General Physics (Phys 1011) Lecture Notes PDF

General Physics (Phys 1011) Lecture Note By Endalkachew Mengistu (PhD) Department of Physics College of Natural and Computational Scien...

General Physics (Phys 1011) Lecture Note By Endalkachew Mengistu (PhD) Department of Physics College of Natural and Computational Sciences Addis Ababa University, Addis Ababa, Ethiopia E-mail:- [email protected] May 15, 2024 1 / 21 By Endalkachew Mengistu (PhD) Department of Physics College ofGeneral Natural Physics and Computational (Phys 1011)Sciences Lecture Addis Note Ababa University, Ad  Chapter 2 - Kinematics in One Dimensions Outlines 1 Distance and Displacement Distance Position and Displacement Speed and Average Speed Velocity and Average Velocity Instantaneous velocity 2 Acceleration Motion with constant acceleration Free fall motion 2 / 21 By Endalkachew Mengistu (PhD) Department of Physics College ofGeneral Natural Physics and Computational (Phys 1011)Sciences Lecture Addis Note Ababa University, Ad  Learning Outcomes At the end of this chapter, the students will be able to: Define kinematic terms such as distance, position, displacement, velocity and acceleration; Identify the d/ce b/n average and instantaneous velocity; Identify the d/ce b/n average and instantaneous acceleration; Describe straight-line motion in terms of average and instantaneous velocity, average and instantaneous acceleration; Interpret graphs of position vs time, velocity vs time, and acceleration vs time for straight-line motion; Drive the equations of motions with constant acceleration; Solve problems involving straight-line motion with constant acceleration, including free-fall problems; and Explain projectile motion and solve problems related to it. 3 / 21 By Endalkachew Mengistu (PhD) Department of Physics College ofGeneral Natural Physics and Computational (Phys 1011)Sciences Lecture Addis Note Ababa University, Ad  Distance and Displacement What is the difference between distance and displacement? Distance, Position and Displacement Distance:- The length of the path followed by the object. Position:- The location of an object with respect to a chosen reference point. Displacement:- The change in position of an object with respect to a given reference frame. For example, when a particle moves along the x-axis from some initial position xi to some final position xf , its displacement is: ∆x = xf − xi. (1) Displacement is a vector quantity. 4 / 21 By Endalkachew Mengistu (PhD) Department of Physics College ofGeneral Natural Physics and Computational (Phys 1011)Sciences Lecture Addis Note Ababa University, Ad  Cont’d... Speed and Average Speed Speed (v): refers to “how fast an object is moving.” the rate at w/c an object covers distance, irrespective of its direction ⇒ a scalar physical quantity. involves both distance and time. expressed in meter m/s, km/hr,... Average speed (vav ) - is the total distance traveled by the object divided by the total elapsed time. Mathematically, this is expressed as: ∆x vav =. (2) ∆t 5 / 21 By Endalkachew Mengistu (PhD) Department of Physics College ofGeneral Natural Physics and Computational (Phys 1011)Sciences Lecture Addis Note Ababa University, Ad  Velocity and Average Velocity Velocity is the rate of change of position of a body in a particular direction. Like speed, it is expressed in m/s, km/s... Average velocity - is the slope of the straight line that connects 2 particular pts on the x (t) curve as shown below: Displacement x(t2 ) − x(t1 ) ∆x vav = = =. (3) Time t2 − t1 ∆t 6 / 21 By Endalkachew Mengistu (PhD) Department of Physics College ofGeneral Natural Physics and Computational (Phys 1011)Sciences Lecture Addis Note Ababa University, Ad  Example: The position of an object moving along an x-axis is given by x = 3t − 4t 2 + t 3 , where x is in meters and t in seconds. Find the position of the object at the following values of t: (a) 1 s, (b) 2 s, (c) 3 s, and (d) 4 s. (e) What is the object’s displacement b/n t = 0 and t = 4 s? (f) What is its vav for the time interval from t = 2 s to t = 4 s? (g) Graph x versus t for 0 ≤ t ≤ 4 s and indicate how the answer for (f) can be found on the graph. ⇒ HW. Solutions: (a) At t = 1 s, x(t = 1 s) = 0 m. (b) At t = 2 s, x(t = 2 s) = - 2 m. (c) At t = 3 s, x(t = 3 s) = 0 m. (d) At t = 4 s, x(t = 4 s) = 12 m. (e) The displacement: ∆x = x(4 s) - x(0 s) = 12 m - 0 m = 12 m. (f) The displacement: ∆x = x(4 s) - x(2 s) = 12 m - (-2 m) = 14 m ⇒ ~vav = ∆~ x 14 ∆t = 2 m/s = 7 m/s. 7 / 21 By Endalkachew Mengistu (PhD) Department of Physics College ofGeneral Natural Physics and Computational (Phys 1011)Sciences Lecture Addis Note Ababa University, Ad  Cont’d... Instantaneous velocity (vinst ) Average velocity/speed of an object do not provide the detail information of the entire motion. So, we may need to know the velocity/speed of an object at a certain instant of time. The slope of the line tangent to the position versus time curve at “a given time” is defined to be the vinst at that time. The vinst is obtained from the average velocity as ∆t → zero: ∆x d~x vinst = lim =. (4) ∆t→0 ∆t dt For uniform motion, vinst = vav. The instantaneous speed of a particle is equal to the magnitude of its vinst. 8 / 21 By Endalkachew Mengistu (PhD) Department of Physics College ofGeneral Natural Physics and Computational (Phys 1011)Sciences Lecture Addis Note Ababa University, Ad  9 / 21 By Endalkachew Mengistu (PhD) Department of Physics College ofGeneral Natural Physics and Computational (Phys 1011)Sciences Lecture Addis Note Ababa University, Ad  Acceleration Average and Instantaneous accelerations When velocity of a particle changes with time, the particle is said to be accelerating. Average acceleration- is the change in velocity ∆~v of an object divided by the time interval ∆t during w/c that change occurs: ∆~v ~vf − ~vi ~aav = =. (5) ∆t tf − ti SI unit: meter per second per second (m/s 2 ) Instantaneous acceleration (~ainst or simply ~a) - is the limit of the ~aav (Eqn 5) as the time interval ∆t goes to zero: ∆~v ~ainst (t) = lim , (6) ∆t→0 ∆t ~v (t + ∆t) − ~v (t) d~v = lim =. (7) ∆t→0 ∆t dt 10 / 21 By Endalkachew Mengistu (PhD) Department of Physics College ofGeneral Natural Physics and Computational (Phys 1011)Sciences Lecture Addis Note Ababa University, Ad  Graphical Representation 11 / 21 By Endalkachew Mengistu (PhD) Department of Physics College ofGeneral Natural Physics and Computational (Phys 1011)Sciences Lecture Addis Note Ababa University, Ad  Cont’d... Example: A baseball player moves in a straight-line path in order to catch a fly ball hit to the outfield. His velocity as a function of time is shown in Figure (a). (i) Find his ~aav at points A, B, and C. (ii) Repeat the problem, using Figure (b) ⇒ HW. 12 / 21 By Endalkachew Mengistu (PhD) Department of Physics College ofGeneral Natural Physics and Computational (Phys 1011)Sciences Lecture Addis Note Ababa University, Ad  Cont’d... Solutions: Acceleration at A: The acceleration at A equals the slope of the line connecting the points (0 s, 0 m/s) and (2.0 s, 4.0 m/s): a = ∆v 4.0−0 2 ∆t = 2.0−0 m/s = 2.0 m/s. 2 Acceleration at B: ∆v = 0, because the segment is horizontal: a = ∆v 4.0−4.0 2 2 ∆t = 3.0−2.0 m/s = 0 m/s. Acceleration at C: The acceleration at C equals the slope of the line connecting the points (3.0 s, 4.0 m/s) and (4.0 s, 2.0 m/s): a = ∆v 2.0−4.0 2 ∆t = 4.0−3.0 m/s = -2.0 m/s. 2 Answer for problem (ii): The accelerations at A, B, and C are 23.0 m/s 2 , 1.0 m/s 2 , and 0 m/s 2 , respectively. 13 / 21 By Endalkachew Mengistu (PhD) Department of Physics College ofGeneral Natural Physics and Computational (Phys 1011)Sciences Lecture Addis Note Ababa University, Ad  Exercise The following figure show the velocity vs. time graph for an object moving along a straight path. (i) Find the ~aav of the object during the time intervals: (a) 0 to 5.0 s, (b) 5.0 s to 15 s, and (c) 0 to 20 s. (ii) Find the ~ainst at (a) 2.0 s, (b) 10 s, and (c) 18 s. 14 / 21 By Endalkachew Mengistu (PhD) Department of Physics College ofGeneral Natural Physics and Computational (Phys 1011)Sciences Lecture Addis Note Ababa University, Ad  Motion with Constant Acceleration The simplest kind of accelerated motion is straight-line motion with constant acceleration. The velocity changes at the same rate throughout the motion. For example, a falling body has a constant acceleration if the effects of the air are not important or neglected. When an object moves with constant acceleration, the ~ainst and ~aav are equal: ~v − ~vo ~ainst = ~aav = ~a = ⇒ ~v = ~vo + ~at, (8) t −0 where ~vo is the velocity at t = 0 and v is the velocity at any later time t. In a similar manner, the ~vav b/n t = 0 and a later time t is: ~x − ~xo ~vav = ⇒ ~x = ~xo + ~vav t, (9) t −0 where ~xo is the position of the particle at t = 0. 15 / 21 By Endalkachew Mengistu (PhD) Department of Physics College ofGeneral Natural Physics and Computational (Phys 1011)Sciences Lecture Addis Note Ababa University, Ad  EXERCISES 1 Show that the equations for motion with constant acceleration are: x − xo = vo t + 12 at 2 , v2 = v2o + 2a(x − xo ), x − xo = vt − 12 at 2 , and x − xo = 12 (vo + v)t. 2 A car accelerates uniformly from rest to a speed of 40.0 km/hr in 12.0 sec. Find (a) the distance the car travels during this time and (b) the constant acceleration of the car. 3 An object moves with constant acceleration 4.00 m/s 2 and over a time interval reaches a final velocity of 12.0 m/s. (a) If its original velocity is 6.00 m/s, what is its displacement during the time interval? (b) What is the distance it travels during this interval? (c) If its original velocity is 26.00 m/s, what is its displacement during this interval? (d) What is the total distance it travels during the interval in part (c)? 16 / 21 By Endalkachew Mengistu (PhD) Department of Physics College ofGeneral Natural Physics and Computational (Phys 1011)Sciences Lecture Addis Note Ababa University, Ad  Free fall motion A freely falling object is any object moving freely under the influence of gravity alone, regardless of its initial motion. Objects thrown upward/downward and those released from rest are all falling freely once they are released. Any freely falling object experiences an acceleration directed downward, regardless of its initial motion. For freely falling bodies the motion is vertical along y-axis so that a is replaced by g and x is replaced by y in the equations of motion for rectilinear motion: 1 vy = voy + gt, y = voy t + gt 2 , & v2y = v2oy + 2gt. (10) 2 ⇒ Free fall is a case of motion with constant acceleration. Example 1 [Answer: (a) voy = 2.94 m/s, (b) t = 0.600 s.] (a) If a flea can jump straight up to a height of 0.440 m, what is its initial speed as it leaves the ground? (b) How long is it in the air? 17 / 21 By Endalkachew Mengistu (PhD) Department of Physics College ofGeneral Natural Physics and Computational (Phys 1011)Sciences Lecture Addis Note Ababa University, Ad  Example 2 Assume that you are on the roof of the physics building, 46.0 m above the ground as shown in the figure below. Your friend, who is 1.80 m tall, is walking alongside the building at a constant speed of 1.20 m/s. If you wish to drop an egg on your friend’s head, where should he be when you release the egg? Assume that the egg is in free fall. 18 / 21 By Endalkachew Mengistu (PhD) Department of Physics College ofGeneral Natural Physics and Computational (Phys 1011)Sciences Lecture Addis Note Ababa University, Ad  Solution First let’s refer to the height of the building and your friend as yb and yh respectively. Again, let the downward direction be positive; i.e., +y be downward. The egg has voy = 0 and ay = g = 9.8 m/s 2. At the height of your friend’s head, the egg has yb − yh = (46.0 − 1.80) m = 44.2 m. r 2(44.2)m q 2(yb −yh ) Then Eqn (10) yields t = g = 9.8m/s 2 = 3s. He walks a distance vox t = (1.20 m/s)(3.00 s) = 3.60 m. Therefore, release the egg when he is 3.60 m from the point directly below you. Exercise: A rock is dropped (from rest) from the top of a 60-m-tall building. (i) How long does it take to fall (a) the first 50 m and (b) the second 50 m? (ii) How far above the ground is the rock 1.2 s before it reaches the ground? 19 / 21 By Endalkachew Mengistu (PhD) Department of Physics College ofGeneral Natural Physics and Computational (Phys 1011)Sciences Lecture Addis Note Ababa University, Ad  Problem 1 A proton moves along the x-axis according to the equation x = 50t + 10t 2 , where x is in meters and t is in seconds. Calculate the (a) ~vav of the proton during the first 3.0 s of its motion, (b) ~vinst of the proton at t = 3.0 s, and (c) ~ainst of the proton at t = 3.0 s. (d) Graph x vs t and indicate how the answer to (a) can be obtained from the plot. (e) Indicate the answer to (b) on the graph. (f) Plot v vs t and indicate on it the answer to (c). Problem 2 The position of a particle as it moves along a y-axis is given by y = (2.0cm) sin( πt4 ), with t in seconds and y in centimeters. Find the (a) ~vav and ~aav of the particle b/n t = 0 & t = 2 s, (b) ~vinst and ~ainst of the particle at t = 0, 1.0, & 2.0 s. Problem 3 Module (pages 65 - 66). 20 / 21 By Endalkachew Mengistu (PhD) Department of Physics College ofGeneral Natural Physics and Computational (Phys 1011)Sciences Lecture Addis Note Ababa University, Ad  Problem 4 Two cars (say, A and B) travel in a straight line. The distance of A from the starting point is given as a function of time by xA (t) = αt + βt 2 , with α = 2.60 m/s and β = 1.2 m/s 2. The distance of B from the starting point is xB (t) = γt 2 − δt 3 , with γ = 2.80 m/s 2 and δ = 0.20 m/s 3. Which car is ahead just after they leave the starting point? At what time(s) (i) are the cars at the same point? (ii) is the distance from A to B neither increasing nor decreasing? (iii) do A and B have the same acceleration? Problem 5 A ball is thrown straight up from the ground with speed v0. At the same instant, a 2nd ball is dropped from rest from a height H, directly above the point where the 1st ball was thrown upward. There is no air resistance. Find the (a) time at w/c the two balls collide, and (b) value of H in terms of v0 and g so that at the instant when the balls collide, the 1st ball is at the highest point of its motion. 21 / 21 By Endalkachew Mengistu (PhD) Department of Physics College ofGeneral Natural Physics and Computational (Phys 1011)Sciences Lecture Addis Note Ababa University, Ad