MASS training of trainers General Physics July 20-25, 2017 GENERAL PHYSICS 1: Classical Mechanics GENERAL PHYSICS 2: Electromagnetism Prof. MARLON FLORES SACEDON PhD in physics (on-going study) Master of Physics BS in Civil Eng’g Visayas State University Baybay City, Leyte

“Physics is the study of interaction between particles.” What is Physics? FORCE “Physics is the study of interaction between particles.” Interaction of objects due to its masses is… Interaction of objects due to its charges is… FORCE Classical electrodynamics Classical mechanics FORCE Thermodynamics Quantum mechanics FORCE Quantum mechanics Relativity FORCE Relativity The result of interaction is FORCE. Gravitational force Chemical force FORCE Electric force etc Magnetic force

This is classical mechanics… Force results motion… QUESTION Time interval, 𝑡 (s) Kinematics of Motion Displacement, 𝑠 (m) 1. How rocket moves? Velocity, 𝑣 (m/s) Acceleration, 𝑎 (m/s2) This is classical mechanics… Dynamics of Motion Force, 𝐹 (N) 2. Why rocket moves? Mass, 𝑚 (kg)

Kinematics of Motion: THE PROJECTILE MOTION

Recall: Uniformly Accelerated Bodies 𝑣 𝑖 𝑣 𝑓 𝑎 𝑠 Physical Quantities Displacement (𝑠) Time interval (𝑡) Average velocity 𝑣 Initial velocity ( 𝑣 𝑖 ) Final velocity ( 𝑣 𝑓 ) Acceleration (𝑎) Formulas Average velocity 𝑣= 𝑠 𝑡 Average velocity 𝑣= 𝑣 𝑖 + 𝑣 𝑓 2 Acceleration 𝑎= 𝑣 𝑓 − 𝑣 𝑖 𝑡 Combined formulas a. 𝑣 𝑓 = 𝑣 𝑖 +𝑎𝑡 b. 𝑠= 𝑣 𝑖 𝑡+ 1 2 𝑎 𝑡 2 c. 𝑣 𝑓 2 = 𝑣 𝑖 2 +2𝑎𝑠

Recall: Uniformly Accelerated Bodies 𝑣 𝑖 𝑣 𝑓 𝑎 𝑠 Formulas Average velocity: 𝑣 𝑎𝑣𝑒 = 𝑠 𝑡 Average velocity: 𝑣 𝑎𝑣𝑒 = 𝑣 𝑖 + 𝑣 𝑓 2 Acceleration: 𝑎= 𝑣 𝑓 − 𝑣 𝑖 𝑡 Combined formulas a. 𝑣 𝑓 = 𝑣 𝑖 +𝑎𝑡 b. 𝑠= 𝑣 𝑖 𝑡+ 1 2 𝑎 𝑡 2 c. 𝑣 𝑓 2 = 𝑣 𝑖 2 +2𝑎𝑠 Problem #1: Horizontal motion Given: 𝑣 𝑖 =3 m/s, 𝑣 𝑓 =9 m/s, & 𝑎=2 m/s2 Calculate: a) Average velocity 𝑣 b) Time interval 𝑡 c) Displacement 𝑠 ANSWERS: a) Average velocity 𝑣 𝑎𝑣𝑒 =6 m/s b) Time interval 𝑡=3 sec c) Displacement 𝑠=18 m

Recall: Uniformly Accelerated Bodies 𝑣 𝑖 𝑣 𝑓 𝑎 𝑠 Formulas Average velocity: 𝑣 𝑎𝑣𝑒 = 𝑠 𝑡 Average velocity: 𝑣 𝑎𝑣𝑒 = 𝑣 𝑖 + 𝑣 𝑓 2 Acceleration: 𝑎= 𝑣 𝑓 − 𝑣 𝑖 𝑡 Combined formulas a. 𝑣 𝑓 = 𝑣 𝑖 +𝑎𝑡 b. 𝑠= 𝑣 𝑖 𝑡+ 1 2 𝑎 𝑡 2 c. 𝑣 𝑓 2 = 𝑣 𝑖 2 +2𝑎𝑠 Problem #2: Horizontal motion Given: 𝑣 𝑖 =3 m/s, 𝑠=20 m, & 𝑎=2 m/s2 Calculate: a) Time interval 𝑡 b) Final velocity 𝑣 𝑓 c) Average velocity 𝑣 𝑎𝑣𝑒 ANSWERS: a) Time interval 𝑡=3.22 sec b) Final velocity 𝑣 𝑓 =9.43 m/s c) Average velocity 𝑣 𝑎𝑣𝑒 =6.22 m/s

Recall: Uniformly Accelerated Bodies 𝑣 𝑖 𝑣 𝑓 𝑎 𝑠 Formulas Average velocity: 𝑣 𝑎𝑣𝑒 = 𝑠 𝑡 Average velocity: 𝑣 𝑎𝑣𝑒 = 𝑣 𝑖 + 𝑣 𝑓 2 Acceleration: 𝑎= 𝑣 𝑓 − 𝑣 𝑖 𝑡 Combined formulas a. 𝑣 𝑓 = 𝑣 𝑖 +𝑎𝑡 b. 𝑠= 𝑣 𝑖 𝑡+ 1 2 𝑎 𝑡 2 c. 𝑣 𝑓 2 = 𝑣 𝑖 2 +2𝑎𝑠 Problem #3: Vertical motion (Free-fall) Given: 𝑣 𝑖 =0, 𝑠=8 m Calculate: a) Time interval 𝑡 b) Final velocity 𝑣 𝑓 c) Average velocity 𝑣 𝑎𝑣𝑒 ANSWERS: a) Time interval 𝑡=1.28 sec b) Final velocity 𝑣 𝑓 =12.53 m/s c) Average velocity 𝑣 𝑎𝑣𝑒 =6.26 m/s

What is projectile motion? Projectile anybody that is given an initial velocity and then allowed to move under the influence of gravity. A body that moves through space usually has a curved path rather than a perfectly straight one. The path followed by a projectile is called its trajectory. x

What is projectile motion? Projectile anybody that is given an initial velocity and then allowed to move under the influence of gravity. A body that moves through space usually has a curved path rather than a perfectly straight one. The path followed by a projectile is called its trajectory. x x

Objectives To understand the principles behind projectile motion. To derive the formulas of projectile. To calculate application problems in projectile.

y x Assumptions Air resistance is neglected (𝐴𝑖𝑟 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛=0) Component of acceleration along x axis is zero ( 𝑎 𝑥 =0). So, motion is uniform. Component of acceleration along y axis is gravitational acceleration ( 𝑎 𝑦 =−𝑔). So, Free-fall. x y 𝑣 𝑦 =0 𝑣= 𝑣 𝑥 𝑣 𝑥 𝑣 𝑦 𝑣 𝑣 𝑥 𝑣 𝑦 𝑣 𝑣 𝑥 −𝑣 𝑦 𝑣 𝜃 2 x y 𝑣 𝑜 𝑣 𝑜𝑥 𝑣 𝑜𝑦 𝜃> 𝜃′ 𝜃

y x Assumptions a. 𝑣 𝑓 = 𝑣 𝑖 +𝑎𝑡 b. 𝑠= 𝑣 𝑖 𝑡+ 1 2 𝑎 𝑡 2 Air resistance is neglected (𝐴𝑖𝑟 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛=0) Component of acceleration along x axis is zero ( 𝑎 𝑥 =0). So, motion is uniform. Component of acceleration along y axis is gravitational acceleration ( 𝑎 𝑦 =−𝑔). So, Free-fall. x y 5. 𝑣= 𝑣 𝑥 2 + 𝑣 𝑦 2 𝑣 𝑦 =0 𝑣= 𝑣 𝑥 𝑣 𝑥 𝑣 𝑦 𝑣 6. 𝑥= 𝑣 𝑜𝑥 𝑡+ 1 2 𝑎 𝑥 𝑡 2 𝑣 𝑥 𝑣 𝑦 𝑣 𝑣 𝑥 −𝑣 𝑦 𝑣 x= 𝑣 𝑜 𝑐𝑜𝑠𝜃.𝑡 x y 7. t= 𝑥 𝑣 𝑜 𝑐𝑜𝑠𝜃 𝑣 𝑜 Eliminate 𝑡 in Eq.7 in Eq.8 to come-up with time-independent equation 𝑣 𝑜𝑥 𝑣 𝑜𝑦 𝜃 8. 𝑦= 𝑣 𝑜𝑦 𝑡+ 1 2 𝑎 𝑦 𝑡 2 = 𝑣 𝑜 𝑠𝑖𝑛𝜃.𝑡− 1 2 𝑔 𝑡 2 1. 𝑣 𝑜𝑥 = 𝑣 𝑜 𝑐𝑜𝑠𝜃 3. 𝑣 𝑥 = 𝑣 𝑜𝑥 = 𝑣 𝑜 𝑐𝑜𝑠𝜃 Time-independent Equation of Projectile 𝑦=𝑥𝑇𝑎𝑛𝜃− 𝑔 𝑥 2 2 𝑣 𝑜 2 𝑐𝑜𝑠 2 𝜃 2. 𝑣 𝑜𝑦 = 𝑣 𝑜 𝑠𝑖𝑛𝜃 4. 𝑣 𝑦 = 𝑣 𝑜𝑦 −𝑔𝑡 = 𝑣 𝑜 𝑠𝑖𝑛𝜃−𝑔𝑡

Formulas for Projectile Motion 1. 𝑣 𝑜𝑥 = 𝑣 𝑜 𝑐𝑜𝑠𝜃 x y 𝑣 𝑜 𝜃 𝑣 𝑜𝑥 𝑣 𝑜𝑦 𝑣 𝑥 −𝑣 𝑦 𝑣 2. 𝑣 𝑜𝑦 = 𝑣 𝑜 𝑠𝑖𝑛𝜃 3. 𝑣 𝑥 = 𝑣 𝑜 𝑐𝑜𝑠𝜃 4. 𝑣 𝑦 = 𝑣 𝑜 𝑠𝑖𝑛𝜃−𝑔𝑡 5. 𝑣= 𝑣 𝑥 2 + 𝑣 𝑦 2 6. 𝑥= 𝑣 𝑜 𝑐𝑜𝑠𝜃.𝑡 7. 𝑡= 𝑥 𝑣 𝑜 𝑐𝑜𝑠𝜃 8. 𝑦= 𝑣 𝑜 𝑠𝑖𝑛𝜃.𝑡− 1 2 𝑔 𝑡 2 The Time-independent Equation of Projectile 9. 𝑦=𝑥𝑇𝑎𝑛𝜃− 𝑔 𝑥 2 2 𝑣 𝑜 2 𝑐𝑜𝑠 2 𝜃

𝑣 𝑜 𝑡 𝑓𝑙𝑖𝑔ℎ𝑡 𝑣 𝑓 Problem Exercise A tennis ball rolls off the edge of a table top 0.75 m above the floor and strikes the floor at a point 1.40 m horizontally from the edge of the table. Ignore air resistance. a). Find the magnitude of the initial velocity. b). Find the time of flight c). Find the magnitude and direction of the velocity of the ball just before it strikes the floor. a). The magnitude of the initial velocity ( 𝑣 𝑜 ) From: The Time-independent Equation of Projectile 9. 𝑦=𝑥𝑇𝑎𝑛𝜃− 𝑔 𝑥 2 2 𝑣 𝑜 2 𝑐𝑜𝑠 2 𝜃 𝑥=1.40𝑚 𝑦=−0.75𝑚 𝑣 𝑜 𝑣 𝑜 =? 𝜃=0 −0.75=1.40𝑇𝑎𝑛0− 9.81( 1.40 2 ) 2 𝑣 𝑜 2 𝑐𝑜𝑠 2 0 𝑣 𝑜 = 9.81( 1.40 2 ) 2(0.75) =3.58 m/s From Formula 7. 𝑡= 𝑥 𝑣 𝑜 𝑐𝑜𝑠𝜃 b). Find the time of flight 𝑡 𝑓𝑙𝑖𝑔ℎ𝑡 = 1.40 3.58𝑐𝑜𝑠0 =0.39 𝑠𝑒𝑐 0.75𝑚 c). Find the magnitude and direction of the velocity of the ball just before it strikes the floor. 𝑡 𝑓𝑙𝑖𝑔ℎ𝑡 3. 𝑣 𝑥 = 𝑣 𝑜 𝑐𝑜𝑠𝜃 4. 𝑣 𝑦 = 𝑣 𝑜 𝑠𝑖𝑛𝜃−𝑔𝑡 𝑣 𝑥 = 𝑣 𝑜 =3.58𝑚/𝑠 1.40𝑚 𝛽 𝑣 𝑓 𝑣 𝑦 =3.58 𝑠𝑖𝑛0 −9.81(0.39)=-3.83m/s 𝑣 𝑓 = 3.58 2 + 3.83 2 =5.24𝑚/𝑠 𝛽= 𝑇𝑎𝑛 −1 3.83 3.58 =46.93𝑜

ACTIVITY #2: A baseball is batted with a speed of 36 m/s at an angle of 37o with the ground. Find how long the ball is in the air, how high the ball goes, how far from the plate it is caught, and the components of the velocity of the ball 4 sec after it is hit. x y 𝑣 𝑜 =36 m/s 𝜃=37𝑜

ACTIVITY #2: A baseball is batted with a speed of 36 m/s at an angle of 37o with the ground. Find how long the ball is in the air, how high the ball goes, how far from the plate it is caught, and the components of the velocity of the ball 4 sec after it is hit. Recall: y= 𝑣 𝑜 𝑠𝑖𝑛𝜃.𝑡− 1 2 𝑔 𝑡 2 x y 0= 𝑣 𝑜 𝑠𝑖𝑛𝜃.𝑡− 1 2 𝑔 𝑡 𝑎𝑖𝑟 2 𝑡 𝑎𝑖𝑟 = 2 𝑣 𝑜 𝑠𝑖𝑛𝜃 𝑔 = 2(36)( sin 37 𝑜 ) 9.81 𝑡 𝑎𝑖𝑟 =4.42 sec 𝑡 𝑎𝑖𝑟 =?

ACTIVITY #2: A baseball is batted with a speed of 36 m/s at an angle of 37o with the ground. Find how long the ball is in the air, how high the ball goes, how far from the plate it is caught, and the components of the velocity of the ball 4 sec after it is hit. Recall: 𝑣 𝑦 = 𝑣 𝑜 𝑠𝑖𝑛𝜃−𝑔𝑡 0= 𝑣 𝑜 𝑠𝑖𝑛𝜃−𝑔 𝑡 𝑝𝑒𝑎𝑘 x y 𝑣 𝑦 =0 𝑡 𝑝𝑒𝑎𝑘 = 𝑣 𝑜 𝑠𝑖𝑛𝜃 𝑔 = (36)( sin 37 𝑜 ) 9.81 𝑡 𝑝𝑒𝑎𝑘 =? 𝑦 𝑚𝑎𝑥 =? 𝑡 𝑝𝑒𝑎𝑘 =2.21 sec Recall: 𝑦= 𝑣 𝑜 𝑠𝑖𝑛𝜃.𝑡− 1 2 𝑔 𝑡 2 𝑦 𝑚𝑎𝑥 =36 𝑠𝑖𝑛37(2.21sec)− 1 2 9.81 (2.21sec) 2 𝑣 𝑜 =36 m/s 𝑦 𝑚𝑎𝑥 =23.92 m 𝑡 𝑎𝑖𝑟 =4.42 sec 𝜃=37𝑜

ACTIVITY #2: A baseball is batted with a speed of 36 m/s at an angle of 37o with the ground. Find how long the ball is in the air, how high the ball goes, how far from the plate it is caught, and the components of the velocity of the ball 4 sec after it is hit. Recall: x= 𝑣 𝑜 𝑐𝑜𝑠𝜃.𝑡 x y 𝑣 𝑦 =0 R= 𝑣 𝑜 𝑐𝑜𝑠𝜃. 𝑡 𝑎𝑖𝑟 𝑡 𝑝𝑒𝑎𝑘 =2.21 sec R=36 sin 37 (4.42 𝑠𝑒𝑐) 𝑦 𝑚𝑎𝑥 =23.92 m R=95.76 m 𝑣 𝑜 =36 m/s 𝑡 𝑎𝑖𝑟 =4.42 sec 𝜃=37𝑜 Range 𝑅=?

ACTIVITY #2: A baseball is batted with a speed of 36 m/s at an angle of 37o with the ground. Find how long the ball is in the air, how high the ball goes, how far from the plate it is caught, and the components of the velocity of the ball 4 sec after it is hit. x y 𝑣 𝑦 =0 𝑡 𝑝𝑒𝑎𝑘 =2.21 sec 𝑦 𝑚𝑎𝑥 =23.92 m 𝑣 𝑜 =36 m/s 𝜃=37𝑜 𝑡 𝑎𝑖𝑟 =4.42 sec Range 𝑅=95.76 m

PROBLEM: A baseball is batted with a speed of 36 m/s at an angle of 37o with the ground. Find how long the ball is in the air, how high the ball goes, how far from the plate it is caught, and the components of the velocity of the ball 4 sec after it is hit. Recall: 𝑣 𝑥 = 𝑣 𝑜 𝑐𝑜𝑠𝜃 x y 𝑣 𝑦 = 𝑣 𝑜 𝑠𝑖𝑛𝜃−𝑔𝑡 𝑣 𝑥 =28.75 m/s 𝑣 𝑦 =−17.57 m/s 𝑣 𝑥 =? −𝑣 𝑦 =? 𝑣 𝑣 𝑜 =36 m/s 𝑡=4 sec 𝑡 𝑎𝑖𝑟 =4.42 sec 𝜃=37𝑜

Assignment #1 𝑣 𝑜 = ?

Assignment #2 𝑣 𝑜 = ? Height cliff ℎ= ?

Assignment #3 Required: Solve for 𝐷

Will the snowball hit the man? Assignment #4 Will the snowball hit the man? 1.90 m

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