Topic 1: INTRO.

Topic 1: INTRO

SI Units Stands for Système international d'unités. It is standard body of measurements, the modern form of the metric system adopted in 1960. ▪The SI system is pretty much the world standard in units. Why use SI units? ▪ universal ▪ easy (metric system) ▪ The fundamental units in the SI system are… Length meter (m) Mass kilogram (kg) Time second (s) Electric Current (I) ampere (A) Temperature kelvin (K) Amount of matter mole Intensity of light/Luminosity candela (cd) ▪ You will also use the gram. In Chemistry. In physics we use the kilogram (SI unit).

1 kg is basic unit of mass, not, I repeat, not 1g !!!!!!!!!!
SI Units Length – 1 meter (1m) is the distance traveled by the light in a vacuum during a time of 1/299,792,458 second. Mass – 1 kilogram (1 kg) is defined as a mass of a specific platinum-iridium alloy cylinder kept at the International Bureau of Weights and Measures at Sevres, France ▪ The International Prototype of the Kilogram was sanctioned in Its form is a cylinder with diameter and height of about 39 mm. It is made of an alloy of 90 % platinum and 10 % iridium. The IPK has been conserved at the BIPM since 1889, initially with two official copies. Over the years, one official copy was replaced and four have been added. 1 kg is basic unit of mass, not, I repeat, not 1g !!!!!!!!!! Time – 1 second (1s) is defined as 9,192,631,770 times the period of one oscillation of radiation from the cesium atom. ▪ One meter is about a yard or three feet. ▪ One kilogram of mass has weight of 9.8 N on earth or about 2.2 pounds in USA -.

Scientific notation and prefixes
EX: The best current estimate of the age of the universe is = × years = 13.7 billion years scientific notation prefix 2. electron mass = kg = 9.1 × kilograms Very large and very small numbers: either scientific notation or prefixes should be used Power of 10 Prefix Name Symbol pico p nano n micro µ milli m centi c kilo k mega M giga G tera T

EX: 1 𝑚 3 = ( 10 2 𝑐𝑚) 3 = 10 6 𝑐𝑚 3 1 𝑚 3 = ( 10 3 𝑚𝑚) 3 = 10 9 𝑚𝑚 3
1 𝑚 3 = ( 10 2 𝑐𝑚) 3 = 𝑐𝑚 3 1 𝑚 3 = ( 10 3 𝑚𝑚) 3 = 𝑚𝑚 3 1 𝑐𝑚 3 = ( 10 −2 𝑚) 3 = 10 −6 𝑚 3 1 𝑚𝑚 3 = ( 10 −3 𝑚) 3 = 10 −9 𝑚 3

EX: 7.2 m3 → mm3 7.2 𝑚 3 = 𝑚𝑚 3 = 7.2 x 𝑚𝑚 3 EX: 100 𝑚𝑚 3 = −3 𝑚 3 = 10 −7 𝑚 3 100 mm3 → m3 EX: 75 𝑔 𝑐𝑚 2 = −3 𝑘𝑔 −2 𝑚 2 =750𝑘𝑔/ 𝑚 2 75 g/cm2 → kg/m2 EX: 20 𝑚 𝑠 = −3 𝑘𝑚 ℎ = 72 km/h 20 m/s → km/h EX: 72 𝑘𝑚 ℎ = 𝑚 3600 𝑠 = 20 m/s 72 km/h → m/s

Uncertainty and error in measurement
No measurement can be "exact". You can never, NEVER get exact value experimentally Error in measurement is expected because of the imperfect nature of our measuring devices. The inevitable uncertainty is inherent in all measurements. It is not to be confused with a mistake or blunder Accuracy is the closeness of agreement between a measured value and a true or accepted value Precision is the degree of exactness (or refinement) of a measurement (results from limitations of measuring device used).

(1) All non-zero digits are significant. 438 g 26.42 m 0.75 cm 3 4 2
SIGNIFICANT FIGURES are reliably known digits + one uncertain (estimate) (1) All non-zero digits are significant. 438 g 26.42 m 0.75 cm 3 4 2 (2) All zeros between non-zero digits are significant. 12060 m cm 4 5 (3) Filler zeros to the left of an understood decimal place are not significant. 220 L 60 g 30. cm 2 1 (4) Filler zeros to the right of a decimal place are not significant. 0.006 L 0.08 g 1 (5) All non-filler zeros to the right of a decimal place are significant. 8.0 L 60.40 g 2 4

Significant figures in calculations
Multiplication and division – round your answer to the same number of significant digits as the quantity with the fewest number of significant digits. Addition and subtraction – round your answer to the same number of decimal places as the quantity with the fewest number of decimal places. Find: calculator result: proper result: (1.2 cm)(2 cm) cm cm2 (2.75 cm) cm cm2 5.350 m/2.752 s m/s m/s ( N)(6 m) Nm Nm 1.2 cm + 2 cm cm cm 2000 m+2.1 m m m m – 2.10 m m m

How to report the measurements
measurement = (best estimate ± absolute uncertainty) units Single Measurement (1 trial) measurement = (reading ± absolute uncertainty) unit 𝑥=(𝑥 𝑟𝑒𝑎𝑑𝑖𝑛𝑔 ± ∆𝑥) 𝑢𝑛𝑖𝑡𝑠 ∆𝑥 𝑖𝑠 𝑖𝑛𝑠𝑡𝑟𝑢𝑚𝑒𝑛𝑡𝑎𝑙 𝑢𝑛𝑐𝑒𝑟𝑡𝑎𝑖𝑛𝑡𝑦 Several repeated measurements measurement = (average ± absolute uncertainty) unit 𝑥 𝑎𝑣𝑔 = 𝑥 𝑖 𝑛 ∆𝑥 = 𝑟𝑎𝑛𝑔𝑒 2 = 𝑥 𝑚𝑎𝑥 − 𝑥 𝑚𝑖𝑛 2 ∆𝑥 𝑥 ▪ 𝐹𝑟𝑎𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑢𝑛𝑐𝑒𝑟𝑡𝑎𝑖𝑛𝑡𝑦 (𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑒𝑟𝑟𝑜𝑟)= ▪ 𝑃𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑢𝑛𝑐𝑒𝑟𝑡𝑎𝑖𝑛𝑡𝑖𝑒𝑠(𝑒𝑟𝑟𝑜𝑟)= ∆𝑥 𝑥 100% The value of best estimate must be expressed to the same precision as the uncertainty Uncertainty has 1 sig. fig.

Propagating uncertainties through calculations
If data are to be multiplied or divided, add the fractional or percentage uncertainty: 𝑦 = 𝑎 · 𝑏 𝑐 ∆𝑦 𝑦 = ∆𝑎 𝑎 + ∆𝑏 𝑏 + ∆𝑐 𝑐 𝑎= 2.3±0.2 𝑚 𝑏= 3.2±0.1 𝑚 𝐴=𝑎∙𝑏 ∆𝐴 𝐴 = ∆𝑎 𝑎 + ∆𝑏 𝑏 ∆𝐴= =0.868 𝐴= 7.4±0.9 𝑚 𝐵= 𝑎 𝑏 ∆𝐵 𝐵 = ∆𝑎 𝑎 + ∆𝑏 𝑏 ∆𝐵= =0.0848 𝑎∙𝑏=7.36 𝑎 𝑏 =0.719 ∆𝑎 𝑎 =0.087 ∆𝑏 𝑏 =0.031 𝐵= 0.7±0.1

Absolute uncertainty in V:
A cylinder has a radius of 1.60 ± 0.01 cm and a height of 11.5 ± 0.1 cm. Find the volume. EX: V = π r2 h = π (1.60) 2 x 11.5 = cm2 = 92 cm2 ∆𝑉 𝑉 = ∆𝑟 𝑟 + ∆𝑟 𝑟 + ∆ℎ ℎ =2 ∆𝑟 𝑟 + ∆ℎ ℎ =2× = Absolute uncertainty in V: ∆𝑉= 𝑉= x cm3 = cm3 V = 92 ± 2 cm3

The power (P) dissipated in a resistor of resistance R carrying a current I is equal to I2R. The value of I has an uncertainty of ±2% and the value of R has an uncertainty of ±10%. The value of the uncertainty in the calculated power dissipation is A. ±8%. B. ±12%. C. ±14%. D. ±20%. P= I2R ∆𝑃 𝑃 = ∆𝐼 𝐼 + ∆𝐼 𝐼 + ∆𝑅 𝑅 = 2% + 2% + 10% = 14%. A student measures the length of a line with a wooden meter stick to be 11 mm  1 mm. What is the percentage error or uncertainty in her measurement? Fractional error = 1 / 11 = 0.09. Percentage error = (1 / 11) ·100% = 9% ▪ Thus 1 mm is the absolute error/uncertainty. ▪ 1 mm is also the precision.

A 9.51  0.15 meter rope ladder is hung from a roof that is
12.56  0.07 meters above the ground. How far is the bottom of the ladder from the ground? ▪ y = a – b = = 3.05 m ▪∆y = ∆a + ∆b = = 0.22 m ▪Thus the bottom is 3.1  0.2 m from the ground. A car travels 64.7  0.5 meters in 8.65  0.05 sec. What is its speed? ▪ v= d/t = 64.7 / 8.65 = 7.48 m s-1 ▪∆v/v = ∆d/d + ∆t/t = 0.5/ /8.65 ▪∆v/v = ▪ ∆v/7.48 = ▪ ∆v = 7.48( ) = 0.10 m s-1. ▪ Thus, the car is traveling at 7.5  0.1 m s-1.

▪ ∆F / F = 0.2 / 10 = = 2%. ▪ ∆m / m = 0.1 / 2 = = 5%. ▪ ∆a / a = ∆F / F + ∆m / m = 2% + 5% = 7%. ▪ ∆r / r = 0.5 / 10 = = 5%. ▪ A = r2. ▪ ∆A/A = ∆r/r + ∆r/r = 5% + 5% = 10%.

Topic 2: Mechanics

Classical Mechanics: - Study of the motion of macroscopic objects and causes of that motion and related concepts of force and energy Kinematics – is concerned with the description of how objects move; their motion is described in terms of displacement, velocity , and acceleration Dynamics – explains why objects change the motion; explains changes using concepts of force and energy.

Displacement – vector (𝑠)=𝑚
Distance = length of the curved line Displacement is the distance moved in a particular direction. It is an object's change in position. Displacement: ∆x = x2 – x1 displacement Q Velocity is the rate of change of displacement. vector (𝑣)=𝑚 𝑠 −1 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦= 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛 Speed is the rate of change of distance Scalar (𝑣)=𝑚 𝑠 −1 𝑠𝑝𝑒𝑒𝑑= 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑐𝑜𝑣𝑒𝑟𝑒𝑑 𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛 Acceleration is the rate of change of velocity. 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛= 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛

a = 3 m/s2 means that velocity changes 3 m/s every second!!!!!!
If an object’s initial velocity is 4 m/s then after one second it will be 7 m/s, after two seconds 10 m/s, …. Acceleration can cause: 1. speeding up 2. slowing down 3. and/or changing direction So beware: both velocity and acceleration are vectors. Therefore 1. if velocity and acceleration are in the same direction, speed of the body is increasing. 2. if velocity and acceleration are in the opposite directions, speed of the body is decreasing. 3. If a car changes direction even at constant speed it is accelerating. Why? Because the direction of the car is changing and therefore its velocity is changing. If its velocity is changing then it must have acceleration. There must be a force acting on the car Velocity vector of a particle moving in a circle with speed 10 m/s at two separate points. The velocity vector is tangential to the circle. Average acceleration is: 𝑎 = ∆ 𝑣 𝑡 𝑎= 14.1 𝑚/𝑠 2𝑠 =7 𝑚/ 𝑠 2 𝑎 =7 𝑚/ 𝑠 2 , (𝑜𝑟 45 0 𝑆𝑊)

Average and Instantaneous velocity and speed
Instantaneous velocity is the velocity of something at a specific time. There is no formula for it using algebra. Only if one knows x(t) , then derivative/slope of it would give v(t). Instantaneous speed is the speed at a specific time. The speedometer of a car reveals information about the instantaneous speed of your car. It shows your speed at a particular instant in time. If direction is included you have instantaneous velocity. Average velocity and average speed is calculated over a period of time. Instantaneous speed is the magnitude of the instantaneous velocity BUT Only if distance traveled is equal to magnitude of displacement, average speed is equal to magnitude of average velocity.

Frames of reference A frame of reference is needed to determine either position or velocity of an object 30 𝑚/𝑠 15 𝑚/𝑠 25 𝑚/𝑠 Both vehicle move forward relative to the stationary tree (when ground is frame of reference) Red car travelling at 30 m/s relative to the tree, travels at + 5m/s relative to silver car, so the gap between cars increases by 5 m/s. On the other hand red car travels at −45 m/s relative to blue car.

What can we find from graphs of motion ?

▪ displacement vs. distance
What is: 1. the total distance travelled by the object during the 10.0 second time interval? 2. the displacement covered by the object during the 10.0 second time interval? Total distance = 8m + 8m + 8m = 24 m Displacement = 8m, away from initial position

Interpreting motion graphs
 The area under a velocity-time graph is the displacement.  The area under an acceleration- time graph is the change in velocity.  Velocity is a slope of displacement – time graph. Average velocity Instantaneous velocity  Acceleration is a slope of velocity – time graph. Average acceleration Instantaneous acceleration

▪ Velocity is gradient of displacement–time graph
instantaneous velocities at P and T 𝑣 𝑃 =2𝑚/𝑠 𝑣 𝑇 =−0.52 𝑚/𝑠 average velocity from P to T 𝑣 𝑎𝑣𝑔 = 24−18 29−8 ≈0.3 𝑚/𝑠

▪ The gradient of a velocity–time graph is the acceleration
1𝑠−5𝑠: 𝑎= 50 5 =10 m /s 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔 5𝑠−7𝑠: 𝑎= 0 2 =125 𝑚 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 7𝑠−8𝑠: 𝑎= −50 1 =−50 m /s 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔 ▪ The area under a velocity-time graph is the displacement. 1𝑠−5𝑠: s= 1 2 ×50×5=125 m 5𝑠−7𝑠: 𝑠=50×2=100 𝑚 7𝑠−8𝑠: 𝑠= 1 2 ×50×1=25 m

▪ The area under a acceleration-time graph is change in velocity.
The acceleration vs. time graph for a car starting from rest. Calculate the velocity of the car and hence draw the velocity vs. time graph. 0 s – 2 s: ∆𝑣= 2 𝑚 𝑠 𝑠 =4 𝑚/𝑠 2 s – 4 s: ∆𝑣=0𝑚/𝑠 4 s – 6 s: ∆𝑣= −2 𝑚 𝑠 𝑠 =−4 𝑚/𝑠 The acceleration had a negative value, which means that the velocity is decreasing. It starts at a velocity of 4 m/s and decreases to 0 m/s.

Kinematic equations of motion for uniform acceleration a is constant
𝑥=𝑢𝑡+ 1 2 𝑎 𝑡 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑐𝑜𝑣𝑒𝑟𝑒𝑑 𝑖𝑛 𝑡𝑖𝑚𝑒 𝑡 𝑣=𝑢+𝑎𝑡 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑎𝑓𝑡𝑒𝑟 𝑡𝑖𝑚𝑒 𝑡 𝑣 𝑎𝑣𝑔 = 𝑢+𝑣 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑥= 𝑢+𝑣 2 𝑡 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑐𝑜𝑣𝑒𝑟𝑒𝑑 𝑖𝑛 𝑡𝑖𝑚𝑒 𝑡 𝑣 2 = 𝑢 2 +2𝑎𝑥 𝑡𝑖𝑚𝑒𝑙𝑒𝑠𝑠 t = the time for which the body accelerates a = acceleration u = the velocity at time t = 0, the initial velocity v = the velocity after time t, the final velocity x = the displacement covered in time t

Free fall is vertical (up and/or down) motion of a body where gravitational force is the only or dominant force acting upon it. (when air resistance can be ignored) Gravitational force gives all bodies regardless of mass or shape, when air resistance can be ignored, the same acceleration. For an object in free fall the speed would decrease by 9.8 m/s every second on the way up, at the top it would reach zero, and increase by 9.8 m/s for each successive second on the way down Possible lab methods to measure velocity in real situations (or acceleration) are: Light gates Device that senses when an object when an object cuts through a beam of light. The time for which beam was broken is recorded. Knowing the length of the object, the average speed of the object through the gate can be calculated. OR: to calculate average speed between to gates, two light gates can be used. Several light gates and a computer can be joined together to make direct calculations of velocity or acceleration. Strobe photography A strobe light gives out very brief flashes of light at fixed time intervals. If a camera is pointed at an object and the only source of light is strobe light, then the developed picture will have captured an object’s motion. Ticker timer A ticker timer can be arranged to make dots on a strip of paper at regular time intervals (typically every fiftieth of a second). If the piece of paper is attached to the object, and the object is in free fall, the dots on the strip will have recorded the distance moved by the object in known time.

Solving problems using equations of motion for uniform acceleration
How far will Pinky and the Brain go in 30.0 seconds if their acceleration is 20.0 m s-2? a = 20 m/s2 Given 𝑥=𝑢𝑡+ 1 2 𝑎 𝑡 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑐𝑜𝑣𝑒𝑟𝑒𝑑 𝑖𝑛 𝑡𝑖𝑚𝑒 𝑡 𝑣=𝑢+𝑎𝑡 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑎𝑓𝑡𝑒𝑟 𝑡𝑖𝑚𝑒 𝑡 𝑣 𝑎𝑣𝑔 = 𝑢+𝑣 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑥= 𝑢+𝑣 2 𝑡 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑐𝑜𝑣𝑒𝑟𝑒𝑑 𝑖𝑛 𝑡𝑖𝑚𝑒 𝑡 𝑣 2 = 𝑢 2 +2𝑎𝑥 𝑡𝑖𝑚𝑒𝑙𝑒𝑠𝑠 t = 30 s Given u = 0 m/s Implicit x = ? 𝑥=𝑢𝑡+ 1 2 𝑎 𝑡 2 x = 9000 m How fast will Pinky and the Brain be going at this instant? v = ? 𝑣=𝑢+𝑎𝑡 v = 600 m s-1 How fast will Pinky and the Brain be going when they have traveled a total of m? v = ? 𝑣 2 = 𝑢 2 +2𝑎𝑥 v = 850 m s-1

A stone is thrown vertically up from the edge of a cliff 35
A stone is thrown vertically up from the edge of a cliff 35.0 m from the ground. The initial velocity of the stone is 8.00 m/s (a) How high will the stone get? (b) When will it hit the ground? (c) What velocity will it have just before hitting the ground? (d) What distance will the stone have covered? (e) What is the average speed and average velocity fro this motion? (f) Make a graph to show the variation of displacement with time. (g) Make a graph to show the variation of velocity with time. Take the acceleration due to gravity to be 10 m/s2. g= – 10 ms-2 v2 = u2 + 2ay ⇒ m from top of cliff (b) y = – 35 m y = ut+(g/2)t ⇒ s (c) v = u + gt v= – 27.6 m/s (d) d = 3.2m m + 35 m = 41.4 m average speed = 11.6 m/s average velocity = m/s

A ball is dropped off of the Empire State Building (381 m tall). How fast is it going when (just before) it hits ground? 𝑥=𝑢𝑡+ 1 2 𝑎 𝑡 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑐𝑜𝑣𝑒𝑟𝑒𝑑 𝑖𝑛 𝑡𝑖𝑚𝑒 𝑡 𝑣=𝑢+𝑎𝑡 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑎𝑓𝑡𝑒𝑟 𝑡𝑖𝑚𝑒 𝑡 𝑣 𝑎𝑣𝑔 = 𝑢+𝑣 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑥= 𝑢+𝑣 2 𝑡 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑐𝑜𝑣𝑒𝑟𝑒𝑑 𝑖𝑛 𝑡𝑖𝑚𝑒 𝑡 𝑣 2 = 𝑢 2 +2𝑎𝑥 𝑡𝑖𝑚𝑒𝑙𝑒𝑠𝑠 a = -10 m/s2 Implicit x = -381 m Given u = 0 m/s Implicit v = ? 𝑣 2 = 𝑢 2 +2𝑎𝑥 v = -87 m s-1 How long does it take to reach the ground? t = ? 𝑥=𝑢𝑡+ 1 2 𝑎 𝑡 2 t = 8.7 s

A cheer leader is thrown up with an initial speed of 7 m s-1. How high does she go? 𝑥=𝑢𝑡+ 1 2 𝑎 𝑡 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑐𝑜𝑣𝑒𝑟𝑒𝑑 𝑖𝑛 𝑡𝑖𝑚𝑒 𝑡 𝑣=𝑢+𝑎𝑡 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑎𝑓𝑡𝑒𝑟 𝑡𝑖𝑚𝑒 𝑡 𝑣 𝑎𝑣𝑔 = 𝑢+𝑣 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑥= 𝑢+𝑣 2 𝑡 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑐𝑜𝑣𝑒𝑟𝑒𝑑 𝑖𝑛 𝑡𝑖𝑚𝑒 𝑡 𝑣 2 = 𝑢 2 +2𝑎𝑥 𝑡𝑖𝑚𝑒𝑙𝑒𝑠𝑠 a = -10 m/s2 Implicit u = 7 m s-1 Given v = 0 m/s Implicit height = ? 𝑣 2 = 𝑢 2 +2𝑎𝑥 x = 2.45 m

A ball is thrown upward at 50 m s-1 from the top of the 300-m Millau Viaduct, the highest bridge in the world. How fast does it hit ground? 𝑥=𝑢𝑡+ 1 2 𝑎 𝑡 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑐𝑜𝑣𝑒𝑟𝑒𝑑 𝑖𝑛 𝑡𝑖𝑚𝑒 𝑡 𝑣=𝑢+𝑎𝑡 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑎𝑓𝑡𝑒𝑟 𝑡𝑖𝑚𝑒 𝑡 𝑣 𝑎𝑣𝑔 = 𝑢+𝑣 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑥= 𝑢+𝑣 2 𝑡 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑐𝑜𝑣𝑒𝑟𝑒𝑑 𝑖𝑛 𝑡𝑖𝑚𝑒 𝑡 𝑣 2 = 𝑢 2 +2𝑎𝑥 𝑡𝑖𝑚𝑒𝑙𝑒𝑠𝑠 a = -10 m/s2 Implicit x = -300 m Implicit u = 50 m/s Given v = ? 𝑣 2 = 𝑢 2 +2𝑎𝑥 v = -90 m s-1 How long does it take to reach the ground? v = -90 m s-1 calculated t = ? 𝑣=𝑢+𝑎𝑡 t = 14 s

1. Dr. Tsimberg, a very strong lady, throws a ball upward with initial speed of 20 m/s.
How high will it go? How long will it take for the ball to come back? Givens: Unknowns: u = 20 m/s t = ? g = - 10 m/s y = ? at the top v = 0

2. Mr. Peterson, hovering in a helicopter 200 m above our school suddenly drops his pen.
How much time will the students have to save themselves? What is the velocity/speed of the pen when it reaches the ground? Givens: u = 0 m/s (dropped) g = 10 m/s2 Unknowns: t = ? v = ?

3. Mrs. Radja descending in a balloon at the speed of 5 m/s above our school drops her car keys from a height of 100 m. How much time will the students have to save themselves? What is the velocity of the keys when they reach the ground? t = ? v = ?

4. Dr. Tsimberg, our very strong lady, goes to the roof and throws a ball upward. The ball leaves her hand with speed 20 m/s. Ignoring air resistance calculate a. the time taken by the stone to reach its maximum height b. the maximum height reached by the ball. c. the height of the building is 60 m. How long does it take for the ball to reach the ground? d. what is the speed of the ball as it reaches the ground? d v = u + gt v = 20 – 10 x 6 = – 40 m/s speed at the bottom is 40 m/s

Comparison of free fall with no air resistance and with air resistance
In vacuum In air Air resistance provides a drag force to objects in free fall. There is a drag force in all fluids ▪ The drag force increases as the speed of the falling object increases resulting in decreasing downward acceleration until it reaches zero. ▪ When the drag force reaches the magnitude of the gravitational force, the falling object will stop accelerating and continue falling at a constant velocity. ▪ This is called terminal velocity/speed.

Constant speed in +x dir.
Projectile motion A projectile is an object that has been given an initial velocity by some sort of short-lived force, and then moves through the air under the influence of gravity and friction. Regardless of the air resistance, the vertical and the horizontal components of the motion are independent. Their combined effects produce unique path - parabola. Speeding up in -y dir. Slowing down in +y dir. ay =g downward ay = -g Constant speed in +x dir. ax = 0

Analysing projectile motion
Separation of motion into horizontal and vertical component ▪ax = 0 in the absence of air resistance. ▪ay = 10m/s2 downward in the absence of air resistance. vy = 0 𝜃 0 𝑢 𝑢 𝑥 =𝑢 cos 𝜃 0 𝑢 𝑦 =𝑢 𝑠𝑖𝑛 𝜃 0 vx =ux vy vx =ux vx =ux vy uy vx =ux ux vy Separation of motion into horizontal and vertical component 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛 𝑟 𝑎𝑡 𝑡𝑖𝑚𝑒 𝑡 : 𝑥= 𝑢 𝑥 𝑡 𝑦= 𝑢 𝑦 𝑡+ 𝑔 2 𝑡 2 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑣 𝑎𝑡 𝑡𝑖𝑚𝑒 𝑡 : 𝑣 𝑥 = 𝑢 𝑥 𝑣 𝑦 = 𝑢 𝑦 +𝑔𝑡 ▪The flight time is depends on vertical motion (uy). ▪ The maximum height is on vertical motion (uy). ▪ The horizontal component of initial velocity (ux ) determines range.

horizontal with constant speed and free fall.
Projectile motion is the combination of two independent motion simultaneously: horizontal with constant speed and free fall. After some time t the object is at the position P(x,y), has displacement 𝑟 from initial position and velocity 𝑣 HORIZONTAL MOTION VERTICAL MOTION 𝑢𝑥 = 𝑢 cos⁡ 𝜃 0 𝑢𝑦 = 𝑢 sin⁡ 𝜃 0 𝑣𝑥 =𝑢𝑥 𝑣𝑦 = 𝑢 𝑦 + 𝑔𝑡 𝑣 𝑦 2 = 𝑢 𝑦 2 +2𝑔𝑦 𝑥=𝑢𝑥t 𝑦= 𝑢 𝑦 𝑡+ 𝑔 2 𝑡 𝑦= 𝑢 𝑦 + 𝑣 𝑦 2 𝑡

▪ New peak below and left. ▪Pre-peak greater than post-peak.
© 2006 By Timothy K. Lund

x = uxt vx = ux y = uyt + 𝑔 2 t 2 vy = uy + gt A cannon fires a projectile with a muzzle velocity of 56 ms-1 at an angle of inclination of 150. (a) What are ux and uy? u = 56 m s-1 uy = u sin   = 150 uy = 56 sin 150  = 150 uy = 15 m s-1. ux = u cos  ux = 56 cos 150 ux = 54 m s-1 (b) What are the equations of motion? (FOR VELOCITY AND POSITION) vx = 54 vy = t x = 54t y = 15t - 5t2

(c) When will the ball reach its maximum height?
At the maximum height, vy = 0. vy = t → 0 = t t = 1.5 s (d) How far from the muzzle will the ball be when it reaches the height of the muzzle at the end of its trajectory From symmetry tup = tdown = 1.5 s so t = 3.0 s. x = 54t → x = 54(3.0) x = 160 m vx = 54 vy = t (e) Sketch the following graphs: a vs. t, vx vs. t, vy vs. t: x = 54t y = 15t - 5t2 t ay The only acceleration is g in the y-direction -10 t vx vx = 54 ms-1 , a constant over time. 54 vy = t linear with a negative gradient and it crosses the time axis at 1.5 s. t vy 15 1.5

A stone is thrown horizontally from the top of a vertical cliff of height 33 m as shown. The initial horizontal velocity of the stone is 18 ms-1 and air resistance may be assumed to be negligible. State values for the horizontal and for the vertical acceleration of the stone.  Horizontal acceleration: ax = 0. Vertical acceleration: ay = -10 ms-2. (b) Determine the time taken for the tone to reach sea level. ▪ Fall time determined by the height: y = uyt - 5t 2 → -33 = 0t - 5t2 t = 2.6 s (c) Calculate the distance of the stone from the base of the cliff when it reaches sea level. ▪ find x at t = 2.6 s: x = uxt → x = 18(2.6) x = 47 m. (d) Calculate the angle that the velocity makes with the surface of the sea. vx=ux=18ms-1 tan  = 26/18 vy = uy – 10t =-26ms-1  = 550

A ball is projected from ground level with a speed of 28 ms-1 at an angle of 300 to the horizontal as shown. There is a wall of height h at a distance of 16 m from the point of projection of the ball. Air resistance is negligible. (a) Calculate the initial magnitudes of (i) The horizontal velocity of the ball 𝑢 𝑥 =𝑢 𝑐𝑜𝑠 =24𝑚 𝑠 −1 (ii) The vertical velocity of the ball 𝑢 𝑦 = 𝑢 𝑠𝑖𝑛 =14m 𝑠 −1 (b) The ball just passes over the wall. Determine maximum height of the wall. ▪ the time to the wall is found from 𝑥 x = uxt → t = 16 / 24 = 0.67s y = uyt – 5t 2 y = 14(0.67) – 5(0.67)2 y = 7.1 m

▪ New peak below and left. A ball is kicked at an angle to the horizontal. The diagram shows the position of the ball every 0.5 s. ▪ Pre-peak greater than post-peak. 3.0 s The acceleration of free fall is g=10 ms-1 . Air resistance may be neglected. 3.0 s (a) Using the diagram determine, for the ball (i) The horizontal component of the initial velocity 𝑥= 𝑢 𝑥 𝑡→ 𝑢 𝑥 =𝑥/𝑡=4𝑚/0.50𝑠 𝑢 𝑥 =8m 𝑠 −1 51 0 (ii) The vertical component of the initial velocity 𝑦= 𝑢 𝑦 𝑡−5 𝑡 2 → 𝑢 𝑦 =𝑦/𝑡+5𝑡 𝑢 𝑦 = 24.5m 𝑠 −1 ≈25m 𝑠 −1 (iii) the displacement after 3.0 s. 𝑟 = 𝜃=𝑎𝑟𝑐 tan = 51 0 𝑚𝑎𝑔. 𝑜𝑓 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡= =38𝑚 (b) On the diagram above draw a line to indicate a possible path for the ball if air resistance were not negligible.

the same range is obtained for two projection angles that add up to 900
750 600 450 Projectile thrown with the same speed at 300 and 600 will have the same range. The one at 300 remains in the air for a shorter time. 300 150 Till now we neglected we either neglected or considered case in which air resistance was negligible. When the air resistance is significant, the range of projectile getting much smaller and the path is not true parabola. It is beyond the scope of this course to mathematically attack this problem. IDEAL PATH ACTUAL PATH Air resistance is particularly significant for fast-moving objects. A batted baseball travels only about 60% as far in air as it would in a vacuum. The path is no longer parabola.

Satellites in circular orbit, such as the moon or space station, fall beneath the paths they would follow if there were no gravity – straight line. During each second the moon travels about one km . In this distance it deviates about one millimeter from a straight line due to the earth’s gravitational pull (dotted line). The moon continually falls toward the earth, as do the planets around the sun. A satellite launched with speeds less than 8 km/s would eventually fall to the Earth. A satellite launched with a speed of 8 km/s would orbit the Earth in a circular path. Launched with a greater speed satellite would orbit the Earth in an elliptical path. If launched with too great of a speed, a satellite/projectile will escape Earth's gravitational influences and continue in motion without actually orbiting the Earth. Such a projectile will continue in motion until influenced by the gravitational influences of other celestial bodies. At 8 km/s atmospheric friction would melt a piece of iron (falling stars). Therefore satellites are launched to altitudes above 150 km. Don’t even try to think they are free of Earth’s gravity. The force of gravity at that altitude is almost as strong as it is at the surface. The only reason we put them there is that they are beyond Earth's atmosphere , not beyond Earth’s gravity.

Inertia is resistance an object has to a change of velocity
Mass is numerical measure of the inertia of a body (kg) Weight is the gravitational force acting on an object . W = mg Force is an influence on an object that causes the object to accelerate • 1 N is the force that causes a 1-kg object to accelerate 1 m/s2. Fnet (resultant force) is the vector sum of all forces acting on an object Free Body Diagram is a sketch of a body and all forces acting on it.

Newton’s first law: An object continues in motion with constant speed in a straight line (constant velocity) or stays at rest unless acted upon by a net external force. Object is in translational equilibrium if  no change in velocity Newton’s second law: • Δp is the change in momentum produced by the net force F in time Δt. • Δp = Δ(mv) 1. velocity changes, mass doesn’t change: Δp = mΔv → F = ma If a net force is acting on an object of mass m, object will acquire acceleration Direction of acceleration is direction of the net force. 2. mass changes, velocity doesn’t change: Δp = vΔm → F = ∆m ∆t v Δm/Δt in (kg/s) Newton’s third law: Whenever object A exerts force on object B, object B exerts an equal in magnitude but opposite in direction force on object A

Normal/Reaction force (Fn or R) is the force which is preventing an object from falling through the surface of another body . That’s why normal force is always perpendicular (normal) to the surfaces in contact. when you have to draw FREE BODY DIAGRAM (object and all forces acting on it), there is a requirement to draw as many normal/reaction forces as there are points of contacts . For example if you have a 2-D car (with two wheels) you have to draw two normal /reaction forces. Each on one wheel. Table with two legs – the same thing. Emu with two legs – the same thing Friction force is the force that opposes slipping (relative motion ) between two surfaces in contact; it acts parallel to surface in direction opposed to slipping. Friction depends on type and roughness of surfaces and normal force.

vertical direction : F sin θ + Fn = mg Horizontal direction: F cos θ – Ffr = ma

direction perpendicular to the incline:
Ffr Fn q mg Ffr Fn mg sin q mg cos mg = direction perpendicular to the incline: Fnet = ma = 0 Fn = mg cos θ direction along the incline direction: Fnet = ma mg sin θ – Ffr = ma

Solid friction 0 ≤ Fs ≤ Fs,max Fd < Fs,max Fd = a constant
▪ Friction acts opposite to the intended direction of motion, and parallel to the contact surface. ▪ Suppose we begin to pull a crate to the right, with gradually increasing force. ▪ We plot the applied force, and the friction force, as functions of time: Force Time static dynamic Fs,max Fd applied force friction force time ▪ During the static phase, the static friction force Fs exactly matches the applied force. ▪ The friction force then almost instantaneously decreases to a constant value Fd, called the dynamic friction force. ▪ Fs increases linearly until it reaches a maximum value Fs,max. Properties of the friction force: ▪ 0 ≤ Fs ≤ Fs,max Fd < Fs,max Fd = a constant Ffr ≤ μs R Ffr = μd R static dynamic 𝑅≡ 𝐹 𝑛

Describing solid friction by coefficients of friction R Ff
FBD, coin x y Describing solid friction by coefficients of friction R Ff EXAMPLE: A piece of wood with a coin on it is raised on one end until the coin just begins to slip. The angle the wood makes with the horizontal is θ is = 15°. What is the coefficient of static friction? mg 15° θ = 15° ∑Fy = 0 ∑Fx = 0 R – mg cos 15° = 0 Ff – mg sin 15° = 0 R = mg cos 15° Ff = mg sin 15° Ff = μs R mg sin 15° = μs mg cos 15° mg sin 15° mg cos 15° μs = = tan 15° = 0.268 Coefficient of static friction between the metal of the coin and the wood of the plank is

Describing solid friction by coefficients of friction R Ff
FBD, coin x y Describing solid friction by coefficients of friction R Ff EXAMPLE: Now suppose the plank of wood is long enough so that you can lower it to the point that the coin keeps slipping, but no longer accelerates (v = 0). If this new angle is 12°, what is the coefficient of dynamic friction? mg 15° θ = 12° ∑Fy = 0 ∑Fx = 0 R – mg cos 12° = 0 Ff – mg sin 12° = 0 R = mg cos 12° Ff = mg sin 12° Ff = μd R mg sin 12° = μd mg cos 12° μd = tan 12° = 0.213 Coefficient of dynamic friction between the metal of the coin and the wood of the plank is

Air Resistance (drag force) and Terminal Velocity
When an object moves through air or any other fluid, the fluid exerts drag force on the moving object. Unlike the friction between surfaces, however, this force depends upon the speed of the object, becoming larger as the speed increases. It also depends upon the size and the shape of the object and the density and kind of fluid. A falling object accelerates due to the gravitational force, mg, exerted on it by the earth. As the object accelerates, however, its speed increases and the drag on it becomes greater and greater until it is equal to the weight of the object. At this point, the net force on the falling object is zero, so it no longer accelerates. Its speed now remains constant; it is traveling at its terminal speed. Terminal speed occurs when the weight force (down) is equaled by the drag force (up). Simplification of something that is pretty complicated: we can write the air resistance force (or drag force) as 𝒇𝒅𝒓𝒂𝒈 = 𝒃𝒗 for very small, slow objects, or 𝒇𝒅𝒓𝒂𝒈 = 𝒃𝒗𝟐 for "human-size objects, depending on the situation. “b” is a constant.

Draw all forces that act on a parachutist. Find 𝐹 and acceleration for
a. parachutist that has just stepped out of the airplane. Σ 𝐹 =𝑚 𝑔 𝑎= 𝑚𝑔 𝑚 mg a = g b. parachutist is falling at increasing speed. Σ 𝐹 =𝑚𝑔− 𝐹 𝑎𝑖𝑟 𝑎= 𝑚𝑔− 𝐹 𝑎𝑖𝑟 𝑚 mg 𝐹 air = a < g Σ 𝐹 the speed is still increasing, and therefore air friction too until c. parachutist is traveling downward with constant velocity (terminal velocity) 𝐹 air Fnet = 0 = a = 0 Σ 𝐹 = 0 mg

Contact Forces Field Forces
Although there are many different contact forces, they are all some form of only four different fundamental field forces existing in the nature. Contact Forces Frictional Force Tension Force Normal Force Air Resistance Force – Drag Force Applied Force Spring Force Field Forces Gravitational Force attraction between objects due to their masses Electromagnetic Force between charges Strong Nuclear Force keeps nucleus together Weak Nuclear Force arise in certain radioactive processes At the atomic level – all contact forces are result of repulsive electromagnetic forces (at very small distances) That means that objects have no actual contact, but their electric fields (outer electrons repel each other)

What is the acceleration of root beer?
Howard, the soda jerk at Bea’s diner, slides a 0.60-kg root beer from the end of the counter to a thirsty customer. A force of friction of 1.2 N brings the drink to a stop right in front of the customer. What is the acceleration of root beer? What is the coefficient of kinetic friction between the glass and the counter? If the glass encounters a sticky patch on the counter, will this spot have a higher or lower coefficient of friction? Vertical direction: Horizontal direction: Resultant force = friction force ΣF = Ffr =1.2 N ΣF = ma 1.2 = 0.60 𝑎 𝑎 = 2.0 m/s2 Fn Vertical acceleration = 0 Vertical force ΣF = 0 Ffr 0.60 kg Fn – mg = 0 Fn = mg = 6.0 N mg = 6N Ffr = 𝜇 Fn 𝜇 = Ffr / Fn = 1.2/6.0 𝜇 = (no units) c. higher

A boy exerts a 36-N horizontal force as he pulls a 52-N sled across a cement sidewalk at constant speed. What is the coefficient of kinetic friction between the sidewalk and the metal sled runners? Ignore air resistance. W = mg = 52 N m = 5.2 kg Vertical direction: Vertical acceleration = 0 Vertical ΣF = 0 Fn–mg = 0 Fn = 52 N Horizontal direction: v is constant, a = 0 and ΣF = 0 Ffr = F = 36 N Ffr = μ Fn 𝜇 = Ffr / Fn = 36/52 𝜇 = 0.69

A force of 40. 0 N accelerates a 5. 0-kg block at 6
A force of 40.0 N accelerates a 5.0-kg block at 6.0 m/s2 along a horizontal surface. a. How large is the frictional force? b. What is the coefficient of friction? m = 5.0 kg F = 40.0 N a = 6.0 m/s2 Vertical direction: a = 0, so ΣF = 0 Fn = mg = 50 N → Ffr = μ Fn = 50 μ horizontal direction: a = 6.0m/s2 ΣF.= ma F – Ffr = ma 40.0 – Ffr = 30 Ffr = 10 N Ffr = μ Fn 𝜇 = Ffr / Fn = 10/50 𝜇 = 0.2

Luke Skywalker starts to pull a sled with Princess Leia across a large ice pond with the force of 100 N at an angle of 30.0° with the horizontal (with nails on his shoes). Find normal force and initial acceleration if the weight of sled and Princess Leia is 800 N and the friction force is 40 N. mg = 800 N m = 80 kg F = 100 N Ffr = 40 N free body diagram components vertical direction : ΣF=0 F sin θ + Fn–mg = 0 50 + Fn = 800 Fn = 750 N Horizontal direction: ΣF= ma F cos θ – Ffr = ma 86.6 – 40 = 80 a a = 0.58 m/s2

Perpendicular direction: ΣF = ma = 0 Fn - mg cos θ = 0
A cute panda, m = 60 kg, is sliding down an iced incline 300. The ice can support up to 550 N. Will bear fall through the ice? If the coefficient of the friction is 0.115, what is the acceleration of the bear? m = 60 kg θ = 300 μ = 0.115 g = 10 m/s2 Perpendicular direction: ΣF = ma = 0 Fn - mg cos θ = 0 Fn = 520 N < 550 N ice can support him, but he should not eat too much Parallel direction: ΣF = ma mg sin θ – Ffr = ma 300 – 60 = 60 a a = 4 m/s2 cute panda is speeding up!!!! Ffr = μ Fn = 60 N

Elevator problem EXAMPLE of FBD Newton’s 3. law:
Question: How does the apparent weight of a person in an elevator depend on the motion of that elevator? EXAMPLE of FBD What will the scale show if the elevator is at rest or moving up with constant speed speeding up (𝑎=3𝑚/ 𝑠 2 ↑ ) slowing down (𝑎=3𝑚/ 𝑠 2 ↓ ) m = 65 kg Newton’s 3. law: Force with which the person acts on the scale (reading of the scale) is equal to the normal force on the person. So, if we find normal force we know the reading of the scale, so called APPARENT WEIGHT

+ apparent weight = weight = 650 N apparent weight > weight Fn
1. draw free body diagram 2. choose upward positive 3. apply Newton’s 2. law : ΣF = ma + Fn 1. elevator is at rest or moving up with constant speed Fn – mg = ma = 0 → Fn = mg = 650 N apparent weight = weight = 650 N Fn 2. elevator is speeding up: (𝑎=3𝑚/ 𝑠 2 ↑ ) Fn – mg = ma → Fn = mg + ma = 845 N apparent weight > weight the scale would show more, and you would feel heavier 3. elevator is slowing down: (𝑎=3𝑚/ 𝑠 2 ↓ ) Fn Fn – mg = ma → Fn = mg + ma = 650 N – 195 N = 455 N apparent weight < weight the scale would show less, and you would feel lighter

𝑇 1 𝑇 2 A system of two cables supports a 150-N ball as shown.
a) What is the tension in the right-and cable? b) What is the tension in the horizontal cable? 𝑇 1 𝑇 2 mg=150N x: ΣF = 0 T1 cos 300 – T2 = T2 = 260 N y: ΣF = 0 T1 sin 300 – 150 = T1 = 300 N

ΣF = ma a is down mg – T= ma 1.1 – T = 0.11a second equation
Two blocks are connected by a string and pulley as shown. Assuming that the string and pulley are massless, find a) the magnitude of the acceleration of each block b) tension force on the blocks the same string – the same tension the same acceleration, except that 110 g accelerate down, and 90 g accelerate up. ΣF = ma a is up T – mg = ma T – 0.9 = 0.09a first equation ΣF = ma a is down mg – T= ma 1.1 – T = 0.11a second equation two equations with two unknowns T – 0.9 = 0.09a (1) 1.1 – T = 0.11a (2) (1) + (2): = 0.2 a ⟹ a = 1 m/s2 T = 0.09a ⟹ T = 0.99 N

a) What is the magnitude of the acceleration of the 40-kg block?
A 10-kg block is connected to a 40-kg block as shown in the figure. The surface on which the blocks slide is frictionless. A force of 50 N pulls the blocks to the right. a) What is the magnitude of the acceleration of the 40-kg block? b) What is the magnitude of the tension T in the rope that connects the two blocks? As these two objects are connected with the same rope, tension is the same and acceleration is the same for two objects. ΣF = ma T = 10a a is to the right 50 – T = 40a a is to the right 2 EQS with 2 unknowns 50 – 10a = 40a a = 1 m/s2 T = 10a T = 10 N

Block A, with a mass of 10 kg, rests on a 30° incline
Block A, with a mass of 10 kg, rests on a 30° incline. The coefficient of kinetic friction is The attached string is parallel to the incline and passes over a massless, frictionless pulley at the top. Block B, with a mass of 8.0 kg, is attached to the dangling end of the string. 80N > 50N, so object B is pulling object A up the incline. Friction force is therefore down the incline.mg cos 𝟖𝟕𝑵 𝑻 𝐹 𝑛 =𝑚𝑔 cos 30 0 𝟓𝟎𝑵 𝑻 17N 𝟖𝟕𝑵 𝐹 𝑛 =87𝑁 30 0 𝟏𝟎𝟎𝑵 𝐹 𝑓𝑟 = 𝜇 𝑑 𝐹 𝑛 =17𝑁 80𝑁 𝐵 𝑜𝑏𝑗𝑒𝑐𝑡 :𝑇=80N 𝐴 𝑜𝑏𝑗𝑒𝑐𝑡 :𝑇−67=10𝑎 The acceleration of B is the same as acceleration of A: 𝑎=1.3𝑚/ 𝑠 2

Linear momentum is defined as the product of an object’s mass and its velocity:
p = mv vector! (kg m/s) Impulse is defined as the product of the net force acting on an object and time interval of action: FΔt vector! (Ns) Impulse F∆t acting on an object will produce the change of its momentum Δp: F∆t = ∆p Δp = mv - mu Ns = kg m/s Achieving the same change in momentum over a longer time requires smaller force, and over a shorter time requires greater force. WHEN YOU TRY TO FIND CHANGE IN MOMENTUM REMEMER TO LABEL VELOCITIES AS POSITIVE OR NEGATIVE The impulse of a time-varying force is represented by the area under the force-time graph.

Law of Conservation of Momentum: The total momentum of a system of interacting particles is conserved - remains constant, provided there is no resultant external force. Such a system is called an “isolated system”. momentum of the system after collision = momentum of the system before collision for isolated system (p = pi ) vector sum REMEMBER TO DRAW A SKETCH OF THE MASSES AND VELOCITIES BEFORE AND AFTER COLLISION. LABEL VELOCITIES AS POSITIVE OR NEGATIVE. Elastic collision: both momentum and kinetic/mechanical energy are conserved. That means no energy is converted into thermal energy Inelastic collision: momentum is conserved but KE is not conserved. Perfectly inelastic collision: the most of KE is converted into other forms of energy when objects after collision stick together. To find how much of KE is converted into thermal energy, subtract KE of the system after collision from KE of the system before the collision. If explosion happens in an isolated system momentum is conserved but KE increases (input of energy from a fuel or explosive material.)

6.7 Ns 𝑝 𝑓𝑖𝑛𝑎𝑙 = 𝑝 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 +𝐹∆𝑡= 5 kg m/s – 6 kg m/s = – 1 kg m/s
A possible force vs. time curve for a ball struck by a bat is shown in the figure. (a) Estimate the impulse delivered to the ball. (b) This 0.25 kg ball was initially moving toward the bat at a speed of 20 m/s. Calculate the exit speed of the ball. 6.7 Ns 𝑝 𝑓𝑖𝑛𝑎𝑙 = 𝑝 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 +𝐹∆𝑡= 5 kg m/s – 6 kg m/s = – 1 kg m/s V = – 4 m/s

A 60. 0-kg astronaut is on a space walk when her tether line breaks
A 60.0-kg astronaut is on a space walk when her tether line breaks. She throws her 10.0-kg oxygen tank away from the shuttle with a speed of 12.0 m/s to propel herself back to the shuttle. What is her velocity? after before 12.0 m/s 60 70 10 v1 = ? u = 0 pbefore = pafter 0 = m1 v1 + m2 v2 0 = 60.0 v (12.0) v1 = − 2.0 m/s moving in the negative direction means toward shuttle

There are two fish in the sea. A 6 kg fish and a 2 kg fish
There are two fish in the sea. A 6 kg fish and a 2 kg fish. The big fish swallows the small one. What is its velocity immediately after lunch? the big fish swims at 1 m/s toward and swallows the small fish that is at rest. before lunch after lunch Net external force is zero. Momentum is conserved. 1 m/s v = ? p before lunch = p after lunch momentum is vector, direction matters; choose positive direction in the direction of big fish. Mu1 + mu2 = (M + m)v + (6 kg)(1 m/s) + (2 kg)(0 m/s) = (6kg + 2 kg) v 6 kg m/s = (8 kg) v v = 0.75 m/s in the direction of the large fish before lunch

+ b. Suppose the small fish is not at rest but is swimming toward the large fish at 2 m/s. before lunch after lunch - 2 m/s 1 m/s v = ? p before lunch = p after lunch Mu1 + mu2 = (M+m)v (6 ) (1 ) + (2 ) (—2 ) = ( ) v 6 — 4 = 8 v v = 0.25 m/s in the direction of the large fish before lunch The negative momentum of the small fish is very effective in slowing the large fish.

+ c. Small fish swims toward the large fish at 3 m/s. before lunch after lunch - 3 m/s 1 m/s v = ? p before lunch = p after lunch Mu1 + mu2 = (M+m)v (6 ) (1 ) + (2 ) (—3 ) = ( ) v 6 — 6 = (8 ) v v = 0 m/s fish have equal and opposite momenta. Zero momentum before lunch is equal to zero momentum after lunch, and both fish come to a halt.

+ d. Small fish swims toward the large fish at 4 m/s. before lunch after lunch - 4 m/s 1 m/s v = ? p before lunch = p after lunch Mu1 + mu2 = (M+m)v (6 ) (1 ) + (2 ) (—4) = (6 + 2 ) v 6 — 8 = 8 v v = — m/s The minus sign tells us that after lunch the two-fish system moves in a direction opposite to the large fish’s direction before lunch.

A red ball traveling with a speed of 2 m/s along the x-axis hits the eight ball. After the collision, the red ball travels with a speed of 1.6 m/s in a direction 37o below the positive x-axis. The two balls have equal mass. At what angle will the eight ball fall in the side pocket? What is the speed of the blue (8th) ball after collision. v2 before collision: after collision: 8 u2 = 0 u1 θ2 8 370 the point of collision pbefore = pafter v1 in x – direction m u1 + 0 = m v1 cos m v2 cos q2 v2 cos q2 = u1 - v1 cos 370 = 0.72 m/s (1) in y – direction = - m v1 sin m v2 sin θ2 v2 sin θ2 = v1 sin 370 = 0.96 m/s (2) direction of v2 ; (2)/(1) tan θ2 = θ2 = 530 (2) → v2 = 0.96 / sin v2 = 1.2 m/s

Work is the product of the component of the force in the direction of displacement and
the magnitude of the displacement. (scalar) W = Fd cos Ѳ (Fd = F cos Ѳ) (Joules) Work done by a varying force F along the whole distance travelled is the area under the graph FcosѲ versus distance travelled. Energy is the ability to do work. Work changes energy. Potential energy is stored energy. (Change in) Gravitational potential energy ∆EP = mg ∆h Elastic potential energy = work is done by external force in stretching/compressing the spring by extension x. EPE = W = ½ kx2 Kinetic energy is the energy an object possesses due to motion EK = ½ mv2

Work done by applied/external force is converted into (changes) potential energy
(when net force is zero, so there is no acceleration). Work – Kinetic energy relationship: work done by net force changes kinetic energy: W = ∆KE = final EK – initial EK = ½ mv2 – ½ mu2 the work done by centripetal force is zero: Wnet = 0 → Wnet = ∆KE → no change in KE → no change in speed; centripetal force cannot change the speed, only direction. Examples: gravitational force on the moon, magnetic force on the moving charge.

Work done by a force F is zero if:
θ = 00 00< θ <900 θ = 900 900< θ <1800 θ = 1800 cos θ = 1 cos θ = + cos θ = 0 cos θ = – cos θ = –1 Work done by a force F is zero if: force is exerted but no motion is involved: no distance moved, no work force is perpendicular to the direction of motion (cos 900 = 0) d F F d F motion normal force tension in the string gravitational force

Example: The graph shows how the length of a spring varies with applied force. i. State the value of the unstretched length of the spring. ii. Use data from the graph to plot another graph of force against extension and from this graph determine the spring constant. i. when the applied force is zero the length of the spring is 20 cm, therefore unstretched length is 20 cm. ii. F/N x/cm

Conservation of energy law: Energy cannot be created or destroyed;
it can only be changed from one form to another. For the system that has only mechanical energy (ME = potential energies + kinetic energy) and there is no frictional force acting on it, so no mechanical energy is converted into thermal energy, mechanical energy is conserved ME1 = ME2 = ME3 = ME mgh1 + ½ mv12 = mgh2 + ½ mv22 = • • • • • • f friction cannot be neglected we have to take into account work done by friction force which doesn’t belong to the object alone but is shared with environment as thermal energy. Friction converts part of kinetic energy of the object into thermal energy. Frictional force has dissipated energy: ME1 – Ffr d = ME (Wfr = – Ffr d)

𝑃= ∆𝐸 𝑡 = 𝑊 𝑡 𝑃=𝐹𝑣 𝑃= 𝑊 𝑡 = 𝐹𝑑 𝑡 =𝐹 𝑑 𝑡
Power is the rate at which work is performed or the rate at which energy is transmitted/converted. scalar (1 W(Watt) = 1 J/ 1s ) 𝑃= ∆𝐸 𝑡 = 𝑊 𝑡 Another way to calculate power 𝑃= 𝑊 𝑡 = 𝐹𝑑 𝑡 =𝐹 𝑑 𝑡 𝑃=𝐹𝑣 Efficiency is the ratio of how much work, energy or power we get out of a system compared to how much is put in.

A father pushes his child on a sled on level ice, a distance 5 m from rest, giving a final speed of 2 m/s. If the mass of the child and sled is 30 kg, how much work did he do? W = ∆ KE = ½ m v2 – 0 = ½ (30 kg)(2)2 = 60 J other way: W = Fd cosθ = Fd F = ma F = 12 N W = 60 J What is the average force he exerted on the child? W = Fd = 60 J, and d = 5 m, so F = 60/5 = 12 N

A 1000-kg car going at 45 km/h. When the driver slams on the brakes, the road does work on the car through a backward-directed friction force. a. How much work must this friction force do in order to stop the car? Instead of slamming on the brakes the work required to stop the car is provided by a tree!!! b. What average force is required to stop a 1000-kg car going at 45 km/h if the car collapses one foot (0.3 m) upon impact? c. What is acceleration

A 1000-kg car going at 45 km/h. When the driver slams on the brakes, the road does work on the car through a backward-directed friction force. How much work must this friction force do in order to stop the car? W = ∆ KE = 0 – ½ m u2 = – ½ (1000 kg) (45 x1000 m/3600 s)2 W = – J = – 78 kJ (the – sign just means the work leads to a decrease in KE) d u Ffr (work done by friction force) v = 0

W = – 78125 J The net force acting on the car is F F = 260 x 103 N
Instead of slamming on the brakes the work required to stop the car is provided by a tree!!! What average force is required to stop a 1000-kg car going at 45 km/h if the car collapses one foot (0.3 m) upon impact? W = – J W = – F d = – ½ m u2 The net force acting on the car is F F = 260 x 103 N corresponds to 29 tons hitting you OUCH Do you see why the cars should not be rigid. Smaller collapse distance, gretaer force, greater acceleration. For half the distance force would double!!!! OUCH, OUCH a = (v2 - u2 )/2d = m/s HUGE!!!!

Dropping down from a pole. As he dives,
PE becomes KE. Total energy is always constant, equal to initial energy. If accounted for air resistance, then how would the numbers change? In presence of air, some energy gets transformed to heat (which is random motion of the air molecules). Total energy at any height would be PE + KE + heat, so at a given height, the KE would be less than in vacuum. What happens when he hits the ground? Just before he hits ground, he has large KE (large speed). This gets transformed into heat energy of his hands and the ground on impact, sound energy, and energy associated with deformation .

h mgh + ½ mu2 = ½ mv2 m cancels
Three balls are thrown from the top of the cliff along paths A, B, and C with the same initial speed (air resistance is negligible). Which ball strikes the ground below with the greatest speed? a. A b. B c. C d. All strike with the same speed h The initial PE + KE of each ball is: mgh + ½ mu2. This amount of energy becomes KE before impact ½ mv2. mgh + ½ mu2 = ½ mv m cancels The speed of impact for each ball is the same: It depends ONLY on HEIGHT and initial SPEED, not mass, not path !!!!!

A child of mass m is released from rest at the top of a water slide, at height h = 8.5 m above the bottom of the slide. Assuming that the slide is frictionless because of the water on it, find the child’s speed at the bottom of the slide. Do you think you could use kinematics equations and Newton’s laws? Try it. Good luck ! mgh = ½ mv2 m cancels on both sides a baby, an elephant and you would reach the bottom at the same speed !!!!! And, by the way, it is the same speed as you were to fall straight down from the same height. v2 = u2 + 2gh = 2gh

A car (toy car – no engine) is at the top of a hill on a frictionless track. What must the car’s speed be at the top of the first hill if it can just make it to the top of the second hill? v2 = 0 PE1 + KE1 = PE2 + KE2 v1 mgh1 + ½ mv12 = mgh2 + ½ mv22 m cancels out ; v2 = 0 gh1 + ½ v12 = gh2 40m 20m v12 = 2g( h2 – h1) v1 = 22 m/s

as shown. Its speed at the lowest point B is:
A simple pendulum consists of a 2.0 kg mass attached to a string. It is released from rest at A as shown. Its speed at the lowest point B is: 1.85 m A B PEA + KEA = PEB + KEB

on the object as it slides down.
● Example: An object of mass 4.0 kg slides 1.0 m down an inclined plane starting from rest. Determine the speed of the object when it reaches the bottom of the plane if a. friction is neglected b. constant friction force of 16 N acts on the object as it slides down. all PE was transformed into KE Friction converts part of KE of the object into heat energy. This energy equals to the work done by the friction. We say that the frictional force has dissipated energy. initial energy = final energy initial energy – Ffr d = final energy Wfr = – Ffr d decreases KE of the object PEA = KEB PEA – Ffr d = KEB mgh = ½ mv mgh – Ffr d = ½ mv2 20 – 16 = 2.5 v2 v = 1.4 m/s

There is another way to calculate power
Calculate the power of a worker in a supermarket who stacks shelves 1.5 m high with cartons of orange juice, each of mass 6.0 kg, at the rate of 30 cartons per minute. P = 45 W There is another way to calculate power Power is equal to force times velocity, providing that both force and velocity are constant and in the same direction. Constant velocity with a force applied ?????? That’s because we are interested in one force only. Net force is obviously zero. Like power of the engine of the car.

How much energy is converted into heat per second.
A car engine has an efficiency of 20 % and produces an average of 25 kJ of useful work per second. How much energy is converted into heat per second. Ein = J heat = 125 kJ – 25 kJ = 100 kJ

Centripetal acceleration causes change in direction of velocity, but doesn’t change speed. It is always directed toward the center of the circle. Centripetal force Period T: time required for one complete revolution/circle speed around circle of radius r:

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