Chapter One: Properties of Matter PDF

Chapter One ; 1. Physical Contents Quantities search I’m feeling lucky 2. Systems of Units 3. Conversion Factors  Physical Quantities Fundamental Quantities Derived Quantities Mass M It is derived in terms of the Length...

Chapter One ; 1. Physical Contents Quantities search I’m feeling lucky 2. Systems of Units 3. Conversion Factors  Physical Quantities Fundamental Quantities Derived Quantities Mass M It is derived in terms of the Length L fundamental quantities. Time T 𝐝𝐢𝐬𝐭𝐚𝐧𝐜𝐞 𝐭𝐫𝐚𝐯𝐞𝐥𝐥𝐞𝐝 𝐥𝐞𝐧𝐠𝐭𝐡 Speed = = 𝐭𝐢𝐦𝐞 𝐭𝐢𝐦𝐞 Other derived quantities Quantity Formula Dimensions 𝐌𝐚𝐬𝐬 𝐌 Density 𝛒= 𝟑 = 𝐌𝐋−𝟑 𝐕𝐨𝐥𝐮𝐦𝐞 𝐋 Velocity 𝐃𝐢𝐬𝐭𝐚𝐧𝐜𝐞 𝐋𝐓 −𝟏 𝐯= 𝐭𝐢𝐦𝐞 Acceleration 𝐯 𝐋𝐓 −𝟏 𝐚= = 𝐋𝐓 −𝟐 𝐭𝐢𝐦𝐞 𝐓 Pressure 𝐅 𝐦𝐚 𝐌𝐋𝐓 −𝟐 𝐏= = = 𝐌𝐋−𝟏 𝐓 −𝟐 𝐀 𝐀 𝐋𝟐 Systems of Units meter (m) centimeter (cm) foot (ft) kilogram (Kg) gram (gm) Slug second (s) second (s) second (s) coulomb (C) coulomb (C) coulomb (C) newton (N) dyne pound (Ib) Slug = 14.59 kg Conversion Factors 1 meter (m) = 100 cm = 39.4 in = 3.28 ft = 6.21 × 10-4 mi 1 inch (in) = 2.54 cm = 0.0254 m 1 foot (ft) = 0.305 m = 30.5 cm 1 mile (mi) = 1610 m = 1.61 km  Example (1) 𝟏 The expression for kinetic energy E = 𝟐 m v2 (where m is the mass of body and v is its speed) and potential energy E = m g h (where g is the acceleration due to gravity and h is the height of the body) look very different but both describe energy. Prove that, the two expressions can be added to each other. Solution 𝟏 𝐋𝐞𝐧𝐠𝐭𝐡 2 Dimension of kinetic energy = m v2 = Mass ( ) = M ( L T-1)2 = M L2 T-2 𝟐 𝐓𝐢𝐦𝐞 𝐋𝐞𝐧𝐠𝐭𝐡 Dimension of potential energy = m g h = Mass ( ) Length = M ( L T-2) L = M L2 T-2 𝐓𝐢𝐦𝐞𝟐 The two expressions have the same dimensions, so they can be added and subtracted from each other. Example (2) In young’s double-slit experiment, the angle θ of constructive interference is related to wavelength λ of the light, the spacing of slit d and the order number n by d sin θ = n λ show that is dimensionally corrected. Solution d sin θ = n λ Length ( dimensionless) = dimensionless ( Length) L*1=1*L L=L So, both sides have dimension of length. Example (3) Hooke’s law states that the force, F in a spring extended by a length x is given by F = - K x. From Newton’s second law F = m a, where m is mass and a is the acceleration. Calculate the dimension of the spring constant K. Solution 𝐅 K= 𝐱 Now, F = m a, so the dimension of the force is given by 𝐋𝐞𝐧𝐠𝐭𝐡 F = mass * (𝐓𝐢𝐦𝐞)𝟐 = M * L* T-2 Therefore, the spring constant has dimension 𝐅 𝐌 ∗ 𝐋∗ 𝐓 −𝟐 K= = = M * T-2 𝐱 𝐋 Example (4) Using the dimensional analysis to set up an expression of the form x ⍺ an tm Where n and m are exponents that must be determined and the symbol ⍺ indicates a proportionality Solution x ⍺ an tm L. H. S. = x = Length = L = L1 T0 𝐋𝐞𝐧𝐠𝐭𝐡 n R. H. S. = an tm = ( ) (Time)m = (L T-2)n (T)m 𝐓𝐢𝐦𝐞𝟐 = Ln T-2n Tm = Ln T-2n+m L. H. S. = R. H. S. L1 T0 = Ln T-2n+m The exponents of L and T must be the same on both sides of the equation. n = 1, -2n + m = 0 -2 * 1 + m = 0 m=2 Example (5) The force attraction between two masses m1 and m2 lying at a distance r is given by 𝒎𝟏 𝒎𝟐 F=G 𝒓𝟐 Where G is the constant of gravitational force. What is the dimension of G? Solution 𝐫𝟐 G=F 𝐦𝟏 𝐦𝟐 𝐋𝐞𝐧𝐠𝐭𝐡 𝐋𝐞𝐧𝐠𝐭𝐡𝟐 = ( Mass * )( ) 𝐓𝐢𝐦𝐞𝟐 𝐌𝐚𝐬𝐬∗𝐌𝐚𝐬𝐬 𝑳𝟐 = M1 L1 T-2 𝐌𝟏 𝑴𝟏 = M-1 L3 T-2 Example (6) Find the law giving the periodic time t of a simple pendulum, if t depends on the pendulum length l, the mass of its bob and g the acceleration due to gravity? Solution Suppose that the required law is given by T = k Lα mβ gγ Dimensions of L.H.S. are [T] = L 0 M0 T 1 Dimensions of R.H.S. are [Lα] [Mβ] [( LT-2) γ] = Lα+ γ Mβ T-2γ Since the dimensions of both sides must be the same L.H.S. = R.H.S. L0 M0 T1 = Lα+ γ Mβ T-2γ α+ γ = 0 ,β=0 , -2γ = 1 And from which we get α = ½ , β =0 , γ = -1/2 𝐋 Then T = k (𝐠)1/2 Example (7) On an interstate highway in a rural region of Wyoming, a car is traveling at a speed of 38 m/s. is the driver exceeding the speed limit of 75 mi/h? 1 mile (mi) = 1610 m 1m = 6.21 × 10-4 mi 38 m = 38 × 6.21 × 10-4 mi = 2.36 × 10-2 mi Speed = 2.36 × 10-2 mi/s 1 h = 60 × 60 s = 3600 s 1 s = (1/3600) h Speed = 2.36 × 10-2 mi/s = 2.36 × 10-2 mi/(1/3600) h = 84.96 mi/h Example (8) Use the dimension analysis to obtain the constants C1 and C2 of the following relationships. Consider x is distance, t is time, F force, W is the work, m is the mass, v is the velocity and E is the energy. m = (2t/ C1)1/2 + m cos( 2 p t / C2 ) t = (2x / C1 + C2 2 t )1/2 v2 = 2 C1 sin( 2p C2 t) W = 2 t2 C1 - 3m x C2 solution 1. m = (2t/ C1)1/2 + m cos( 2 p t / C2 ) M = ( T/C1)1/2 = M *cos ( T/C2) solution solution