37 sin 53.1 y = (v 0x ) t ½ g t 2 v x = v 0x v y = v 0y g t 47 Your Review Card (Back) 48

I The Same Problem? Example 3.3. Question: How fast must the motorcycle leave the cliff-top? 49 II 50

The Same Problem? Example 2: A projectile is launched from ground level to the top of a cliff which is R = 195 m away and H = 155 m high. The projectile lands on top of the cliff T = 7.60 s after it is fired. Use 2sin cos = sin2 if necessary. The acceleration due to gravity is g = 9.80 m/s 2 pointing down. Ignore air friction. a. Find the initial velocity of the projectile (magnitude v 0 and direction ). b. Find a formula of tan in terms of g, R, H and T. II t = 7.6 s Kinematics (2D) 51 Intentionally left with blank page 52

Firing at a More Complex Target A moving target presents a real-life scenario. It is possible to solve a falling body as the target. This problem is a classic on standardized exams. http://www.youtube.com/watch?v=cxvshnrxljw III 53 Magic? or Physics? Can you explain? III Giancoli s Textbook 3 rd Ed. We see An apple in motion (x direction) tends to stay in motion (x direction). Motion with constant velocity in x direction. http://link.brightcove.com/services/player/bcpid36804639001?bckey=aq~~,aaaaci JPQzk~,qiwYyUrE_-dz5lglGrCClkfJDM1jW3zH&bclid=0&bctid=109459228001 54

55 Prob. 3.06 56

Prob. 3.06 Alike A runner covers one lap of a circular track 40.0m in diameter in 63.1s. P 2 P 1 P 3 a) Average speed and magnitude of average velocity for one lap? b) Average speed and magnitude of average velocity for the first half lap in 28.7s? c) [What If] Average speed and magnitude of average velocity for the first 3-quarter lap in 47.3s? 57 Prob. 3.36 Alike Workbook 58

Average Acceleration? 59 Intentionally left with blank page 60

Modified Example 3.3 A hunter aims directly at a target (on the same level) 65.0 m away. Note that the magnitude of gravitational acceleration on the Earth is g = 9.80 m/s 2. (a) (10 pts) If the bullet leaves the gun at a speed of 145 m/s, by how much will it miss the target? (b) (15 pts) At what angle should the gun be aimed so the target will be hit? Kinematics (2D) 61 Modified Example 3.3 Workbook 62

Modified Example 3.3 Answer Key (a) d = 0.985 m (b) = 0.5 x sin 1 (9.80 x 65.0/145 2 ) = 0.868 Note that the following is a wrong approach for part (b): = tan 1 (0.985/65.0) = 0.868 even though the answer numerically agrees with the correct one. Below is an exercise of two approaches by changing the magnitude of the velocity (v 0 ), but using the same distance (R) of 65.0 m, where you see a larger discrepancy as v 0 decreases. 63 Reflection - Modified Example 3.3(b) A hunter aims directly at a target (on the same level) 65.0 m away. Note that the magnitude of gravitational acceleration on the Earth is g = 9.80 m/s 2. (a) (10 pts) If the bullet leaves the gun at a speed of 145 m/s, by how much will it miss the target? (b) (15 pts) At what angle should the gun be aimed so the target will be hit? Step 1: Draw a diagram (or picture) of the situation, with coordinate axes. y R = 65.0 m t = 0, v 0 = 145 m/s t = T x 64

Reflection - Modified Example 3.3(b) Step 2: Setting up and solving a system of equations. R = 65.0 m x = R y = 0 Kinematics (2D) 65 Reflection - Modified Example 3.3(b) COMMON MISTAKE d tan = d / R, where d is from part (a). R Accidentally, you get the same answer! ( = 0.868 o ) Your claim: I should get a full credit because the final answer is correct. My response: No What is wrong with this? 66

Reflection - Modified Example 3.3(b) COMMON MISTAKE Wrong True (W-T)/(T) R v 0 [m] [m/s] [deg] [deg] [%] 65.0 145 0.867886 0.868086 0.023% 65.0 80.0 2.84901 2.85609 0.25% 65.0 40.0 11.2583 11.7305 4.0% 65.0 26.0 25.2277 35.2214 28% 65.0 25.3 26.4543 42.1839 37% 67 Intentionally left with blank page 68

P.3-23 & P.3-41 Kinematics (2D) 69 Example 2 Uniform circular motion applied to a daring carnival ride. Referred to the worked example on page 88. + v = D / T = 2 R / T P3-38 Hint: A jet plane comes in for a downward dive as shown in the figure. The bottom part of the path is a quarter circle having a radius of curvature of R = 330m. At this trajectory (= a part of circular motion), pilots lose consciousness at an acceleration of 5.1g. Translation: a rad = 5.10 g = 5.10 x 9.80 for R = 330 m. Setting-up a rad = v^2 / R à calculate v 70

2 71 The four equations shown to the right apply to any straight-line motion with constant acceleration a x. Problem-Solving Strategy 72

A motorcycle with constant acceleration Problem-Solving Strategy for an accelerating motorcycle. Given! Recap From Chap 2 73 A motorcycle with constant acceleration Problem-Solving Strategy for an accelerating motorcycle. Given! 74

Direction of Acceleration? A car is parked in its driveway facing away from the street. Its owner backs the car out onto the street. As the car starts from rest and gains speed up to 5 mph backing out of the driveway, how is the velocity and acceleration pointed? 1) Velocity points toward the rear of the car, and its acceleration points toward its rear. 2) Velocity points toward the front of the car, and its acceleration points toward its front. 3) Velocity points toward the front of the car, while its acceleration points toward its rear. 4) Velocity points toward the rear of the car, while its acceleration points toward its front. 5) Velocity points toward the rear of the car, and its acceleration is zero. [ ] [ ] [ ] [ ] [ a=0] 75 Acceleration (with Calculus)? 76

Prob. 2.74 77 Prob. 2.74 Workbook Introduction 78

Example 2.10 A 2-euro coin is dropped from the Leaning Tower of Pisa. It starts from rest at height 50.0 m and falls freely. a) Compute its position after 2.00 s. b) Find the velocity after 2.00 s. c) Find the time at which the coin hits the ground. d) Find the velocity just before the coin hits the ground. e) Sketch a y-t graph. 50 m 0 m Kinematics (1D) 79 a y = 9.80 980m/s 2 y 0 = 50.0 m v y0 = 0.00 m/s Workbook y = y 0 + v y0 t + ½ a y t 2 y = 50.0 + ½ ( ) t 2 (1) v y = v y0 + a y t v y = ( ) t (2) v y 2 = v y0 2 + 2a y (y y 0 ) v y 2 = 2 ( ) (y 50.0) (3) Kinematics (1D) 80

Example 2.10 [What If] A 2-euro coin is dropped from the Leaning Tower of Pisa. It starts from rest at height 50.0 m and falls freely. a) Compute its position after 2.00 s. b) Find the velocity after 2.00 s. c) Find the time at which the coin hits the ground. d) Find the velocity just before the coin hits the ground. e) Sketch a y-t graph. 0 m 50 m Kinematics (1D) 81 a y = +9.80 980m/s 2 y 0 = 0.0 m v y0 = 0.00 m/s Workbook y = y 0 + v y0 t + ½ a y t 2 y = 0.0 + ½ ( ) t 2 (1) v y = v y0 + a y t v y = ( ) t (2) v y 2 = v y0 2 + 2a y (y y 0 ) v y 2 = 2 ( ) (y 0.0) (3) Kinematics (1D) 82

Old Exam 1 for Practice (II) a) 15.8 m/s b) 23.4 m/s c) Kinematics (1D) *) 15 min. *) similar to the HW problems *) a similar figure was used. *) modified to have 3 parts. 83 Workbook 84

Workbook If v 0 is not zero, 85 Intentionally left with blank page 86

1 87 http://hyperphysics.phy-astr.gsu.edu/hbase/vect.html A x = cos (180+45) = 0707 A y = sin(180+45) = 0.707 OR A x = cos (45) = 0707 A y = sin(45) = 0.707 You add - by hands. A 88

B/A or A/B? 89 Angle between A = 4 ^ i 4 ^ j and the x-axis is: 180+45 45 180-45 A http://www.regentsprep.org/regents/math/algtrig/att3/referenceangles.htm 90

91 Workbook 92

Find the magnitude of F 2 and its direction relative to F 1. F F F F 1300 N Force as Vectors R 1 2 90 o 1300 N Vector direction x-y coordinates Newton s Laws of Motion 93 Workbook Find the magnitude of F 2 and its direction relative to F 1. F F F F R 1 2 1300 N 90 o 1300 N y x Newton s Laws of Motion 94

[Quick Quiz 1] Find: (a)a+b (b)a B (c) A+B+C (d) A+B C (e) B C (f) A C 95 [Quick Quiz 1] Find: (a)a+b (b)a B (c) A+B+C (d) A+B C (e) B C (f) A C 96

Adding vectors using their components 97 Intentionally left with blank page 98

Displacement (Vector) y x Introduction 99 Direction: North Magnitude: 2.6 km y x Introduction 100

Direction: East Magnitude: 4.0 km y x Introduction 101 Direction: 45 o North of East Magnitude: 3.1 km y x Introduction 102

Direction:? Magnitude:? D 2 D 3 D 1 D R D 1 D 2 D 3 x y Introduction D R 103 Direction: 37.7 o North of East Magnitude: 7.83 km D 2 D 3 D 1 D R Introduction x y D 1 0 km 2.6 km D 2 4.0 km km D 3 km km D R 6.19 km km 104

53 54 Prob. 1-54 5.80 km 4.00 km 2.00 km 50 o 5.50 km 105 Prob. 1-62 Workbook 106

56 Prob. 1-56 Kinematics (1D) 107 Prob. 1-56 Workbook 108

Old Exam 1 for Practice (I) *) 15 min. *) similar to the HW problems *) a textbook figure was used. *) modified to have 3 parts. a) 228 km, 29.6 o b) 189 km, 10.5 o c) No 109 Workbook 110

Velocity as Vectors 111 Workbook P.3-36 Alike.. 112

Two bodies with different accelerations Follow the figure in which the police officer and motorist have different accelerations. Two lines 113 Intentionally left with blank page 114

Motion with Constant Acceleration y v y Diagnostic Test v y -t graph t A B C 115 Motion with Constant Acceleration y v y Diagnostic Test v y -t graph t A B C 116

Diagnostic Test Motion with Constant Acceleration y a y a y -t graph t A B C 117 Diagnostic Test Motion with Constant Acceleration y a y a y -t graph t a y = 9.8 m/s 2 A B C 118

Diagnostic Test Motion with Constant Acceleration a y a y -t graph t y A B C 119 Diagnostic Test Motion with Constant Acceleration a y a y -t graph a y = +9.8 m/s 2 t y A B C 120

Problem 1 [What If] A 2-euro coin is dropped from the Leaning Tower of Pisa. It starts from rest at height 50.0 m and falls freely. a) Compute its position after 2.00 s. b) Find the velocity after 2.00 s. c) Find the time at which the ball hits the ground. d) Find the velocity just before the ball hits the ground. e) Sketch a y-t graph. See Appendix Kinematics (1D) 121 ISEE Problem 1 Solution (Cont d) D.A.D. Identify unknowns! a y = 9.80 m/s 2 y 0 = 0.0 m v y0 = 0.00 m/s y =? v y =? @ t = 2.00 s 0 m 50.0 m y 2 unknowns 2 equations if t is given. Kinematics (1D) 122

ISEE y = y 0 + v y0 t + ½ a y t 2 y = 0.0 + ½ ( 9.80) t 2 (1) v y = v y0 + a y t v y = ( 9.80) t (2) v y 2 = v y0 2 + 2a y (y y 0 ) v y 2 = 2 ( 9.80) (y 0.0) (3) (a)eq. 1 (b)eq. 2 (c) Eq. 1 (d)eq. 3 OR Eq. 2 Kinematics (1D) 123 ISEE y = y 0 + v y0 t + ½ a y t 2 y = 0.0 + ½ ( 9.80) t 2 (1) v y = v y0 + a y t v y = ( 9.80) t (2) v y 2 = v y0 2 + 2a y (y y 0 ) v y 2 = 2 ( 9.80) (y 0.0) (3) (a)eq. 1 t = 2.00 s, thus y = 19.6 m (was 30.4 m) (b)eq. 2 v y = 19.6 m/s (was 19.6 m/s) (c) Eq. 1 50.0 = 0.0 + ½ ( 9.80) t 2, thus, t = 3.19 s (d)eq. 3 y = 50.0 v y = 31.3 x m/s or +31.3 m/s OR Eq. 2 v y = ( 9.80)(3.19) = 31.3 m/s Kinematics (1D) y direction] 124

Example 1: A Hunter A hunter aims directly at a target (on the same level) 65.0 m away. Note that the magnitude of gravitational acceleration on the Earth is g = 9.80 m/s 2. (a) (10 pts) If the bullet leaves the gun at a speed of 145 m/s, by how much will it miss the target? (b) (15 pts) At what angle should the gun be aimed so the target will be hit? Kinematics (2D) 125 Your Review Card y R = 65.0 m x t = 0, v 0 = 145 m/s d t = T x = (v 0 cos 0 ) t = tan -1 (v y /v x ) v 0x = v 0 cos 0 v 0y = v 0 sin 0 y = (v 0x ) t ½ g t 2 v x = v 0x v y = v 0y g t 126

Example 1(a) - Cont d R = 65.0 m d T = R/v 0 [x] Motion with constant velocity y = ½ g (R / v 0 ) 2 Kinematics (2D) 127 Example 1(b) Cont d y COMMON MISTAKE #2 How to find T? t = 0, v 0 = 145 m/s R = 65.0 m x t = T (a) = R/v 0 = 0.448 s Question: What s wrong with T (a) = R/v 0? Hint #1: Remember vector kinematics Hint #2: Remember the velocity is a vector quantity Hint #3: v 0x =? [145 m/s is a wrong choice.] Kinematics (2D) 128

Kinematics (2D) 129 Was: What is the acceleration? v 0x = 28 m/s x x 0 = 55 m v x 2 = v 0x 2 + 2 a x (x x 0 ) Newton s Laws of Motion 130

Was: What is the acceleration? Now: What is the net force? m = 1500 kg v 0x = 28 m/s F x net? x x 0 = 55 m v x 2 = v 0x 2 + 2 a x (x x 0 ) F x net = m a x Newton s Laws of Motion 131 m = 1500 kg F x net? Was: What is the acceleration? Now: What is the net force? v 0x = 28 m/s x x 0 = 55 m v 2 x = v 2 0x + 2 a x (x x 0 ) a x = 7.1 m/s 2 F x = m a x (1500 kg) x ( 7.1 m/s 2 ) v 2 x = v 2 0x + 2 a x (x x 0 ) F net x = m a x Newton s Laws of Motion 132