Physics G10 Notes PDF

CONTENTS 1. GENERAL PHYSICS ······················································································ 1 1.1. Quantity and Unit ··········································································· 1 1.2. Length and Time ·········································...

CONTENTS 1. GENERAL PHYSICS ······················································································ 1 1.1. Quantity and Unit ··········································································· 1 1.2. Length and Time ············································································ 2 1.3. Speed, Velocity and Acceleration ························································· 4 1.4. Mass and Weight ··········································································· 10 1.5. Volume and Density ······································································· 12 1.6. Force························································································· 13 1.7. Moment ····················································································· 16 1.8. Work, Energy and Power ································································· 18 1.9. Simple Machines··········································································· 22 2. THERMAL PHYSICS ···················································································· 25 2.1. Kinetic theory ·············································································· 25 2.2. Thermal properties········································································· 27 2.3. Gas laws ···················································································· 30 2.4. Transfer of thermal energy································································ 32 3. PROPERTIES OF WAVES··············································································· 36 3.1. General waves ·············································································· 36 3.2. Sound ························································································ 38 3.3. Light ························································································· 39 4. ELECTRICITY ···························································································· 46 4.1. Static electricity ············································································ 46 4.2. Electric circuit ·············································································· 47 4.3. Practical electricity circuit ································································ 52 5. MAGNETISM ····························································································· 55 5.1. Simple phenomenon of magnetism ······················································ 55 5.2. Electromagnetic effect ···································································· 57 6. INTRODUCTORY ELECTRONICS ··································································· 62 6.1. Electron ····················································································· 62 6.2. C.R.O. ······················································································· 62 7. ATOMIC PHYSICS ······················································································· 64 7.1. Nuclear atom ··············································································· 64 7.2. Radioactivity ··············································································· 65  1. GENERAL PHYSICS 1.1. Quantity and Unit Physical quantities There are many physical quantities in Physics. These Physical quantities can be divided into two types as shown below; Type of quantities Base quantity Derived quantity Mass Speed Length Volume Example Time Area Current Force Temperature They have only one SI unit. They can be expressed by combining Explanation suitable base quantities. SI unit The value of a physical quantity is written as a number by a suitable unit. The International System of Units is adopted in Physics. The following table shows some of SI units. Physical quantity SI unit Symbol for unit Example Length metre m 100m Mass kilogram kg 60kg Time second s 30s Current Ampere A 15A Temperature Kelvin K 150K (Basically, units of quantities are shown as SI units or derived SI units in this textbook.) Prefixes Sometime a physical quantity is too big or too small to be conveniently expressed in SI units. Then some symbols are used as the prefixes instead of Zeros or many places. Prefixes are multiples or decimals of ten. The following table shows some prefixes Prefixes Symbol Exponent Meaning Example Mega M 106 1,000,000 3Mm = 3,000,000m kilo K 103 1,000 5km = 5,000m centi C 10-2 1/100 (=0.01) 2cm = 0.02m Milli M 10-3 1/1000 (=0.001) 6mm = 0.006m Micro Μ 10-6 1/1000000 (=0.000001) 7μm = 0.000007m Scalar and Vector Definition A scalar is a quantity having magnitude only. (e.g. mass, length, area, volume, density, time, distance, speed, energy, temperature, current, voltage...) Definition A vector is a quantity having both magnitude and direction. (e.g. weight, displacement, velocity, acceleration, force, moment...) 1 1.2. Length and Time Length Definition Length is defined as the measurement of something from one end to the other. SI unit is metre. The symbol of unit is m. Instruments for measuring length Instrument Uses for length Accuracy Example of measured objects Measuring tape Long length 1mm Length of classroom, Height of building Ruler Medium length 1mm Width of paper, Length of pen Vernier calipers Short length 0.1mm Diameter of pen, Internal diameter of tube Micrometer Very short length 0.01mm Diameter of hair, Thickness of razor blade screw gauge Correct (1.3cm) Ruler and Measuring tape Wrong (1.2cm) Wrong (1.4cm) How to use the ruler or the measuring tape 0 mark (1) Put the 0 mark on the end of the object. 0 2 (2) Read the mark at the other end of the object. Caution to use; Object The eye must be placed vertically above the mark on the scale. Inside jaws Vernier calipers A pair of vernier calipers is shown in the diagram. Stem The outside jaws are usually used to measure lengths of something such as external diameter. And inside jaws are used to measure internal diameter of a tube or cylinder. Main scale Vernier scale How to use the vernier calipers (1) Put an object to be measured between jaws. Outside jaws (2) Read the main scale before the 0 mark of vernier scale. (3) Look at the vernier scale and find a marking on the vernier scale that is in line with the main scale. (Commonly, the reading on the vernier scale is for the 2nd place of decimal in centimetre.) (4) Add the main scale reading and the vernier scale reading. [Example] Step (2) Main scale reading = 1.2cm (before the 0 mark of vernier scale.) 1 2 0 5 Step (3) Step (1) Vernier scale reading = 0.05cm Measured object (in line with the main scale) Step (4) Main scale reading + vernier scale reading = 1.2 + 0.05 = 1.25 cm 2  anvil Measured sleeve thimble Micrometer screw gauge object A micrometer screw gauge is shown as the diagram. How to use the micrometer screw gauge spindle ratchet Main scale (1) Thimble is turned until the object is Circular scale gripped between the anvil and the spindle very gently. frame (2) Read the main scale on the sleeve before the edge of thimble. (3) Look at the circular scale on the thimble. Find a marking on the circular scale that is in line with the horizontal line of the main scale. (Commonly, each division of circular scale represents a length of 0.01mm) (4) Add the main scale reading and the circular scale reading. [Example] Circular scale 15 Step (3) 1 2 3 4 Circular scale reading = 0.12 mm Main scale (in line with the horizontal line of Step (2) 10 the main scale.) Main scale reading = 4.5 mm (before the edge of thimble.) Step (4) Main scale reading + circular scale reading = 4.5 + 0.12 = 4.62 mm Time SI unit: second (The symbol is s) Other units: minute, hour, day, month, year, century Conversion of the unit 1year = 365days = 8760hours = 525600minutes = 31536000s 1 day = 24hours = 1440minutes = 86400s 1hour = 60minutes = 3600s 1minute = 60s Instrument for measuring the time → Clock, Watch, Stopwatch, Pendulum Simple pendulum The diagram below shows a simple pendulum with length l. The length l should be from the ceiling to the centre of the bob. Definition Period (T) is defined as the time taken for one complete oscillation. (time taken from A to C and back again to A.) l Formula t T= n A C T: Period [s] n: number of oscillation B t: time taken for n oscillation [s] Experiment : What changes the period of pendulum? One complete oscillation ①standard ②shorter arc ③heavier bob ④shorter string 3  n: number of t: time taken T: period [s] oscillations [s] (=t/n) ① standard ② shorter arc ③ heavier bob ④ shorter string Conclusion: Period of pendulum depends on... → the length of the pendulum (l). → acceleration due to gravity (g). [Example] The diagram shows a pendulum oscillating between position A and C. It takes 3s to go from A to C and back to B. What is its period? DATA Solution 3 t 3 4 A C t = 3s n = (three quarter oscillations) T = = t÷n = 3÷ = 3× = 4s 4 n 4 3 B [EXERCISE] (1) Vernier callipers are used to measure wooden cubes (a) and (b) as shown below. What is the width of the cubes? (a) 10 [cm] 11 (b) [cm] 6 0 5 0 5 (2) Read the measurements of micrometer screw gauge below. (a) 40 (b) (c) 30 3 4 5 11 12 13 5 10 11 35 25 [mm] [mm] 0 [mm] 30 (3) Find the period of a pendulum if it oscillates 15 times completely for 45 seconds. (4) The diagram shows a pendulum oscillating between positions A to C. A C (a) It takes 3s to go from A to C. What is its period? B (b) How long does it take for 12 times complete oscillations? 1.3. Speed, Velocity and Acceleration Distance and Displacement Definition Distance is defined as the total length taken between two points. → It is a scalar. Definition Displacement is defined as the change of position of a point in a particular direction. → It is a vector. SI units of both distance and displacement are metre [m]. [Example 1] A car moves 5km to the East and 3km to the North. N What is the distance and the displacement of the car? 3km → Distance of the car is 8km (= 5km +3km). W E 5km → Displacement of the car is 5km East and 3km North. Car Car S 4 [Example 2] The circumference of a roundabout is 10m and the car turns it once. What is the distance and the displacement of the car? → Distance of the car is 10m. → Displacement of the car is 0m because it came back to the starting position. Car [Example 3] Your distance You walk forward 15m and backward 5m. 15m What is your distance and your displacement? → Your distance is 20m (= 15m + 5m). 5m Your displacement → Your displacement is 10m (= 15m - 5m) forward. Speed Definition Speed is defined as the rate of change of distance traveled with time. Various speeds (m/s) → It is a scalar. -man’s walking: 1.5 The unit of speed is metre per second [m/s]. -100m runner: 10 Formula Distance traveled -Zoom bus: 30 Speed = [m/s] Time taken -baseball thrown by a pitcher: 42 Total distance traveled Average Speed = [m/s] -sound in air (20℃): 340 Total time taken -Light: 300000000 [Example] A car travels a distance of 540km from Lusaka to Katete in 10 hours. Find the average speed in km/hr and m/s. DATA Solution Total distance traveled = 540km Total distanve traveled 540km = 540,000m Average speed in km/hr = = = 54km/hr Total time taken 10hrs Total time taken = 10Hrs 540000m Average speed in m/s = = 15m/s = 36000s 36000s Velocity Definition Velocity is defined as the rate of change of displacement with time. → It is a vector. The unit of velocity is metre per second [m/s]. (It is the same unit as speed.) [Example] N What are their speeds and their velocities? 10m/s → They have the same speeds of 10m/s but they have different velocities. Car2 W E 10m/s → Car1 has the velocity of 10m/s East. Car1 S → Car2 has the velocity of 10m/s North. 5 Acceleration Definition Acceleration is defined as the rate of change of velocity with time. → It is a vector. The unit of acceleration is metre per second squared [m/s2]. Formula v-u a= t a: Acceleration [m/s2] v: final velocity [m/s] u: initial velocity [m/s] t: time taken [s] [Example 1] A car starting from rest increases its velocity uniformly to 15m/s in 5s. What is its acceleration? DATA Solution 5s u = 0m/s v-u 15 - 0 a= = = 3m/s2 a=? v = 15m/s t 5 rest (u=0m/s) v = 15m/s t = 5s a=? [Example 2] If a car slows down from 72km/hr and stops in 10s, calculate the acceleration. DATA Solution 10s u = 72km/hr = 20m/s v-u 0 - 20 v = 0m/s a= = = -2m/s2 a=? Stop (v=0m/s) t 10 u = 72km/hr t = 10s a=? 72km 72000m 20m When the velocity reduces, the acceleration becomes a negative number. 72km/hr = = = = 20m/s The acceleration is called the retardation or deceleration. 1hr 3600s 1s Uniformly accelerated liner motion If a body moves with a uniform acceleration (the acceleration is constant), three important equations are given below. Formula v = u + at 1 x = ut + at2 2 2 2 v = u + 2ax a: Acceleration [m/s2] v: final velocity [m/s] u: initial velocity [m/s] t: time taken [s] x: distance covered [m] [Example 1] A car traveling at 10m/s accelerates at 2m/s2 for 3s. What is its final velocity? DATA Solution 3s u = 10m/s a = 2m/s2 v = u +at = 10 + 2×3 a = 2m/s2 t = 3s v=? = 16m/s u = 10m/s v=? [Example 2] A motorcycle starting from rest acquires a velocity of 72km/hr in 5s. (a) What is its acceleration? (b) How far does it travel during this time? 6  7200m 3600s (a) DATA Solution v = 72km/hr = 20m/s v-u 20 - 0 a= = = 4m/s2 5s u = 0m/s t 5 t = 5s a=? rest (u=0m/s) a = ? (a) v = 72km/hr (b) DATA Solution x = ? (b) u = 0m/s 1 2 1 x = ut + at = 0×5+ ×4×52 t = 5s 2 2 a = 4m/s2 = 50m x=? Don’t use the formula: D=S×T. Because there is acceleration. [Example 3] The velocity of an object is uniformly reduced from 50m/s to 30m/s. If the deceleration is –4m/s2, how much is the distance of the body decelerating? DATA Solution 2 2 u = 50m/s v = u +2ax a = -4m/s2 v = 30m/s v2 - u2 302 - 502 -1600 u = 50m/s v = 30m/s x= = = a = -4m/s2 2a 2×(-4) -8 x=? x=? = 200m Acceleration due to gravity All objects accelerate uniformly towards the earth if air resistance is ignored. It is called acceleration due to gravity. It is represented by the symbol ‘g’. g = 9.8m/s2 ≈ 10m/s2 If a stone is dropped from the top of a tall building, it accelerates uniformly downwards. Free fall If you release a stone without applying force, it starts from rest. It is called free fall. → Free fall Stop momentarily u = 0m/s a = g = 10m/s2 If you throw up a stone, the stone decelerates to the top. Throwing Free fall Then it stops momentarily at the top. And then it starts falling freely. up → Throwing up v = 0m/s a = -g = -10m/s2 [Example 1] A body falls freely from rest. Air resistance is ignored. (g = 10m/s2) (a) What is its velocity after 1s? (b) How far does it reach in 1s? (a) DATA Solution rest (u = 0m/s) u = 0m/s v = u + at Free fall a = g = 10m/s2 = 0 + 10×1 t = 1s = 10m/s x =?(b) 1s 10m/s2 v=? (b) DATA Solution u = 0m/s Free fall 1 1 v = ?(a) x = ut + at2 = 0×1+ ×10×12 a = g = 10m/s2 2 2 t = 1s = 5m x=? Try to solve for x using the formula : v2 = u2 +2ax 7  [Example 2] A stone is thrown upward with an initial velocity of 20m/s. Air resistance is ignored. (g = 10m/s2) (a) How far does it take to reach the top? (b) How long does it take to reach the top? (c) What is its velocity just before reaching the ground? (d) How long does it take to reach the ground? (a) DATA Solution u = 20m/s v2 = u2 +2ax v = 0m/s u = 0m/s v = 0m/s throwing v2 - u2 02 - 202 -400 a = -g = -10m/s2 x= = = up 2a 2×(-10) -20 Throwing x=? = 20m up Free fall (b) DATA Solution x = ?(a) u = 20m/s v = u + at t = ?(b) v = 0m/s throwing v-u 0-20 a = -g = -10m/s2 t = = up a -10 u = 20m/s v = ?(c) t=? = 2s Total time taken =? (d) (c) DATA Solution 2 2 x = 20m v = u +2ax u = 0m/s Free fall = 02 + 2×10×20 a = g = 10m/s2 =400 v=? v = 20m/s Total time taken (t) =t1 +t2 (d) DATA Solution t1: time taken from the ground to the top u = 0m/s v = u + at2 t2: time taken from the top to the ground a = g = 10m/s2 free fall v-u 20-0 v = 20m/s t2 = = = 2s a 10 t1 = 2s, t2 = ?, t = ? t = t1 + t2 = 2+2 = 4s Speed (velocity) – time graph – Speed (velocity) – time graphs tell stories about the movement of an object. – The gradient of the speed – time graph is equal to the acceleration of the object. – The area under the speed – time graph represents the distance traveled by the object. The diagrams below show the speed – time graphs for different kinds of motion. (a) Constant speed (no acceleration) (b) Uniform acceleration speed speed Horizontal straight line Straight line sloping upward to the right time time (c) Uniform deceleration (d) Non-uniform acceleration (or deceleration) speed speed Straight line sloping downward to the right Curved line time time [Example] A car moving from rest acquires a velocity of 20m/s with uniform acceleration in 4s. It moves with this velocity for 6s and again accelerates uniformly to 30m/s in 5s. It travels for 3s at this velocity and then comes to rest with uniform deceleration in 12s. (a) Draw a speed – time graph. (b) Calculate the total distance covered. 8 (c) Calculate the average speed. constant constant accelerate speed accelerate speed 30m/s decelerate rest 20m/s 20m/s 30m/s stop 4s 6s 5s 3s 12s (a) Speed 30 20 10 A B C D E time 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 (b) To find the total distance covered, calculate the area under the speed – time graph. 1 1 Area A = triangle = bh = ×4×20 = 40 Turn it ! 2 2 Area B = rectangle = lb = 6×20 = 120 20 C 30 20 ←a 1 1 Area C = trapezium = (a+b)h = (20 +30)×5 = 125 2 2 5 h→ 5 C Area D = rectangle = lb = 3×30 = 90 1 1 30 ←b Area E = triangle = bh = ×12×30 = 180 2 2 Total distance covered = Total area = 40 + 120 + 125 + 90 + 180 = 555m (c) DATA Solution Total distance covered = 555m Total distance covered 555 Total time taken = 30s Average speed = = = 18.5m/s Total time taken 30 Average speed = ? [EXERCISE] (1) You walk from Chisale to St. Francis, a distance of 6km, in 50 minutes. Find the average speed in m/s. (2) A man rides on a bicycle. He accelerates from rest to 8m/s in 5s. What is his acceleration? (3) A man drives a car at 54km/hr. He brakes and it stops in 3s. Calculate the deceleration. (4) A car starting from rest accelerates uniformly at 5m/s2 in 3s. (a) Calculate the final velocity. (b) Calculate the distance covered. (5) A car accelerated uniformly from 10m/s to 20m/s. It traveled a distance of 50m during this time. (a) What was the acceleration of the car? (b) How long did it take to travel this distance? (6) A stone released from the top of a building takes 3s to reach the ground. The air resistance is ignored. (a) What is the final velocity? (b) How tall is the building? (7) A car starting from rest accelerates uniformly to 20m/s in 5s. And it accelerates more to 40m/s in 2s. And then it decelerates until it stops 8s later. (a) Draw the speed–time graph. (b) Calculate the deceleration. (c) Calculate the total distance traveled. (d) Calculate the average speed. [TRY] Explain the reason why a piece of paper falls more slowly than a stone, although both of them are on the Earth and supposed to have the same acceleration: 10m /s2. 9  Beam balance 1.4. Mass and Weight Mass Known mass Measured object object Definition Mass is defined as the quantity of matter in a substance. → The mass of an object is the same everywhere. The unit of mass is kilogram [kg]. Read the mass of known mass object when Instrument for measuring the mass → Beam balance the beam is balanced. Weight Spring balance Definition Weight is defined as the attractive force exerted on an object by gravity. → The weight of an object varies from place to place. (e.g. from the earth to the moon) Pointer The unit of weight is Newton [N]. Instrument for measuring the weight → Spring balance, Bathroom scale Scale Spring Relationship between mass and weight The weight of an object is directly proportional to its mass. Measured object Formula w = mg w: weight [N] Read the scale at the pointer. m: mass [kg] g: acceleration due to gravity[m/s2]  Gravity of the moon is only about 1/6 that of the earth. 1 g = 10×6 m/s2 (≈ 1.67 m/s2) on the moon  No gravity in outer space. g = 0 m/s2 in the outer space The following table shows the changes to the mass and weight of an astronaut when he travels from the earth to the moon in a spacecraft. Earth Moon Outer space Mass 60kg 60kg 60kg Weight 600N 100N 0N [Example] The mass of a man is 70kg. What is his weight on the earth? DATA Solution m = 70kg, g = 10m/s2, w = ? w = mg = 70×10 = 700N Centre of gravity Definition The center of gravity of a body is defined as the point through which its whole weight appears to act. Balanced Pen Center of gravity Hand (finger) 10 Experiment To find the centre of gravity. Apparatus: Paper, Pen, String Procedure: (1) Fold the paper tightly and make the shape which you want to find the centre of gravity. ( It must be flat.) (2) Make three holes near the edge. Three holes are as far as possible. Short string (3) Cut the string into two, short one(5cm) and long one(15cm). (4) Make the plumb by the long string and a pen. Hole (5) Thread a hole with the short string. Paper (6) Hang the plumb on the short string. (7) Draw the line of the plumb on the paper after it is steady. (The paper must be free to move at the hanging point.) Long string (8) Repeat (5) to (7) for other holes. Pen (9) Find the intersection of three lines. (10) Make sure that the paper balances by putting a finger at the point of intersection. Conclusion: Since the centre of gravity lies on each of the lines, the intersection locates the centre of gravity. Stability Definition The stability of an object is defined as the ability of an object to regain its original position after it has been displaced slightly. The position of centre of gravity affects the stability of a body. [Example 1] Which is more stable? (2) (2) (1) (1) Higher centre Lower centre of gravity of gravity → (1) is more stable than (2) because it has a lower centre of gravity. [Example 2] Which is more stable? (1) (2) (1) (2) Smaller base area Larger base area → (2) is more stable than (1) because it has a larger base area. To increase the stability of an object,  the centre of gravity should be as low as possible.  the base area should be as large as possible. [EXERCISE] (1) The mass of a man is 1200g. What is his weight on the earth and on the moon? The gravity on the moon is 1/6 that on the earth (g earth=10m/s2). (2) The weight of an object is 300N on the earth. (a) What is its mass on the earth? (b) What is its weight and mass on the moon? The gravity on the moon is 1/6 that on the earth (g earth =10m/s2). (c) What is its weight and mass in the outer space? [TRY] Why is it not advisable to put heavy luggage on the roof of a minibus? 11 1.5. Volume and Density Volume Definition Volume is defined as the amount of space an object occupies. The unit of volume is Cubic metre [m3]. Formula Height (h) Volume of cuboid = l×b×h l: length b: breadth h: height Breadth (b) Length (l) Instrument for measuring volume of liquid → Measuring cylinder (1) (2) How to use the measuring cylinder 25cm3 (1) Pour the measured liquid into the measuring cylinder. Wrong (2) Read the scale at the flat surface of liquid. Caution to use; Correct 22 cm3 – Put the measuring cylinder on the horizontal surface. – Place the eye level with the flat surface of liquid. 20cm3 Wrong (The surface of liquid is curved where it meets the glass. This surface is called the meniscus.) Horizontal surface Density Definition Density is defined as mass per unit volume. The unit of density is kilogram per cubic metre [kg/m3]. (Also gram per cubic centimetre [g/c m3] is frequently used.) Formula m D= V D: Density [kg/m3] or [g/cm3] m: mass [kg] or [g] V: volume [m3] or [cm3] [Example 1] A material has a mass of 450g and a volume of 50cm3. What is its density? DATA Solution m = 450g m 450 D= = = 9 g/cm3 V = 50cm3 V 50 D =? [Example 2] A body of mass 500g was suspended in 100cm3 of water by a piece of cotton. The water level rises to 150 cm3. What is the density of the body? 150cm3 DATA Solution 100cm3 m = 500g, V = 50cm3 , D =? m 500 D= = = 10 g/cm3 V 50 Volume of the body = Total volume – volume of water = 150 – 100 = 50 cm3 [EXERCISE] (1) A metal has a mass of 255g and a volume of 30cm3. What is its density? (2) A cube with side 2m long has a mass of 8kg. What is its density? (3) A container of mass 200g contains 160cm3 of liquid. The total mass of the container and liquid is 520g. What is the density of the liquid? 12 [TRY] (1) How many g/c m3 is equal to 1kg/m3? (2) Try to find out if an egg will sink or float in (a) pure water and (b)salt water. Suggest the reason for it. 1.6. Force Force → Force is a pull or a push. Ability of force → The unit of force is Newton [N]. → Force can change the size or shape of a body. → It is a vector. →Weight is a kind of force. → Force can change the motion of a body. | → Force can be measured by a spring balance, as weight is measured. Resultant force If two forces or more act on an object, they can be simplified. It is called a resultant force. [Example 1] Resultant force 8N 3N 5N 8N Compound Resultant 5N force arrows Resultant force 8N-5N Resultant force [Example 2] Resultant force 4N 5N 2N 4N 11N Compound Resultant force 5N arrows 11N 11N-(5N+4N) [Example 3] Resultant force is 0N 1N 3N 3N 2N Compound Resultant force 4N 2N arrows (3N+2N)-(4N+1N) 4N 1N If a resultant force is 0N on an object, then the forces are balanced on the object. Newton’s first law of motion Law If all the forces are balanced on a body,  if it is at rest, it will continue to stay at rest  if it is moving, it will keep moving at a constant speed in a straight line. The property of a body that resists a change to its motion is called Inertia. Inertia depends on the mass of an object. If something has a high resistance to change of motion (a high mass), it is said to have high inertia. [Example] Which is easier to move, a wire car or a truck? Truck → The wire car is easier to move than the truck Wire car because the wire car has lower inertia. (The truck has higher inertia.) low inertia high inertia Newton’s second law of motion Law Unbalanced forces acting on a body produce acceleration in the direction of the force. 13  This acceleration is directly proportional to the force and inversely proportional to the mass of the body. Formula F = ma F: Force (Resultant force) [N] m: mass [kg] a: acceleration [m/s2] [Example 1] A force of 350N acts on an object of mass 10kg. Find the acceleration. DATA Solution 10kg 350N a =? F = 350N, m = 10kg, a =? F 350 a= = = 35 m/s2 m 10 [Example 2] A man pushes an 8kg luggage on the smooth floor. It starts from rest and reaches 15m/s in 5s. (a) What is its acceleration? (b) What is the acting force on the luggage? (a) DATA Solution Rest (u =0m/s) v = 15m/s u = 0m/s, v = 15m/s v-u 15 - 0 a= = = 3m/s2 a =? (a) t = 5s, a=? t 5 8kg 5s F =? (b) (b) DATA Solution m = 8kg, a = 3m/s2, F =? F = ma = 8×3 = 24N Friction Definition Friction is a force which acts to stop the motion of two touching surfaces. → Friction acts in the opposite direction of the motion or tendency of motion. → The larger friction exists between rougher surfaces. For example, if you push a luggage, friction is caused between the luggage and the floor. Its direction is opposite to pushing force and it resists moving. If you push it on a smooth floor like ice, friction reduces and it is easy to push. Pushing force Friction Application of friction Consequence of friction – We can walk without slipping. – Friction causes energy lost and – Vehicle (car) can move and stop. reduces the efficiency of machines. – We can grip (hold) something – It causes rapid wear and tear. (e.g. pen) in our hands. – Nails hold something tight. ↓ To reduce friction, lubricants (oil, grease) are added to machines. Centripetal force The force of circular motion is always at right angle to the motion. The direction of the force is always towards the centre of circular motion. This force is called the centripetal force. Orbit of Moon the moon Centripetal force Earth Pulling force of gravity (Centripetal force) Direction of motion The moon is in a circular orbit round the earth because the earth pulls the moon by the force of gravity. 14 Load and Extension Experiment To find the relationship between loads and extensions on a spring. Apparatus: spring, loads Procedure: (1) Hold an end of a spring (the other end must be free) and measure the length of the spring. It is called the original length. (2) Hang a load on the free end. (3) Measure the length of the spring. (4) Calculate the extension by [Extension = length of spring – original length] (5) Repeat (2) to (4) adding loads. Extension (6) Calculate load (7) Plot a graph of Extension against load. (Use a scale of 2cm to 1N and 2cm to 2cm.) Result: (Samples of results are shown in the brackets.) Graph (Samples of results are plotted as below.) Extension 14 Extension [cm] Load Length of spring Extension load 12 0N (10.5cm) (0cm) (–) 10 8 1N (12.5cm) (2.0cm) (2.0) 6 2N (15.0cm) (4.5cm) (2.25) 4 3N (17.5cm) (7.0cm) (2.33) 2 4N (19.5cm) (9.0cm) (2.25) 0 0 1 2 3 4 5 6 5N (21.5cm) (11.5cm) (2.30) Load [N] Conclusion: The extension of a loaded spring is directly proportional to the load (force) applied. It is called Hooke’s Law. Formula Extension = Constant load But springs have not been permanently stretched. They have the limitations. The limitation is called Elastic limit. [Example] A load of 4N extends a spring by 10mm. What load would extend it by 15mm? DATA Solution Extension 1 = 10mm Extension 1 Extension 2 Load 1= 4N = Load 1 Load 2 Extension 2 = 15mm 10mm Load 1×Extension 2 4×15 60 Load 2=? Load 2 = = = 15mm Extension 1 10 10 4N = 6N ? [EXERCISE] (1) An 8kg object accelerates at 2m/s2. What is the acting force on it? (2) A force of 20N acts on a mass of 5kg. What is the acceleration? (3) If an object accelerates at 6m/s2 by a force of 24N, What is the mass? (4) A pupil pushes a wheelbarrow carrying 30kg sand. It accelerates from rest to 2m/s in the distance of 1m. (b) What is the acceleration? (c) What is the acting force on the wheelbarrow? (5) A load of 1N extends a spring by 5mm. What load would extend it by 1cm? (6) Calculate the extension of a spring that would be produced by a 20N load if a 15N load extends the spring by 3cm? 15 1.7. Moment Moment Definition Moment of a force about a pivot is defined as the product of the force and the perpendicular distance of its line of action from pivot. → Moment is a turning effect of a force about a certain point. Formula M = Fd M: Moment [Nm] F: Force [N] d: perpendicular distance [m]  Perpendicular distance must be a distance from the pivot to the force.  Perpendicular distance must be at right angle to the force. Force Pivot In this case, there is a moment because Force is perpendicular to the bar. Perpendicular distance The force can produce the turning effect. Pivot Force In this case, there is no moment because Force is in the same direction of distance. Not perpendicular distance The force doesn’t produce the turning effect. M=0 [Example] Calculate the moment of the force at the pivot. 3N Pivo DATA Solution F=3N, d=2m, M=? M = Fd = 3×2 = 6Nm 2m Principle of moment Law For a body to be in equilibrium (balanced), the sum of clockwise moments about any point is equal to the sum of anticlockwise moments about the same point. ↓ In other words If a body is balanced, then Formula Total clockwise moment = Total anticlockwise moment Experiment To verify the principle of moment. Apparatus: long ruler (30cm or more), 3 strings, Loads Procedure: (1) Hang a ruler by a string at the cntre of mass and make it balanced. (2) Hang some loads (e.g. 2N) at a certain point (e.g. 10cm from pivot). (3) Find the position where other loads (e.g. 4N) are hung to balance the ruler and find the distance from the pivot to the position. (4) Calculate the clockwise moment and the anticlockwise moment. (5) Repeat (2) to (4) with different pairs of loads and distances. 16  d1 d2 Ruler Result: (Samples of results are shown in the brackets.) Anticlockwise Clockwise F1 d1 M1(=F1d1) F2 d2 M2(=F2d2) F1 F2 M1 M2 (2N) (10cm) (0.2Nm) (4N) (5cm) (0.2Nm) Loads Conclusion: If a body is balanced, then the total clockwise moment is equal to the total anticlockwise moment. [Example 1] Calculate the force F if it is balanced. Data & Solution 0.5m 0.4m Anticlockwise Clockwise F D M F d M F 100N F 0.5m 0.5F 100N 0.4m 40Nm Total anticlockwise moment = Total clockwise moment 0.5F = 40 F = 80N [Example 2] Calculate the distance d if it is balanced. Data & Solution 2m Anticlockwise Clockwise d 1m F d M F d M 5N d 5d 3N 1m 3Nm 5N 3N 1N 1N 2m 2Nm Total anticlockwise moment = Total clockwise moment 5d = 3 + 2 d = 1m [EXERCISE] (1) Find the moment and its direction. 0.5N 5m (a) Pivot (b) Pivot 10N 2m (c) 2m 5m Pivot 3N (2) Find the force F if it is in equilibrium. (a) (b) 3m (c) 2m 2m 6m 3m 1m 1m 1m F 3N 2N F F 3N 5N 3N (3) Find the distance d if it is balanced. (a) (b) 3m (c) d d 4m d 1m 1m 1m 4N 2N 2N 1N 6N 3N 5N 3N (4) A metre ruler hangs by a string at the 80cm mark and a mass of 140g hangs at the 95cm mark. The weight of the ruler appears on the centre of mass. (a) Where is the pivot? (b) What is the weight of the 140g mass? (c) Calculate the weight of the ruler W (d) Calculate the mass of the ruler. 17 1.8. Work, Energy and Power Work Definition Work is defined as the product of the force and the distance moved in the direction of the force. The unit of work is Joule [J]. Formula W = Fd W: Work [J] F: Force [N] d: distance moved in the direction of the force [m] If a man pushes an object on the floor, he does the work on the object because the distance is in the same direction of the force. Force Force Work done distance But if a woman carries a container on her head, she does no work on the container because the distance is in the different direction of the force. (in this case perpendicular to the force) Force Force No Work distance [Example] A force of 5N acts on a 3kg brick, moving it 8m horizontally from rest. Find the work done by the force. 5N 5N DATA Solution F = 5N W = Fd = 5×8 = 40J 8m d = 8m W =? [EXERCISE] (1) A man pushes the big stone through 10m and exerts the force of 25N. Find the work done by the man. (2) A crane lifts a weight of 200N through 50m. Find the work done by the crane. (3) A crane lifts a car of mass 500kg through 5m. Find the work done by the crane. (g = 10m/s2) (4) A car of mass 1000kg is accelerated at 2m/s2 from rest in 20s. Find (a) the force acting on the car. (b) the distance travelled by the car in this period. (c) the work done by the car in this period. 18 Energy Definition Energy is defined as the ability to do work. The unit of energy is Joule [J]. Potential energy Definition Potential energy is defined as the energy by the position or state of an object. The potential energy due to height is called gravitational potential energy. Formula PE = mgh Box This object has the ability PE: Potential energy [J] to do work of: m: mass [kg] h g: acceleration due to gravity [m/s2] m F×d = mg×h = mgh h: height [m] h [Example] A 20kg object is raised to a height of 5m. What is its potential energy? DATA Solution 20kg m =20kg h = 5m PE = mgh = 20×10×5 g = 10m/s2 PE=? = 1000J = 1kJ 5m Kinetic energy Definition Kinetic energy is defined as the energy due to the motion of an object. Formula 1 2 KE = mv 2 KE: Kinetic energy [J] m: mass [kg] v: velocity [m/s] [Example] A 2kg stone is thrown with a velocity of 5m/s. What is its kinetic energy? DATA Solution 2kg 5m/s m =2kg 1 2 1 v = 5m/s PE = mv = ×2×52 2 2 KE=? = 25J Conservation of energy Law Energy can be changed from one form to another, but cannot be created or destroyed. [Example 1] If a ball falls freely from a certain point, how does the energy change? (1) P.E. = 1000J (1) Before a ball is released, its potential energy is 1000J and the kinetic energy is K.E. = 0J 0J because it doesn’t move. Total energy=1000J P.E. = 500J (2) At the midpoint of its journery, the potential energy drops to 500J but the (2) K.E. = 500J kinetic energy increses to 500J. Total energy is still 1000J. Total energy=1000J P.E. = 0J (3) Just before hitting the ground, the potential energy becomes 0 J but the kinetic (3) energy increases to 1000J. There is no change in the total energy throughout its K.E. = 1000J falling. Total energy=1000J 19  [Example 2] How does the energy change on the pendulum? (1) The pendulum bob is pulled to position (1). Before it is released, its potential energy is 200J and kinetic energy is 0J because it is at rest. (2) As the bob moves from (1) to (2), it loses potential energy and gains kinetic energy to 200J because of (1) (3) reducing the height and increasing the velocity. It has P.E. = 200J maximum velocity at the bottom (2). P.E. = 200J K.E. = 0J K.E. = 0J (2) (3) Moving from (2) to (3), the bob slows down, losing Total energy=200J Total energy=200J kinetic energy but gaining poteintial energy. If air P.E. = 0J K.E. =200J resistance is ignored, the height of (1) is the same as the height of (3) because these potential energies must be Total energy=200J the same. Conservation of energy → Each energy can be changed but total energy is constant. →When there are only PE and KE, P.E. + K.E. = constant Use of energy (Source, Transducer, Use) When energy is used for our life,first energy comes from energy sources, and the energy forms are changed by a transducer. Then useful energy is provided. The diagrams below show some examples of sources of energy, transducers and useful energy provieded. Source of energy Transducer Useful energy Food You Movement of your body [Chemical energy] [Kinetic energy] Sun light Solar system Electricity [Solar(light) energy] [Electrical energy] Wind Generator Electricity [Kinetic energy] [Electrical energy] Electricity Electrical motor Movement of motor [Electrical energy] [Kinetic energy] Environmental effect If sources of energy are used in our life, it may affect on the environment. The followings show some cases of effect to enviroment. We must prevent such kind of bad influence. [Case 1] Effect of a car Fuel Movement Car [Chemical energy] [Kinetic energy] – Sound causes the noise pollution. Sound (Noise) Heat – Heat causes global warming. Exhaust gas – Exhaust gas causes the air pollution. [Case 2] Effect of a hydraulic Plant Stocked water Movement of water Hydro- Electricity Lake Kariba [Potential energy] [Kinetic energy] generator [Electrical energy] Hydrauric plants are clearn. It doesn’t cause any pollution. However it may changes natural ecosystem. When the dam is built, very large space is needed. Many trees are cut. It is a deforestration. And also many living things (animals and insects) lose their living places. 20 [Case 3] Effect (risk) of an atomic Plant Atomic power Radioactive substance Electricity station [Nuclear energy] [Electrical energy] If an atomic power station is damaged, the radiation leaks. It brakes cells or gene of living things. As a result, it causes some diseases such as cancer. Power Definition Power is defined as the rate of doing work. The unit of power is Watt [W]. Formula W E P= = t t P: Power [W] W: work done [J] t: time taken [s] E: Energy changed [J] [Example] A crane can lift 200kg to a height of 100m in 20s. What is the useful power of the crane? Step (1) Find the weight. Step (2) Find the work done. DATA Solution (1) Step (3) Find the power. m = 200kg W = mg = 200×10 = 2,000N g = 10m/s2 W =? DATA Solution (2) F = Weight = 2000N Work = Fd = 2000×100 = 200,000J d = 100m Work =? 100m DATA Solution (3) Work done 200000 Work done = 200,000J t = 20s P= = t 20 200kg P =? = 10,000W = 10kW [EXERCISE] (1) A 10kg rock is on the hill of 50m. What is its potential energy? (2) A book which has a mass of 1.2kg is put on the desk. the height of desk is 0.8m. Calculate the potential energy. (3) A car of mass 500kg moves with a velocity of 20m/s. What is its kinetic energy? (4) A 3kg brick is released from the top of building. Calculate; (a) the velocity after one second. (b) the kinetic energy after one second. (5) A rock of mass 200kg is dropped from a height of 200m. What is the potential energy and the kinetic energy at; (a) 0 second (b) 2seconds (c) 4seconds (d) just before it hits the ground. (6) A force of 1000N is needed to push a mass of 30kg through a distance of 40m to raise an inclined plane to a height of 5m. Calculate; (a) the weight of an object. (b) the mechanical advantage. (c) the velocity ratio. (d) the efficiency of the inclined plane. (e) the energy at the height of 5m. (f) the work done by the force of 1000N. (g) the power developed if the force is exerted for 20s. (7) A 60kg pupil runs for 600m in 1minute uniformly. (a) Calculate his velocity. (b) Calculate his kinetic energy. (c) If eating one banana enables a pupil to perform about 3kJ of work, how may banana should he eat? 21 1.9. Simple Machine Machine Definition Machine is any device by means of which a force applied at one point can be used to overcome a force at a different point. → The applied force is called Effort Rock → The force which effort overcomes is called Load Effort (The force which an object pulls or pushes on a machine is called Load.) Load Types of simple machine (1) Lever (2) Pulley (Examples) – Wheelbarrow – Borehole – Scissors – Opener – Slasher – Hoe – Shovel – Cooking stick Single fixed pulley Single moving pulley (3) Gear (4) Inclined plane Mechanical Advantage (M.A.) Definition The mechanical advantage of a machine is defined as the ratio of the load to the effort. Formula Load M.A. = Effort Velocity Ratio (V.R.) Definition The velocity ratio of a machine is defined as the ratio of the distance moved by effort to the distance moved by the load in the same time. Formula distance moved by effort(dE) V.R. = distance moved by load in the same time(dL) Efficiency Definition The efficiency of a machine is defined as the ratio of the useful work done by the machine to the total work put into the machine. Useful work done Work done by load Efficiency = = Total work put into Work done by effort Since, W=Fd, Load and Effort are forces Load×dL Load dL Load dE Efficiency = ×100 = × ×100 = ÷ ×100 = M.A.÷V.R.×100 Effort×dE Effort dE Effort dL Formula M.A. Efficiency = ×100 [%] (Efficiency  100%) V.R. [Example 1] Calculate the mechanical advantage of the diagrams below. 22  (1) Lever DATA Solution 20N Effort = 20N Load 100 M.A. = = =5 Load = 100N Effort 20 100N M.A. =? (2) Single moving Pulley 15N DATA Solution Effort = 15N Load 30 Load = 30N M.A. = = =2 Effort 15 M.A. =? 30N (3) Inclined Plane DATA Solution 5N Effort = 5N Load 35 Load = weight = 35N M.A. = = =7 Effort 5 M.A. =? 35N [Example 2] Calculate the velocity ratio of the diagrams below. (1) Lever DATA Solution distance from pivot to effort dE V.R. = 50cm 1m = 1m dL distance from pivot to load distance from pivot to effort = 50cm = 0.5m = distance from pivot to load V.R. =? 1 = =2 0.5 (2) Inclined Plane DATA Solution 100m Height = 40m dE Length of slope V.R. = = Length of slope= 100m dL height 40m V.R. =? 100 = = 2.5 40 [Example 3] The diagram below shows a pulley system. An effort of 1000N is required to raise a load of 4500N. (a) Find the mechanical advantage. (b) Find the velocity ratio. (c) Find the efficiency. (a) DATA Solution Effort = 1000N Load 4500 Load = 4500N M.A. = = = 4.5 M.A. =? Effort 1000 (b) To find the velocity ratio of Solution the pulley system, count the number of lines connected 6 lines connected to moving pulleys. to moving pulley. V.R. = 6 1000N (c) DATA Solution M.A. = 4.5, V.R. = 6 M.A. 4.5 4500N Efficiency = ×100 = ×100 = 75% Efficiency =? V.R. 6 23 [EXERCISE] (1) Calculate the mechanical advantage, the velocity ratio and the efficiency of the diagrams below. (a) (b) (c) 40m 20N 40N 2m 120N 80N 20cm 80cm 340N 192N (2) The system of levers with a velocity ratio of 25 overcomes a resistance of 3300N when an effort of 165N is applied to it. (b) Calculate the mechanical advantage of the system. (c) Calculate its efficiency. [TRY] Prove that M.A.  V.R. 24 2. THERMAL PHYSICS 2.1. Kinetic theory State of matter Matter exists in the three states of Solid, Liquid and Gas. The physical difference between the three states of matter depends on the arrangement and behaviour of the molecules in each particular state. This difference can be explained in trems of the Kinetic Theory, model which states that;  Matter is made up of very small particles called molecules.  These molecules are not stationary but are constantly moving.  The degree of movement of the molecules depends on their temperature. Solid (1) Properties – Fixed shape and volume (3) Molecular model (Solid model) – Normally hard and rigid – Incompressible – Large force needed to change shape – High density (2) Arrangement and movement of particles – The particles are close together. – They are arranged in a regular pattern. – The attractive forces between them are strong. (The attractive force is called Cohesive force.) – They vibrate to and fro at the fixed positions. (They cannot change positions.) Liquid (1) Properties – Fixed volume but does not have a fixed shape (2) Molecular model (Liquid model) – Not compressible – High density (2) Arrangement and movement of particles – The particles are close together but they have wider space than those in solid. – The attractive forces between them are weaker than those in solid. – They move vigorously. – They can move from one position to another. Gas (1) Properties – No fixed shape or volume (3) Molecular model (Gas model) – Compressible – Low density (2) Arrangement and movement of particles – The particles are very far apart. – They can hardly attract each other. – They move randomly with a high speed. Brownian motion Brownian motion provides an evidence of the continuous random motion of the molecules in the air.  Experimental Set up A microscope is used to look into a smoke cell that contains some smoke particles (as well as air molecules) as shown below. 25  light source converging microscope light lens smoke cell smoke  Observation When the light strikes the smoke particles, they are observed as bright specks of light. The path of They move in a random zig-zag path as the smoke particle diagram in the circle.  Explanation smoke particle The zig-zag movement is due to the collisions of the smoke paticles with invisible air molecules View of microscope that move about randomly in the smoke cell. This is called Brownian motion. Diffusion Definition Diffusion is defined as the process by which different substances mix as a result of the randum motions of their molecules. → The substances move freely from a region of high concentration to a region of low concentration at their own pace. → The rate of diffusion depends on the temperature and the density of the substances involved. → It supports the kinetic theory, since the particles must be moving to mix, and gases can be seen to diffuse faster than liquids. [Case 1] Diffusion of ink in water The colour of water is changed by the diffusion of ink paticles. (without mixing) [Case 2] Diffusion of mosquito coil The smell of mosquito coil reaches to your place by the diffusion of its smoke paticles. Change of state The names of changes of states are indicated in the diagram below. Evaporation Solid Melting Liquid Boiling Gas (ice) Solidification (water) Condensation (steam) (freezing) The temperature is unchanged during the change of state of a substance. All the energy supplied to the substance is used for breaking the strong forces between molecules.  The melting point is the temperature at which a solid melts.  The boiling point is the temperature at which a liquid boils. 26  Temperature Boiling point 100 oC Melting point 0 oC Time solid solid and liquid liquid and gas Evaporation liquid gas Definition Evaporation is defined as the change of a liquid into a gas at the surface. It occurs at any temperature but occurs more rapidly at higher temperature because heat gives more kinetic energy to the molecules and they escape from the surface faster. If temperature increases, molecules move faster Increasing temperature and they have larger kinetic energy. Then Some molecules escape from the they escape from the surface of liquid at any temperature. surface faster. Increased gas pressure on the surface of the liquid Gas (air) molecules have higher speed reduces the rate of evaporation because more because of increasing of pressure. collisions occur between the evaporating liquid molecules and the gas molecules, and some of the More gas (air) molecules collide evaporated liquid molecules bounce back into the with liquid molecules. liquid. (See the right diagram.) Liquid molecules bounce on the surface. The molecules that have the largest kinetic energy escape from the liquid. Then, the average kinetic energy of molecules in the liquid is reduced, and also the temperature of liquid reduces. This is called the cooling effect of evaporation. [The difference between evaporation and boiling] Evaporation Boiling Occurs at any temperature Occurs at the boiling point Occurs on the surface of liquid Occurs within liquid No bubbles Bubbles appear [TRY] How do you feel when spirits are applied on your hand? Why? 2.2. Thermal properties Thermal expansion When a body of something is heated, the body increases in size. It is called Thermal expansion. When molecules get heat energy, they have more kinetic energy. They move or vibrate more. Then they need larger spaces between them. Applications of thermal expansion  Thermometer See after next section.  Bimetallic strip This is a compound bar made from two metals riveted together. When it is heated, it bends because the metals expand differently. 27  [Bimetallic strip of copper and iron] Copper Heating Copper expands more than iron when they are headed. It causes Iron the bimetallic strip to bend. This is used as a thermostat which is a switch to keep the temperature in electrical circuits e.g. a pressing iron, refrigerator and fire alarm. [Electric pressing iron] When a bimetallic strip is bent by heat, the circuit is disconnected. This system controls the heat on the iron. It is called a thermostat.  Rivets Rivets are a form of nail used to hold two metal plates tightly together. If hot rivet is used, the rivet contracts after cooling. It can fix plates tightly. hot rivet The rivet is still hot. It holds two The rivet contracts after cooling. metal plates. Then it can fix them tightly.  Wheel fitting A slightly larger axles does not fit into the wheel. However it can fit into the wheel by cooling because it is contracted. As it warms up again, it will expand back to its natural size. It causes a tight fit b