PHY 101 (100 Level) First Semester 20242025 Physics PDF

# PHYSICS. ## 3rd December 24 Physics is the scientific study of the natural world around us. It includes the study of matter, energy and the fundamental laws that govern the behaviour of the physical Universe. - Space and time. Space is the three-dimensional extent within entities that exist a...

# PHYSICS. ## 3rd December 24 Physics is the scientific study of the natural world around us. It includes the study of matter, energy and the fundamental laws that govern the behaviour of the physical Universe. - Space and time. Space is the three-dimensional extent within entities that exist and have a physical relationship to one another. Space is a three-dimensional Continuum containing positions and directions. Time is defined as the measure of a change in a physical quantity or a magnitude used to quantify the duration of events. Time is considered absolute and one-dimensional (because it flows at the same rate everywhere), flowing at the same rate everywhere. ## Units Units are the standard of measurement used to express physical quantities. There are several Systems of Units but the most commonly used System in physics is the International System of Units (SI). - **Basic Units of fundamental Units are:** - Meter (m) - Unit of length. - Kilogram (kg) - mass - Time - Seconds (s) - Temperature - Kelvin (K) - Electric Current - Ampere (A) - Amount of Substance - Mole (mol) - Luminous Intensity - Candela (Cd) - **Derive Unit** Derived Units are units that can be expressed in terms of base units. (combination of two or more fundamental units). Examples includes: - **Newton (N) - Unit of force** - N = kgm/s^2 - **Joules (J) - Unit of energy/workdone** - J = force x distance - J = kgm/s^2 x m - J = kgm^2/s^2 - **Watt (W) - Unit of power** - W = force x distance x time - W = kgm^2/s^2 x s - W = kgm^2/s^3 ## **Dimensions** Dimensions are the fundamental characteristics of physical quantities. There are several dimension including: - Length ( L) - NB: Any quantity that can be measured in meters is represented with (L) eg width, breadth, height - Mass (M) - Time (T) - Temperative (θ) [K] - Electric Current (I) - Luminous Intensity (J). ## **Dimensional Analysis** Dimensional Analysis is the process of analysing the dimensions of physical quantities to determine their relationship. This can be useful to: 1. Checking the validity of equations. 2. Deriving new equations. 3. Analysing the behaviour of physical systems. ## Solved Examples. 1. Speed = distance/time - L/T = LT^−1 2. Acceleration = Displacement/time - L/T^2 = LT^−2 3. Force = Mass x acceleration - M x LT^−2 = MLT^−2 4. Energy = Force x distance - MLT^−2 x L = MLT^−2 5. Pressure = force/area - MLT^−2 / LxL = ML^−1T^−2 6. Power = force x velocity - MLT^−2 x LT^−1 = ML^2T^−3 ## **Assignment.** Find the dimensions of the following: 1. Density 2. Viscosity. 3. Surface tension 4. Strain 5. Stress. ## Solutions. 1. Density = mass/volume - M/L^3 = ML^−3 2. Viscosity is the ratio of shear stress to shear rate. Shear stress is force per unit area and shear rate is the rate of change of velocity with respect to distance. - Viscosity = Shear Stress/Shear rate - force x area/time - MLT^−2/L^2/T - MLT^−2L^2/T - MLT^−1/ T 3. Surface tension is the force per unit length acting on the surface of a liquid. - Surface tension = Force/length - MLT^−2/L - MT^−2 4. Strain is the ratio of length change to original length. - Strain = Length/Length - L/L - Therefore, Strain is dimensionless. 5. Stress is the force per unit area - Stress = Force/area - MLT^−2/L^2 = ML^−1T^−2 ## Displacement, Velocity and Acceleration. ### Displacement Displacement is the change in position of an object from one point to another. It is a vector quantity which means it has both magnitude and direction. Change in displacement can be represented by Δx. ### Vector Vector is the rate of change of displacement in respect to time. It is also represented by letter V. ### **Acceleration.** Acceleration is the rate of change of Velocity in respect to time. It is also a vector velocity. Often represented by letter A. - NB: Kinematics is the study. ## **A car travels from rest to a speed of 60km/h in 10 seconds, what is its acceleration?** - U=0 - V = 60km/h = 60 x 1000/3600 - t = 10 sec - a = ? - a = (V - U)/t - a = (60 x 1000/3600 - 0)/10 - a = 1.67 m/s^2 ## **An object is dropped from a height of 100m, what is its velocity after 2 sec (g = 9.8m/s^2)?** - V^2 = u^2 + 2 as - V^2 = 0^2 + 2 x 9.8 x 100 - V^2 = 1960 - V = 14 m/s ## **A particle moves along a straight line with an initial velocity of 20m/s. If it accelerates at a rate of 5m/s^2 for 4 seconds what is its final velocity?** - U = 20m/s - a = 5m/s - t = 4 sec - V = U + at - V = 20 + (5 x 4) V = 40 m/s ## **An Object is thrown upward from the ground with an initial velocity of 25m/s. If it takes 5 sec to reach its maximum height, what is its maximum height?** - U = 25m/s - t = 5 sec - a = -9.8m/s^2 - S = ut + (1/2)at^2 - S = 25 x 5 + (1/2)(-9.8) x 5^2 - S = 125 - 245 x 1/2 - S = 125 -122.5 - S = 2.5m ## **A particle moves along a straight line with a constant acceleration of 2m/s^2. If it starts from rest and travels a distance of 100m, what is the final velocity?** - V = ? - S = 100 m - a = 2 m/s^2 - U = 0 - V^2 = u^2 + 2as - V^2 = 0^2 + 2 x 2 x 100 - V^2 = 400 - V = 20 m/s ## **Relative Motion** Relative motion refers to the motion of an object with respect to another object or a reference frame. In other words, it is the motion of an object as seen by an observer who is also moving. There are several types of relative motion which includes translational motion, rotational motion and circular motion. Newton's Law of motion can be applied to relative motion to describe the motion of objects with respect to each other. ### Relative Velocity. Relative velocity of two objects is the velocity of one object with respect to the other. ### Relative acceleration Relative acceleration involves the acceleration of one object with respect to the other. Relative motion can be used to solve collision problems where two objects collide, and they change momentum. ## **Example:** Two cars are moving in the same direction on a straight road. Car A is moving at a velocity of 60km/hr and Car B is moving at 80km/hr. What is the relative velocity of Car B with respect to Car A? - B to A - 80 - 60 = 20 km/h ## **A car is accelerating from rest to a velocity of 30m/s in 6 sec. What is the relative acceleration of the car with respect to the road?** - Speed = distance/time - a = (v-u)/t - a = 30/6 - a = 5m/s^2 ## **A car of mass 1500kg is moving at a velocity of 20m/s. It collides inelastically with a truck of mass 3000kg moving at a velocity of 10m/s. What is the relative velocity of the car and truck after the collision?** - m1 = 1500kg - v1 = 20 m/s - m2 = 3000kg - v2 = 10m/s - m1*v1 + m2*v2 = (m1+m2)U - 1500 x 20 + 3000 x 10 = (3000 + 1500)U - 30000+30000 = 4500U - 60000 = 4500U - U = 60000/4500 - U = 13.33 m/s ## **Two cars are moving in opposite directions on a straight road. Car A is moving at 60km/hr and Car B at 80km/hr. What is the relative velocity?** - 80 - (-60) = 140 km/hr ## **A boat is moving with a speed of 5m/s in still water. The river has a current of 3m/s. Find the resultant velocity of the boat if it trying to move perpendicular to the river current.** - R^2 = Vb^2 + Vr^2 - R^2 = 3^2 + 5^2 - R^2 = 9 + 25 - R^2 = 34 - R = 5.83 m/s ## **Conservation Principles.** Conservation has to do with preservation of a particular physical quantity during a process. This includes: - a. Conservation of energy. - b. Conservation of linear momentum - c. Conservation of charge - d. Conservation of angular momentum. ## **Conservative forces & Non-Conservative forces** Conservative forces are those in which the work done is independent of the path taken but only on the initial and final position. Eg gravitational force, electric forces, work done by elastic spring, etc. - **Work done by Conservative forces has 4 properties** - i. It is reversible - ii. It is independent of the path taken, but only on the starting and final point. - iii. When the starting and final points are the same, the total work done is 0 (zero) - iv. In the absence of non-conservative forces, the total mechanical energy is constant. - Et = Ept + Ek - Ep = Potential energy - Ek = Kinetic energy - Et = Total energy - **for Non-conservative forces, the work done is path dependent and is mostly irreversible** ## **Example:** Consider an object of mass 10kg at position A 100m above the ground. The object drops to a position B 30m above the ground. Calculate the final speed of position B, and show that the work-done is conservative g = 9.8 m/s^2 - V = 0 m/s - Vf = Vo^2 - 2g (xf-xo) - Vf = 0^2 - 2(9.8) (30-100) - Vf = 1372 - Vf = √1372 - Vf = 37.04 m/s^-1 **At position A** - Ep = mgh - Ep = 10 x 9.8 x 100 - Ep = 9800J - Ek = (1/2)mu^2 = 0 **At position B** - Ep = mgh - Ep = 10 x 9.8 x 30 - Ep = 2940J - Ek = (1/2)mu^2f - Ek = (1/2) x 1372 x 10 - Ek = 6860J **Total energy at position A** - ET = Ep + Ek - ET = 9800 + 0 - ET = 9800J **Total energy at position B** - ET = Ep + Ek - ET = 2940 + 6860 - ET = 9800J **Conclusion** => Work done is conservative - ΔME = O **For conservative forces there is no change in mechanical energy.** ## **Conservation Of Linear Momentum.** Momentum = mass of an object x velocity Momentum is the quantity of motion and is obtained by the product of the mass of an object and its velocity. It has units of Kgms^-1 or N.s. - **The law of conservation of linear momentum states that in a system of colliding objects, the total momentum remains constant provided that there is no net external force acting on the system.** ## **Types Of Collision** ### Elastic Collision - a. momentum is Conserved. - Σmju; = Σmf Uf - b. Ek is Conserved - Σ'1/2m; u;^2 = Σ'1/2m; u;^2 ### Inelastic Collision - a. momentum is Conserved - b. Ek is not conserved - Σ'1/2m; u^2; ≠ Σ'1/2m; u^2f - 1.e total kinetic energy before collision is not equal to kinetic energy after collision. **For completely inelastic collisions, the two bodies stick together after collision, and move as one body.** ## **Moments Of Inertia.** 7th January 2025 Consider a rigid body made up of large numbers of particles with masses m1, m2, m3, ... Mn at perpendicular distances R1, R2, R3, ..., Rn from the axis of rotation The moment of inertia I of the rigid body is defined as I = m1r1^2 + m2r2^2... mnrn^2= Σmiri^2 ## **Moment of Inertia Of Some Common Shapes or objects.** Consider a uniform bar or rod of length (l). **Case 1: Axis of rotation at one end.** - a = l - b = l - I = (1/3) m (a^3+b^3)/(a+b) - I = (1/3) m (l^3+l^3)/(l+l) - I = (1/3) m (l^3+l^3)/2l - I = (1/3) m (2l3)/2l - I = (1/3) m l^2 - I = (1/3) m(l^2/4) - I = (1/12)ml^2 **Case 2: Axis of rotation about the middle** - a = l/2 - b = l/2 - I = (1/3)m (a^3+b^3)/(a+b) - I = (1/3) m (l/2)^3+(l/2)^3)/(l/2+l/2) - I = (1/3)m (l^3/8+ l^3/8)/l - I = (1/3) m (l^3/4)/l - I = (1/3)ml^2/4 - I = (1/12)ml^2 **Case 3: Axis of rotation about one end** - a = 0 - b = l - I = (1/3)m (a^3+b^3)/(a+b) - I = (1/3)m (0^3+ l^3)/(0+l) - I = (1/3) m (l^3)/l - I = (1/3)ml^2 2. Circular ring with axis of rotation through its Center - I = mr^2 3. Circular disk about the Center - I = (1/2)mr^2 4. Hollow Cylinder - I = (1/2) m (R1^2 + R2^2) 5. Solid Cylinder - I = (1/2) mr^2 6. Thin walled hollow cylinder - I = mr^2 7. Sphere about a diameter - I = (2/5) mr^2 8. Thick walled hollowed sphere - I = (2/3)mr^2 9. Rectangular plate with axis of rotation through the Center - I = (1/12)m (a^2+b^2) 10. Axis of rotation along one edge - I = (1/3)ma^2 ## **Precession Of A Gyroscope.** 9th January 2025 A gyroscope is a spinning disk in which the axis of rotation is free to assume any rotation. The precession of a gyroscope can be illustrated by a biro or pen cover if it is placed on a flat surface near the surface of the earth at an angle to the vertical, and it is not spinning. It will fall over. But if the pen cover is spinning as its axis, rather than fall over due to its torque, it processes about the vertical. ## **Gravitation** A force exists between two objects whose magnitude is given by the product of the two masses divided by the square of the distance between them. Newton's Law of gravitation can be expressed as: F12 = - G(m1*m2)/r^2 * r^12 where G = Gravitational constant. i.e 6.67 x 10^-11 N.m^2/kg^2. - F12 = F12 is the force on the object 1 exerted by object 2 - r^12 = r^12 is the unit vector that points from object 1 towards object 2 ## **Gravitational Potential Energy** Consider a particle of mass 'm' in displace between two points P and Q above the earth's surface. The change in the gravitational potential energy as regards the displacement is defined as the negative of the work done by the gravitational force during that displacement. - ΔU = Uf - ui = -∫P^Q F(r)dr The gravitational force can now be expressed as: - F(r) = -G(m*me)/r^2 * r^2 - Substinate r^2 in exo where F(r) = -G(m*me)/r^2 * r^2 - Ui - Ui = -∫ri^rf F(r)dr - = -∫ri^rf -G(m*me)/r^2 * r^2 dr - = -∫ri^rf -G(m*me)/r^2 dr - = -∫ri^rf -G(m*me)/r^2 * (1/r^2) dr - = -G(m*me) *∫ri^rf 1/r^2 dr - = -G(m*me)* (-1/r)|ri^rf - = -G(m*me) * (1/rf - 1/ri) Taking the post ri = ∞ be equals infinity ie as the reference point, we obtained the gravitational potential energy as - G(m*me)/r This expression is valid provide that r ≥ Re. Also, the gravitational potential energy of a pair of bodies of masses m1 and m2 separated by distance. - U = -G(m1*m2)/r This expression shows that the gravitational potential energy for a pair of particles varies as 1/r, whereas the force varies as 1/r^2. - U α 1/r - F α 1/r^2 ## **Satellite Motion and Orbit** Consider a satellite of mass 'm' moving with speed 'v' in an orbit around the earth of mass 'me'. The Centripetal force required to keep the satellite in orbit is supplied by the gravitational attraction between the satellite and the earth. - F = (m*u^2)/r = G(m*me)/r^2 Hence, V = √Gme/r Where r= Re + h - Re = radius of the earth - h= altitude above the earth surface ## **Velocity Of Escape** 14TH JANUARY 2025 If an object of mass 'm' is projected from the earth's surface at point B with an initial speed 'v', to enable it escape from the gravitational influence of the earth, we have the kinetic energy of the object as: - K = (1/2)*mu^2 - So, v= √2Gme/R Hence, gravitational field strength 'g'. - g = G*me/R^2 ## **Kepler's Law.** Keplar discovered 3 empirical laws that accurately describe motion of the planet. Kepler's law states that: 1. Each planet moves in an elliptical orbit with the sun located at a focus of the Ellipse. - The line joining the sun and other planets sweeps equal areas in equal times. - The squares of the period of the revolution of the planets are proportional to the cubes of their mean distances to the sun. ## **Example 1** A piece of iron is placed 23 cm away from a piece of Nickel that has a mass of 46 kg. Given that the force of gravity between them is 2.9 x 10^-8 N. determine the mass of the piece of iron. Take universal gravitational Constant G = 6.67 x 10^-11 Nm^2/kg^2 - r = 23cm = 0.23m = 23 x 10^-2 - m1 = 46kg - F = 2.9 x 10^-8 - F = G(m1*m2)/r^2 - m2 = Fr^2/G*m1 - m2 = 2.9 x 10^-8* (23 x 10^-2)^2 / (6.67 x 10^-11) x 46 - m2 = 0.025 kg - m2 = 25g ## **Example 2: What is mass of the planet given that the acceleration due to gravity at its surface is 6.003m/s^2, its radius is 2.4km and g = 6.67 x 10^-11 N/m^2/kg^2?** - g = G*M/r^2 - M = gr^2/G - r = 2.4km = 2.4 x 10^3 m - M = 6.003 * (2.4 x 10^3)^2 / 6.67 x 10^-11 - M = 5.184 x 10^24 kg ## **Co-efficient Of Restitution** 16th January 2025 e = (V2 - V1)/(U1 - U2) - e=1 Perfectly elastic collision - e>1 Energy is lost - e=0 Perfectly inelastic collision. ## **Example:** Two bodies of mass 2kg each collides. The velocities before collision are UA = 15i + 30j, UB = -10i + 5j and VA = -5i + 20j after collision. Calculate the final velocity UB: - Σm1u1 = Σm1v1 - m1v1a + m1v1b = m1v2a + m2 v2b - 2(15i+30j) + 2 (- 10i + 5j) = 2(-5i + 20j) + 2 UB - 30i + 60j - 20i + 10j = -10i + 40j + 2 UB - 20i + 30j = 2 UB - UB = (20i + 30j)/2 UB = (10i + 15j) m/s^2 ## **A 1.0 kg ball moving at 12m/s collides with a 2.0 kg ball moving in the opposite direction at 24 m/s. Determine the velocities of the ball after impact if:** - a. e= 1/2 - b. the balls stick together after collision - c. the collision is perfectly elastic. **Solution.** - Using: m1u1a + m1u1b = m1v2a + m2 v2b where m1 = 1 kg m2 = 2 kg u1 = 12 m/s u2 = -24m/s - 1 * 12 + 2 * (-24) = 1 * (v1) + 2 * (v2) - 12 - 48 = v1 + 2 v2 - 36 = - v1 + 2 v2 - V1 + 2V2 = -36..... equation * **a. e = 1/2 V1 + 2V2 = -36.... equation *** - e = (V2 - V1)/(U1 - U2) - 1/2 = (V2 -V1)/(12 - (-24) - 2 * (V2 - V1) = 3 * (V1 - V2) - 2 * V2 - 48 = 3V1 - 3V2 - 5 * V2 - 3 * V1 = 48 - V2 = (3 * V1 + 48)/5 **b. the balls stick together after collision** - V2 = V1 **Sub. V2=V1 into equation *** - V1 + 2 V1 = -36 - 3 * V1 = -36 - V1 = -12 m/s - V2 = -12 m/s **c. the collision is perfectly elastic** - e = 1 **Sub. e = 1 in equation *** - V2 - V1 = U1 - U2 - V2 - V1 = 12 + 24 - V2 - V1 = 36 - V2 = V1 + 36 **Sub. V2 = V1 + 36 into equation** - V1 + 2(V1 + 36) = -36 - V1 + 2V1 + 72 = -36 - 3V1 = -36 -72 - 3V1 = - 108 - V1 = -36m/s - V2 = -36 + 36 - V2 = 0 m/s ## **Moment of Vector, Torque.** The moments of a Vector A about a point O is another vector with a magnitude equal to the product of the magnitude of A and the perpendicular distance of O from the direction of A. Moment of C - IAI x distance from O If the vector is a force, its moment is called *torque*. TORQUE is the rotational effect of a force on a body - T = F x l Where l = distance - T = Irl |F|sin Ө Where T is the position vector of the point from the origin through which the force is acting. Ө is the angle between the line r, and the force vector. - T = rFsinӨ ## **Mechanical Equilibrium** A body is said to be in equilibrium if all forces acting on it is balanced. Types of Mechanical Equilibrium: **A. Translational equilibrium** ΣF = 0 This means that: 1. the body remains at rest or in motion with a constant velocity. **B. Rotational Equilibrium** ΣT = 0 This means that the sum of the clockwise moments about a point on the body is equal to the sum of anticlockwise moment about the same point. This is known as the principle of moment. This means that the body is rotationally at rest or rotating with constant angular velocity. ## **Example:** What are the magnitude and direction of the torque about the origin on a particle located at coordinates (0, -4, 3m) due to: - a) force F1, with F1x = 2N, and F1y = F1z= 0 - b) force F2, with F2x = 0, and F2y = 2N and F2z = 4N ### Solutions #### a) - T = r x F - T = (-4J + 3K) x (2I) - T = (0 - 0) - J(0-6) + K(0+8) - T = 8K - |T| = √(6^2 + 8^2) - ITI = 10 Nm - Direction y-z plane #### b) - T = r x F - T = (-4J + 3K) x (2J +4K) - T = (1(16-6) - J(0-0) + K( 0 -0)) - T = -22I - |T| = √(-22)^2 = 22 Nm - Direction: negative x-axis. ## **Example 2: Find the magnitude and direction of the torque about a point with coordinates (2m, 0, -3m) due to Fi (Force) ** - T = r x F - Frelative= (0, 5, 4m) - (2m, 0, -3) - Frelative = (-2, -4, 6) - T = r x F - T = (-2, -4, 6) x (3, 2, -1) - T = (1(0+0) - J( 0 - (12) + K(0 - 8)) - T = 12j + 8k - |T| = √(12^2 + 8^2) - |T| = √208 Nm - Direction y-z plane ## **Angular Momentum for a particle** The angular momentum of a particle of mass 'm' with linear momentum P = mv as it passes though a point A, a position vector r from the origin, is given by - L = r x P - L = r x mv - |L| = rmv sin Ө Where Ө is the angle between the position vector r and the direction of the velocity vector v From the 2nd law - F = d(P)/dt - F = d(mv)/dt = net force - Fnet = d(P)/dt - Fnet = d(rxmv)/dt - d(L)/dt = d(r x mv)/dt - d(L)/dt = mv(dr/dt + r x d(v)/dt) - d(L)/dt = mv x v + r x m * d(v)/dt - d(L)/dt = r x Fnet - d(L)/dt = Fnet This implies that the rate of change of angular momentum of a particle with time is equal to the vector sum of all the torque of the particle. ## **For System of Particles.** - L = ΣLi = L1 + L2 + ... + LN - d(L)/dt = Σi Tneti = Tnet This implies that the rate of change of angular momentum of a system of particles is equal to the vector sum of all the torques on its individual particles. ## **For a rigid body** - The average distance between the particles are constant throughout the motion. - Li = angular momentum of the ith particle - Li = rmi V i sin Ө - Li = rmi (ri*w) * Sin Ө - Li = mri^2*w*Sin 90° - Li = I iw. where I= moment of inertia - for all the particles: L = I*w . - Recall: (dL)/dt = Tnet - Tnet = dL/dt = d(Iw)/dt - Tnet = Idw/dt Tnet = I * α where α = Angular acceleration ## **Conservation Of Angular Momentum** The principle of Conservation of α states that provided that the net external torque acting on a system is zero(0), the total angular momentum of the system is constant, i.e conserved. - d(L)/dt = Tnet - if Tnet=0 - then dL/dt=0 - dL/dt = 0 - ∫dL = ∫0 - L1 = L2 - L= Constant - L1 = Lf - I1w1 = I2wf ## Example 1. **A turbine fan in a jet engine has a moment of inertia of 2.5kgm^2 about its axis of rotation. As the turbine is starting up, its angular velocity is a function of time is: - w = (40 rad/s^2)t^2. **Find:** - a) the fans angular momentum as a function of time and its value at t= 3 secs. - b) the net torque (Tnet) acting on the fan as a function of time. 2. **A solid cylinder of mass 10kg and radius 0.5m is rotating about its central axis with an angular velocity of 3rad/sec. Find its angular momentum.** 3. ** A 2kg particle has a position vector r = 4.0i - 2.0j relative to the origin. Its velocity is given by v = - 6.0t^2 j. Find: ** - a) the particle's angular momentum. - b) the torque acting on the particle. - c) the angular momentum of the particle about a point with coordinates (-2,-3,0) 4. ** A figure skater is spinning with an angular velocity of 2 rad/sec. If she brings her arm closer to her body, reducing her moment of inertia by a factor of 2. What is her new angular velocity?** ## Solutions ### 1 a) - I = 2.5kgm^2 - L = Iw = 2.5 x (40 rad/s^2)t^2 - L= (100t^2) rad/s^3 - L (t=3) - L = 100 x 3^2 - L = 9