PHY1210 Mechanics Lecture Notes PDF

BAYERO UNIVERSITY, KANO COLLEGE OF NATURAL AND PHARMACEUTICAL SCIENCES (FACULTY OF PHYSICAL SCIENCES) DEPARTMENT OF PHYSICS LECTURE NOTES ON...

BAYERO UNIVERSITY, KANO COLLEGE OF NATURAL AND PHARMACEUTICAL SCIENCES (FACULTY OF PHYSICAL SCIENCES) DEPARTMENT OF PHYSICS LECTURE NOTES ON PHY1210 (Mechanics) BY Yusuf A. Musa Copyright © Yusuf A. Musa, 2021 – All rights reserved. For more notes, questions and answers visit: www.aymusa.com Email: [email protected] 1|P H Y 1 2 1 0 ( M e c h a n i c s ) – L e c t u r e N o t e s b y Y u s u f A. M u s a , w w w. a y m u sa. c o m Course Outline Physical Quantities and Unit System Dimensional Analysis Vectors and Scalar Quantities Kinematics Force, Momentum and Impulse Collision Theory Circular Motion Work, Energy and Power Rotational Dynamics Universal Gravitation and Simple Machines Grading of the Course C.A 30 Examination 70 Pass Mark for the Course: 40 Recommended Textbooks/Resources Schaum’s outline of College Physics, 11th edition Nelkon and Parker, Advanced Level Physics, 7th edition Halliday, Resnick and walker, Fundamentals of Physics, 9th edition www.aymusa.com 2|P H Y 1 2 1 0 ( M e c h a n i c s ) – L e c t u r e N o t e s b y Y u s u f A. M u s a , w w w. a y m u sa. c o m 1.0 PHYSICAL QUANTITIES AND UNITS SYSTEM 1.1 Definition of Mechanics Mechanics can be defined as a branch of physics that deals with the study of the behaviour of objects in motion or at rest. Mechanics can be broadly classified in three sub-branches namely: statics, dynamics and kinematics. Statics deals with the study of objects at rest which may be under the influence of forces. Dynamics on the other hand, deals with the study of objects in motion and the forces causing such motion. Kinematics is the branch of mechanics that studies the behaviour of objects in motion without considering the force acting on them. 1.2 Physical Quantities Most observations in physics which are quantitative and descriptive are usually imprecise. A physical quantity is any quantity that can be measured and consists of a magnitude and unit. Examples of physical quantities are: mass, time, temperature, pressure, distance, density etc. All physical quantities are classified into two types namely: fundamental quantities and derived quantities. Fundamental quantities are those physical quantities from which other physical quantities can be expressed. Physical quantities that can be expressed in terms of the fundamental quantities are called derived quantities. The fundamental physical quantities in physics are: length, time, mass, temperature, electric current, amount of substance and luminous intensity. Table 1.0 below shows some derived quantities with their corresponding fundamental quantities. Table 1.0: derived quantities with their corresponding fundamental quantities S/N Derived Quantity Fundamental Quantities 1 Area Length 2 Velocity Length, Time 3 Density Length, Mass 4 Pressure Length, Time, Mass 5 Work done Length, Time, Mass 6 Frequency Time 7 Specific heat capacity Length, Time, Mass, Temperature 1.3 Units and Measurement Units of measure are vital parts of any physical quantities in physics. Just as a person is known by his/her name, the same way physical quantities are known by their units of measurement. The unit of measurement is defined as the standard used for the measurement of that physical quantity. The standard unit is always universal accepted and it does not change with time. There are large number of physical quantities, and each of them should have the units of measurement. However, it is not necessary to define separate unit for each and every physical quantity. So, there is need to have a unit system of measurement. The units of measurement defined for the fundamental quantities are called as the fundamental or basic units of measurement. There are four categories of these units of measurement namely: CGS (centimetre, gram, second) system. MKS (metre, kilogram, second) system, FPS (foot, pound, second) and SI system of units of measurement. The SI system or the international system of units has now replaced all the systems of measurement. Table 2.0 shows the fundamental quantities with their corresponding unit system. 3|P H Y 1 2 1 0 ( M e c h a n i c s ) – L e c t u r e N o t e s b y Y u s u f A. M u s a , w w w. a y m u sa. c o m  Table 2.0: derived quantities with their corresponding fundamental quantities S/N Fundamental Quantity CGS unit MKS unit SI unit 1 Time second (s) second (s) seconds (s) 2 Length centimetre (cm) metre (m) metre (m) 3 Mass gram (g) kilogram (kg) kilogram (kg) 4 Temperature --------- --------- Kelvin (0K) 5 Electric Current --------- --------- Ampere (A) 6 Luminous Intensity --------- --------- Candela (Cd) 7 Amount of Substance --------- --------- mole (mol) ---------- Not defined In addition to the SI units of the seven fundamental quantities, two supplementary units are also defined as shown in table 3.0 below. Table 3.0: supplementary quantity with their corresponding SI units S/N Supplementary Quantity Name of Unit Symbol of Unit 1 Pane angle radian rad 2 Solid angle steradian sr Sometimes some SI derived units have special names with the SI base unit equivalents as shown in table 4.0. Table 4.0: Some special units with their corresponding SI base units S/N Name Symbol SI Base Unit Equivalent 1 Newton N Kgms-2 2 Joule J Kgm2s-2 3 Pascal Pa Kgm-1s-2 4 Watt W Kgm2s-3 5 Coulomb C As 6 Volt V Kgm2s-3A-1 7 Farad F Kg-1m-2s4A2 8 Ohm Ω Kgm2s-3A-2 9 Tesla T Kgs-2A-1 10 Henry H Kgm2s-2A-2 11 Hertz Hz s-1 1.4 Definitions of SI Base Units Metre: One metre is defined as the length of the path travelled in aa vacuum during a time interval of 1/299,793,458 of a second. Second: One second is the duration of 9162631770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium -133 atoms. Kilogram: One kilogram is the mass equal to the mass of standard platinum-iridium alloy cylinder (90% platinum and 110% iridium) kept at the international Bureau of weights and measurements in Paris, France. The cylinder is about 4cm in height and 4cm in diameter. 4|P H Y 1 2 1 0 ( M e c h a n i c s ) – L e c t u r e N o t e s b y Y u s u f A. M u s a , w w w. a y m u sa. c o m Ampere It is the current which when flows through two infinitely long straight conductors of negligible cross- section placed at a distance of 1 m in vacuum produces a force of 2 × 10−7 N/m between them. Kelvin: One kelvin is defined as 1 / 273.16 part of the thermodynamic temperature of triple point of water. Mole It is the amount of substance of a system which contains as many elementary particles (atoms, molecules, ions etc.) as there are atoms in 12 g of carbon-12. 1 Candela It is luminous intensity in a perpendicular direction of a surface of ( 600000 ) 𝑚2 of a black body at the temperature of freezing platinum under a pressure of 1.013 × 105 N/m2. 1.5 SI Unit Prefix Magnitudes of physical quantities range from very large to very small. For example, mass of the sun is 10 30 kg and mass of the electron is 10-31 kg. Hence, prefix is used to describe these magnitudes of measurement. SI units prefix represents multiples or sub-multiples of the SI base units. Table 5.0 shows some of the most commonly used SI unit prefix. Table 5.0: SI units Prefix S/N Name of Unit Symbol of Prefix Value 1 kilo K 103 2 mega M 106 3 giga G 109 4 tera T 1012 5 peta P 1015 6 deca da 101 7 deci d 10-1 8 centi cm 10-2 9 milli m 10-3 10 micro 𝜇 10-6 11 nano 𝑛 10-9 12 pico p 10-12 Units conversions for some units are shown in table 6.0. Table 6.0: Units Conversions Units Conversion 1 day = 24 hours 1 mile = 1.609344 km 1 cm3 = 10-6 m2 1 hour = 60 minutes 1 pound = 0.453592 kg 1 mm3 = 10-9 m2 1 minute = 60 seconds 1 tonne = 1000 kg 1 litre = 0.001 m3 1 year = 365 days 1 mg = 10-6 kg 1 dm3 = 0.001 m3 1 feet = 0.3048 m 1 kg = 1000 g 1 oil barrel = 158.987 litres 1 yard = 0.9144 m 1 hectare = 10000 m2 1 calorie = 4.184 Joules 1 inch = 0.0254 m 1 cm2 = 10-4 m2 1 dyne = 10-5 Newton 1 kilometre = 1000 m 1 mm2 = 10-6 m2 1 horse power = 735.5 Watts 5|P H Y 1 2 1 0 ( M e c h a n i c s ) – L e c t u r e N o t e s b y Y u s u f A. M u s a , w w w. a y m u sa. c o m EXAMPLE 1 Convert the following to m/s: (a) 21 cm/hr (b) 30 miles/day (c) 35 km/min EXAMPLE 2 Convert the following to m3/s: (a) 5 L/min (b) 7 cm3/week (c) 12 cL/day EXAMPLE 3 The speed of sound is measured to be 342 m/s on a certain day. What is this measurement in kilometres per hour? EXAMPLE 4 The density of aluminium is 2.7 g/cm3. What is the density in kilograms per cubic meter? EXAMPLE 5 Mount Everest, at 29028 ft, is the tallest mountain on Earth. What is its height in kilometres? EXAMPLE 6 A nurse gives a patient a painkiller at 8:32am. At 2.12pm the patient complains that the pain has return and the nurse administers another painkiller to the patient. How long did the original painkiller last? 2.0 DIMENSIONAL ANALYSIS It is well understood that it does not make sense to add together a distance of, say 2 km and a time of 15 seconds. The same applies to subtraction. Distance/length and time are different from one another in a fundamental way but it is possible to divide length by time to get a velocity. It is also possible to multiply quantities, for example length multiplied by length gives area. Dimensional analysis is a means of simplifying a physical quantity in terms of the fundamental quantities. The dimension of a physical quantity is defined as the power to which the fundamental quantities are raised to express that quantity. Dimensional analysis allows us to: i. convert from one unit system to another ii. establish the relation among various physical quantities iii. check the correctness of equations and iv. non-dimensionalizing of ordinary differential equations. We use the symbol [ ] to indicate the ‘dimension of’ a physical quantity: [time] = 𝑇, [length] = 𝐿, [mass] = 𝑀 and [temperature] = 𝜃 The dimensions of other quantities can be written in terms of 𝑀, 𝐿 , 𝑇 and 𝜃 using the definition of the particular quantity. The dimension of any angle has dimension of 1 (dimensionless). For any transcendental functions such as sine, cosine, tangent, exponential etc., the dimension is 1 (dimensionless). The dimensions of physical quantities 𝐴, 𝐵, 𝐶, ….. 𝑍 obey the following rules of algebra: 𝐴 [𝐴] [𝐴𝐵𝐶 ⋯ 𝑍] = [𝐴][𝐵][𝐶] ⋯ [𝑍], [𝐴𝑛 ] = [𝐴]𝑛 , [ ] = , and [𝑛] = 1 (1) 𝐵 [𝐵] Where 𝑛 is any real number. The principle of dimensional homogeneity holds the view that all additive terms in a physical equation must have the same dimensions. Dimensional homogeneity is a useful tool for checking formulae. If 𝐴, 𝐵, 𝐶, ….. 𝑍 represent some physical quantities such that 𝐴 = 𝐵 ± 𝐶 ± 𝐷 ± 𝐸 ± ⋯± 𝑍 (2) By the principle of dimensional homogeneity, 6|P H Y 1 2 1 0 ( M e c h a n i c s ) – L e c t u r e N o t e s b y Y u s u f A. M u s a , w w w. a y m u sa. c o m  [𝐴] = [𝐵] = [𝐶] = [𝐷] = [𝐸] = ⋯ = [𝑍] (3) EXAMPLE 7 Find the dimension for each of the following physical quantities: (a) volume (b) density (c) velocity (d) acceleration (e) force (f) momentum (g) work done (h) kinetic energy (i) power (j) angular velocity (k) pressure (l) specific het capacity (m) coefficient off thermal expansion EXAMPLE 8 Use the principle of dimensional homogeneity to check whether the following formulae are dimensionally correct where 𝑢 and 𝑣 are velocities, 𝑔 is an acceleration, 𝑥 is a distance, 𝑡 is a time and 𝜃 is the angle. 1 𝑣 𝑣2 (a) 𝑣 = 2𝑎𝑡 (b) 𝑥 = 𝑢𝑡 + 2 𝑎𝑡 2 (c) 𝑠𝑖𝑛𝜃 = 𝑥 (d) 𝑡 = √2𝑎𝑥 EXAMPLE 9 The period of oscillation T of a simple pendulum depends on its mass m, length of the string l and the value of the acceleration due to gravity g. Show that 𝑙 𝑇 = 𝑘√ 𝑔 Where 𝑘 is a real constant number. EXAMPLE 10 Poiseuille found that the volume of a liquid flowing through a horizontal cylindrical tube per second 𝑄 𝑃 depends upon the pressure gradient 𝑙 , radius 𝑟 of the tube and dynamic viscosity 𝜂. The SI unit of viscosity is kgm-1s-1. 𝑙 is the length of the tube. Use dimensional analysis to show that 𝑘𝑝𝑟 4 𝑄= 𝜂𝑙 Where 𝑘 is a real constant number. EXAMPLE 11 Determine the dimensions of the constants 𝐴 and 𝐵 so that the following expressions are dimensionally correct where 𝑥 is a distance, 𝑡 is a time and 𝑣 is velocity. 1 𝐴 (a) 𝑥 = 𝐴𝑠𝑖𝑛(2𝜋𝐵𝑡) (b) 𝑣 = 2 𝐵𝑒 −𝐴𝑡 (c) 𝑣 = √ 𝑥 − 𝐵𝑡 2 EXAMPLE 12 Consider the model for the angle of a nonlinear simple pendulum 𝑑2𝜃 𝑔 2 = − 𝑠𝑖𝑛𝜃 𝑑𝑡 𝑙 𝑡 By using the transformation 𝛼 = 𝑙 in the above equation, show that the transformed equation obtained √𝑔 is dimensionless. What is the significance of the transformation? 7|P H Y 1 2 1 0 ( M e c h a n i c s ) – L e c t u r e N o t e s b y Y u s u f A. M u s a , w w w. a y m u sa. c o m 3.0 VECTORS AND SCALAR QUANTITIES A scalar quantity is any quantity which has only magnitude but no direction. Scalars may or may not have units associated with them. Examples: mass, volume, energy, money etc. A vector quantity on the other hand, is any quantity which has both magnitude and direction. The magnitude of a vector is a scalar. Examples: displacement, velocity, acceleration, electric field, magnetic field etc. 3.1 Vector Notation Vectors are denoted as a letter with an arrow over the top or a bold letter such as 𝐴⃗ or A. The magnitude of a vector A is represented by |𝑨| or 𝐴 while its direction by 𝑨 ̂. Thus, 𝐴⃗ = 𝐴𝐴̂ (4) 3.2 Vector Diagrammatic Representation Diagrammatically, vectors are represented by an arrow pointing in the direction of the vector. The length of the vector represents the magnitude of the vector. Vectors always have tail and head as shown in figure 1.0 below Head Tail Figure 1.0: Diagrammatic representation of a vector The length of the arrow does not necessarily represent a length but rather the magnitude of the vector quantity under consideration. In 2– dimensional cartesian coordinates, a vector 𝑨 can be broken down into its three components x and y i.e. 𝐴⃗ = 𝐴𝑥 𝑖̂ + 𝐴𝑦 𝑗̂ (5) 𝐴⃗ 𝑉𝑦 𝜃 𝑉𝑥 Figure 2.0: Diagrammatic representation of a vector in 2 - dimension 𝐴𝑥 = 𝐴𝑐𝑜𝑠𝜃 and 𝐴𝑦 = 𝐴𝑠𝑖𝑛𝜃 Where: 𝐴𝑥 is the component of the vector A along the x – direction (horizontal) 𝐴𝑦 is the component of the vector A along the y – direction (vertical) 𝑖̂ represents direction along the x – axis 𝑗̂ represents direction along the y – axis The magnitude of the vector represented by |𝑨| or 𝐴 is thus given by |𝑨| = √𝐴2𝑥 + 𝐴2𝑦 (6) 𝑖̂ and 𝑗̂ are called units vector and has magnitude of 1. The angle between 𝑖̂ and 𝑗̂ is 900. 8|P H Y 1 2 1 0 ( M e c h a n i c s ) – L e c t u r e N o t e s b y Y u s u f A. M u s a , w w w. a y m u sa. c o m 3.3 Vector Addition/Subtraction Diagrammatically, two or more vectors can be added graphically to give a resultant vector by arranging the vectors in a head to tail fashion. The resultant vector is drawn from the tail of the first to the head of the last vector. Subtracting two or more vectors graphically can also be done by flipping one or more vectors then proceed to add the vectors. Algebraically, given two vectors 𝐴⃗ = 𝐴𝑥 𝑖̂ + 𝐴𝑦 𝑗̂ ⃗⃗ = 𝐵𝑥 𝑖̂ + 𝐵𝑦 𝑗̂, and 𝐵 The two vectors can be added/subtracted as follows: 𝐴⃗ + 𝐵⃗⃗ = 𝐴𝑥 𝑖̂ + 𝐴𝑦 𝑗̂ + 𝐵𝑥 𝑖̂ + 𝐵𝑦 𝑗̂ = (𝐴𝑥 + 𝐵𝑥 )𝑖̂ + (𝐴𝑦 + 𝐵𝑦 )𝑗̂ 𝐴⃗ − 𝐵 ⃗⃗ = 𝐴𝑥 𝑖̂ + 𝐴𝑦 𝑗̂ − (𝐵𝑥 𝑖̂ + 𝐵𝑦 𝑗̂) = (𝐴𝑥 − 𝐵𝑥 )𝑖̂ + (𝐴𝑦 − 𝐵𝑦 )𝑗̂ If 𝝀 is any scalar number, multiplying a vector A by n does not change its magnitude but only its direction. Thus, 𝐴⃗ = 𝜆(𝐴𝑥 𝑖̂ + 𝐴𝑦 𝑗̂) = (𝜆𝐴𝑥 )𝑖̂ + (𝜆𝐴𝑦 )𝑗̂ 3.4 Vector Dot/Scalar Multiplication The result of a dot product of two vectors is a scalar. The vector dot or scalar product of two vectors 𝐴⃗ and ⃗⃗ is denoted by 𝐴⃗ ∙ 𝐵 𝐵 ⃗⃗ and is given by 𝐴⃗ ∙ 𝐵 ⃗⃗ = 𝐴𝐵𝑐𝑜𝑠𝜃 Where 𝜃 is the angle between the two vectors. From the above definition it can be seen that: 𝑖̂ ∙ 𝑖̂ = |𝑖̂||𝑖̂|𝑐𝑜𝑠𝜃 = 1 × 1 × cos (0) = 1 𝑗̂ ∙ 𝑗̂ = |𝑗̂||𝑗̂|𝑐𝑜𝑠𝜃 = 1 × 1 × cos (0) = 1 𝑖̂ ∙ 𝑗̂ = |𝑖̂||𝑗̂|𝑐𝑜𝑠𝜃 = 1 × 1 × cos (90) = 0 𝑗̂ ∙ 𝑖̂ = |𝑗̂||𝑖̂|𝑐𝑜𝑠𝜃 = 1 × 1 × cos (90) = 0 Thus, 𝐴⃗ ∙ 𝐵 ⃗⃗ = (𝐴𝑥 𝑖̂ + 𝐴𝑦 𝑗̂) ∙ (𝐵𝑥 𝑖̂ + 𝐵𝑦 𝑗̂) = 𝐴𝑥 𝐵𝑥 𝑖̂ ∙ 𝑖̂ + 𝐴𝑥 𝐵𝑦 𝑖̂ ∙ 𝑗̂ + 𝐴𝑦 𝐵𝑥 𝑗̂ ∙ 𝑖̂ + 𝐴𝑦 𝐵𝑦 𝑗̂ ∙ 𝑗̂ = 𝐴𝑥 𝐵𝑥 + 𝐴𝑦 𝐵𝑦 ∴ 𝐴⃗ ∙ 𝐵 ⃗⃗ = 𝐴𝑥 𝐵𝑥 + 𝐴𝑦 𝐵𝑦 3.5 Vector Cross Product/Multiplication The result of a cross product of two vectors is a vector. The vector cross product of two vectors 𝐴⃗ 𝑎𝑛𝑑 𝐵 ⃗⃗ is denoted by 𝐴⃗ × 𝐵 ⃗⃗ and is defined as 𝐴⃗ × 𝐵 ⃗⃗ = 𝑛̂ 𝐴𝐵𝑐𝑜𝑠𝜃 And its magnitude is given by |𝐴⃗ × 𝐵 ⃗⃗ | = 𝐴𝐵𝑠𝑖𝑛𝜃 Where 𝜃 is the angle between the two vectors. 9|P H Y 1 2 1 0 ( M e c h a n i c s ) – L e c t u r e N o t e s b y Y u s u f A. M u s a , w w w. a y m u sa. c o m From the above definition it can be seen that: 𝑖̂ × 𝑖̂ = 𝑗̂ × 𝑗̂ = 𝑘̂ × 𝑘̂ = 𝑛̂ × 1 × 1 × 𝑠𝑖𝑛0 = 0 𝑖̂ × 𝑗̂ = 𝑘̂ × 1 × 1 × 𝑠𝑖𝑛90 = 𝑘̂ 𝑗̂ × 𝑘̂ = 𝑖̂ × 1 × 1 × 𝑠𝑖𝑛90 = 𝑖̂ 𝑘̂ × 𝑖̂ = 𝑗̂ × 1 × 1 × 𝑠𝑖𝑛90 = 𝑗̂ 𝑗̂ × 𝑖̂ = 𝑘̂ × 1 × 1 × sin(−90) = −𝑘̂ 𝑘̂ × 𝑗̂ = 𝑖̂ × 1 × 1 × sin(−90) = −𝑖̂ 𝑖̂ × 𝑘̂ = 𝑗̂ × 1 × 1 × sin(−90) = −𝑗̂ Thus in 2 - dimension, 𝐴⃗ × 𝐵 ⃗⃗ = (𝐴𝑥 𝑖̂ + 𝐴𝑦 𝑗̂) × (𝐵𝑥 𝑖̂ + 𝐵𝑦 𝑗̂) = 𝐴𝑥 𝐵𝑥 𝑖̂ × 𝑖̂ + 𝐴𝑥 𝐵𝑦 𝑖̂ × 𝑗̂ + 𝐴𝑦 𝐵𝑥 𝑗̂ × 𝑖̂ + 𝐴𝑦 𝐵𝑦 𝑗̂ × 𝑗̂ = 𝐴𝑥 𝐵𝑦 𝑖̂ × 𝑗̂ + 𝐴𝑦 𝐵𝑥 𝑗̂ × 𝑖̂ = 𝐴𝑥 𝐵𝑦 𝑘̂ − 𝐴𝑦 𝐵𝑥 𝑘̂ ∴ 𝐴⃗ × 𝐵 ⃗⃗ = 𝐴𝑥 𝐵𝑦 − 𝐴𝑦 𝐵𝑥 Example 13 Given the vectors as shown below ⃗⃗ 𝐵 𝐴⃗ 𝐶⃗ Show diagrammatically for each of the expressions: (a) 𝐴⃗ + 𝐵 ⃗⃗ (b) 𝐴⃗ − 𝐵 ⃗⃗ (c) 𝐴⃗ − 𝐶⃗ + 𝐵 ⃗⃗ (d) 𝐴⃗ − 𝐵 ⃗⃗ − 𝐶⃗ Example 14 If 𝐴⃗ = 3𝑖̂ − 4𝑗̂ and 𝐵 ⃗⃗ = −𝑖̂ + 2𝑗̂, find : (a) |𝐴⃗ | (b) 𝐴̂ (c) 𝐴⃗ + 2𝐵 ⃗⃗ (d) 𝐴⃗ ∙ 𝐵 ⃗⃗ (e) 𝐴⃗ × 𝐵 ⃗⃗ 10 | P H Y 1 2 1 0 ( M e c h a n i c s ) – L e c t u r e N o t e s b y Y u s u f A. M u s a , w w w. a y m u s a. c o m EXCERCISES 1 Q1. A car is traveling at a speed of 33 m/s. (a) What is its speed in kilometres per hour? (b) Is it exceeding the 90 km/ h speed limit? Q2. Find the dimension for each of the following physical quantities: (a) electric potential (b) permittivity of free space (c) magnetic field (d) electric charge (e) angular momentum (f) gravitational constant (g) surface tension (h) moment of inertia (i) angular velocity (j) modulus of elasticity (k) coefficient of viscosity (l) Boltzmann constant (m) electric dipole moment Q3. The frequency 𝑓 of a stretched string depends upon the tension 𝐹 (dimensions of force), length 𝑙 of the string and the mass per unit length 𝑚 of string. Derive the formula for frequency using dimensional analysis. Q4. The centripetal force 𝐹 acting on a particle moving uniformly in a circle may depend upon mass 𝑚, velocity 𝑣 and radius 𝑟 of the circle. Derive the formula for 𝐹 using the method of dimensions. Q5.The air bubble formed by explosion inside water performs oscillations with time period 𝑇 which depends on pressure p, density 𝜌 and on energy due to explosion 𝐸. Establish relation between 𝑇, 𝑝, 𝐸 and 𝜌 using dimensional analysis. Q6. The velocity 𝑣 of particle depends upon the time 𝑡 according to the equation 𝑐 𝑣 = √𝑎𝑡 + 𝑏𝑡 + 𝑑+𝑡 Determine the dimension of 𝑎, 𝑏, 𝑐 and 𝑑. Q7. The Van der Waals equation of state of a real gas of volume 𝑉, pressure 𝑃 and temperature 𝑇 is give by 𝑛 2 𝑉 (𝑃 + ( ) ) ( − 𝑏) = 𝑅𝑇 𝑉 𝑛 Where 𝑛 is the number of moles of the gas and 𝑎 and 𝑏 are constants. Determine the dimensions of 𝑎 and 𝑏 in the above equation. Q8. The frictional force 𝐹 acting on a small spherical bob in a fluid depends on its radius 𝑟, the viscosity 𝜂 of the fluid and its speed 𝑣. Show by dimensional analysis that 𝐹 = 𝑘𝜂𝑣𝑟 Where 𝑘 is a constant number. Q9. Taking the x-axis is horizontal and points to the right, resolved the vectors given in the following figure to their scalar components and express them in vector component form. (a) 𝐴⃗ (b) 500 250 𝐶⃗ 11 | P H Y 1 2 1 0 ( M e c h a n i c s ) – L e c t u r e N o t e s b y Y u s u f A. M u s a , w w w. a y m u s a. c o m 4.0 KINEMATICS Kinematics is a branch of mechanics that deals with the study of the motion of a body in terms in terms of space and time while ignoring the forces that caused the motion. Kinematics basically means studying how things are moving, not why they're moving. It includes concepts such as distance or displacement, speed or velocity, and acceleration, and it looks at how those values vary over time. Kinematics can be studied in one dimension, like a car moving along a straight road; in two dimensions, like a stone thrown into air ; and even in three dimensions. Things just get a little more complicated as you add dimensions. 4.1 KINEMATICS IN ONE DIMENSION In this section we are going to consider when the motion of the body is always longa a straight path either along the x-direction, y-direction or z-direction. 4.1.1 Position, Distance and Displacement Position(x): the position of a body is the location of the body with respect to a chosen reference point or frame that can be consider to be the origin of the coordinates system (cartesian, cylindrical or spherical). Distance (x): Distance is the length of path followed by a body measures from the reference point or frame of reference. Distance is a scalar quantity with SI unit of metre (m). Displacement (∆𝒙): Displacement is the change in position of a body in a specified direction. The shortest distance between any two point is the displacement. Displacement is a vector quantity with SI unit of metre (m). ∆𝑥 = 𝑥𝑓 − 𝑥0 Where 𝑥𝑓 is the final position of the body and 𝑥0 is its initial position. 4.1.2 Speed, Velocity and Acceleration Speed (v): average speed is the total distance covered by a body divided by the total time taken. Speed is a scalar is scalar quantity with SI unit of m/s. total distance travelled speed = total time taken Velocity (v): Average velocity is the change of displacement of a body divided by the change in time interval during the motion. Velocity is a scalar is a vector quantity with SI unit of m/s. displacement 𝑥𝑓 − 𝑥0 velocity = = change of time 𝑡𝑓 − 𝑡0 Where 𝑡𝑓 and 𝑡0 are the final time and initial time of the motion of the body respectively. 4.1.3 Instantaneous Speed, Velocity and Acceleration Instantaneous Velocity: Instantaneous velocity is the velocity of a body at any given instant of time. Mathematically, ∆𝑥 𝑑𝑥 Instataneous velocity = lim = ∆𝑡→0 ∆𝑡 𝑑𝑡 Instantaneous Speed: instantaneous speed of a body is the speed of a body at any given instant of time as it is given as the magnitude of the instantaneous velocity at the same instant of time. 12 | P H Y 1 2 1 0 ( M e c h a n i c s ) – L e c t u r e N o t e s b y Y u s u f A. M u s a , w w w. a y m u s a. c o m Instantaneous Acceleration: Instantaneous acceleration is the acceleration of a body at any instant of time. Mathematically, ∆𝑣 𝑑𝑣 Instataneous acceleration = lim = ∆𝑡→0 ∆𝑡 𝑑𝑡 For motion with constant/uniform acceleration, the rate with which velocity changes with time is constant. Thus, 𝑣𝑓 + 𝑣0 𝑣 + 𝑢 Averaage velocity = = 2 2 If on the other hand the body moves with constant/uniform velocity, then its acceleration will be zero. For motion with constant/uniform acceleration, average and instantaneous acceleration are equal. 4.1.4 Equations of Uniformly Accelerated Motion A very common and simple type of one-dimensional motion is that in which the acceleration is constant (uniform). In this type of motion, we are going to consider an object which moves from an initial position 𝑥0 , at time 𝑡0 = 0, to a position 𝑥𝑓 , at time 𝑡𝑓 = 𝑡 with initial velocity 𝑢 and final velocity 𝑣 with constant acceleration a. Thus, 𝑣−𝑢 𝑎= 𝑡 𝑣 = 𝑢 + 𝑎𝑡 (7) We know that 𝑣+𝑢 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑥𝑓 − 𝑥0 𝑥𝑓 − 𝑥0 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 = = = = 2 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑡𝑖𝑚𝑒 𝑡−0 𝑡 𝑣 + 𝑢 𝑥𝑓 − 𝑥0 = 2 𝑡 2(𝑥𝑓 − 𝑥0 ) 𝑣+𝑢 = (8) 𝑡 Putting equation (7) into equation (8), we get 2(𝑥𝑓 − 𝑥0 ) 𝑢 + 𝑎𝑡 + 𝑢 = 𝑡 2(𝑥𝑓 − 𝑥0 ) 2𝑢 + 𝑎𝑡 = 𝑡 2𝑢𝑡 + 𝑎𝑡 2 = 2(𝑥𝑓 − 𝑥0 ) 1 𝑢𝑡 + 𝑎𝑡 2 = 𝑥𝑓 − 𝑥0 2 1 𝑥𝑓 − 𝑥0 = 𝑢𝑡 + 𝑎𝑡 2 (9) 2 From equation (7), 𝑣−𝑢 (10) 𝑡= 𝑎 13 | P H Y 1 2 1 0 ( M e c h a n i c s ) – L e c t u r e N o t e s b y Y u s u f A. M u s a , w w w. a y m u s a. c o m Putting equation (10) into equation (9), we get 𝑣−𝑢 1 𝑣−𝑢 2 𝑣−𝑢 1 𝑣−𝑢 𝑣−𝑢 𝑥𝑓 − 𝑥0 = 𝑢 [ ] + 𝑎 [𝑡 ] = 𝑢( )+ 𝑎( )( ) 𝑎 2 𝑎 𝑎 2 𝑎 𝑎 𝑣−𝑢 1 𝑣−𝑢 𝑣−𝑢 𝑣 𝑢 𝑣−𝑢 𝑣 𝑢 𝑣−𝑢 𝑣+𝑢 𝑥𝑓 − 𝑥0 = ( ) [𝑢 + 𝑎 ( )] = ( ) [𝑢 + − ] = ( )( + ) = ( )( ) 𝑎 2 𝑎 𝑎 2 2 𝑎 2 2 𝑎 2 1 1 2 𝑥𝑓 − 𝑥0 = (𝑣 − 𝑢)(𝑣 + 𝑢) = (𝑣 − 𝑢2 ) 2𝑎 2𝑎 2𝑎(𝑥𝑓 − 𝑥0 ) = 𝑣 2 − 𝑢2 𝑣 2 = 𝑢2 + 2𝑎(𝑥𝑓 − 𝑥0 ) (11) Therefore, the equations for uniformly accelerate motion are: 𝑣 = 𝑢 + 𝑎𝑡 1 𝑥𝑓 − 𝑥0 = 𝑢𝑡 + 𝑎𝑡 2 and 2 𝑣 2 = 𝑢2 + 2𝑎(𝑥𝑓 − 𝑥0 ) If a body travels along a straight line with uniform acceleration, the distance travelled in the n th time will be given by 𝑥𝑛 = distance travvelled for the first 𝒏 seconds − distance travelled for last (𝒏 − 𝟏) seconds 1 1 1 1 𝑥𝑛 = 𝑢𝑛 + 𝑎𝑛2 − [𝑢(𝑛 − 1) + 𝑎(𝑛 − 1)2 ] = 𝑢𝑛 + 𝑎𝑛2 − 𝑢𝑛 + 𝑢 − 𝑎(𝑛2 − 2𝑛 + 1) 2 2 2 2 1 1 1 1 = 𝑎𝑛2 + 𝑢 − 𝑎𝑛2 + 𝑛𝑎 − 𝑎 = 𝑢 + 𝑎(𝑛 − ) 2 2 2 2 1 𝑥𝑛 = 𝑢 + 𝑎 (𝑛 − ) (12) 2 4.1.5 Free Fall Motion (Vertical Downward Motion) When an object falls towards the ground, the force that causes the objects to fall is called gravity. The acceleration caused by gravity is typically written as 𝒈 and is called acceleration due to gravity. Any object, be it a ball or an elephant, experiences the same acceleration 𝑔 when it is dropped or thrown downward. The acceleration due to gravity 𝑔 does not dpends on the nature of the material and it normally has the value of 9.81𝑚𝑠 −2 and acts vertically downward. As a body moves vertically downward its velocity/speed increases which resulted in making the body to bounced off when it strikes the surface. 14 | P H Y 1 2 1 0 ( M e c h a n i c s ) – L e c t u r e N o t e s b y Y u s u f A. M u s a , w w w. a y m u s a. c o m  m 𝑢=0 Body of mass, 𝑚 at rest 𝑔 = 9.81𝑚𝑠−2 𝑣 ℎ Ground surface Figure 3.0: Diagrammatic representation of a body falling from rest vertically downward. For a vertically free fall motion, the equations of motion are 𝑣 = 𝑔𝑡 1 2 ℎ= 𝑔𝑡 and 𝑣 2 = 2𝑔ℎ 2 4.1.6 Vertical Upward Motion When an object is thrown vertically upward, gravity acts in opposite direction and brings it back to the ground surface so that at its maximum height, its final velocity is zero as shown in the figure below. The velocity of the ball tends to decrease more slowly and slowly until it comes momentarily to a stop, and it then falls back faster and faster till it hits the plane of projection. For a vertical upward motion, the acceleration due to gravity has the value of −9.81𝑚𝑠 −2. 𝑣=0 𝑔 = − 9.81𝑚𝑠−2 ℎ 𝑢 Body of mass, 𝑚 at rest m Ground surface Figure 4.0: Diagrammatic representation of a body thrown vertically upward. For a vertically upward motion, the equations of motion are 𝑢 = 𝑔𝑡 1 ℎ = 𝑢𝑡 − 𝑔𝑡 2 and 𝑢2 = 2𝑔ℎ 2 15 | P H Y 1 2 1 0 ( M e c h a n i c s ) – L e c t u r e N o t e s b y Y u s u f A. M u s a , w w w. a y m u s a. c o m 4.1. 7 Relative Motion Motion does not happen in isolation. For example, if you are travelling in a train and the train is moving at a speed of 100 km/hr, then your speed according to another passenger sitting on that train is zero. According to him, you are not moving. But if someone observes you from outside the train, standing on the ground, according to him, you are moving with 100 km/hr as you are on the train and the train is moving with 100 km/hr. Relative motion is a motion where more than one body is in motion and we are in some way examining the motion of one moving body relative to another moving body. The concept of reference frames was first introduced to discuss relative motion in one or more dimensions. When we say an object has a certain velocity, then this velocity is with respect to some frame that is known as the reference frame. In everyday life, when we measure the velocity of an object, the reference frame is taken to be the ground or the earth. The relative velocity of an object 𝑨 with respect to object 𝑩 (denoted by 𝑉𝐴𝐵 )is the rate of change of position of the object A with respect to object B. If 𝑉𝐴 and 𝑉𝐵 be the velocities of objects A and B with respect to the ground, then The relative velocity of A with respect to B is 𝑉𝐴𝐵 = 𝑉𝐴 − 𝑉𝐵 The relative velocity of B with respect to A is 𝑉𝐵𝐴 = 𝑉𝐵 − 𝑉𝐴 EXAMPLE 15 A car travels up a hill at a constant speed of 37 km/h and returns down the hill at a constant speed of 66 km/h. Calculate the average speed of the car for the whole trip. EXAMPLE 16 A goat and a dog are in a road race to defend the honour of their breed. The goat moves the entire 1000 meters at a speed of 0.2 m/s. The dog runs the first 200 meters at 2 m/s, stops to take a nap for 45 minutes, and awakens to finish the last 800 meters with an average speed of 3 m/s. Who wins the race and by how much time? EXAMPLE 17 An airplane accelerates down a runway at 3.20 m/s2 for 32.8s until is finally lifts off the ground. Determine the distance travelled before take-off. EXAMPLE 18 A bike accelerates uniformly from rest to a speed of 7.10 m/s over a distance of 35.4 m. Determine the acceleration of the bike. EXAMPLE 19 An airplane lands with an initial velocity of 70.0 m/s and then decelerates at 1.50 m/s 2 for 40.0 s. What is its final velocity? EXAMPLE 20 You are driving along the street at the speed limit (80km/h) and 50 metres before reaching a traffic light, you notice it becoming yellow. You accelerate to make the traffic light within the 3 seconds it takes for it to turn red. What is your speed as you cross the intersection? Assume that the acceleration is constant and there is no air resistance. EXAMPLE 21 A tarin covers 60 miles between 2:00pm to 5:00pm. How fast the train is going after 3:00pm.? 16 | P H Y 1 2 1 0 ( M e c h a n i c s ) – L e c t u r e N o t e s b y Y u s u f A. M u s a , w w w. a y m u s a. c o m EXAMPLE 22 A body moving with uniform acceleration covers 6m in the 3rd second and 16m in the 5th second. Calculate the initial velocity, acceleration and distance moved in the 6th second. EXAMPLE 23 An object starting from rest moves with uniform acceleration of 4 m/s 2. Find its displacement (i) 5seconds later (ii) in the 5th second (iii) in the 8th second. EXAMPLE 24 A tennis ball is thrown vertically downward from a height of 30m to hit the top of a man’s head. If the height of the man is 3.5m tall, what is the time taken for the ball to hit the top of the man’s head?. (Take g=10m/s2) EXAMPLE 25 A boy drops a ball from the roof of a house which takes 3 seconds to hit the ground. Calculate the velocity before the ball crashes to the ground. (Take g=10m/s2) EXAMPLE 26 Calculate the time taken for a body to reach a height of 45m vertically upward if it is projected with an initial velocity of 30m/s. (Take g=10m/s2) EXAMPLE 27 A body is thrown vertically upward with an initial velocity of 20m/s. Find the highest point that it will reach and the time to reach this height. (Take g=10m/s2) EXAMPLE 28 A ball was dropped from the top of a building of height 200m, a stone was projected vertically upward at an initial velocity of 80m/s at the same time. Find the time and height of the two bodies when they are passing each other. (Take g=10m/s2) EXAMPLE 29 Two bodies A and B are travelling with the same speed 100 km/hr in opposite directions. Find the relative velocity of body A with respect to body B and relative velocity of body B with respect to body A. EXAMPLE 30 A truck is traveling south at a speed of 70 km/h toward an intersection. A car is traveling east toward the intersection at a speed of 80 km/h. What is the velocity of the car relative to the truck? EXAMPLE 31 A boat heads north in still water at 4.5 m/s directly across a river that is running east at 3.0 m/s. What is the velocity of the boat with respect to the river? EXAMPLE 32 A pilot must fly his plane due north to reach his destination. The plane can fly at 300 km/h in still air. A wind is blowing out of the northeast at 90 km/h. (a) What is the speed of the plane relative to the ground? (b) In what direction must the pilot head her plane to fly due north? 17 | P H Y 1 2 1 0 ( M e c h a n i c s ) – L e c t u r e N o t e s b y Y u s u f A. M u s a , w w w. a y m u s a. c o m 4.2 KINEMATICS IN TWO DIMENSIONS (PROJECTILE MOTION) In the previous section, we investigated one-dimensional motion in which the acceleration is constant. Let us now turn now to a two-dimensional motion in which the acceleration remains constant in both magnitude and direction. The most common type of a two - dimensional motion with constant acceleration is Projectile motion. A projectile is any object thrown into air and allowed to move on its own under the influence of gravity and air resistance. Examples of projectile are: a kicked football, a thrown javelin, a launched missile etc. The motion of a projectile is called projectile motion. The path followed by a projectile is called its trajectory. In analysing a projectile motion, we normally neglected the effect of air resistance. A projectile motion is considered to be two independent motions: one in the horizontal direction acceleration with zero acceleration and the other in the vertical direction with constant acceleration as shown in the diagram below. y 𝑢 = initial velocity of projection 𝜃 = angle of projection 𝑢𝑥 = 𝑢𝑐𝑜𝑠𝜃, 𝑢𝑦 = 𝑢𝑠𝑖𝑛𝜃 𝑔𝑥 = 0, 𝑔𝑦 = −𝑔 A 𝑢 Trajectory H 𝜃 B O x R Figure 5.0: Diagrammatic representation of the motion of a projectile from ground level. From the above diagram, the projectile at point O is projected at an angle of 𝜃 to the horizontal with initial velocity 𝑢. The body reaches its maximum height at a point A and later strikes the ground surface at point B with final velocity 𝑣. The horizontal distance covered by the projectile from the point of projection to the point where it strikes thee ground surface is called the range of projection, 𝑅 and the total time taken to achieve the range of projection is called the time of flight. H is the maximum height reached by the projectile. The equations of motion of the projectile in the x-direction are: 𝑣𝑥 = 𝑢𝑐𝑜𝑠𝜃 (13) 𝑥 = (𝑢𝑐𝑜𝑠𝜃)𝑡 (14) Similarly, the equations of motion of the projectile in the y-direction are: 𝑣𝑦 = 𝑢𝑠𝑖𝑛𝜃 − 𝑔𝑡 (15) 1 𝑦 = (𝑢𝑠𝑖𝑛𝜃)𝑡 − 𝑔𝑡 2 (16) 2 𝑣𝑦2 = 𝑢2 sin2 𝜃 − 2𝑔𝑦 (17) 18 | P H Y 1 2 1 0 ( M e c h a n i c s ) – L e c t u r e N o t e s b y Y u s u f A. M u s a , w w w. a y m u s a. c o m 4.2.1 Trajectory of Projectile Motion From equation (13), we get 𝑥 𝑡= (18) 𝑢𝑐𝑜𝑠𝜃 Putting equation (18) into equation (16) we will get 𝑥 1 𝑥 2 𝑔 𝑦 = (𝑢𝑠𝑖𝑛𝜃) − 𝑔( ) = 𝑥𝑡𝑎𝑛𝜃 − ( 2 2 ) 𝑥 2 𝑢𝑐𝑜𝑠𝜃 2 𝑢𝑐𝑜𝑠𝜃 2𝑢 cos 𝜃 𝑔 (19) ∴ 𝑦 = 𝑥𝑡𝑎𝑛𝜃 − ( 2 2 ) 𝑥 2 2𝑢 cos 𝜃 From equation (19), it can be seen that the trajectory of the projectile is a parabola. 4.2.2 Time to Reach Maximum Height (𝑡ℎ ) At the maximum height (point A in figure 5.0), 𝑣𝑦 = 0 and 𝑡 = 𝑡𝐻 From equation (15), we will have 0 = 𝑢𝑠𝑖𝑛𝜃 − 𝑔𝑡𝐻 𝑢𝑠𝑖𝑛𝜃 ∴ 𝑡𝐻 = 𝑔 4.2.3 Maximum Height Reached by Projectile (𝐻) At the maximum height (point A in figure 5.0), 𝑣𝑦 = 0 and 𝑦 = 𝐻 From equation (17), we will have 0 = 𝑢2 sin2 𝜃 − 2𝑔𝐻 𝑢2 sin2 𝜃 ∴𝐻= 2𝑔 Thus, greatest height will be obtained when 𝜃 = 900 4.2.4 Time of Flight of the Projectile (𝑇) When the projectile strikes the ground (point B in figure 5.0), 𝑦 = 0 and 𝑡 = 𝑇 From equation (16), we will have 1 0 = (𝑢𝑠𝑖𝑛𝜃)𝑇 − 𝑔𝑇 2 2 2𝑠𝑖𝑛𝜃 ∴𝑇= = 2𝑡𝐻 𝑔 4.2.5 Horizontal Range Covered by the Projectile (𝑅) When the projectile strikes the ground (point B in figure 5.0), 𝑥 = 𝑅 and 𝑡 = 𝑇 From equation (14), we will have 2𝑢𝑠𝑖𝑛𝜃 𝑢2 (2𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃) 𝑢2 𝑠𝑖𝑛2𝜃 𝑅 = (𝑢𝑐𝑜𝑠𝜃) = = 𝑔 𝑔 𝑔 𝑢2 𝑠𝑖𝑛2𝜃 ∴𝑅= 𝑔 Thus, maximum range will be obtained when 𝜃 = 450 19 | P H Y 1 2 1 0 ( M e c h a n i c s ) – L e c t u r e N o t e s b y Y u s u f A. M u s a , w w w. a y m u s a. c o m The velocity with which the projectile strikes the ground will be given by 𝑣 = √𝑣𝑥2 + 𝑣𝑦2 = √𝑢2 cos2 𝜃 + (𝑢𝑠𝑖𝑛𝜃 − 𝑔𝑇)2 ∴ 𝑣 = √𝑢2 cos2 𝜃 + (𝑢𝑠𝑖𝑛𝜃 − 𝑔𝑇)2 Let us now consider the case when the projectile is launched at a height of 𝒉 above the ground surface as shown in figure 6.0. 𝑢 = initial velocity of projection y 𝜃 = angle of projection 𝑢𝑥 = 𝑢𝑐𝑜𝑠𝜃, 𝑢𝑦 = 𝑢𝑠𝑖𝑛𝜃 𝑔𝑥 = 0, 𝑔𝑦 = −𝑔 C Trajectory H 𝜃 O ℎ ℎ x D R Figure 6.0: Diagrammatic representation of the motion of a projectile at a height above ground level. The equation of motion along the y-direction will be given by 1 𝑦 − ℎ = (𝑢𝑠𝑖𝑛𝜃)𝑡 − 𝑔𝑡 2 2 1 𝑦 = ℎ + (𝑢𝑠𝑖𝑛𝜃)𝑡 − 𝑔𝑡 2 (20) 2 Putting equation (18) into equation (20), we will have 𝑔 𝑦 = ℎ + 𝑥𝑡𝑎𝑛𝜃 − ( ) 𝑥2 (21) 2𝑢2 cos 2 𝜃 Equation (21) is the equation of a projective which projected into the air at a height of h above the ground level. The maximum height reached by the projectile will be given by 𝑢2 sin2 𝜃 (22) Maximum height reched = ℎ + 𝐻 = ℎ + 2𝑔 Which should always take note that the value of ℎ is negative when the body goes down and positive when it goes up. 20 | P H Y 1 2 1 0 ( M e c h a n i c s ) – L e c t u r e N o t e s b y Y u s u f A. M u s a , w w w. a y m u s a. c o m EXAMPLE 33 A ball is launched upward at a velocity of 60 m/s in a direction making an angle of 250 with the horizontal. Take 𝑔 = 9.81 𝑚/𝑠 2. Calculate (a) the maximum height and horizontal range covered by the projection (b) what is the total time spent by the ball in air? (c) what is the total time spent by the ball in air? (d) how high up or down is the ball after 2.2s? (e) what is the magnitude of the velocity of the ball just before it hits the ground? EXAMPLE 34 Find the angle of projection such that the maximum height reached by the projectile equals the horizontal range covered. EXAMPLE 35 The trajectory of a projectile launched from the ground is given by the equation 𝑦 = −0.025𝑥 2 + 0.5𝑥, where 𝑥 and 𝑦 are the coordinates of the projectile on a rectangular system of axes. Find the initial velocity and the angle at which the projectile was launched. EXAMPLE 36 A stone is to be launched at an angle of 300 so that it falls beyond the pond of length 20 metres as shown in the diagram below. Take 𝑔 = 9.81 𝑚/𝑠 2. What is the range of values of the initial velocity 𝑢 so that the projectile falls between points M and N? Trajectory 𝑢 Pond 0 30 M N O 10 m 10 m 20 m EXAMPLE 37 A basketball player throws a ball with an initial velocity of 6.5 𝑚/𝑠 at an angle of 490 above the horizontal to a top of the basket which is located a horizontal distance 𝐿 from the point of release as shown in the diagram below. What is the required value of 𝐿 if the basketball is to reach the basket? Take 𝑔 = 9.81 𝑚/𝑠 2. The effect of air resistance can be ignored. L 6.5 𝑚/𝑠 490 1.01 m Basketball player 2.3 m 21 | P H Y 1 2 1 0 ( M e c h a n i c s ) – L e c t u r e N o t e s b y Y u s u f A. M u s a , w w w. a y m u s a. c o m EXAMPLE 38 A stone was thrown upward from the top of a cliff at an angle of 370 to the horizontal with initial speed of 65 m/s. If the height of the cliff is 125 m, how long is it before the stone hits the ground? Take 𝑔 = 10 𝑚/𝑠 2. EXAMPLE 39 A plan is flying horizontally with a constant speed of 75 m/s at an altitude of 914 m. If the pilot drops a package with same horizontal speed of 76 m/s, determine the angle 𝜃 at which he must sight the target so that when the package is released it falls and strikes the target. Take 𝑔 = 10 𝑚/𝑠 2. EXAMPLE 40 An object is projected horizontally at 8.0 m/s from the top of a 122.5 m building. How far from the base of the building will the object strike the ground? EXAMPLE 41 A football player kicks a ball at a 2.4m tall goalpost from a distance of 30m. He kicks the ball at an angle of 10 degrees above the horizontal and with a velocity of 25 m/s. Will he score a goal? EXAMPLE 42 A daredevil is attempting to jump his motorcycle over a line of buses parked end to end by driving up a 32 0 ramp at a speed of 40.0 m/s (144 km/h). How many buses can he clear if the top of the take-off ramp is at the same height as the bus tops and the buses are 20.0 m long? EXAMPLE 43 Suppose a soccer player kicks the ball from a distance 30 m toward the goal. Find the initial speed of the ball if it just passes over the goal, 2.4 m above the ground, given the initial direction to be 40 0 above the horizontal. EXAMPLE 44 A basketball player is running at 5.00 m/s directly toward the basket when he jumps into the air to dunk the ball. He maintains his horizontal velocity. What vertical velocity does he need to rise 0.750 m above the floor? 22 | P H Y 1 2 1 0 ( M e c h a n i c s ) – L e c t u r e N o t e s b y Y u s u f A. M u s a , w w w. a y m u s a. c o m EXCERCISES 2 Q1. The position of an object moving in a straight line is given by 𝑥 = 3𝑡 − 4𝑡 + 𝑡 3 , where 𝑥 is in metres and 𝑡 in seconds. (a) Calculate the position of the object at t=1.0, 2.0, 3.0, and 4.0 s. (b) Calculate the object’s displacement between t = 0 and t = 4.0 s. (c) Calculate the average velocity for the time interval from t = 2.0 to t = 4.0 s. (d) Calculate the time at which the object will be brought to rest. (e) What is the acceleration of the object at any time t? Q2. An automobile travels on a straight road for 40 km at 30 km/h. It then continues in the same direction for another 40 km at 60 km/h. Calculate the average speed of the car during this 80 km trip. Q3. Two trains, each having a speed of 30 km/h, are headed toward each other on the same straight track. A bird that can fly 60 km/h flies off the front of one train when they are 60 km apart and heads directly for the other train. On reaching the other train it flies directly back to the first train, and so forth. Calculate the total distance the bird travels before the trains crash. Q4. A particle had a velocity of 18 m/s and 2.4s later its velocity was 30 m/s in the opposite direction. Calculate the average acceleration of the particle during this 2.4s interval. Q5. An object has a constant acceleration of 3.2 m/s2. At a certain clock reading its velocity is +9.6 m/s. Determine its velocity (a) 2.5s earlier and (b) 2.5s later. Q6. A car moving with constant acceleration covers the distance between two points 60 m apart in 6.0 s. Its speed as it passes the second point is 15 m/s. (a) Calculate the speed at the first point. (b) Calculate the acceleration. (c) At what prior distance from the first point was the car at rest? Q7. An object falls from a bridge that is 45 m above the water. It falls directly into a small boat moving with constant velocity that was 12 m from the point of impact when the object was released. Calculate the speed of the boat. Q8. A stone is dropped into the water from a bridge 144 ft. above the water. Another stone is thrown vertically down 1.0 s after the first is dropped. Both stones strike the water at the same time. Calculate the initial speed of the second stone. (g =32 ft/s2). Q9. A projectile is launched at ground level with an initial speed of 50.0 m/s at an angle of 30 0 above the horizontal. It strikes a target above the ground 3.00 seconds later. What are the x and y distances from where the projectile was launched to where it lands? Q10. A ball is kicked with an initial velocity of 16 m/s in the horizontal direction and 12 m/s in the vertical direction. (a) At what speed does the ball hit the ground? (b) For how long does the ball remain in the air? (c)What maximum height is attained by the ball? Q11. Serving at a speed of 170 km/h, a tennis player hits the ball at a height of 2.5 m and an angle θ below the horizontal. The service line is 11.9 m from the net, which is 0.91 m high. What is the angle θ such that the ball just crosses the net? Will the ball land in the service box, whose outline is 6.40 m from the net? Q12. Can a goalkeeper at her/ his goal kick a soccer ball into the opponent’s goal without the ball touching the ground? The distance will be about 95 m. A goalkeeper can give the ball a speed of 30 m/s. 23 | P H Y 1 2 1 0 ( M e c h a n i c s ) – L e c t u r e N o t e s b y Y u s u f A. M u s a , w w w. a y m u s a. c o m 5.0 FORCE, MOMENTUM AND IMPULSE 5.1 Force Force is any agent that can change the state of motion of an object. When you kick a ball a force is applied and the ball changes its position or state of motion. Force is a vector quantity having both magnitude and direction. The SI unit of force is Newton (N). 5.2 Mass and Weight Mass and weight are the most confusing concepts in physics. Sometimes we use them interchangeably in daily life; however, in physics we must be careful while using them. Mass is the quantity of matter in a body which is constant everywhere. The SI unit of mass is kilogram (kg). weight is of course related to the mass of a body, but it has a little bit different definition. Weight is the force of gravity acting upon an object. Since it is a force, weight is directly proportional to the acceleration due to gravity, 𝑔. Weight is vector quantity which changes from place to place on the earth. Thus, 𝑤 = 𝑚𝑔 (23) Where 𝑤 is the weight of the body in Newtons (N) and m is the mass of the body in kilograms (kg). Mass is the measure of the inertia of a body. Inertia is the measure of the reluctance of a body at rest to move or its reluctance to stop while in motion. The more the mass of a body the higher its inertia. 5.3 Newton’s First law of motion In his first law of motion, Newton stated that all objects save their state of motion. In other words, if an object is at rest it will continue to be at rest and if it is in motion, it will continue to be in motion unless a non-zero force is applied on it. Newton’s first law of motion is sometimes called the law of inertia which stated that, objects want to continue their state of motion whether they are at rest or in motion. 5.4 Newton’s Second Law of Motion According to the Newton’s second law of motion, force (F) is linearly proportional to product of mass (m) and acceleration of a body. Thus, 𝐹 = 𝑚𝑎 (24) If the mass is constant, when we increase the force, the object gains acceleration with the same amount or, if the force is constant, when we decrease the mass, the acceleration increases with the same amount. 5.5 Newton’s Third Law of Motion Newton’s third law states that when two bodies interact, they apply forces (action and reaction) to one another that are equal in magnitude and opposite in direction. The third law is also known as the law of (25) action and reaction. This law is important in analyzing problems of static equilibrium, where all forces are balanced, but it also applies to bodies in uniform or accelerated motion. 𝑎𝑐𝑡𝑖𝑜𝑛 = −𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 5.6 Momentum and Impulse Any object can have momentum. This is because momentum is mass in motion. The way we determine an object’s momentum is fairly straightforward. Momentum (p) is the object’s mass (m) times its velocity (v). Thus, 𝑝 = 𝑚𝑣 (26) Momentum is proportional to both mass and velocity, meaning that a change in one will cause the same amount of change in the other. So, if you increase the mass of an object, you also increase its momentum. 24 | P H Y 1 2 1 0 ( M e c h a n i c s ) – L e c t u r e N o t e s b y Y u s u f A. M u s a , w w w. a y m u s a. c o m The same is true for velocity. Increase or decrease the object’s velocity and you increase or decrease its momentum by the same amount. To understand how a change in momentum affects an object, we look to impulse. Force is an important factor, but time also counts. Specifically, when we are interested in knowing how long the force acts. For example, if you push a box across the floor for just a few seconds, the time interval is very short. But if you push a box across the same floor and you do so with the same force as before, but this time for several minutes, you’ve increased the amount of time the force acts. This longer time interval leads to a grater change in momentum. This change in momentum is called impulse. Impulse has the same SI unit of kgm/s as that of momentum. 𝑣−𝑢 Impulse = change in momentum = 𝑚(𝑣 − 𝑢) = 𝑚 ( ) 𝑡 = (𝑚𝑎)𝑡 = 𝐹𝑡 𝑡 Impulse = 𝑚(𝑣 − 𝑢) = 𝐹𝑡 (27) EXAMPLE 45 A net force of 6N acts upon an object with a mass of 2kg. if the object were initially moving with velocity of 1m/s when the force was applied, what will be its displacement after 5s? EXAMPLE 46 A 1.5kg ball drops vertically onto a floor, hitting the floor with a speed of 25 m/s. The ball was in contact with the floor for 0.020s and then rebounds to a height of 11.5m. What is the magnitude of the average force of the floor on the ball? EXAMPLE 47 A ball of mass 0.500kg moving with a speed 15.0 m/s collides with a wall and bounces back with a speed of 10.5m/s. If the motion is in a straight line, calculate the initial and final momenta of the ball. If the ball exerted an average force of 1Kn on the wall, how long did the collision last? 25 | P H Y 1 2 1 0 ( M e c h a n i c s ) – L e c t u r e N o t e s b y Y u s u f A. M u s a , w w w. a y m u s a. c o m 6.0 WORK, ENERGY AND POWER 6.1 Work (W) Work is done when force is exerted on an object and the object moves from one place to another. Work is as a result of force acting over a certain distance called displacement. If the force F, and the displacement x, are in the same direction, then work done is given by the formula Work done = 𝐹𝑥 𝑊 = 𝐹𝑥 (28) The SI unit of work done is Newton-metre (Joule). 1𝑁𝑚 = 1𝐽. Equation (28), depends on force applied and the displacement being in the same direction. There are many situations in which the force is not in the same direction as the displacement. In a case like this, only the component of the force that is along the path of the displacement contributes to the work done. If we treat the force and displacement as vectors, then 𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 = 𝐹⃗ ∙ 𝑥⃗ = 𝐹𝑥𝑐𝑜𝑠𝜃 (29) Where 𝜃 is the angle between the force and the displacement vectors. 6.2 ENERGY Energy is the capacity or ability for doing work. You must have energy to accomplish work. To do 100 Joules of work, you must expend 100 Joules of energy. There basically two main types of energy, kinetic and potential energy. Energy is measure in Joules (J). 6.2.1 Kinetic Energy (K) Kinetic energy is energy of motion. The kinetic energy of an object is the energy it possesses because of its motion. The kinetic energy of a body of mass 𝑚, moving with velocity 𝑣 is given by 1 (30) Kinetic energy, K = 𝑚𝑣 2 2 Kinetic energy is an expression of the fact that a moving object can do work on anything it hits; it quantifies the amount of work the object could do as a result of its motion. 6.2.2 Potential Energy (P) Potential energy is energy that is stored. There are various types of stored or potential energy. Chemical energy from a battery is a form of potential energy, elastic energy in a stretched rubber band is a form of potential energy too. The most commonly referred to form of potential energy in physics is that of a gravitational potential energy. Gravitational potential energy is the energy that is stored due to an object’s position. It is dependent on the mass 𝑚 of the object, the height ℎ of the object above the ground surface, and the acceleration due to gravity, 𝑔. Thus, Potential energy, P = 𝑚𝑔ℎ (31) 6.2.3 Mechanical Energy (E) In the process of doing work, the object that is doing the work exchanges energy with the object upon which the work is done. When the work is done upon the object, that object gains energy. The energy acquired by the objects upon which work is done is known as mechanical energy. Mechanical energy is the energy that is possessed by an object due to its motion and/or due to its position. Mechanicl energy, E = kinetic energy + potential energy 26 | P H Y 1 2 1 0 ( M e c h a n i c s ) – L e c t u r e N o t e s b y Y u s u f A. M u s a , w w w. a y m u s a. c o m  𝐸 =𝐾+𝑃 (32) the principle of conservation of mechanical energy sates that the total mechanical energy in any given system remains constant as long as the forces acting are conservative forces. Thus, 1 1 𝑚𝑣𝑖2 + 𝑚𝑔ℎ𝑖 = 𝑚𝑣𝑓2 + 𝑚𝑔ℎ𝑓 (33) 2 2 A conservative force is a force which does not change the total mechanical energy. Magnetic, electrical, gravitational forces are examples of conservative forces. 6.2.4 Work -Energy Theorem If you push a stationary car, the car will start moving. That means, the car has gained some velocity. We say that the car has gained some kinetic energy. According to the work energy theorem, this work is converted into the kinetic energy of the car. The work – energy theorem gives the relationship between work done and kinetic energy. For the sake of simplicity, we will consider the case in which the resultant force 𝐹 is constant in both magnitude and direction and is parallel to the velocity of the particle as shown in figure 7.0 below. 𝑣𝑖 𝑣𝑓 𝐹 𝐹 m m 𝑥 Figure 7.0: Diagrammatic representation of work -energy theorem. The particle is moving with constant acceleration 𝑎 along a straight line. The relationship between the net force and the acceleration is given by the equation 𝐹 = 𝑚𝑎 (Newton’s second law), and the particle’s displacement 𝑥, can be determined from the equation: 𝑣𝑓2 = 𝑣𝑖2 + 2𝑎𝑥 𝑣𝑓2 − 𝑣𝑖2 𝑥= (34) 2𝑎 The work of the net force is calculated as the product of its magnitude (𝐹 = 𝑚𝑎) and the particle’s displacement 𝑥. Substituting equation (34) above equations yields: 𝑣𝑓2 − 𝑣𝑖2 1 1 Work done = 𝐹𝑥 = 𝑚𝑎 ( ) = 𝑚𝑣𝑓2 − 𝑚𝑣𝑖2 = change in kinetic energy 2𝑎 2 2 1 1 𝑊 = 𝑚𝑣𝑓2 − 𝑚𝑣𝑖2 (35) 2 2 Thus, the work done by the net force on an object equals the change in the object’s kinetic energy. 27 | P H Y 1 2 1 0 ( M e c h a n i c s ) – L e c t u r e N o t e s b y Y u s u f A. M u s a , w w w. a y m u s a. c o m 6.3 POWER We can define power as the rate of doing work, and it is the amount of energy consumed per unit of time. Therefore, it can be calculated by dividing work done by time. The formula for power is given below. work done 𝑊 (36) Power, P = = time 𝑡 Where, 𝑃 is the power, 𝑊 is the work done and 𝑡 is the time taken. As power doesn’t have any direction, it is a scalar quantity. The SI unit of power is Joules per Second (J/s), which is termed as Watt. Watt can be defined as the power needed to do one joule of work in one second. The unit Watt is dedicated in honour of Sir James Watt, the developer of the steam engine. One common unit of energy is the kilowatt-hour (kWh). If we are using one kW of power, a kWh of energy will last one hour. Horsepower is another unit of measurement. One horsepower is equivalent to 735.7 Watts. From the definition of work done, 𝑊 = 𝐹𝑥 𝐹𝑥 ⇒ Power = = 𝐹𝑣 𝑡 Where 𝑣 is the velocity of the body. When the force 𝐹 acting on the body is not in the direction of motion, the power can be expressed as 𝑃𝑜𝑤𝑒𝑟 = 𝐹⃗ ⋅ 𝑣⃗ = 𝐹𝑣𝑐𝑜𝑠𝜃 (37) Where 𝜃 is the angle between the force and velocity vectors. EXAMPLE 48 How much work is done when a weightlifter lifts a 200N barbell from the floor to a height of 2m? EXAMPLE 49 An ice skater with a mass of 50 kg is gliding across the ice at a speed of 8 m/s when her friend comes up from behind and gives her a push, causing her speed to increase to 12 m/s. How much work did the friend do on the skater? EXAMPLE 50 A body of mass 10kg at rest is subjected to a force of 16N. Find the kinetic energy at the end of 10 s. EXAMPLE 51 A particle moves along 𝑥 − 𝑎𝑥𝑖𝑠 from 𝑥 = 0 to 𝑥 = 8 under the influence of a force given by 𝐹 = 3𝑥 2 − 4𝑥 + 5. Find the work done in the process. EXAMPLE 52 A garage hoist lifts a truck up 2 meters above the ground in 15 seconds. Find the power delivered to the truck. The mass of the truck is 1000 kg. EXAMPLE 53 A body of mass 5kg is thrown up vertically with a kinetic energy of 1000 J. If acceleration due to gravity is 10ms-2, find the height at which the kinetic energy becomes half of the original value. 28 | P H Y 1 2 1 0 ( M e c h a n i c s ) – L e c t u r e N o t e s b y Y u s u f A. M u s a , w w w. a y m u s a. c o m EXCERCISES 3 Q1. A 0.450-kg hammer is moving horizontally at 7.00 m/s when it strikes a nail and comes to rest after driving the nail 1.00 cm into a board. Assume constant acceleration of the hammer-nail pair. Calculate the duration of the impact. What was the average force exerted on the nail? Q2. A particle of mass 5 kg moving with a velocity. (a) What is the momentum of the particle as a function of time? (b) What is the net force acting on this particle? Q3. Calculate the final speed of a 110-kg rugby player who is initially running at 8.00 m/s but collides head- on with a padded goalpost and experiences a backward force of 1.76 × 104 N for 5.5 × 10−2 𝑠 Q4. A gun fires 8 bullets per second into a target X. If the mass of each bullet is 3 g and its speed 600 m/s. Then, calculate the power delivered by the bullets. Q5. A box is pulled with a force of 25 N to produce a displacement of 15 m. If the angle between the force and displacement is 300, find the work done by the force. Q6. If an object of mass 2 kg is thrown up from the ground reaches a height of 5 m and falls back to the Earth (neglect the air resistance). Calculate: (a) The work done by gravity when the object reaches 5 m height (b) The work done by gravity when the object comes back to Earth. Q7. A weight lifter lifts a mass of 250 kg with a force 5000 N to the height of 5 m. (a) What is the work done by the weight lifter? (b) What is the work done by the gravity? (c) What is the net work done on the object? Q8. Calculate the energy consumed in electrical units when a 75 W fan is used for 8 hours daily for one month (30 days). 1 electrical unit = 1kWh. Q9. A vehicle of mass 1250 kg is driven with an acceleration 0.2 m/s2 along a straight level road against an external resistive force 500 N. Calculate the power delivered by the vehicle’s engine if the velocity of the vehicle is 30 m/s. Q10. force 𝑓 = 𝑖̂ + 2𝑗̂ + 3𝑘̂ acts on a particle and displays it through a distance 𝑥 = 4𝑖̂ + 6𝑗̂. Calculate the work done on the particle. 29 | P H Y 1 2 1 0 ( M e c h a n i c s ) – L e c t u r e N o t e s b y Y u s u f A. M u s a , w w w. a y m u s a. c o m 7.0 COLLISION THEORY 7.1 What is a Collision? Collision means two objects coming into contact with each other for a very short period. In other words, collision is a reciprocative interaction between two masses for a very short interval wherein the momentum and energy of the colliding masses changes. You might have seen two billiard balls colliding with each other in the course of the game. This forceful coming together of two separate bodies is called collision. Collision normally involves two masses 𝑚1 and 𝑚2. 7.2 One-Dimensional Collision Between Two Objects One dimensional sudden interaction of masses is that collision in which both the initial and final velocities of the masses lie in one line. All the variables of motion are contained in a single dimension. Consider a one- dimensional elastic collision between two objects moving in the x - direction. One object, with mass 𝑚1 and initial velocity 𝑢1 , collides with an object of mass 𝑚2 and initial velocity 𝑢2. No forces other than the interaction force between the objects act during the collision. After the collision, the final velocities are 𝑣1 , f and 𝑣2 respectively. Because there are no external forces in the x -direction, momentum is constant in the x - direction. Equating the momentum before and after the collision gives the relation 𝑚1 𝑢1 + 𝑚2 𝑢2 = 𝑚1 𝑣1 + 𝑚2 𝑣2 (38) The total kinetic energy before and after the collision may not always be the same as some amount energy is sometimes lost as in form of another energy such as heat energy. Equation (38) is the statement of the law of conservation of linear momentum. 7.3 Characterizing Collisions In a collision, the ratio of the magnitudes of the initial and final relative velocities is called the coefficient of restitution and denoted by the symbol e, relative speed after collison 𝑣𝑎 𝑣2 − 𝑣1 𝑒=| |=| |=| | relative speed before collion 𝑣𝑏 𝑢1 − 𝑢2 Where 𝑣𝑏 and 𝑣𝑎 are the relative velocities of the body before and after the collision respectively. If the magnitude of the relative velocity does not change during a collision, 𝑒 = 1, then the change in kinetic energy is zero. Collisions in which there is no change in kinetic energy (∆𝐾) are called perfectly elastic collisions, Thus, for a perfectly elastic collision, 𝑒 = 1 If the magnitude of the final relative velocity is less than the magnitude of the initial relative velocity, 𝑒 < 1, then the change in kinetic energy is negative. Collisions in which the kinetic energy decreases are called inelastic collisions, Thus, for an inelastic collision, 𝑒< 1 If the two objects stick together after the collision, then the relative final velocity is zero, 𝑒 = 0. Such collisions are called totally inelastic. Thus, for a totally inelastic collision, 𝑒= 0 30 | P H Y 1 2 1 0 ( M e c h a n i c s ) – L e c t u r e N o t e s b y Y u s u f A. M u s a , w w w. a y m u s a. c o m If the magnitude of the final relative velocity is greater than the magnitude of the initial relative velocity, e > 1, then the change in kinetic energy is positive. Collisions in which the kinetic energy increases are called super elastic collisions, Thus, for a super elastic collision, 𝑒> 1 7.4 One-Dimensional Totally Inelastic Collision Between Two Objects For a totally inelastic collision, 𝑒 = 0 so that 𝑣1 = 𝑣2 = 𝑣 = common velocity. Thus, equation (38) reduces to 𝑚1 𝑢1 + 𝑚2 𝑢2 = (𝑚1 + 𝑚2 )𝑉 𝑚1 𝑢1 + 𝑚2 𝑢2 ∴𝑣= (39) 𝑚1 + 𝑚2 The change or lost in kinetic energy before and after the collision will be given by 1 1 1 ∆𝐾 = (𝑚1 + 𝑚2 )𝑣 2 − (𝑚1 𝑢12 + 𝑚2 𝑢22 ) = [(𝑚1 + 𝑚2 )𝑣 2 − 𝑚1 𝑢12 − 𝑚2 𝑢22 ] 2 2 2 1 ∴ ∆𝐾 = [(𝑚1 + 𝑚2 )𝑣 2 − 𝑚1 𝑢12 − 𝑚2 𝑢22 ] (40) 2 7.5 One-Dimensional Perfectly Elastic Collision Between Two Objects For a perfectly elastic collision, 𝑒 = 1. 𝑣2 − 𝑣1 = 𝑢1 − 𝑢2 𝑣2 = 𝑢1 − 𝑢2 + 𝑣1 (41) Putting equation (41) into equation (38), we will have 𝑚1 𝑢1 + 𝑚2 𝑢2 = 𝑚1 𝑣1 + 𝑚2 (𝑢1 − 𝑢2 + 𝑣1 ) 𝑚1 𝑢1 + 𝑚2 𝑢2 − 𝑚2 𝑢1 + 𝑚2 𝑢2 = 𝑚1 𝑣1 + 𝑚2 𝑣1 (𝑚1 +𝑚2 )𝑣1 = (𝑚1 − 𝑚2 )𝑢1 + 2𝑚2 𝑢2 (𝑚1 − 𝑚2 )𝑢1 + 2𝑚2 𝑢2 ∴ 𝑣1 = (42) (𝑚1 +𝑚2 ) Putting equation (42) into equation (41), we will have (𝑚1 − 𝑚2 )𝑢1 + 2𝑚2 𝑢2 (𝑢1 − 𝑢2 )(𝑚1 +𝑚2 ) + (𝑚1 − 𝑚2 )𝑢1 + 2𝑚2 𝑢2 𝑣2 = 𝑢1 − 𝑢2 + = (𝑚1 +𝑚2 ) (𝑚1 +𝑚2 ) 𝑚1 𝑢1 +𝑚2 𝑢1 − 𝑚1 𝑢2 − 𝑚2 𝑢2 + 𝑚1 𝑢1 − 𝑚2 𝑢1 + 2𝑚2 𝑢2 = (𝑚1 +𝑚2 ) 2𝑚1 𝑢1 − 𝑚1 𝑢2 + 𝑚2 𝑢2 (𝑚2 − 𝑚1 )𝑢2 + 2𝑚1 𝑢1 = = (𝑚1 +𝑚2 ) (𝑚1 +𝑚2 ) (𝑚2 − 𝑚1 )𝑢2 + 2𝑚1 𝑢1 ∴ 𝑣2 = (43) (𝑚1 +𝑚2 ) Equations 42 and 43 are the final velocities of the two bodies after the perfectly elastic collision. Consider what happens in the limits 𝑚1 = 𝑚2 in Equations 42 and 43. Then 31 | P H Y 1 2 1 0 ( M e c h a n i c s ) – L e c t u r e N o t e s b y Y u s u f A. M u s a , w w w. a y m u s a. c o m  0 + 2𝑚2 𝑢2 0 + 2𝑚1 𝑢1 𝑣1 = = 𝑢2 and 𝑣2 = = 𝑢1 (𝑚2 +𝑚2 ) (𝑚1 +𝑚1 ) Thus, if the two objects are identical, or have the same mass the objects will exchange their velocities, and unless we could somehow distinguish the objects, we might not be able to tell if there was a collision at all. If on the other hand, we consider what happens in the limits 𝑚1 >> 𝑚2 and 𝑚1 is at rest in Equations 42 and 43. Then (𝑚1 − 𝑚2 )0 + 2𝑚2 𝑢2 0 𝑚2 𝑣1 = = +2 ≈0 (𝑚1 ) 𝑚1 𝑚1 and (𝑚2 −𝑚1)𝑢2 + 0 𝑚1 𝑢2 𝑣2 = ≈− + 0 = −𝑢2 (𝑚1 ) 𝑚1 Thus, the less massive object rebounds with the same speed relative to the more massive object which barely changed its speed. 7.6 Two-Dimensional Collision Between Two Objects A collision in two dimensions obeys the same rules as a collision in one dimension: Total momentum in each direction is always the same before and after the collision Total kinetic energy is the same before and after for an elastic collision The kinetic energy is not calculated for each direction separately, but depends on the magnitude of the total velocity of each object. When the objects are free to move in two dimensions, we can continue to use the same law of conservation of linear momentum here as follows: 𝑚1 𝑢1𝑥 + 𝑚2 𝑢2𝑥 = 𝑚1 𝑣1𝑥 + 𝑚2 𝑣2𝑥 (44) and 𝑚1 𝑢1𝑦 + 𝑚2 𝑢2𝑦 = 𝑚1 𝑣1𝑦 + 𝑚2 𝑣2𝑦 (45) For elastic collisions, 1 1 1 1 𝑚1 𝑢12 + 𝑚2 𝑢22 = 𝑚1 𝑣12 + 𝑚2 𝑣22 (45) 2 2 2 2 Where: 𝑢1𝑥 𝑎𝑛𝑑 𝑢1𝑦 are the initial velocities of object of mass 𝑚1 in the x and y directions respectively. 𝑢2𝑥 𝑎𝑛𝑑 𝑢2𝑦 are the initial velocities of object of mass 𝑚2 in the x and y directions respectively. 𝑣1𝑥 𝑎𝑛𝑑 𝑣1𝑦 are the final velocities of object of mass 𝑚1 in the x and y directions respectively. 𝑣2𝑥 𝑎𝑛𝑑 𝑣2𝑦 are the final velocities of object of mass 𝑚2 in the x and y directions respectively. 32 | P H Y 1 2 1 0 ( M e c h a n i c s ) – L e c t u r e N o t e s b y Y u s u f A. M u s a , w w w. a y m u s a. c o m EXAMPLE 54 What is the velocity of a 900 kg car initially moving at 30.0 m/s, just after it hits a 150 kg deer initially running at 12.0 m/s in the same direction? Assume the deer remains on the car. EXAMPLE 55 Train cars are coupled together by being bumped into one another. Suppose two loaded train cars are moving toward one another, the first having a mass of 150,000 kg and a velocity of 0.300 m/s, and the second having a mass of 110,000 kg and a velocity of 0.120 m/s. What is their final velocity? EXAMPLE 56 A 0.240 kg billiard ball that is moving at 3.00 m/s strikes the bumper of a pool table and bounces straight back at 2.40 m/s (80% of its original speed). The collision lasts 0.0150 s. (a) Calculate the average force exerted on the ball by the bumper. (b) How much kinetic energy in joules is lost during the collision? (c) What percent of the original energy is left? EXAMPLE 57 A basketball with an initial speed of 3 meters per second impacts a hard floor at an angle of 600 to the floor. If the collision has a coefficient of restitution of 0.8, what is the expected speed and angle of the basketball after the impact? EXAMPLE 58 A 10.0 g bullet is stopped in a block of wood of mass 5.00kg. The speed of the bullet–plus–wood combination immediately after the collision is 0.600 m/s. What was the original speed of the bullet? EXAMPLE 59 An unstable nucleus of mass 17 × 10-27 kg initially at rest disintegrates into three particles. One of the particles, of mass 5.0 × 10-27 kg, moves along the y-axis with a speed of 6.0 × 106 m/s. Another particle of mass 8.4 × 10-27 kg, moves along the x-axis with a speed of 4.0 × 106 m/s. Find (a) the velocity of the third particle and (b) the total energy given off in the process. EXAMPLE 60 A billiard ball moving at 5.00 m/s strikes a stationary ball of the same mass. After the collision, the first ball moves at 4.33 m/s at an angle of 300 with respect to the original line of motion. Assuming a perfectly elastic collision (and ignoring friction and rotational motion), find the struck ball’s velocity. EXAMPLE 60 Let a ball be dropped from height ℎ to a fixed horizontal plane. If 𝑢 be the velocity of the sphere just before striking the plane. If 𝑣 be the vertically upward velocity with which the ball rebounds to a height 𝐻. Show that 𝐻 = 𝑒 2ℎ Where is the coefficient of restitution between the material of the ball and the horizontal plane. 33 | P H Y 1 2 1 0 ( M e c h a n i c s ) – L e c t u r e N o t e s b y Y u s u f A. M u s a , w w w. a y m u s a. c o m 8.0 CENTER OF MASS 8.1 What is Center of Mass (C.M)? The center of mass of a body or system of a particle is defined as a point where the whole of the mass of the body or all the masses of a set of particles appeared to be concentrated. When we are studying the dynamics of the motion of the system of a particle as a whole, then we need not worry about the dynamics of individual particles o