Uploaded by
Ammar Ahmed Thakur
Physics Notes & Worksheet: Kinematics, Dynamics, Forces
advertisement
Table of Contents 00 SF NOTES 1 00 SF NOTES.pdf 1 worksheet sf.pdf 3 worksheet sf 2.pdf 4 MS worksheet sf.pdf 5 MS worksheet sf 2.pdf 6 01a PHYSICAL QUANTITIES 7 01 PHYSICAL QUANTITIES AND UNITS 7 Physical quantity 7 Base units 7 Derived units 7 Homogeneity of physical equations 8 Homogenous Equation: 8 EXAMPLE 1 9 EXAMPLE 2 9 EXAMPLE 3 10 EXAMPLE 4 11 EXAMPLE 5 12 EXAMPLE 6 13 EXAMPLE 7 14 EXAMPLE 8 15 QUESTION 1 16 QUESTION 2 17 QUESTION 3 18 QUESTION 4 19 ESTIMATIONS 20 QUESTION 5 21 QUESTION 6 21 QUESTION 7 22 QUESTION 8 22 QUESTION 9 23 QUESTION 10 23 Prefixes 24 QUESTION 11 25 QUESTION 12 25 QUESTION 13 25 Scalars and vectors 25 Add and subtract coplanar vectors. 26 By using head to tail or tip to tail rule EXAMPLE 14 By using two perpendicular components EXAMPLE 15 26 26 27 27 Coplanar vectors in equilibrium triangle of forces 28 Vector subtraction 29 QUESTION 16 30 QUESTION 17 31 QUESTION 18 32 QUESTION 19 32 QUESTION 20 34 01b ERRORS AND UNCERTAINTIES 36 Uncertainty 36 Systematic Errors 37 Zero error 37 Random Error 38 Question 1 38 Question 2 38 Question 3 39 Precision and Accuracy 39 Example 1 40 Question 4 41 Question 5 41 Question 6 42 Question 7 44 Fractional and percentage uncertainty 45 Combining uncertainties 45 Adding or subtracting uncertainties 45 Multiplying or dividing uncertainties 45 If there is a power on one or both then 45 Multiplying or dividing uncertainties in percentages 46 If there is a power on one or both, then 46 Example 2 47 Example 3 48 Question 8 49 Example 4 51 Example 5 52 Question 9 53 Question 10 54 Question 11 55 01c CRO AND CALIBRATION 56 CRO and Calibration 56 Using Cathode Ray Oscilloscope 56 Question 1 57 Question 2 57 Question 3 58 Question 4 58 Question 5 59 Question 6 60 CALIBRATION 61 Calibration curve 61 Question 7 62 Question 8 63 02 KINEMATICS 65 Kinematics of linear motion 65 Definition of Velocity 65 Average velocity 65 Definition of Acceleration 65 Case 1: Linear motion with uniform (constant) velocity 65 Case 2: Linear motion with uniform acceleration 67 Case 3: Linear motion with uniform (constant) deceleration 69 Equations of motion 71 Question 1 73 Question 2 73 Question 3 74 Question 4 74 Question 5 76 Combining Case 2 and Case 3 (Vertical motion under gravity) 78 Question 6 81 Question 7 83 Question 8 85 Question 9 85 Question 10 85 Question 11 85 Question 12 86 Question 13 86 Rebounding or Bouncing 88 Acceleration time graph 88 Velocity-time graph 88 Displacement-time graph 89 Speed-time graph 89 Question 14 90 Question 15 90 Question 16 91 Question 17 91 Question 18 92 Question 19 92 Question 20 95 Projectile motion 96 Question 21 98 Question 22 99 Question 23 100 Half-projectile motion 101 Question 24 103 Question 25 104 03 DYNAMICS 106 Dynamics 106 Definition of mass 106 Weight 106 First law of motion 106 Example 1 107 Second law of Newton 107 Momentum 108 A special case of Newton’s second law of motion 108 Force-time graph 109 Defining the newton 109 Examples question for Newton’s 2nd law 109 Example 2 109 Example 3 110 Example 4 110 Example 5 111 Question 1 111 Example 6 112 Example 7 113 Example 8 114 Question 2 115 Question 3 117 Question 4 118 Question 5 119 Objects on an inclined plane 120 Acceleration down the slope (frictionless surface) 120 Example 8 121 Question 6 122 Question 7 123 Question 8 124 Show that K.E =,, - .- . Question 9 124 125 Frictional and Viscous drag forces 125 Falling under gravity with air resistance (terminal velocity) 125 Question 10 Newton’s third law 126 129 Question 11 130 Question 12 131 Single body collision/collision with a rigid surface (wall) 131 Perfectly Elastic collisions 131 Inelastic Collison 131 Inelastic collision (comes to a halt after Collison) 131 Principle/law of conservation of momentum 132 Types of collisions 132 Elastic collision 132 Question 13 133 Inelastic collisions 134 Example 9: 135 Question 14 135 Question 15 136 Question 16 137 Question 17 137 Using law of conservation of momentum to prove third law 137 Using 2nd and 3rd law to prove law of conservation of momentum 138 Question 18 138 Oblique collisions/2D collisions 140 Question 19 141 Question 20 142 Question 21 143 04 FORCES DENSITY PRESSURE 144 Forces, Density and Pressure 144 Center of gravity 144 Turning effect of force / Moment of a force 144 Line of action 144 Find the perpendicular distance with respect to the force 145 Find the perpendicular component of force with respect to the distance 145 Principle of moments 145 Equilibrium 146 Torque of a couple 146 Find the resultant moment for the following: 146 Question 1 146 Question 2 147 Question 3 147 Question 4 148 Question 5 148 Quesiton 6 149 Example 1 150 Question 7 151 Question 8 152 Question 9 153 Density 154 Pressure 154 Derivation of hydrostatic pressure P=ρgh 154 Question 10 155 Question 11 157 Upthrust = (Archimedes’ principle) 158 Question 12 159 Question 13 159 Question 14 159 Question 15 160 Question 16 160 Question 17 161 Question 18 163 Question 19 164 Question 20 165 Question 21 165 Question 22 166 Question 23 166 Question 24 167 Question 25 168 05 WORK POWER ENERGY 169 WORK POWER AND ENERGY 169 Work done 169 Work done against friction or drag 169 Work transforming into other forms 169 Work done into KE 170 Question 1 170 Question 2 171 Work done into GPE 171 Question 3 171 Question 4 172 Power 172 Derivation P=F.v 172 Question 5 173 Question 6 173 Question 7 173 Question 8 174 Efficiency 174 Question 9 174 Question 10 175 Question 11 175 Question 12 175 Law of conservation of energy 176 Question 13 176 Question 14 176 Question 15 177 Question 16 178 Question 17 179 Question 18 180 Question 19 181 Question 20 183 Question 21 185 Work done by a gas 188 Question 22 189 Question 23 189 Question 24 189 Force-displacement graphs 190 06 DEFORMATION 191 Deformation 191 Tensile and compressive force 191 Hooke’s law 191 Force-extension graph (Hooke’s law) 192 Question 1 193 Question 2 193 Limit of proportionality 193 Elastic limit 194 Elastic deformation 194 Question 3 194 Question 4 196 Finding Work done using Force-extension graphs 197 Elastic potential energy or strain energy 197 Change in strain energy or elastic potential energy 198 Question 5 199 Question 6 199 Question 7 200 Question 8 200 Question 9 201 Question 10 202 Stress 203 Strain 203 Young Modulus 203 Graph of stress-strain 204 Strain energy per unit volume 205 Question 11 206 Question 12 206 Question 13 207 Question 14 207 Question 15 208 Question 16 208 Question 17 208 Experiment to determine the Young Modulus 209 Method 209 Reducing uncertainty 210 Measurements to determine Young modulus 210 Question 18 212 Question 19 212 Question 20 213 Question 21 215 Effective spring constant 217 Question 22 218 Question 23 218 Question 24 218 Question 25 219 Question 26 219 Question 27 219 Question 28 220 Elastic and plastic deformation 220 Question 29 221 Question 30 222 Question 31 222 Increase in resistance due to change in length 223 SIR SAMEER UZ ZAMAN · · · · · · · WORKED EXAMPLE 1 of 223 In addition and subtraction round off the answer to the lowest no of decimal places in the raw data for e.g. 12.015 + 13.9 = 25.9 12.567 + 13.9 = 26.5 In multiplication and division the answer should be rounded off to the lowest no of significant figure in the raw data For example: 2.31 x 9 = 20.79 rounded off to 20, one sf, because 2.31 is three sf and 9 is one sf, hence the answer is rounded off to one sf Note: for multiplication and division the answer could be given to one sf more than the lowest one. In that case the answer for the above example would be 21 2 of 223 SIR SAMEER UZ ZAMAN USING APPROPRIATE NO OF SF IN CALCULATIONS Significant Figures Worksheet 1. Indicate how many significant figures there are in each of the following measured values. 246.32 1.008 700000 107.854 0.00340 350.670 100.3 14.600 1.0000 0.678 0.0001 320001 2. Calculate the answers to the appropriate number of significant figures. 32.567 135.0 + 1.4567 246.24 238.278 + 98.3__ 658.0 23.5478 + 1345.29__ 3. Calculate the answers to the appropriate number of significant figures. a) 23.7 x 3.8 = e) 43.678 x 64.1 = b) 45.76 x 0.25 = f) 1.678 / 0.42 = c) 81.04 x 0.010 = g) 28.367 / 3.74 = d) 6.47 x 64.5 = h) 4278 / 1.006 = 3 of 223 Solve the Problems and Round Accordingly. 1 ) 161 x 0.071 = __________ 11) 2002 ÷ 48.0 = __________ 2 ) 303 x 0.8 = __________ 12) 0.004 x 0.085 x 5003 = __________ 3 ) 37.7 x 48 x 170 = __________ 13) 51.0 x 9 x 2100 = __________ 4 ) 80 x 16.33 x 4002 = __________ 14) 5.8 x 0.37 = __________ 5 ) 8070 ÷ 7.0 = __________ 15) 300 x 6.124 x 4009 = __________ 6 ) 19 x 4.1 = __________ 16) 70 ÷ 5.58 = __________ 7 ) 90 ÷ 7.276 = __________ 17) 0.007 x 0.6 = __________ 8 ) 606 ÷ 3.7 = __________ 18) 0.02 x 0.5 x 90 = __________ 9 ) 42 x 9.14 = __________ 19) 6006 ÷ 1.4 = __________ = __________ 20) 28.69 x 80 = __________ 10) 1500 ÷ 6.49 4 of 223 Significant Figures Worksheet Key 1. Indicate how many significant figures there are in each of the following measured values. 246.32 5 sig figs 1.008 4 sig figs 700000 1 sig fig 107.854 6 sig figs 0.00340 3 sig figs 350.670 6 sig figs 100.3 4 sig figs 14.600 5 sig figs 1.0000 5 sig figs 0.678 3 sig figs 0.0001 1 sig fig 320001 6 sig figs 2. Calculate the answers to the appropriate number of significant figures. 32.567 135.0 + 1.4567 169.0 246.24 238.278 + 98.3__ 582.8 658.0 23.5478 + 1345.29__ 2026.8 3. Calculate the answers to the appropriate number of significant figures. a) 23.7 x 3.8 = 90. e) 43.678 x 64.1 = 2.80 x 103 b) 45.76 x 0.25 = 11 f) 1.678 / 0.42 = 4.0 c) 81.04 g x0.010 = 0.81 g) 28.367 / 3.74 = 7.58 d) 6.47 x 64.5 = 417 h) 4278 / 1.006 = 4252 5 of 223 Solve the Problems and Round Accordingly. 1 ) 161 x 0.071 11 = __________ 11) 2002 ÷ 48.0 41.7 = __________ 2 ) 303 x 0.8 200 = __________ 12) 0.004 x 0.085 x 5003 2 = __________ 3 ) 37.7 x 48 x 170 310,000 = __________ 13) 51.0 x 9 x 2100 1,000,000 = __________ 4 ) 80 x 16.33 x 4002 5,000,000 = __________ 14) 5.8 x 0.37 2.1 = __________ 5 ) 8070 ÷ 7.0 1,200 = __________ 15) 300 x 6.124 x 4009 7,000,000 = __________ 6 ) 19 x 4.1 78 = __________ 16) 70 ÷ 5.58 10 = __________ 7 ) 90 ÷ 7.276 10 = __________ 17) 0.007 x 0.6 0.004 = __________ 8 ) 606 ÷ 3.7 160 = __________ 18) 0.02 x 0.5 x 90 0.9 = __________ 9 ) 42 x 9.14 380 = __________ 19) 6006 ÷ 1.4 4,300 = __________ 230 = __________ 20) 28.69 x 80 2,000 = __________ 10) 1500 ÷ 6.49 6 of 223 01 PHYSICAL QUANTITIES AND UNITS Physical quantity A quantity with a magnitude and a unit is a physical quantity. Base units All units in science are derived from seven base units. These are as follows: Mass Distance Time Current Amount of substance Temperature Light Intensity kilogram meter second ampere mole Kelvin candela kg m s A mol K cd Derived units There are many other units that we use, but all of these are derived by multiplication or division of some combinations of the base units. Here are some of the derived units: SIR SAMEER UZ ZAMAN Notes: 7 of 223 F=ma Homogeneity of physical equations Homogenous Equation: An equation in which the base units of all quantities added or subtracted is the same on either side of the equation. Meaning that the units of the quantity on one side of the equation is equal to the units on the other side. SIR SAMEER UZ ZAMAN Notes: 8 of 223 EXAMPLE 1 EXAMPLE 2 Given that the equation is homogenous. SIR SAMEER UZ ZAMAN Notes: 9 of 223 EXAMPLE 3 Given that the equation is homogenous. SIR SAMEER UZ ZAMAN Notes: 10 of 223 EXAMPLE 4 Given that the equation is homogenous. SIR SAMEER UZ ZAMAN Notes: 11 of 223 EXAMPLE 5 SIR SAMEER UZ ZAMAN Notes: 12 of 223 EXAMPLE 6 Given that the equation is homogenous. SIR SAMEER UZ ZAMAN Notes: 13 of 223 EXAMPLE 7 Given that the equation is homogenous. SIR SAMEER UZ ZAMAN Notes: 14 of 223 EXAMPLE 8 SIR SAMEER UZ ZAMAN Notes: 15 of 223 QUESTION 1 SIR SAMEER UZ ZAMAN Notes: 16 of 223 QUESTION 2 SIR SAMEER UZ ZAMAN Notes: 17 of 223 QUESTION 3 SIR SAMEER UZ ZAMAN Notes: 18 of 223 QUESTION 4 SIR SAMEER UZ ZAMAN Notes: 19 of 223 ESTIMATIONS When making an estimate, it is only reasonable to give the figure to 1 or at most 2 significant figures since an estimate is not very precise. Some examples are below: SIR SAMEER UZ ZAMAN Notes: 20 of 223 QUESTION 5 QUESTION 6 SIR SAMEER UZ ZAMAN Notes: 21 of 223 QUESTION 7 QUESTION 8 SIR SAMEER UZ ZAMAN Notes: 22 of 223 QUESTION 9 QUESTION 10 SIR SAMEER UZ ZAMAN Notes: 23 of 223 Prefixes Now you have units, you often need to group these into larger or smaller numbers to make them more manageable. For example, you don't say that you are going to see someone who lives 100,000 m away from you; you say they live 100 km away from you. Here is a quick list of the common quantities used: Name Symbol Scaling factor Tera T Giga G Mega 12 Common example 10 1,000,000,000,000 109 1,000,000,000 Large computer hard drives can be terabytes in size. Computer memories are measured in gigabytes. M 106 1,000,000 A power station may have an output of 600 MW (megawatts). kilo k 103 1,000 Mass is often measured in kilograms (i.e. 1000 grams). deci d 10-1 0.1 Fluids are sometimes measured in deciliters (i.e. 0.1 liter). centi c 10-2 0.01 Distances are measured in centimeters (i.e. 100th of a meter). milli m 10-3 0.001 Time is sometimes measured in milliseconds. micro μ 10-6 1,000,000th nano n 10-9 pico p 10-12 micrometer is often used to measure wavelengths of electromagnetic waves. nanometer is used to measure atomic spacing. Picometre is used to measure atomic radii. SIR SAMEER UZ ZAMAN Notes: 24 of 223 QUESTION 11 QUESTION 12 QUESTION 13 Scalars and vectors A vector is a quantity, which has both a magnitude and a direction. Examples of vectors are displacement, velocity, acceleration, moments (or torque), momentum, force, electric field etc. SIR SAMEER UZ ZAMAN Notes: A scalar is a quantity, which has magnitude (numerical size) only. Examples of scalars are Distance, speed, mass, time, temperature, work done, pressure, power, electric charge, energy, volume, and temperature etc. these quantities can be added or subtracted by adding or subtracting their magnitudes. 25 of 223 Add and subtract coplanar vectors. By using head to tail or tip to tail rule EXAMPLE 14 SIR SAMEER UZ ZAMAN Notes: 26 of 223 By using two perpendicular components EXAMPLE 15 SIR SAMEER UZ ZAMAN Notes: 27 of 223 Coplanar vectors in equilibrium triangle of forces If system is in equilibrium that mean there is no net force on it. If we add up all the forces, by using head to tail rule, it would form a closed vector triangle (meaning there would be no place to draw a resultant vector). In this triangle the arrows for the forces follow each other. SIR SAMEER UZ ZAMAN Notes: 28 of 223 Vector subtraction The easiest way to think about vector subtraction is in terms of adding a negative vector. What’s a negative vector? It’s the same vector as its positive counterpart, only pointing in the opposite direction. As shown below A – B, then, is the same thing as A + (–B). For instance, let’s take the two vectors A and B: To subtract B from A, take a vector of the same magnitude as B, but pointing in the opposite direction, and add that vector to A, using either the tip-to-tail method or the parallelogram method. SIR SAMEER UZ ZAMAN Notes: 29 of 223 QUESTION 16 SIR SAMEER UZ ZAMAN Notes: 30 of 223 QUESTION 17 SIR SAMEER UZ ZAMAN Notes: 31 of 223 QUESTION 18 QUESTION 19 SIR SAMEER UZ ZAMAN Notes: 32 of 223 SIR SAMEER UZ ZAMAN Notes: 33 of 223 QUESTION 20 SIR SAMEER UZ ZAMAN Notes: 34 of 223 SIR SAMEER UZ ZAMAN Notes: 35 of 223 Uncertainty Natural variations in measurements are called uncertainties. They may come about for the following reasons: ● No instrument is exactly precise. ● Different people may use different instruments. ● Sometimes instruments are wrongly read. ● The instrument’s adjustments may change. The uncertainty in a set of readings is the maximum deviation from the mean value. To illustrate this point, let us consider the following example; An experiment was conducted to determine the free-fall acceleration. The results obtained were as follows: g (in ms−2) = 9.81, 9.80, 9.78, 9.81, 9.83, 9.81, 9.82, 9.81, 9.84, 9.79, 9.80, 9.83, 9.81, 9.79, 9.80, 9.82, 9.81, 9.82. The mean value (to 3 sig fig.) is 9.81 ms−2. If N represents the number of times each value is obtained, the graph of N against g will be as follows: 7 The absolute uncertainty is the maximum deviation from the mean value. ∆𝑔𝑔 = (9.84 − 9.81) 𝑚𝑚𝑚𝑚 −2 = 0.03 𝑚𝑚𝑚𝑚 −2 Note: Absolute uncertainty should always be stated to 1 significant figure. N 6 5 4 3 2 1 0 9.76 9.78 9.80 9.82 9.84 The different types of errors that cause the uncertainties are classified as systematic and random errors. 9.86 g SIR SAMEER UZ ZAMAN Notes: Summary: 36 of 223 Systematic Errors ⎯ It is an error that is consistent in magnitude as well as in sign. ⎯ It cannot be reduced by taking average of many readings. Zero error is an example of systematic errors. Systematic errors can be reduced by the following methods: ⎯ By fixing the instrument ⎯ By using a different instrument ⎯ By using a different method The figure on the right is an example of systematic error. All the measured values are away from the true value by the same amount. Zero error Zero error is defined as the condition where a measuring instrument registers a reading when there should not be any reading. In case of vernier calipers it occurs when a zero on main scale does not coincide with a zero on vernier scale. The zero error may be of two types: when the scale is towards numbers greater than zero it is positive; else negative. The method to correct for a zero error is to use the formula 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑟𝑟𝑟𝑟𝐴𝐴𝑟𝑟𝑟𝑟𝑟𝑟𝑔𝑔 = 𝑀𝑀𝑟𝑟𝐴𝐴𝑚𝑚𝐴𝐴𝑟𝑟𝑟𝑟𝑟𝑟 𝑟𝑟𝑟𝑟𝐴𝐴𝑟𝑟𝑟𝑟𝑟𝑟𝑔𝑔 − (𝑧𝑧𝑟𝑟𝑟𝑟𝑧𝑧 𝑟𝑟𝑟𝑟𝑟𝑟𝑧𝑧𝑟𝑟) If the zero error is positive, then it is subtracted from the measured reading and if the error is negative it is added to the measured reading. SIR SAMEER UZ ZAMAN Notes: Summary: 37 of 223 Random Error ⎯ Random errors are neither consistent in magnitude nor in sign. ⎯ They can be reduced by taking an average of many readings. Errors caused by environmental factors e.g., air currents, variations in temperature or pressure, human reaction time etc. are examples of random errors. The figure on the right is an example of random error. The measurements are evenly scattered around the true graph. Question 1 Question 2 SIR SAMEER UZ ZAMAN Notes: Summary: 38 of 223 Question 3 Precision and Accuracy A set of reading is said to have good precision if all the readings are close to each other (even if the mean value is far from the true value). • • The smaller the random error the higher the precision. Precision in a reading is judged by the number of decimal places. For e.g., 5.21 ±0.01 is more precise than 5.2 ±0.1. A set of readings is said to have good accuracy if the mean value is close to the true value (even if the readings are not close to each other). • • The smaller the systematic error more accurate the readings A large systematic error will shift all the readings away and therefore the mean value will be very far from the true value. SIR SAMEER UZ ZAMAN Notes: Summary: 39 of 223 Example 1 1. Steel rule can be read to the nearest millimetre it is used to measure the length of a bar whose true length is 895 mm repeated measurements give the following readings: length /mm 892, 891, 892, 891, 891, 892 a) Are the readings accurate and precise to within 1 mm? b) Are the readings precise to within 1 mm? c) Are the readings accurate to within 1 mm? Solution: To check the accuracy, we see how far is the mean value from the true value. Mean value = 891.5 mm True value = 895 mm Difference = 895 mm − 891.5 mm = 3.5 mm Therefore, the readings are not accurate to within 1 mm. To check the precision, we see the maximum deviation of the reading from the mean value. The mean value is 891.5 mm Max value = 892mm Mean value = 891.5 mm Max deviation = 892 mm − 891.5 mm = 0.5 mm Therefore, the readings are precise to 1 millimetre, so the correct answer is b. SIR SAMEER UZ ZAMAN Notes: Summary: 40 of 223 Question 4 Question 5 SIR SAMEER UZ ZAMAN Notes: Summary: 41 of 223 Question 6 SIR SAMEER UZ ZAMAN Notes: Summary: 42 of 223 SIR SAMEER UZ ZAMAN Notes: Summary: 43 of 223 Question 7 SIR SAMEER UZ ZAMAN Notes: Summary: 44 of 223 Fractional and percentage uncertainty If ∆𝑥𝑥 is the absolute uncertainty in x, then 𝐹𝐹𝑟𝑟𝐴𝐴𝐴𝐴𝐴𝐴𝑟𝑟𝑧𝑧𝑟𝑟𝐴𝐴𝐴𝐴 𝐴𝐴𝑟𝑟𝐴𝐴𝑟𝑟𝑟𝑟𝐴𝐴𝐴𝐴𝑟𝑟𝑟𝑟𝐴𝐴𝑢𝑢 𝑟𝑟𝑟𝑟 𝑥𝑥 = ∆𝑥𝑥 𝐴𝐴𝑟𝑟𝑟𝑟 𝑥𝑥 𝑝𝑝𝑟𝑟𝑟𝑟𝐴𝐴𝑟𝑟𝑟𝑟𝐴𝐴𝐴𝐴𝑔𝑔𝑟𝑟 𝐴𝐴𝑟𝑟𝐴𝐴𝑟𝑟𝑟𝑟𝐴𝐴𝐴𝐴𝑟𝑟𝑟𝑟𝐴𝐴𝑢𝑢 𝑟𝑟𝑟𝑟 𝑥𝑥 = ∆𝑥𝑥 × 100 𝑥𝑥 Combining uncertainties The following rules should be applied: ⎯ ⎯ For a sum or a difference, add the absolute uncertainties. For a product or a quotient, add the fractional (or the percentage) uncertainties. Adding or subtracting uncertainties 𝐴𝐴 = 𝑥𝑥 + 𝑢𝑢 𝑂𝑂𝑂𝑂 𝐴𝐴 = 𝑥𝑥 − 𝑢𝑢 ∆𝐴𝐴 = ∆𝑥𝑥 + ∆𝑢𝑢 Multiplying or dividing uncertainties 𝐴𝐴 = 𝑥𝑥𝑢𝑢 𝑂𝑂𝑂𝑂 𝐴𝐴 = 𝑥𝑥 𝑢𝑢 ∆𝐴𝐴 ∆𝑥𝑥 ∆𝑢𝑢 = + 𝑥𝑥 𝑢𝑢 𝐴𝐴 If there is a power on one or both then 𝐴𝐴 = 𝑥𝑥 𝑚𝑚 𝑢𝑢 𝑛𝑛 𝑂𝑂𝑂𝑂 𝐴𝐴 = 𝑥𝑥 𝑚𝑚 𝑢𝑢 𝑛𝑛 ∆𝑥𝑥 ∆𝑢𝑢 ∆𝐴𝐴 = 𝑚𝑚 � � + 𝑟𝑟 � � 𝑥𝑥 𝑢𝑢 𝐴𝐴 For example: SIR SAMEER UZ ZAMAN Notes: Summary: 45 of 223 𝑥𝑥 2 𝐴𝐴 = 𝑥𝑥 2 �𝑢𝑢 𝑂𝑂𝑂𝑂 𝐴𝐴 = �𝑢𝑢 ∆𝐴𝐴 ∆𝑥𝑥 1 ∆𝑢𝑢 = 2� � + � � 𝐴𝐴 𝑥𝑥 2 𝑢𝑢 Multiplying or dividing uncertainties in percentages 𝐴𝐴 = 𝑥𝑥𝑢𝑢 𝑂𝑂𝑂𝑂 𝐴𝐴 = 𝑥𝑥 𝑢𝑢 %∆𝐴𝐴 = (%∆𝑥𝑥 + %∆𝑢𝑢) If there is a power on one or both, then 𝑥𝑥 𝑚𝑚 𝑢𝑢 𝑛𝑛 𝐴𝐴 = 𝑥𝑥 𝑚𝑚 𝑢𝑢 𝑛𝑛 𝑂𝑂𝑂𝑂 𝐴𝐴 = %∆𝐴𝐴 = (𝑚𝑚 × %∆𝑥𝑥) + (𝑟𝑟 × %∆𝑢𝑢) For example: 𝐴𝐴 = 𝑥𝑥 2 �𝑢𝑢 𝑂𝑂𝑂𝑂 𝐴𝐴 = 𝑥𝑥 2 �𝑢𝑢 1 %∆𝐴𝐴 = (2 × %∆𝑥𝑥) + � × %∆𝑢𝑢� 2 Important note: if the power is negative then ignore the negative sign when calculating the uncertainty. For example 1 𝐴𝐴 = 𝑥𝑥 2 �𝑢𝑢 𝑂𝑂𝑂𝑂 𝐴𝐴 = 𝑥𝑥 2 𝑢𝑢 −2 ∆𝐴𝐴 ∆𝑥𝑥 1 ∆𝑢𝑢 = 2� � + � � 𝐴𝐴 𝑥𝑥 2 𝑢𝑢 Or if we are calculating percentage uncertainty then 1 𝐴𝐴 = 𝑥𝑥 2 �𝑢𝑢 𝑂𝑂𝑂𝑂 𝐴𝐴 = 𝑥𝑥 2 𝑢𝑢 −2 1 %∆𝐴𝐴 = (2 × %∆𝑥𝑥) + � × %∆𝑢𝑢� 2 In any case the uncertainties are always added. SIR SAMEER UZ ZAMAN Notes: Summary: 46 of 223 Example 2 2. A thermometer can be read to an accuracy of ±0.5 °C this thermometer is used to measure a temperature rise from 40 degrees to 100 degrees. What is the percentage uncertainty in the measurement of the temperature rise? a) 0.5% b) 0.8% c) 1.3% d) 1.7% Solution: Initial temperature = 40 ± 0.5 ℃ final temperature = 100 ± 0.5 ℃ value of rise = 100℃ − 40℃ = 60 ℃ % uncertainty in rise = 1 × 100 60 = 1.67 ≈ 1.7% Therefore, the correct choices d SIR SAMEER UZ ZAMAN Notes: Summary: 47 of 223 Example 3 3. Resistance of an unknown resistor is found by measuring the potential difference V across the resistor and the current through it. And by using the 𝑉𝑉 formula 𝑂𝑂 = 𝐼𝐼 .The voltmeter reading has a 3% uncertainty and ammeter has 2% uncertainty what is the uncertainty in the calculated resistance a) 1.5% b) 3% c) 5% d) 6% Solution: For products of quotient, we at the percentage uncertainties. 𝑂𝑂 = 𝑉𝑉 𝐼𝐼 %𝐴𝐴𝑟𝑟𝐴𝐴𝑟𝑟𝑟𝑟𝐴𝐴𝐴𝐴𝑟𝑟𝑟𝑟𝐴𝐴𝑢𝑢 𝑟𝑟𝑟𝑟 𝑂𝑂 = % 𝐴𝐴𝑟𝑟𝐴𝐴𝑟𝑟𝑟𝑟𝐴𝐴𝐴𝐴𝑟𝑟𝑟𝑟𝐴𝐴𝑢𝑢 𝑟𝑟𝑟𝑟 𝑉𝑉 + % 𝐴𝐴𝑟𝑟𝐴𝐴𝑟𝑟𝑟𝑟𝐴𝐴𝐴𝐴𝑟𝑟𝑟𝑟𝐴𝐴𝑢𝑢 𝑟𝑟𝑟𝑟 𝐼𝐼 =3+2 =5 Correct choice is c SIR SAMEER UZ ZAMAN Notes: Summary: 48 of 223 Question 8 SIR SAMEER UZ ZAMAN Notes: Summary: 49 of 223 SIR SAMEER UZ ZAMAN Notes: Summary: 50 of 223 Example 4 4. The density of material of a rectangular block is determined by measuring the linear dimensions of the block. The table shows the results obtained together with their uncertainties. mass = (25.0 ± 0.1) g length = (5.00 ± 0.01) cm breadth = (2.00 ± 0.01) cm height = (1.00 ± 0.01) cm The density is calculated to be 2.50 gcm−3 . What is the uncertainty in the result? a) ±0.01 gcm−3 b) ±0.02 gcm−3 c) ±0.05 gcm−3 d) ±0.13 gcm−3 Solution 𝐷𝐷𝑟𝑟𝑟𝑟𝑚𝑚𝑟𝑟𝐴𝐴𝑢𝑢 = 𝜌𝜌 = 𝑚𝑚𝐴𝐴𝑚𝑚𝑚𝑚 𝐴𝐴𝑟𝑟𝑟𝑟𝑔𝑔𝐴𝐴ℎ × 𝑏𝑏𝑟𝑟𝑟𝑟𝐴𝐴𝑟𝑟𝐴𝐴ℎ × ℎ𝑟𝑟𝑟𝑟𝑔𝑔ℎ𝐴𝐴 𝑚𝑚 25 = = 2.50 𝐴𝐴 × 𝑏𝑏 × ℎ 5 × 2 × 1 ∆𝜌𝜌 ∆𝑚𝑚 ∆𝐴𝐴 ∆𝑏𝑏 ∆ℎ = + + + 𝑚𝑚 𝐴𝐴 𝑏𝑏 ℎ 𝜌𝜌 0.1 0.01 0.01 0.01 ∆𝜌𝜌 = + + + 2.50 25.0 5.00 2.00 1.00 ∆𝜌𝜌 = 0.021 2.50 ∆𝜌𝜌 = 0.05 Correct choice is c SIR SAMEER UZ ZAMAN Notes: Summary: 51 of 223 Example 5 5. In an experiment a radio-controlled car takes 2.50 ± 0.05 s to travel 40.0 ± 0.1 m. What is the car's average speed and uncertainty in this value? a) 16 ±1 ms −1 b) 16.0 ±0.2 ms−1 c) 16.0 ±0.4 ms−1 d) 16.00 ±0.36 ms−1 Solution 𝐴𝐴𝑎𝑎𝑟𝑟𝑟𝑟𝐴𝐴𝑔𝑔𝑟𝑟 𝑚𝑚𝑝𝑝𝑟𝑟𝑟𝑟𝑟𝑟 = 𝑎𝑎 = 𝑟𝑟𝑟𝑟𝑚𝑚𝐴𝐴𝐴𝐴𝑟𝑟𝐴𝐴𝑟𝑟 𝐴𝐴𝑟𝑟𝑚𝑚𝑟𝑟 𝑚𝑚 40.0 = = 16.0 (3 𝑚𝑚𝑟𝑟𝑔𝑔. 𝑓𝑓𝑟𝑟𝑔𝑔. ) 𝐴𝐴 2.50 ∆𝑎𝑎 ∆𝑚𝑚 ∆𝐴𝐴 = + 𝑎𝑎 𝑚𝑚 𝐴𝐴 0.1 0.05 ∆𝑎𝑎 = + 16.0 40.0 2.50 ∆𝑎𝑎 = 0.36 ≈ 0.4 (1 𝑚𝑚𝑟𝑟𝑔𝑔. 𝑓𝑓𝑟𝑟𝑔𝑔) SIR SAMEER UZ ZAMAN Notes: Summary: 52 of 223 Question 9 SIR SAMEER UZ ZAMAN Notes: Summary: 53 of 223 Question 10 SIR SAMEER UZ ZAMAN Notes: Summary: 54 of 223 Question 11 SIR SAMEER UZ ZAMAN Notes: Summary: 55 of 223 CRO and Calibration Using Cathode Ray Oscilloscope The time interval is equal to the number of divisions from the peak of one spike to the peak of the other multiplied by the time base settings. 𝐴𝐴𝑟𝑟𝑚𝑚𝑟𝑟 = 𝑟𝑟𝑧𝑧 𝑧𝑧𝑓𝑓 (𝑟𝑟𝑟𝑟𝑎𝑎 𝑧𝑧𝑟𝑟 𝐴𝐴𝑚𝑚) × 𝐴𝐴𝑟𝑟𝑚𝑚𝑟𝑟 𝑏𝑏𝐴𝐴𝑚𝑚𝑟𝑟 𝑚𝑚𝑟𝑟𝐴𝐴𝐴𝐴𝑟𝑟𝑟𝑟𝑔𝑔 𝐴𝐴𝑟𝑟𝑚𝑚𝑟𝑟 𝑝𝑝𝑟𝑟𝑟𝑟𝑟𝑟𝑧𝑧𝑟𝑟 = 𝑟𝑟𝑧𝑧 𝑧𝑧𝑓𝑓 (𝑟𝑟𝑟𝑟𝑎𝑎 𝑧𝑧𝑟𝑟 𝐴𝐴𝑚𝑚) 𝑟𝑟𝑟𝑟 𝑧𝑧𝑟𝑟𝑟𝑟 𝐴𝐴𝑢𝑢𝐴𝐴𝐴𝐴𝑟𝑟 × 𝐴𝐴𝑟𝑟𝑚𝑚𝑟𝑟 𝑏𝑏𝐴𝐴𝑚𝑚𝑟𝑟 𝑚𝑚𝑟𝑟𝐴𝐴𝐴𝐴𝑟𝑟𝑟𝑟𝑔𝑔 The relationship between time period and frequency is given by = 1 𝑓𝑓 SIR SAMEER UZ ZAMAN Notes: Summary: 56 of 223 Question 1 Question 2 SIR SAMEER UZ ZAMAN Notes: Summary: 57 of 223 Question 3 Question 4 SIR SAMEER UZ ZAMAN Notes: Summary: 58 of 223 Question 5 SIR SAMEER UZ ZAMAN Notes: Summary: 59 of 223 Question 6 SIR SAMEER UZ ZAMAN Notes: Summary: 60 of 223 CALIBRATION The act of comparing an instrument to a known standard is calibrating it. For e.g. if you take a long wooden stick and use a standard metre rule to mark graduations on it, that wooden stick is now calibrated to measure length. Calibration curve These can be used to find out an unknown quantity by using another quantity. For e.g., the rotation of a dial (in degrees) can be used to measure pressure. The fig. below shows the calibration curve for that dial. In this particular case the advantage of this curve is that, this dial will be more sensitive at lower pressures and its disadvantage is that it is less sensitive at higher pressure. In other word the needle would rotate more at lower pressures (let’s say going from 0 to 1 pascals) and less at higher pressures (let’s say going from 8 to 9 pascals). SIR SAMEER UZ ZAMAN Notes: Summary: 61 of 223 Question 7 SIR SAMEER UZ ZAMAN Notes: Summary: 62 of 223 Question 8 SIR SAMEER UZ ZAMAN Notes: Summary: 63 of 223 SIR SAMEER UZ ZAMAN Notes: Summary: 64 of 223 Kinematics of linear motion Definition of Velocity It is defined as ‘The rate of change of displacement.’ 𝑎𝑎 = 𝑚𝑚 𝐴𝐴 (1) Average velocity If you have an initial velocity ‘u’ and a final velocity ‘v’ then the average velocity is given by the equation: 𝑎𝑎 = 𝑎𝑎 + 𝐴𝐴 2 (2) Definition of Acceleration it is defined as ‘The rate of change of velocity.’ Mathematically, 𝐴𝐴 = 𝑎𝑎 − 𝐴𝐴 𝐴𝐴 (3) Case 1: Linear motion with uniform (constant) velocity Since the velocity is constant, the acceleration is zero. The body covers the same distance in each second. Since the direction of motion is not changing, the velocitytime graph is the same as the speed-time graph. The magnitude of the velocity is constant. SIR SAMEER UZ ZAMAN Notes: Summary: 65 of 223 Since the velocity is the gradient of the displacement-time graph, a constant velocity means that the gradient of the displacement-time graph is constant, i.e. the displacement increases at a steady rate. Since there is no change in direction of motion of the body, the displacement-time graph is the same as the distance time graph. SIR SAMEER UZ ZAMAN Notes: Summary: 66 of 223 Case 2: Linear motion with uniform acceleration Uniform acceleration means it remains constant with time We know that acceleration is the gradient of the velocity-time graph. So if acceleration is constant, velocity-time graph has a constant gradient i.e. it will be a straight line with a constant gradient. SIR SAMEER UZ ZAMAN Notes: Summary: 67 of 223 SIR SAMEER UZ ZAMAN Notes: Increasing velocity means the gradient of displacement-time graph increases. i.e. the displacement increases at an increasing rate. Note that the displacement-time graph is flat near t=0, i.e. the gradient is zero at t=0. In case the velocity is not zero at t=0, the velocity-time graph is shown below Summary: 68 of 223 The displacement-time graph will not have a zero gradient at t=0 as shown below SIR SAMEER UZ ZAMAN Notes: Case 3: Linear motion with uniform (constant) deceleration When a body is decelerating, the velocity and acceleration have opposite directions, and therefore they have a opposite signs. If we take velocity as positive, we must take acceleration as negative. But if we decide to take acceleration as positive, then the velocity must be negative. In short, the important thing to remember is that in case of deceleration, velocity and acceleration have opposite signs. Summary: 69 of 223 Here we are taking velocity as positive and acceleration as negative. Thus acceleration graph is a horizontal negative straight line. Constant deceleration means the gradient of the velocity-time graph is constant, which means that the speed decreases at a steady rate. SIR SAMEER UZ ZAMAN Notes: Summary: 70 of 223 Decreasing velocity means the gradient of the displacement-time graph becomes less and less steep with time. In other words, since the object is slowing down its displacement increases at a dropping rate. Since the object is moving in a straight line without any change in direction, the displacement-time graphs are the same similarly the velocity-time graph and the speed-time graphs are identical. Equations of motion These equations are valid only for linear motion with constant acceleration, i.e. for case 2 and case 3. Let’s consider case 2 again. SIR SAMEER UZ ZAMAN Notes: Summary: 71 of 223 The velocity time diagram for this case is as follows: The initial velocity is u, the final velocity is v. The acceleration of the body is given by 𝐴𝐴 = 𝑎𝑎 − 𝐴𝐴 𝐴𝐴 (1) The area under the velocity-time graph is the displacement. 𝑚𝑚 = 𝐴𝐴𝑟𝑟𝑟𝑟𝐴𝐴 𝐴𝐴𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝐴𝐴ℎ𝑟𝑟 𝑎𝑎 − 𝐴𝐴 𝑔𝑔𝑟𝑟𝐴𝐴𝑝𝑝ℎ 𝑚𝑚 = 1 (𝐴𝐴 + 𝑎𝑎)𝐴𝐴 2 (2) Substituting v from eq (1) into eq (2) we get 𝑚𝑚 = 1 (𝐴𝐴 + 𝐴𝐴 + 𝐴𝐴𝐴𝐴)𝐴𝐴 2 𝑚𝑚 = 1 (2𝐴𝐴 + 𝐴𝐴𝐴𝐴)𝐴𝐴 2 1 𝑚𝑚 = 𝐴𝐴𝐴𝐴 + 𝐴𝐴𝐴𝐴 2 2 (3) SIR SAMEER UZ ZAMAN Notes: Summary: 72 of 223 (4) We know that 𝑎𝑎 − 𝐴𝐴 𝐴𝐴 𝑎𝑎 − 𝐴𝐴 𝐴𝐴 = 𝐴𝐴 𝐴𝐴 = Substituting this expression for t into eq (2) we get 𝑚𝑚 = 1 (𝑎𝑎 − 𝐴𝐴) (𝐴𝐴 + 𝑎𝑎 ) 2 𝐴𝐴 𝑚𝑚 = (𝑎𝑎 + 𝐴𝐴)(𝑎𝑎 − 𝐴𝐴) 2𝐴𝐴 𝑚𝑚 = (𝑎𝑎 2 + 𝐴𝐴2 ) 2𝐴𝐴 2𝐴𝐴𝑚𝑚 = 𝑎𝑎 2 − 𝐴𝐴2 Question 1 Question 2 (5) SIR SAMEER UZ ZAMAN 1 𝑚𝑚 = 𝑎𝑎𝐴𝐴 − 𝐴𝐴𝐴𝐴 2 2 Notes: Summary: 73 of 223 Question 3 Question 4 SIR SAMEER UZ ZAMAN Notes: Summary: 74 of 223 SIR SAMEER UZ ZAMAN Notes: Summary: 75 of 223 Question 5 SIR SAMEER UZ ZAMAN Notes: Summary: 76 of 223 SIR SAMEER UZ ZAMAN Notes: Summary: 77 of 223 Combining Case 2 and Case 3 (Vertical motion under gravity) A cricket ball is thrown vertically upwards. It reaches the highest point and returns to the throwers hand. Considering the effect of air resistance to be negligible, sketch the s-t the, v-t and a-t graphs for the motion of the ball. (take upward positive.) Since the upward direction is taken as positive, the acceleration due to gravity will be negative because it always points downwards. The magnitude of the acceleration is constant because the air resistance is negligible. The velocity is positive in the upward motion, and negative in the downward motion. The speed decreases in the upward motion and increases in the downward motion. Since the acceleration is constant, the gradient of the velocity-time graph will be constant i.e. the velocity time graph will be a straight line with a gradient equal to -9.81 𝑚𝑚𝑚𝑚 2 . SIR SAMEER UZ ZAMAN Notes: Summary: 78 of 223 The ball is thrown vertically upwards at t = 0, with a velocity u. At t = t1 , the ball reaches the highest point where it momentarily comes to rest. From t = 0 to t = t1 , the speed decreases until it becomes zero at the maximum height. After t = t1 the body moves downwards with an increasing speed. The direction of the velocity is downwards, and therefore the sign of the velocity is negative. At t= t2 , the ball is caught by the thrower. From t = 0 to t = t1 , the ball is slowing, therefore the displacement increases at a decreasing rate. In the downward motion from t = t1 to t = t2 , the displacement decreases at an increasing rate. Since the direction of motion is changing, speed time graph is different from the velocity time graph, and the distance time graph is different from the displacement time graph. SIR SAMEER UZ ZAMAN Notes: Summary: 79 of 223 SIR SAMEER UZ ZAMAN Notes: Summary: 80 of 223 Question 6 SIR SAMEER UZ ZAMAN Notes: Summary: 81 of 223 SIR SAMEER UZ ZAMAN Notes: Summary: 82 of 223 Question 7 SIR SAMEER UZ ZAMAN Notes: Summary: 83 of 223 SIR SAMEER UZ ZAMAN Notes: Summary: 84 of 223 Question 8 Question 9 Question 10 Question 11 SIR SAMEER UZ ZAMAN Notes: Summary: 85 of 223 Question 12 Question 13 SIR SAMEER UZ ZAMAN Notes: Summary: 86 of 223 SIR SAMEER UZ ZAMAN Notes: Summary: 87 of 223 Rebounding or Bouncing A steel ball is dropped from a height above the hard horizontal surface. The ball bounces several times before coming to rest. Ignoring the effect of air resistance, sketch the acceleration-time graph, velocity-time graph, displacement-time and speed-time graphs for the motion of the ball. (Take upwards as positive). Acceleration time graph When the ball is not in contact with the surface, the acceleration of the ball is the acceleration due to gravity which always points downwards and has a constant magnitude. thus, when the ball is not in contact with the surface, its acceleration is -9.81 𝑚𝑚𝑚𝑚 −2 regardless of how the ball is moving. When the ball hits the surface, it experiences a large upward force from the surface which momentarily changes the direction (and magnitude) of acceleration. So we see “spikes” in the graph for those short periods when the ball is in contact with the surface. At other times, the acceleration remains constant at -9.81 𝑚𝑚𝑚𝑚 −2 . Velocity-time graph As the ball is dropped at t=0 (with initial velocity zero) it gains a downward velocity. Thus, the sign of the velocity is negative and the speed increases at a steady rate (because the acceleration is constant). At t=t1 the ball hits the surface and the direction of the velocity reverses suddenly. The ball then begins to rise and the speed decreases at a steady rate util it becomes zero at t=t2 when the ball reaches maximum height after the first bounce. The ball then starts to move down. The SIR SAMEER UZ ZAMAN Notes: Summary: 88 of 223 velocity of the ball becomes negative, and the speed increases until it hits the surface for the second time at t=t3 Notice that all slant lines are parallel as they all have the same gradient -9.81𝑚𝑚𝑚𝑚 −2 Displacement-time graph Speed-time graph Let’s sketch first the velocity time graph for one bounce only. Since the speed is always positive, the speed-time graph can be obtained by flipping the negative part of the velocity-time graph. SIR SAMEER UZ ZAMAN Notes: Summary: 89 of 223 Question 14 Question 15 SIR SAMEER UZ ZAMAN Notes: Summary: 90 of 223 Question 16 Question 17 SIR SAMEER UZ ZAMAN Notes: Summary: 91 of 223 Question 18 Question 19 SIR SAMEER UZ ZAMAN Notes: Summary: 92 of 223 SIR SAMEER UZ ZAMAN Notes: Summary: 93 of 223 SIR SAMEER UZ ZAMAN Notes: Summary: 94 of 223 Question 20 ` SIR SAMEER UZ ZAMAN Notes: Summary: 95 of 223 Projectile motion motion of the object is shown below. (Effect of air resistance is negligible). If the effect of air resistance is negligible, then projectile motion is a twodimensional motion. Moreover, the motion is also symmetrical, which means that SIR SAMEER UZ ZAMAN Notes: Summary: 96 of 223 the time for the object to move from the ground to the highest point is equal to the time for it to move from the highest point back to the ground. Since the motion is not linear, we can’t use the equations of motion directly. The important thing to note is that as the object moves it covers the horizontal distance as well as vertical distance. In other words, projectile motion is a combination of horizontal and vertical motions. In the absence of air resistance, the horizontal part of the motion and the vertical part of the motion are independent of each other, and therefore can be studied separately using equations of motion. Horizontal component of the motion Vertical component of the motion 1- There is no acceleration in the horizontal direction, therefore the horizontal component of velocity stays constant. 𝑎𝑎 = 𝐴𝐴 = 𝐴𝐴 cos 2- The horizontal displacement can be found using the equation 𝑥𝑥 = (𝐴𝐴 cos )𝐴𝐴 1- There is a constant downward acceleration a=g=-9.81 𝑚𝑚𝑚𝑚 −2 . The vertical component of velocity at any time can be calculated by using the equations. 𝑎𝑎 = 𝐴𝐴 + 𝐴𝐴𝐴𝐴 𝑎𝑎 = 𝐴𝐴 sin +(−9.81)𝐴𝐴 Also, 𝑎𝑎 2 = 𝐴𝐴 2 + 2𝐴𝐴𝑢𝑢 𝑎𝑎 2 = 𝐴𝐴 2 + 2(−9.81)𝑢𝑢 2- The vertical displacement at any time can be found using the equation 1 𝑢𝑢 = 𝐴𝐴 𝐴𝐴 + 𝐴𝐴𝐴𝐴 2 2 1 𝑢𝑢 = (𝐴𝐴 sin )𝐴𝐴 + (−9.81)𝐴𝐴 2 2 Once vx and vy are found, we can calculate the magnitude and direction of the resultant velocity by 𝑎𝑎 2 = 𝑎𝑎 2 + 𝑎𝑎 2 𝑎𝑎 = 2 + 𝑎𝑎 2 𝑎𝑎 SIR SAMEER UZ ZAMAN Notes: Summary: 97 of 223 And we can calculate the angle with tan Question 21 = 𝑎𝑎 𝑎𝑎 SIR SAMEER UZ ZAMAN Notes: Summary: 98 of 223 Question 22 SIR SAMEER UZ ZAMAN Notes: Summary: 99 of 223 Question 23 SIR SAMEER UZ ZAMAN Notes: Summary: 100 of 223 Half-projectile motion An object is projected horizontally from a height with a velocity “u”. The motion of the object is shown below. (Effect of air resistance is negligible). In the absence of air resistance, the horizontal part of the motion and the vertical part of the motion are independent of each other, and therefore can be studied separately using equations of motion. For the vertical motion we can consider the object to be falling from rest i.e., its initial vertical velocity is going to be zero. SIR SAMEER UZ ZAMAN Notes: Summary: 101 of 223 Horizontal component of the halfprojectile motion 1- There is no acceleration in the horizontal direction, therefore the horizontal component of velocity stays constant. 𝑎𝑎 = 𝐴𝐴 2- The horizontal displacement can be found using the equation 𝑥𝑥 = 𝐴𝐴 . 𝐴𝐴 Since time for both components stay the same and 𝐴𝐴 = 2𝑢𝑢 𝑔𝑔 So 𝑥𝑥 = 𝐴𝐴 2𝑢𝑢 𝑔𝑔 Vertical component of the halfprojectile motion 1- There is a constant downward acceleration a=g=-9.81 𝑚𝑚𝑚𝑚 −2 . The vertical component of velocity at any time can be calculated by using the equations. 𝑎𝑎 = 𝐴𝐴 + 𝐴𝐴𝐴𝐴 Since 𝐴𝐴 = 0 𝑎𝑎 = (−9.81)𝐴𝐴 Also, 𝑎𝑎 2 = 𝐴𝐴 2 + 2𝐴𝐴𝑢𝑢 Since 𝐴𝐴 = 0 𝑎𝑎 2 = 2(−9.81)𝑢𝑢 2- The vertical displacement at any time can be found using the equation 1 𝑢𝑢 = 𝐴𝐴 𝐴𝐴 + 𝐴𝐴𝐴𝐴 2 2 Since 𝐴𝐴 = 0 1 𝑢𝑢 = (−9.81)𝐴𝐴 2 2 3- Time can be found using 𝐴𝐴 = 2𝑢𝑢 𝑔𝑔 SIR SAMEER UZ ZAMAN Notes: Summary: 102 of 223 Question 24 SIR SAMEER UZ ZAMAN Notes: Summary: 103 of 223 Question 25 SIR SAMEER UZ ZAMAN Notes: Summary: 104 of 223 SIR SAMEER UZ ZAMAN Notes: Summary: 105 of 223 Dynamics Definition of mass Mass is the property of a body that resists change in motion. Weight It is the force with which gravity attracts an object. This force is considered to act on the center of gravity of the object. Mathematically it is the product of mass and the acceleration of freefall. = 𝑚𝑚𝑔𝑔 First law of motion An object at rest will stay at rest and an object moving with constant velocity will continue to move at constant velocity, unless acted upon by an external force. - This law describes object with no net force or where all forces on the object are balanced. - Since there are no net forces acting on the object it will not accelerate. SIR SAMEER UZ ZAMAN Notes: Summary: 106 of 223 Example 1 Second law of Newton The resultant force acting on an object is equal to the rate of change of its momentum. The resultant force and the change in momentum are in the same direction. This statement defines what we mean by a force. Force is an interaction that causes an object’s momentum to change. So, if an object’s momentum is changing, there must be a force acting on it. 𝐹𝐹𝑧𝑧𝑟𝑟𝐴𝐴𝑟𝑟 = 𝑟𝑟𝐴𝐴𝐴𝐴𝑟𝑟 𝑧𝑧𝑓𝑓 𝐴𝐴ℎ𝐴𝐴𝑟𝑟𝑔𝑔𝑟𝑟 𝑧𝑧𝑓𝑓 𝑚𝑚𝑧𝑧𝑚𝑚𝑟𝑟𝑟𝑟𝐴𝐴𝐴𝐴𝑚𝑚 𝐹𝐹𝑛𝑛 The unit for force is Newton = ∆ ∆𝐴𝐴 SIR SAMEER UZ ZAMAN Notes: Summary: 107 of 223 Momentum It is defined as the product of mass and velocity. This is known as Linear momentum or momentum. = 𝑚𝑚𝑎𝑎 Where is momentum, 𝑎𝑎 is velocity & 𝑚𝑚 is mass. The unit of momentum is kg m s-1. It is a vector quantity, and it has the same direction as the object’s velocity. A special case of Newton’s second law of motion SIR SAMEER UZ ZAMAN Notes: Imagine an object of constant mass ‘𝑚𝑚’ acted upon by a resultant force 𝐹𝐹𝑛𝑛 . The force will change the momentum of the object. According to Newton’s second law of motion, we have: ∆ ∆𝐴𝐴 𝑚𝑚𝑎𝑎 − 𝑚𝑚𝐴𝐴 = ∆𝐴𝐴 𝐹𝐹𝑛𝑛 𝐹𝐹𝑛𝑛 = Where ‘𝐴𝐴’ is the initial velocity of the object, ‘𝑎𝑎’ is the final velocity of the object and ‘ 𝐴𝐴’ is the time taken for the change in velocity. The mass m of the object is constant; hence the above equation can be rewritten as: 𝐹𝐹𝑛𝑛 = 𝑚𝑚 𝑎𝑎 − 𝐴𝐴 ∆𝐴𝐴 since 𝐴𝐴 = 𝑎𝑎 − 𝐴𝐴 ∆𝐴𝐴 𝐹𝐹𝑛𝑛 = 𝑚𝑚𝐴𝐴 So, The equation we used above, F = ma, is a simplified version of Newton’s second law of motion. For a body of constant mass, its acceleration is directly proportional to the resultant force applied to it. Summary: 108 of 223 Force-time graph The area under force-time graph represents the change in momentum or impulse. Mathematically; (𝑟𝑟𝑚𝑚𝑝𝑝𝐴𝐴𝐴𝐴𝑚𝑚𝑟𝑟) ∆ = 𝐹𝐹 × 𝐴𝐴 Defining the newton One newton is the force that will give a 1 kg mass an acceleration of 1 𝑚𝑚 𝑚𝑚 −2 in the direction of the force. Examples question for Newton’s 2nd law Example 2 SIR SAMEER UZ ZAMAN Notes: Summary: 109 of 223 Example 3 Example 4 SIR SAMEER UZ ZAMAN Notes: Summary: 110 of 223 Example 5 Question 1 SIR SAMEER UZ ZAMAN Notes: Summary: 111 of 223 Example 6 SIR SAMEER UZ ZAMAN Notes: Summary: 112 of 223 Example 7 SIR SAMEER UZ ZAMAN Notes: Summary: 113 of 223 Example 8 SIR SAMEER UZ ZAMAN Notes: Summary: 114 of 223 Question 2 SIR SAMEER UZ ZAMAN Notes: Summary: 115 of 223 SIR SAMEER UZ ZAMAN Notes: Summary: 116 of 223 Question 3 SIR SAMEER UZ ZAMAN Notes: Summary: 117 of 223 Question 4 (iv) calculate the force exerted by the floor. SIR SAMEER UZ ZAMAN Notes: Summary: 118 of 223 Question 5 SIR SAMEER UZ ZAMAN Notes: Summary: 119 of 223 Objects on an inclined plane When an object is placed on an inclined plane it will naturally slide down the slope, and the greater the slope the faster it slides down. This is due to the weight of the object being pulled down due to gravity. But not all of its weight is responsible for this sliding motion, only a component of it is responsible. This can be illustrated by resolving the weight of the object with respect to the inclined plane. The weight of the object act vertically downward we can resolve the weight into two components, component of weight that is perpendicular to the slope. = 𝑚𝑚𝑔𝑔 cos And component of weight that is parallel to the slope also known as component of weight down the slope. = 𝑚𝑚𝑔𝑔 𝑚𝑚𝑟𝑟𝑟𝑟 𝑚𝑚𝑔𝑔 cos is balanced by the normal reaction force would not contribute toward its motion. applied by the slope so it Acceleration down the slope (frictionless surface) the ground the net force on it is 𝐹𝐹𝑛𝑛 = 𝑚𝑚𝑔𝑔 sin 𝑚𝑚𝐴𝐴 = 𝑚𝑚𝑔𝑔 sin Hence the acceleration down the slope is giver by 𝐴𝐴 = 𝑔𝑔 sin The greater the slope, the greater the object’s acceleration when sliding down the slope. SIR SAMEER UZ ZAMAN Notes: Summary: 120 of 223 Example 8 SIR SAMEER UZ ZAMAN Notes: Summary: 121 of 223 Question 6 SIR SAMEER UZ ZAMAN Notes: Summary: 122 of 223 Question 7 SIR SAMEER UZ ZAMAN Notes: Summary: 123 of 223 Question 8 Show that K.E = From the definition of momentum, we have = 𝑚𝑚𝑎𝑎 By squaring both sides, we get 2 2 = 𝑚𝑚2 𝑎𝑎 2 = 𝑚𝑚. 𝑚𝑚. 𝑎𝑎 2 2 𝑚𝑚 = 𝑚𝑚. 𝑎𝑎 2 Dividing by 2 on both sides, we get 2 2𝑚𝑚 = 1 𝑚𝑚. 𝑎𝑎 2 2 = 1 𝑚𝑚𝑎𝑎 2 2 We know that So, 2 2𝑚𝑚 = SIR SAMEER UZ ZAMAN Notes: Summary: 124 of 223 Question 9 Frictional and Viscous drag forces When an object moves on a surface friction acts on it and when an object moves through a fluid for e.g., air or water drag force acts on it. Both of these forces act in the opposite direction of motion. Drag forces increase the faster an object moves, so an object accelerating through a fluid will eventually reach terminal velocity. As the drag force increases and becomes equal to the forward accelerating force. Falling under gravity with air resistance (terminal velocity) As an object falls down its acceleration is ‘g’ however, the object is falling under air resistance so as the velocity increases the acceleration decreases because the net force (𝐹𝐹𝑛𝑛 = 𝑟𝑟𝑟𝑟𝑔𝑔ℎ𝐴𝐴 − 𝐴𝐴𝑟𝑟𝑟𝑟 𝑟𝑟𝑟𝑟𝑚𝑚𝑟𝑟𝑚𝑚𝐴𝐴𝐴𝐴𝑟𝑟𝐴𝐴𝑟𝑟) decreases. As the object continues to increase its velocity the air resistance also continues to increase. Eventually the air resistance becomes as large as the weight of the object and net force on it becomes zero and it reaches terminal velocity. SIR SAMEER UZ ZAMAN Notes: Summary: 125 of 223 Question 10 SIR SAMEER UZ ZAMAN Notes: Summary: 126 of 223 SIR SAMEER UZ ZAMAN Notes: Summary: 127 of 223 SIR SAMEER UZ ZAMAN Notes: Summary: 128 of 223 Newton’s third law When two bodies interact, the forces they exert on each other are equal and opposite. These two forces are sometimes described as action and reaction, but this is misleading as it sounds as though one force arises as a consequence of the other. In fact, the two forces appear at the same time and we can’t say that one caused the other. The two forces which make up a ‘Newton’s third law pair’ have the following characteristics: SIR SAMEER UZ ZAMAN Notes: They act on different objects. They exist for the same amount of time. They are equal in magnitude. They are opposite in direction. They are forces of the same type. Same type of forces can mean that: Two objects may attract each other because of the gravity of their masses – these are gravitational forces. Two objects may attract or repel because of their electrical charges – electrical forces. Two objects may touch – contact forces. Two objects may be attached by a string and pull on each other – tension forces. Two objects may attract or repel because of their magnetic fields – magnetic forces. Summary: 129 of 223 For example, if there are two objects A and B. If these objects collide with each other the object A will exert a force on object B and object B will exert a force on object A. (these forces are contact forces) These forces will be equal in magnitude, opposite in direction and will exist for the same amount of time (the time that they are in contact with each other). In the above example an object A is colliding with another object B, as they collide with each other they exert a force on one another during this collision. FAB is the force on object A applied by object B and FBA is the force on object B applied by object A. These forces exist only during the collision hence they exist for the same amount of time, these forces are on different objects and equal in magnitude. Note: We can use the third law to prove law of conservation of momentum or use the law of conservation of momentum to prove third law. Question 11 SIR SAMEER UZ ZAMAN Notes: Summary: 130 of 223 Question 12 Single body collision/collision with a rigid surface (wall) Perfectly Elastic collisions If an object moving with velocity v collides with a surface it will bounce back with its velocity reversed Inelastic Collison In this case object will bounce back with its velocity less than v Inelastic collision (comes to a halt after Collison) In this case the object will be at rest with its velocity to be zero SIR SAMEER UZ ZAMAN Notes: Summary: 131 of 223 Principle/law of conservation of momentum 1. In a closed system (i.e., no external force is acting on the system) 2. The total momentum always remains constant. This can also apply to colliding bodies, in that case total momentum before the collision is equal to the total momentum after the collision. Types of collisions For all types of collisions Momentum and total energy are always conserved. Elastic collision These collisions have the following properties: • • Kinetic energy is conserved (KE before and after the collision is the same) Speed of approach is equal to the speed of separation. Speed of approach and speed of separation is determined as follows: • • If the objects are moving in opposite direction, then their speeds are added. If the objects are moving in the same direction, then their speeds are subtracted. To check if the collision is elastic or not, we can use two methods. Method 1 Find total KE before the collision and compare it to the total KE after the collision. If they are equal, then the collision is elastic otherwise it’s not. Method 2 Find speed of approach and the speed of separation if they are equal, then the collision is elastic otherwise it’s not. SIR SAMEER UZ ZAMAN Notes: Summary: 132 of 223 Question 13 SIR SAMEER UZ ZAMAN Notes: Summary: 133 of 223 Inelastic collisions • • Kinetic energy is not conserved. Speed of approach is not equal to the speed of separation. SIR SAMEER UZ ZAMAN Notes: Summary: 134 of 223 Example 9: Interactions where one object splits into two or two objects that stick together to form one single object will always be inelastic. Question 14 SIR SAMEER UZ ZAMAN Notes: Summary: 135 of 223 Question 15 SIR SAMEER UZ ZAMAN Notes: Summary: 136 of 223 Question 16 Question 17 SIR SAMEER UZ ZAMAN Notes: Using law of conservation of momentum to prove third law Since law of conservation of momentum states that in a closed system momentum is conserved. During Collison momentum lost by one object must be equal to the momentum gained by the other object so if PA is the momentum change for object A after collision and PB is the momentum change for object B. Since momentum lost by one object is equal to the momentum gained by the other. ∆ = −∆ And ∆P = Ft from second law of motion 𝐹𝐹 𝐴𝐴 = −𝐹𝐹 𝐴𝐴 Summary: 137 of 223 Since t = t (during the collision they remain in contact for the same amount of time) 𝐹𝐹 = −𝐹𝐹 Using 2nd and 3rd law to prove law of conservation of momentum According to second law 𝐹𝐹 = According to third law ∆ 𝐴𝐴 𝐹𝐹 = −𝐹𝐹 And time for which they remain in contact is the same so, 𝐴𝐴 = 𝐴𝐴 = 𝐴𝐴 Therefore, ∆ ∆ =− 𝐴𝐴 𝐴𝐴 Since t = t = t ∆ = −∆ Question 18 SIR SAMEER UZ ZAMAN Notes: Summary: 138 of 223 SIR SAMEER UZ ZAMAN Notes: Summary: 139 of 223 Oblique collisions/2D collisions m1 has momentum, and this is in the forward direction. During the collision, this momentum is shared between the two masses. We can see this because each has a component of velocity in the forward direction. At the same time, each ball gains momentum in the y-direction, because each has a y-component of velocity. These must be equal in magnitude and opposite in direction, otherwise we would conclude that momentum had been created out of nothing. The mass m1 moves at a greater angle, but its velocity is less than that of the mass m2. Since linear momentum is always conserved. For this type of collision, the momentum in x-direction and y-direction is conserved. For x 𝑚𝑚1 𝐴𝐴 = 𝑚𝑚1 𝑎𝑎1 𝐴𝐴𝑧𝑧𝑚𝑚 1 + 𝑚𝑚2 𝑎𝑎2 𝐴𝐴𝑧𝑧𝑚𝑚 2 For y (the object is moving horizontally so it has no y-component before the impact) 0 = 𝑚𝑚1 𝑎𝑎1 𝑚𝑚𝑟𝑟𝑟𝑟 1 0 = 𝑚𝑚1 𝑎𝑎1 𝑚𝑚𝑟𝑟𝑟𝑟 𝑚𝑚1 𝑎𝑎1 𝑚𝑚𝑟𝑟𝑟𝑟 1 + 𝑚𝑚2 (−𝑎𝑎2 𝑚𝑚𝑟𝑟𝑟𝑟 1 − 𝑚𝑚2 𝑎𝑎2 𝑚𝑚𝑟𝑟𝑟𝑟 = 𝑚𝑚2 𝑎𝑎2 𝑚𝑚𝑟𝑟𝑟𝑟 2 2) 2 SIR SAMEER UZ ZAMAN Notes: Summary: 140 of 223 Question 19 SIR SAMEER UZ ZAMAN Notes: Summary: 141 of 223 Question 20 SIR SAMEER UZ ZAMAN Notes: Summary: 142 of 223 Question 21 SIR SAMEER UZ ZAMAN Notes: Summary: 143 of 223 Forces, Density and Pressure Center of gravity The centre of gravity of an object is defined as the point where its entire weight appears to act. Turning effect of force / Moment of a force The quantity which tells us about the turning effect of a force is its moment. The moment of a force depends on two quantities: • • The magnitude of the force (the bigger the force, the greater its moment) The perpendicular distance of the force from the pivot (the further the force acts from the pivot, the greater its moment). The moment of a force is defined as follows: The moment of a force is equal to the product of force and perpendicular distance of the pivot from the line of action of the force. Mathematically, 𝑀𝑀𝑧𝑧𝑚𝑚𝑟𝑟𝑟𝑟𝐴𝐴𝑚𝑚 = 𝐹𝐹. 𝑟𝑟 Moment is a vector quantity. SI unit is Nm. Line of action It is a geometric representation of how the force is applied. It is the line through the point at which the force is applied in the direction of force. If the force is not perpendicular to the surface, then there are two methods to find moment SIR SAMEER UZ ZAMAN Notes: Summary: 144 of 223 Find the perpendicular distance with respect to the force For this condition, first we draw the line of action of the force and then we find the perpendicular distance by drawing a perpendicular line from the pivot to the line of action of the force. Then the moment is given by force multiplied by the perpendicular distance. 𝑀𝑀𝑧𝑧𝑚𝑚𝑟𝑟𝑟𝑟𝐴𝐴 = 𝐹𝐹 × 𝑟𝑟𝑚𝑚𝑟𝑟𝑟𝑟 Find the perpendicular component of force with respect to the distance We can also find the component of the force that is perpendicular to the distance. 𝑀𝑀𝑧𝑧𝑚𝑚𝑟𝑟𝑟𝑟𝐴𝐴 = 𝐹𝐹 sin × 𝑟𝑟 = 𝐹𝐹. 𝑟𝑟𝑚𝑚𝑟𝑟𝑟𝑟 Principle of moments It states that in equilibrium the sum of all moments is zero or net moments are zero Or the sum of all clockwise moments is equal to the sum of all anti-clockwise moments 𝐴𝐴𝑚𝑚 𝑧𝑧𝑓𝑓 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝐴𝐴𝑧𝑧𝐴𝐴 𝑟𝑟𝑚𝑚𝑟𝑟 𝑚𝑚𝑧𝑧𝑚𝑚𝑟𝑟𝑟𝑟𝐴𝐴𝑚𝑚 = 𝐴𝐴𝑚𝑚 𝑧𝑧𝑓𝑓 𝐴𝐴𝐴𝐴𝐴𝐴 𝐴𝐴𝑟𝑟𝐴𝐴𝑟𝑟𝐴𝐴𝐴𝐴𝑧𝑧𝐴𝐴 𝑟𝑟𝑚𝑚𝑟𝑟 𝑚𝑚𝑧𝑧𝑚𝑚𝑟𝑟𝑟𝑟𝐴𝐴𝑚𝑚 SIR SAMEER UZ ZAMAN Notes: Summary: 145 of 223 Equilibrium An object is said to be an equilibrium if these two conditions are met 1- Sum of all forces acting on it is zero or there is no resultant force 2- Sum of all clockwise moments is equal to sum of all anticlockwise moments Torque of a couple A pair of equal and anti-parallel forces are called couple, and they produce rotation only; that’s torque of a couple. It is the product of one of the forces and the perpendicular distance between the forces. 𝑧𝑧𝑟𝑟 𝐴𝐴𝑟𝑟 ( ) = 𝐹𝐹 × 𝑟𝑟 Find the resultant moment for the following: Question 1 SIR SAMEER UZ ZAMAN Notes: Summary: 146 of 223 Question 2 Question 3 SIR SAMEER UZ ZAMAN Notes: Summary: 147 of 223 Question 4 Question 5 SIR SAMEER UZ ZAMAN Notes: Summary: 148 of 223 Quesiton 6 SIR SAMEER UZ ZAMAN Notes: Summary: 149 of 223 Example 1 If line of action for all forces passes through a single point the object will be in equilibrium because the distance from P for all the force would be zero hence net moment about p would also be zero 𝐹𝐹𝐴𝐴𝑧𝑧𝑚𝑚 + (− 𝐴𝐴𝑧𝑧𝑚𝑚 ) = 0 𝐹𝐹𝑚𝑚𝑟𝑟𝑟𝑟 + 𝑚𝑚𝑟𝑟𝑟𝑟 + (− ) = 0 2 × 𝐴𝐴 = 𝑚𝑚𝑟𝑟𝑟𝑟 × 𝐴𝐴 3 SIR SAMEER UZ ZAMAN Notes: Summary: 150 of 223 Question 7 SIR SAMEER UZ ZAMAN Notes: Summary: 151 of 223 Question 8 SIR SAMEER UZ ZAMAN Notes: Summary: 152 of 223 Question 9 SIR SAMEER UZ ZAMAN Notes: Summary: 153 of 223 Density Mass per unit volume = m V P= F A Pressure It is defined as Force per unit area Derivation of hydrostatic pressure P=ρgh P= F A consider a column of water with a height h and base area A Since we know that the force exerted will be due to the weight so, P= W A P= mg A We know that W=mg, For the definition of density m= P= Vg A P= Ahg A And P = gh Static fluid pressure does not depend on the shape or total mass of the fluid. SIR SAMEER UZ ZAMAN Notes: Summary: 154 of 223 Question 10 SIR SAMEER UZ ZAMAN Notes: Summary: 155 of 223 SIR SAMEER UZ ZAMAN Notes: Summary: 156 of 223 Question 11 SIR SAMEER UZ ZAMAN Notes: Summary: 157 of 223 Upthrust = (Archimedes’ principle) It is a force exerted by fluid when an object is immersed in it completely or partially. This force is exerted because of the difference in pressure at the top and bottom. Pressure on the top is less than the pressure at the bottom, because top is closer to the surface it experiences less pressure than the bottom which is at a greater depth. The pressure on the sides do not matter because the pressure on either side is equal so they cancel out. So, the upthrust is caused by the difference of the two forces. 𝐹𝐹 And we know that = 𝐹𝐹 𝑚𝑚 − 𝐹𝐹 = and 𝐹𝐹 = 𝐴𝐴 𝐹𝐹 = 𝐹𝐹 =( 𝐹𝐹 2 𝐴𝐴 2 − − = 1 𝐴𝐴 1 )𝐴𝐴 𝐴𝐴 We also know that ∆ = 𝜌𝜌𝑔𝑔ℎ2 − 𝜌𝜌𝑔𝑔ℎ1 ∆ = 𝜌𝜌𝑔𝑔∆ℎ So, 𝐹𝐹 = 𝜌𝜌𝑔𝑔∆ℎ𝐴𝐴 𝐹𝐹 = 𝜌𝜌𝑔𝑔 𝑉𝑉 SIR SAMEER UZ ZAMAN Notes: the volume of the object that is immersed in it. Summary: 158 of 223 The formula above shows that the Upthrust is dependent on the volume of the object immersed (fully or partially) in the fluid. If the object is fully immersed, then at any depth (be it closer to the surface or at the very bottom) the Upthrust remains the same. Upthrust is simply equal to the weight of the liquid displaced by the immersed object. Question 12 Question 13 Question 14 SIR SAMEER UZ ZAMAN Notes: Summary: 159 of 223 Question 15 Question 16 SIR SAMEER UZ ZAMAN Notes: Summary: 160 of 223 Question 17 SIR SAMEER UZ ZAMAN Notes: Summary: 161 of 223 SIR SAMEER UZ ZAMAN Notes: Summary: 162 of 223 Question 18 SIR SAMEER UZ ZAMAN Notes: Summary: 163 of 223 Question 19 SIR SAMEER UZ ZAMAN Notes: Summary: 164 of 223 Question 20 Question 21 SIR SAMEER UZ ZAMAN Notes: Summary: 165 of 223 Question 22 Question 23 SIR SAMEER UZ ZAMAN Notes: Summary: 166 of 223 Question 24 SIR SAMEER UZ ZAMAN Notes: Summary: 167 of 223 Question 25 SIR SAMEER UZ ZAMAN Notes: Summary: 168 of 223 WORK POWER AND ENERGY Work done It is the product of force and displacement in the direction of force. If the force and displacement are already parallel to one another as shown above = 𝐹𝐹. 𝑚𝑚 When force and displacement are at an angle as shown on the right, then we have to either resolve the force or the displacement in order to find work. = 𝐹𝐹. 𝑚𝑚 cos If force and displacement are perpendicular to one another then work done is zero. Work done against friction or drag Work done in overcoming friction or any sort of resistive force such as drag produces heat. Which means work done against friction or to overcome it transforms into heat energy. 𝑛𝑛 𝑛𝑛 = 𝑟𝑟𝐴𝐴𝐴𝐴 = 𝐹𝐹 . 𝑚𝑚 Work transforming into other forms Work can be changed into other forms of energy such as KE, PE, Heat, etc. SIR SAMEER UZ ZAMAN Notes: Summary: 169 of 223 Work done into KE If a force F is applied on an object, it will accelerate and cover some displacement. During this acceleration velocity will go from ‘u’ to ‘v’ as it travels a displacement ‘s’ If the surface is smooth then =∆ Since W=F.s 𝐹𝐹. 𝑚𝑚 = ∆ And F=ma 𝑚𝑚𝐴𝐴𝑚𝑚 = ∆ Form equations of motion we know that 2𝐴𝐴𝑚𝑚 = 𝑎𝑎 2 − 𝐴𝐴2 𝑧𝑧𝑟𝑟 𝐴𝐴 = 𝑎𝑎 2 − 𝐴𝐴2 2𝑚𝑚 Substituting acceleration, we get 1 𝑚𝑚(𝑎𝑎 2 − 𝐴𝐴2 ) = ∆ 2 If u=0 then, = 1 𝑚𝑚𝑎𝑎 2 2 The equations above show that a constant force produces a change in KE. Note: if velocity is constant there is no acceleration hence net force is zero so there will be no change in KE Question 1 SIR SAMEER UZ ZAMAN Notes: Summary: 170 of 223 Question 2 Work done into GPE If an object is raised form a height of ℎ1 to ℎ2 against gravity with constant velocity, then 𝑧𝑧𝑟𝑟 𝑟𝑟𝑧𝑧𝑟𝑟𝑟𝑟 = 𝐹𝐹. 𝑚𝑚 = In this case the displacement is vertical so it is the change in height ℎ2 − ℎ1 = h 𝐹𝐹. (ℎ2 − ℎ1 ) = 𝐹𝐹. h = GPE Since force needed to lift the object with constant velocity is equal to 𝑚𝑚𝑔𝑔 ℎ = This is only valid when there is no work done against friction or drag. Question 3 = 𝑚𝑚𝑔𝑔 SIR SAMEER UZ ZAMAN Notes: Summary: 171 of 223 Question 4 Power It is defined as the Rate of change of energy = 𝐴𝐴 𝑧𝑧𝑟𝑟 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑔𝑔𝑢𝑢 𝐴𝐴 Derivation P=F.v = = Since 𝑎𝑎𝑟𝑟𝐴𝐴𝑧𝑧𝐴𝐴𝑟𝑟𝐴𝐴𝑢𝑢 = 𝑚𝑚𝑛𝑛 𝑚𝑚 𝐴𝐴 𝐹𝐹. 𝑚𝑚 𝐴𝐴 = so, = 𝐹𝐹. 𝑎𝑎 SIR SAMEER UZ ZAMAN Notes: Summary: 172 of 223 Question 5 Question 6 Question 7 SIR SAMEER UZ ZAMAN Notes: Summary: 173 of 223 Question 8 Efficiency efficiency of a system is the ratio of useful energy output from the system to the total energy input 𝑟𝑟𝑓𝑓𝑓𝑓𝑟𝑟𝐴𝐴𝑟𝑟𝑟𝑟𝑟𝑟𝐴𝐴𝑢𝑢 = Question 9 𝐴𝐴𝑚𝑚𝑟𝑟𝑓𝑓𝐴𝐴𝐴𝐴 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑔𝑔𝑢𝑢 𝑧𝑧𝐴𝐴𝐴𝐴𝑝𝑝𝐴𝐴𝐴𝐴 𝐴𝐴𝑧𝑧𝐴𝐴𝐴𝐴𝐴𝐴 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑔𝑔𝑢𝑢 𝑟𝑟𝑟𝑟𝑝𝑝𝐴𝐴𝐴𝐴 SIR SAMEER UZ ZAMAN Notes: Summary: 174 of 223 Question 10 Question 11 Question 12 SIR SAMEER UZ ZAMAN Notes: Summary: 175 of 223 Law of conservation of energy Energy can neither be created nor be destroyed it merely changes from one form to the other. Or Energy at the beginning = Energy at the end Question 13 Question 14 An object of mass 2 kg initially at rest moves down a frictionless slope at an angle of 30° and reaches a velocity v the distance travelled down the slope is 12 m find the velocity and also sketch the diagram. SIR SAMEER UZ ZAMAN Notes: Summary: 176 of 223 Question 15 An object of mass 2 kg at rest initially slides down a slope with an angle 30° and the slope has a constant frictional force of 2N. The mass reaches a velocity v when distance travelled down the slope is 12 m find the velocity and sketch the diagram SIR SAMEER UZ ZAMAN Notes: Summary: 177 of 223 Question 16 An object of mass 2 kg at rest is pulled up a frictionless slope with a force of 30N the slope makes an angle 30° and the object reaches a velocity v. The distance travelled up the slope is 12 m find the velocity and sketch the diagram. SIR SAMEER UZ ZAMAN Notes: Summary: 178 of 223 Question 17 An object of mass 2 kg at rest is pulled up a slope, with a force of 30 N. The slope makes an angle 30° and has a constant friction force of 4 N. The object reaches a velocity v at the end of the journey as it moves a distance 12 m up the slope. Find the velocity and sketch the diagram SIR SAMEER UZ ZAMAN Notes: Summary: 179 of 223 Question 18 SIR SAMEER UZ ZAMAN Notes: Summary: 180 of 223 Question 19 SIR SAMEER UZ ZAMAN Notes: Summary: 181 of 223 SIR SAMEER UZ ZAMAN Notes: Summary: 182 of 223 Question 20 SIR SAMEER UZ ZAMAN Notes: Summary: 183 of 223 SIR SAMEER UZ ZAMAN Notes: Summary: 184 of 223 Question 21 SIR SAMEER UZ ZAMAN Notes: Summary: 185 of 223 SIR SAMEER UZ ZAMAN Notes: Summary: 186 of 223 SIR SAMEER UZ ZAMAN Notes: Summary: 187 of 223 Work done by a gas Consider a gas contained in a cylinder by means of a frictionless piston with a crosssectional area A the pressure outside and inside the cylinder is the same and equal to the atmospheric pressure since = 𝐹𝐹 𝐴𝐴 And 𝐹𝐹 = . 𝐴𝐴 When the gas expands with a constant pressure the piston moves up by a distance ℎ as shown so, = 𝐹𝐹. ℎ = . 𝐴𝐴. ℎ We know that 𝐴𝐴 ℎ is the change in volume 𝑉𝑉 So, = 𝑉𝑉 SIR SAMEER UZ ZAMAN Notes: Summary: 188 of 223 Question 22 Question 23 Question 24 SIR SAMEER UZ ZAMAN Notes: Summary: 189 of 223 Force-displacement graphs SIR SAMEER UZ ZAMAN Notes: The area under the force displacement graph represents the work done by the force. 𝑧𝑧𝑟𝑟 𝑟𝑟𝑧𝑧𝑟𝑟𝑟𝑟 = 1 𝐹𝐹. 𝑟𝑟 2 Summary: 190 of 223 Deformation Change in shape, volume, or size due to an applied force is called deformation. These forces can be compressive or tensile (caused by tension). Tensile and compressive force A pair of forces that act away from each other. This is known as tension and causes extension. A pair of forces the act toward each other. This is known as compression and causes compression. Hooke’s law When a force is applied on an object it may extend or compress. According to Hooke’s law the force applied is proportional to the extension, provided that the object is within its limit of proportionality. 𝐹𝐹 𝑟𝑟 𝐹𝐹 = 𝑟𝑟 Where k is the spring constant (also known as force constant) i.e., force per unit extension or length. SIR SAMEER UZ ZAMAN Notes: Summary: 191 of 223 Force-extension graph (Hooke’s law) Gradient of F-e (force-extension) graph is equal to the spring constant. The higher values of k represent a rigid spring and lower values of k represent a flexible spring. SIR SAMEER UZ ZAMAN Notes: Summary: 192 of 223 Question 1 Question 2 Limit of proportionality It is defined as the point beyond which force is no longer directly proportional to the extension. The limit of proportionality lies within the elastic limit. SIR SAMEER UZ ZAMAN Notes: Summary: 193 of 223 Elastic limit It is defined as the point beyond which the object will permanently deform. For example, if you stretch a spring beyond a certain point, it will become stretched and will not return to its original length. Elastic deformation Elastic deformation happens when force is applied it changes the length, shape, size or dimensions of the object, but when the force is removed the object returns to its original length, shape, size or dimensions. Question 3 SIR SAMEER UZ ZAMAN Notes: Summary: 194 of 223 SIR SAMEER UZ ZAMAN Notes: Summary: 195 of 223 Question 4 SIR SAMEER UZ ZAMAN Notes: Summary: 196 of 223 Finding Work done using Force-extension graphs The area under the force-extension graph represents the work done by the force. 𝑧𝑧𝑟𝑟 𝑟𝑟𝑧𝑧𝑟𝑟𝑟𝑟 = 1 𝐹𝐹. 𝑥𝑥 2 𝑧𝑧𝑟𝑟 𝑟𝑟𝑧𝑧𝑟𝑟𝑟𝑟 = 1 𝐹𝐹. 𝑟𝑟 2 For a spring (or any other object) force applied may cause extension or compression so the work done to extend or compress the object is stored in it as elastic potential energy also known as strain energy. Elastic potential energy or strain energy The elastic potential energy of a material deformed within its limit of proportionality from the area under the force–extension graph. strain energy = 1 𝐹𝐹𝑟𝑟 2 substitute 𝐹𝐹 = 𝑟𝑟 in above equation we get, strain energy = 1 2 𝑟𝑟 2 SIR SAMEER UZ ZAMAN Notes: Summary: 197 of 223 Change in strain energy or elastic potential energy A spring is extended by a force of 𝐹𝐹1 and an extension of 𝑥𝑥1 , then the force is increased to 𝐹𝐹2 and the extension becomes 𝑥𝑥2 . As shown on the graph on the right The change in elastic potential energy can be derived as follows, DERIVATION: = 𝑓𝑓𝑟𝑟𝑟𝑟𝐴𝐴𝐴𝐴 𝑚𝑚𝐴𝐴𝑟𝑟𝐴𝐴𝑟𝑟𝑟𝑟 𝑟𝑟𝑟𝑟𝑟𝑟𝑔𝑔𝑢𝑢 − 𝑟𝑟𝑟𝑟𝑟𝑟𝐴𝐴𝑟𝑟𝐴𝐴𝐴𝐴 𝑚𝑚𝐴𝐴𝑟𝑟𝐴𝐴𝑟𝑟𝑟𝑟 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑔𝑔𝑢𝑢 = 1 1 𝐹𝐹2 𝑥𝑥2 − 𝐹𝐹1 𝑥𝑥1 2 2 According to Hooke’s law 𝐹𝐹 = 𝑥𝑥 Substituting 𝐹𝐹1 = 𝑥𝑥1 𝐴𝐴𝑟𝑟𝑟𝑟 𝑚𝑚𝑧𝑧, 𝐹𝐹1 = 𝑥𝑥1 𝐴𝐴𝑟𝑟𝑟𝑟 𝐹𝐹2 = 𝑥𝑥2 𝐹𝐹2 = 𝑥𝑥2 we get = 1 1 ( 𝑥𝑥2 )𝑥𝑥2 − (𝑥𝑥1 )𝑥𝑥1 2 2 = 1 2 1 2 𝑥𝑥 − 𝑥𝑥 2 2 2 1 = 1 (𝑥𝑥22 − 𝑥𝑥12 ) 2 SIR SAMEER UZ ZAMAN Notes: Summary: 198 of 223 Question 5 Question 6 SIR SAMEER UZ ZAMAN Notes: Summary: 199 of 223 Question 7 Question 8 SIR SAMEER UZ ZAMAN Notes: Summary: 200 of 223 Question 9 SIR SAMEER UZ ZAMAN Notes: Summary: 201 of 223 Question 10 SIR SAMEER UZ ZAMAN Notes: Summary: 202 of 223 Stress It is defined as Force per unit area 𝐴𝐴𝑟𝑟𝑟𝑟𝑚𝑚𝑚𝑚 ( ) = 𝐹𝐹 𝐴𝐴 Where ( ) (𝑚𝑚𝑟𝑟𝑔𝑔𝑚𝑚𝐴𝐴) is stress F is force and A is the area or cross-sectional area Stress and pressure have the same SI unit of Pa (Pascal) Strain It is defined as the ratio of extension to the original length 𝐴𝐴𝑟𝑟𝐴𝐴𝑟𝑟𝑟𝑟 ( ) = 𝑟𝑟 𝐴𝐴 Where, 𝐴𝐴 is the original length and e is the extension or compression. It is a dimensionless quantity, so it has no units. Young Modulus In the region where stress Is proportional to strain then, 𝐴𝐴𝑟𝑟𝑟𝑟𝑚𝑚𝑚𝑚 𝐴𝐴𝑟𝑟𝐴𝐴𝑟𝑟𝑟𝑟 𝐴𝐴𝑟𝑟𝑟𝑟𝑚𝑚𝑚𝑚 = × 𝐴𝐴𝑟𝑟𝐴𝐴𝑟𝑟𝑟𝑟 𝐴𝐴𝑟𝑟𝑟𝑟𝑚𝑚𝑚𝑚 𝐴𝐴𝑟𝑟𝐴𝐴𝑟𝑟𝑟𝑟 = The constant E is known as the young modulus. Hence Young modulus is defined as the ratio of stress to strain. As long as the object remains within its limit of proportionality. 𝐹𝐹 = 𝐴𝐴 𝑟𝑟 𝐴𝐴 = 𝐹𝐹𝐴𝐴 𝐴𝐴𝑟𝑟 SIR SAMEER UZ ZAMAN Notes: Summary: 203 of 223 Young modulus is a material property, this is same for same material regardless of its length, cross-sectional area, shape etc. it is in essence the measure of the material’s rigidity or in other words how easily the material can be bended or stretched. Since it is a material property, we can use it to compare different materials and see which one is stiffer or flexible. Its unit is Pa pascals. Graph of stress-strain The gradient is equal to the young modulus (E). SIR SAMEER UZ ZAMAN Notes: Summary: 204 of 223 Strain energy per unit volume The area under the stress-strain graph is equal to strain energy per unit volume 𝐴𝐴𝑟𝑟𝑟𝑟𝐴𝐴 𝐴𝐴𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝐴𝐴ℎ𝑟𝑟 𝑔𝑔𝑟𝑟𝐴𝐴𝑝𝑝ℎ = 1 𝑚𝑚𝐴𝐴𝑟𝑟𝑟𝑟𝑚𝑚𝑚𝑚 × 𝑚𝑚𝐴𝐴𝑟𝑟𝐴𝐴𝑟𝑟𝑟𝑟 2 𝐴𝐴𝑟𝑟𝑟𝑟𝐴𝐴 𝐴𝐴𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝐴𝐴ℎ𝑟𝑟 𝑔𝑔𝑟𝑟𝐴𝐴𝑝𝑝ℎ = 1 2 1 𝐹𝐹 𝑟𝑟 × × 2 𝐴𝐴 𝐹𝐹𝑟𝑟 is strain energy and 𝐴𝐴 is volume so, 𝐴𝐴𝑟𝑟𝑟𝑟𝐴𝐴 𝐴𝐴𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝐴𝐴ℎ𝑟𝑟 𝑔𝑔𝑟𝑟𝐴𝐴𝑝𝑝ℎ = 𝑚𝑚𝐴𝐴𝑟𝑟𝐴𝐴𝑟𝑟𝑟𝑟 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑔𝑔𝑢𝑢 𝑎𝑎𝑧𝑧𝐴𝐴𝐴𝐴𝑚𝑚𝑟𝑟 SIR SAMEER UZ ZAMAN Notes: Summary: 205 of 223 Question 11 Question 12 SIR SAMEER UZ ZAMAN Notes: Summary: 206 of 223 Question 13 Question 14 SIR SAMEER UZ ZAMAN Notes: Summary: 207 of 223 Question 15 Question 16 Question 17 SIR SAMEER UZ ZAMAN Notes: Summary: 208 of 223 Experiment to determine the Young Modulus • To measure the Young Modulus of a metal in the form of a wire requires a clamped horizontal wire over a pulley (or vertical wire attached to the ceiling with a mass attached) as shown in the diagram below • A reference marker is needed on the wire. This is used to accurately measure the extension with the applied load The independent variable is the load The dependent variable is the extension • • Method 1. Measure the original length of the wire using a metre ruler and mark this reference point with tape 2. Measure the diameter of the wire with micrometer screw gauge or digital calipers 3. Measure or record the mass or weight used for the extension e.g. 300 g 4. Record initial reading on the ruler where the reference point is 5. Add mass and record the new scale reading from the metre ruler 6. Record final reading from the new position of the reference point on the ruler 7. Add another mass and repeat method SIR SAMEER UZ ZAMAN Notes: Summary: 209 of 223 Reducing uncertainty • SIR SAMEER UZ ZAMAN Notes: To reduce the uncertainty in the final answer, take the following precautions when measuring o o o o o Take pairs of readings of the diameter right angles to each other, to ensure the wire is circular Six to ten readings altogether is enough to get an average value Remove the load and check the wire returns to the original limit after each reading. A little 'creep' is acceptable but a large amount indicates that the elastic limit has been exceeded Take several readings with different loads and find average Use a Vernier scale to measure the extension of the wire Measurements to determine Young modulus 1. Determine extension x from final and initial readings 2. Plot a graph of force against extension and draw line of best fit Summary: 210 of 223 3. Determine gradient of the force v extension graph 4. Calculate cross-sectional area from 𝐴𝐴 = 𝑟𝑟 2 4 5. Calculate Young modulus from 𝑧𝑧𝐴𝐴𝑟𝑟𝑔𝑔 𝑀𝑀𝑧𝑧𝑟𝑟𝐴𝐴𝐴𝐴𝐴𝐴𝑚𝑚 = 𝐹𝐹𝐴𝐴 𝐴𝐴𝑥𝑥 Since gradient is 𝑧𝑧𝐴𝐴𝑟𝑟𝑔𝑔 𝑚𝑚𝑧𝑧𝑟𝑟𝐴𝐴𝐴𝐴𝐴𝐴𝑚𝑚 = 𝐴𝐴𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝐴𝐴 × 𝐴𝐴 𝑥𝑥 SIR SAMEER UZ ZAMAN Notes: Summary: 211 of 223 Question 18 Question 19 SIR SAMEER UZ ZAMAN Notes: Summary: 212 of 223 Question 20 SIR SAMEER UZ ZAMAN Notes: Summary: 213 of 223 SIR SAMEER UZ ZAMAN Notes: Summary: 214 of 223 Question 21 SIR SAMEER UZ ZAMAN Notes: Summary: 215 of 223 SIR SAMEER UZ ZAMAN Notes: Summary: 216 of 223 Effective spring constant In Series: let’s consider two springs in series, force on both the springs remain the same since they are supporting the same weight. The total extension in series is the sum of extension of each spring, since the weight ‘W’ causes the spring 1 to extend by an extension of ‘e’ and in the same way the weight ‘W’ causes spring 2 extends by an extension of ‘e’ as well. So, the total extension in this case is 2e The combined spring constant in fig(A) is ½ k (it is the reciprocal of the total extension). The combined spring constant in series is given by: 1 = 1 + 1 1 + 2 In Parallel: Force is equally divided among both the springs. The total extension in parallel is equal to the individual extension of a spring. This is because the weight W is divided among the two springs so each one of them supports only ½ W so the extension in this case it is ½ e for both of them and the total extension of the system is also ½ e. The combined spring constant in (B) is 2k (the reciprocal of the total extension). In general, the combined spring constant is given by = 1 + 2 + NOTE: if load on is constant then, 1 𝑟𝑟 So 𝑟𝑟 1 Total extension is inversely proportional to the total spring constant of the system. SIR SAMEER UZ ZAMAN Notes: Summary: 217 of 223 Question 22 Question 23 Question 24 SIR SAMEER UZ ZAMAN Notes: Summary: 218 of 223 Question 25 Question 26 Question 27 SIR SAMEER UZ ZAMAN Notes: Summary: 219 of 223 Question 28 Elastic and plastic deformation Elastic deformation happens when force is applied it changes the length, shape, size or dimensions of the object, but when the force is removed the object returns to its original length, shape, size or dimensions. The figure on the right shows a stressstrain graph for an elastic deformation. Note while loading and the energy is stored as Strain Energy and while unloading it is released back. The loading and unloading curve/line follows the same path, showing that all the energy stored was released. SIR SAMEER UZ ZAMAN Notes: Summary: 220 of 223 Plastic deformation occurs when force is removed and the object does not return to its original length, there is a permanent deformation in its length, shape, size or dimensions. The fig on the right shows that loading and unloading curves do not follow the same path. Meaning that energy stored is not equal to the energy released. The difference in the energies is the work done in order to cause permanent deformation. stress strain NOTE: The slope of the unloading line is same as the loading line. This is because the slope represents Young’s modulus (only in the region where stress strain) and it is a material property hence it will not change even after there has been a permanent deformation of the material. Question 29 SIR SAMEER UZ ZAMAN Notes: Summary: 221 of 223 Question 30 Question 31 SIR SAMEER UZ ZAMAN Notes: Summary: 222 of 223 Increase in resistance due to change in length Since resistance of a wire depends on the length of the conductor. When it is stretched the resistance of the wire will change. Resistance of a conductor is given by R = -sectional area and l is the length of the conductor Since is constant and assuming that when the wire is stretched the crosssectional area of the wire doesn’t change, resistance will be directly proportional to the length. 𝑂𝑂 𝐴𝐴 so, the change is resistance will also be proportional to the change in length ∆𝑂𝑂 ∆𝐴𝐴 It can be said that the percentage change in R is equal to the percentage change in l Mathematically, ∆𝐴𝐴 ∆𝑂𝑂 × 100 = × 100 𝐴𝐴 𝑂𝑂 Since, ∆ SIR SAMEER UZ ZAMAN Notes: = strain, then the above equation can also be written as ∆𝑂𝑂 = 𝑚𝑚𝐴𝐴𝑟𝑟𝐴𝐴𝑟𝑟𝑟𝑟 𝑂𝑂 Summary: 223 of 223
0
advertisement
Download
advertisement
Add this document to collection(s)
You can add this document to your study collection(s)
Sign in Available only to authorized usersAdd this document to saved
You can add this document to your saved list
Sign in Available only to authorized users