Physics

LessonAbout 1.5 h2 of 2 in Physics

Physics

Physics is the fundamental branch of physical science that studies matter, energy, motion, force, and their mutual interactions in the physical universe.

1. Physical Quantities, Vectors, and Units

1.1 Units and Physical Constants

Standard physical constants establish the relationship between mass, force, and gravitational acceleration across measurement systems.

Weight Formula

\[ W = mg \]

ME Board October 1994

Example. The weight of a mass of \(10\ \text{kg}\) at a location where the acceleration of gravity is \(9.77\ \text{m/s}^2\) is

Solution. Calculate the gravitational force acting on the mass.

\[ W = mg \]
\[ W = (10)(9.77) = 97.7\ \text{N} \]

Answer: 97.7 N

1.2 Vector and Scalar Quantities

Physical quantities are classified based on directional characteristics:

  • Scalars have magnitude only without directional orientation.
  • Vectors have both magnitude and spatial direction.
  • Vector addition requires resolving orthogonal components.

Resultant Vector Magnitude

\[ R = \sqrt{R_x^2 + R_y^2 + R_z^2} \]

ME Board March 1998

Example. Assume the three force vectors intersect at a single point: \(\mathbf{F}_1 = 4\mathbf{i} + 2\mathbf{j} + 5\mathbf{k}\), \(\mathbf{F}_2 = -2\mathbf{i} + 7\mathbf{j} - 3\mathbf{k}\), \(\mathbf{F}_3 = 2\mathbf{i} - \mathbf{j} + 6\mathbf{k}\). What is the magnitude of the resultant force vector, R?

Solution. Sum the respective directional components and find the magnitude.

\[ \mathbf{R} = (4 - 2 + 2)\mathbf{i} + (2 + 7 - 1)\mathbf{j} + (5 - 3 + 6)\mathbf{k} = 4\mathbf{i} + 8\mathbf{j} + 8\mathbf{k} \]
\[ |\mathbf{R}| = \sqrt{(4)^2 + (8)^2 + (8)^2} = 12 \]

Answer: 12 units

2. Gravitation

2.1 Newton’s Laws of Motion

Classical dynamics are governed by Newton’s fundamental laws:

  • First Law: Inertia maintains constant velocity without net force.
  • Second Law: Unbalanced force causes proportional mass acceleration.
  • Third Law: Interacting bodies exert equal and opposite forces.

Newton’s Second Law of Motion

\[ F = ma \]

ME Board October 1997

Example. Compute the acceleration in \(\text{ft/s}^2\) of a \(10\ \text{lbm}\) object acted upon by an unbalanced force of \(4.4\ \text{lbf}\).

Solution. Apply Newton’s second law relating force, mass, and gravitational acceleration.

\[ F = \dfrac{W}{g}a \]
\[ 4.4 = \dfrac{10}{32.2}a \]
\[ a = 14.168\ \text{ft/s}^2 \]

Answer: 14.2 ft/s\(^2\)

2.2 Friction and Frictional Forces

Friction is the resistive force opposing relative tangential sliding between contacting surfaces in contact.

Static Friction Force

\[ F_f = \mu_s N \]

Force on Inclined Plane

\[ P = W \sin\theta + \mu W \cos\theta \]

EE Board October 1997

Example. A \(250\ \text{lb}\) block is initially at rest on a flat surface that is inclined at \(30^\circ\). If the coefficient of kinetic friction is \(0.30\) and the coefficient of static friction is \(0.40\), find the force required to start the block moving up the plane.

Solution. Determine normal force perpendicular to the plane and sum forces along the incline using static friction.

\[ N = W \cos\theta = 250 \cos 30^\circ = 216.506\ \text{lb} \]
\[ P = W \sin\theta + \mu_s N \]
\[ P = 250 \sin 30^\circ + 0.40(216.506) = 212\ \text{lb} \]

Answer: 212 lb

2.3 Universal Gravitation

The Law of Universal Gravitation states that mutual gravitational attraction force is proportional to the mass product and inversely proportional to squared separation distance.

Law of Universal Gravitation

\[ F = \dfrac{G m_1 m_2}{s^2} \]

Example. Two steel balls of masses \(500\ \text{kg}\) and \(50\ \text{kg}\), respectively are placed with their centers \(0.5\ \text{m}\) apart. The two balls attract with a force of

Solution. Apply Newton’s law of universal gravitation using constant \(G = 6.67 \times 10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2\).

\[ F = \dfrac{G m_1 m_2}{s^2} \]
\[ F = \dfrac{(6.67 \times 10^{-11})(500)(50)}{(0.5)^2} = 6.67 \times 10^{-6}\ \text{N} \]

Answer: 6.67 \(\times 10^{-6}\) N

3. Work, Energy, and Power

3.1 Work and Energy Principle

Work is performed when force produces displacement along its line of action. The work-energy principle equates net performed work to kinetic energy change.

Work Formula

\[ W = F s \cos\theta \]

Work-Energy Principle

\[ F s = \dfrac{1}{2} m v^2 \]

Example. What average force is necessary to stop a .45 caliber bullet of mass of \(15\ \text{grams}\) and speed of \(300\ \text{m/s}\) as it penetrates a block to a distance of \(5\ \text{cm}\)?

Solution. Equate the work done by the retarding force to the initial kinetic energy of the bullet.

\[ F s = \dfrac{1}{2} m v^2 \]
\[ F(0.05) = \dfrac{1}{2}(0.015)(300)^2 \]
\[ F = 13,500\ \text{N} = 13.5\ \text{kN} \]

Answer: 13.5 kN

3.2 Conservation of Energy

Gravitational potential energy expresses energy stored due to vertical elevation position above a reference datum.

Gravitational Potential Energy

\[ PE = m g h \]

ME Board April 1995

Example. A \(10\ \text{kg}\) block is lifted vertically through a height of \(3\ \text{meters}\). Calculate the change in potential energy.

Solution. Compute potential energy from mass, gravitational acceleration, and vertical lift distance.

\[ \Delta PE = m g h \]
\[ \Delta PE = (10)(9.81)(3) = 294.3\ \text{J} \approx 294\ \text{J} \]

Answer: 294 J

3.3 Power

Power represents the time rate at which work is accomplished or energy is transferred.

Mechanical Power

\[ P = \dfrac{W}{t} \]
\[ P = \dfrac{\Delta KE}{\Delta t} \]

ME Board October 1997

Example. A rocket is moving through a vacuum. It changes its velocity from \(9020\ \text{ft/sec}\) to \(5100\ \text{ft/sec}\) in \(48\ \text{seconds}\). How much power is required to accomplished this if the rocket’s mass is \(13,000\ \text{slugs}\)?

Solution. Compute change in kinetic energy over time and convert to horsepower (\(1\ \text{hp} = 550\ \text{ft}\cdot\text{lb/s}\)).

\[ \Delta KE = \dfrac{1}{2} m (v_1^2 - v_2^2) = \dfrac{1}{2}(13,000)[(9020)^2 - (5100)^2] = 3.597 \times 10^{11}\ \text{ft}\cdot\text{lb} \]
\[ P = \dfrac{3.597 \times 10^{11}}{48} = 7.49 \times 10^9\ \text{ft}\cdot\text{lb/s} \]
\[ P = \dfrac{7.49 \times 10^9}{550} = 1.36 \times 10^7\ \text{hp} \]

Answer: 1.36 \(\times 10^7\) hp

4. Fluid Mechanics

4.1 Density, Specific Weight, and Pressure

Fluid properties define mass and force concentration characteristics per volume unit:

  • Water mass density is 1000 kg/m\(^3\) in SI metric units.
  • Water weight density is 62.4 lb/ft\(^3\) in customary units.
  • Water specific weight is 9.81 kN/m\(^3\) in force units.

Mass Density

\[ \rho = \dfrac{m}{V} \]

Specific Weight

\[ D = \rho g \]

ME Board October 1994

Example. The mass of air in a room \(3\ \text{m} \times 5\ \text{m} \times 20\ \text{m}\) is known to be \(350\ \text{kg}\). Find its density.

Solution. Compute total volume of the room and divide mass by volume.

\[ V = (3)(5)(20) = 300\ \text{m}^3 \]
\[ \rho = \dfrac{m}{V} = \dfrac{350}{300} = 1.167\ \text{kg/m}^3 \]

Answer: 1.167 kg/m\(^3\)

4.2 Hydrostatics and Archimedes Principle

Archimedes’ Principle establishes that an immersed body experiences an upward buoyant force equal to the weight of fluid displaced.

Archimedes Buoyant Force

\[ BF = \rho g V_s \]

EE Board April 1996

Example. Determine the submerged depth of a cube of steel \(0.3\ \text{m}\) on each side floating in mercury. The specific gravities of steel and mercury are \(7.8\) and \(13.6\) respectively.

Solution. Equate the weight of the steel cube to the buoyant force of displaced mercury.

\[ W = BF \]
\[ (7.81)(9.81)(0.3)^3 = (13.6)(9.81)(0.3)^2 d \]
\[ d = 0.172\ \text{m} \]

Answer: 0.172 m

5. Waves, Sound, and Oscillations

5.1 Periodic Motion and Harmonic Oscillations

Hooke’s Law dictates linear elastic restoring force proportional to displacement in simple harmonic motion systems.

Hooke’s Law

\[ F = k s \]

EE Board October 1995

Example. A \(50\ \text{g}\) mass hangs at the end of the spring. When \(20\ \text{grams}\) more are added to the end of the spring, it stretches \(7\ \text{cm}\) more. Find the spring constant.

Solution. Apply Hooke’s law relating additional force to spring elongation.

\[ F = k s \]
\[ 20 = k (7) \]
\[ k = 2.857 \approx 2.8 \]

Answer: 2.8

5.2 Wave Motion, Sound, and Doppler Effect

Transverse wave velocity along flexible strings depends directly on internal tension and mass distribution per unit length.

Transverse Wave Velocity

\[ v = \sqrt{\dfrac{T}{\mu}} \]

REE September 2001

Example. A steel wire \(2\ \text{meters}\) long has a mass of \(20\ \text{grams}\) and is stretched with a tension of \(1000\ \text{newtons}\). What is the velocity of propagation of a transverse wave in the wire?

Solution. Compute mass per unit length \(\mu\) and solve wave propagation velocity.

\[ \mu = \dfrac{20\ \text{g}}{2\ \text{m}} = 10 \times 10^{-3}\ \text{kg/m} \]
\[ v = \sqrt{\dfrac{T}{\mu}} = \sqrt{\dfrac{1000}{10 \times 10^{-3}}} = 316.23\ \text{m/s} \approx 316\ \text{m/s} \]

Answer: 316 m/s

6. Light and Optics

6.1 Reflection, Refraction, and Snell’s Law

Light behavior across media boundaries exhibits distinct optical phenomena:

  • Refraction angle changes when light enters a new medium.
  • Frequency remains constant during medium light transition.
  • Critical angle occurs when refraction angle equals \(90^\circ\).

Snell’s Law of Refraction

\[ n_1 \sin\theta_1 = n_2 \sin\theta_2 \]

Example. A beam of light enters a lake at an angle of incidence of \(40^\circ\). Find the angle of refraction if the index of refraction of water is \(1.33\).

Solution. Apply Snell’s law of refraction using air refractive index \(n_1 = 1.00\).

\[ n_1 \sin\theta_1 = n_2 \sin\theta_2 \]
\[ (1.00) \sin 40^\circ = 1.33 \sin\theta_2 \]
\[ \sin\theta_2 = 0.4833 \implies \theta_2 = 28.9^\circ \approx 29^\circ \]

Answer: 29 deg

6.2 Lenses and Optical Phenomena

Spherical thin lenses form real or virtual images by refracting incident light rays through defined focal points.

Thin Lens Equation

\[ \dfrac{1}{f} = \dfrac{1}{d_o} + \dfrac{1}{d_i} \]

Example. A jeweler examines a diamond with a loupe with a focal length of \(8.0\ \text{cm}\). If the gem is positioned so that its image is at the normal near point, \(25\ \text{cm}\), how far from the lens must the diamond be held?

Solution. Apply the thin lens equation with virtual image distance \(d_i = -25\ \text{cm}\).

\[ \dfrac{1}{f} = \dfrac{1}{d_o} + \dfrac{1}{d_i} \]
\[ \dfrac{1}{8} = \dfrac{1}{d_o} - \dfrac{1}{25} \]
\[ \dfrac{1}{d_o} = \dfrac{33}{200} \implies d_o = 6.06\ \text{cm} \approx 6.1\ \text{cm} \]

Answer: 6.1 cm

7. Modern and Relativistic Physics

7.1 Lorentz Factor

The Lorentz factor is the fundamental mathematical multiplier calculating space, time, and mass transformations at relativistic speeds.

Lorentz Factor

\[ \gamma = \dfrac{1}{\sqrt{1 - \dfrac{v^2}{c^2}}} \]

Where: \(\gamma\) = Lorentz factor

Example. A particle accelerator propels an electron to a velocity of \(0.80c\) relative to the laboratory. Calculate the exact Lorentz factor (\(\gamma\)) for this electron.

Solution. Substitute velocity into the Lorentz factor equation.

\[ \gamma = \dfrac{1}{\sqrt{1 - \dfrac{v^2}{c^2}}} \]
\[ \gamma = \dfrac{1}{\sqrt{1 - (0.80)^2}} = \dfrac{1}{\sqrt{1 - 0.64}} = \dfrac{1}{\sqrt{0.36}} \]
\[ \gamma = \dfrac{1}{0.60} = \dfrac{5}{3} \]

Answer: \(5/3\)

7.2 Time Dilation

Time dilation causes moving clocks to tick measurably slower relative to a stationary reference frame.

Time Dilation

\[ \Delta t' = \gamma \Delta t \]

Where: \(\Delta t'\) = dilated time, \(\Delta t\) = proper time

Example. A subatomic particle moves past a sensor in a laboratory at a speed of \(0.90c\). If the particle has a measured lifetime of \(2.00 \times 10^{-6}\ \text{s}\) in its own rest frame (proper time), what is the lifetime of the particle as measured by the stationary laboratory sensor?

Solution. Compute the Lorentz factor and multiply by proper rest time.

\[ \gamma = \dfrac{1}{\sqrt{1 - (0.90)^2}} = \dfrac{1}{\sqrt{1 - 0.81}} = \dfrac{1}{\sqrt{0.19}} = 2.294 \]
\[ \Delta t' = \gamma \Delta t = (2.294)(2.00 \times 10^{-6}\ \text{s}) = 4.59 \times 10^{-6}\ \text{s} \]

Answer: \(4.59 \times 10^{-6}\ \text{s}\)

7.3 Length Contraction

Length contraction describes the physical foreshortening of moving objects along their axis of motion.

Length Contraction

\[ L = \dfrac{L_0}{\gamma} \]

Where: \(L_0\) = proper length

Example. An aerospace engineer is tracking a spacecraft that has a proper length of \(100.0\ \text{m}\) (measured when completely at rest). If the spacecraft passes an Earth observation station at a constant velocity of \(0.60c\), what is the contracted length of the spacecraft as measured by the station’s radar?

Solution. Multiply proper length by inverse Lorentz factor.

\[ L = L_0 \sqrt{1 - \dfrac{v^2}{c^2}} \]
\[ L = (100.0\ \text{m})\sqrt{1 - (0.60)^2} = (100.0\ \text{m})\sqrt{0.64} = (100.0\ \text{m})(0.80) = 80.0\ \text{m} \]

Answer: \(80.0\ \text{m}\)

7.4 Relativistic Velocity Addition

Relativistic velocity addition combines frame velocities ensuring total speed never exceeds the speed of light.

Relativistic Velocity Addition

\[ u = \dfrac{v + u'}{1 + \dfrac{vu'}{c^2}} \]

Where: \(u\) = total combined velocity, \(u'\) = velocity relative to moving frame

Example. A deep-space cruiser is traveling at a speed of \(0.70c\) relative to Earth. The cruiser fires a drone forward (in its direction of travel) at a speed of \(0.50c\) relative to the cruiser itself. Calculate the velocity of the drone as measured by an observer on Earth.

Solution. Substitute forward velocities into the relativistic velocity addition formula.

\[ u = \dfrac{v + u'}{1 + \dfrac{vu'}{c^2}} \]
\[ u = \dfrac{0.70c + 0.50c}{1 + (0.70)(0.50)} = \dfrac{1.20c}{1 + 0.35} = \dfrac{1.20c}{1.35} = 0.889c \]

Answer: \(0.889c\)

Example. A deep-space exploration vessel is traveling away from Earth at a velocity of \(0.80c\). The vessel launches a probe directly backward (toward Earth) at a speed of \(0.60c\) relative to the vessel itself. Calculate the velocity and direction of the probe as measured by an observer on Earth.

Solution. Assign positive sign away from Earth and negative sign for backward launch direction.

\[ u = \dfrac{v + u'}{1 + \dfrac{vu'}{c^2}} \]
\[ u = \dfrac{0.80c + (-0.60c)}{1 + (0.80)(-0.60)} = \dfrac{0.20c}{1 - 0.48} = \dfrac{0.20c}{0.52} = 0.385c \]

Answer: \(0.385c\)

7.5 Escape Velocity

Escape velocity is the minimum ballistic velocity required to permanently break free from gravitational attraction.

Escape Velocity

\[ v_e = \sqrt{\dfrac{2GM}{R}} \]

Where: \(v_e\) = minimum escape velocity

Example. Calculate the minimum escape velocity required for a module to launch from the surface of the Moon and never fall back. The mass of the Moon is \(7.35 \times 10^{22}\ \text{kg}\), its radius is \(1.74 \times 10^6\ \text{m}\), and \(G = 6.674 \times 10^{-11}\ \text{m}^3\text{kg}^{-1}\text{s}^{-2}\).

Solution. Substitute lunar mass, lunar radius, and universal gravitational constant into the formula.

\[ v_e = \sqrt{\dfrac{2GM}{R}} \]
\[ v_e = \sqrt{\dfrac{2(6.674 \times 10^{-11})(7.35 \times 10^{22})}{1.74 \times 10^6}} \]
\[ v_e = \sqrt{\dfrac{9.811 \times 10^{12}}{1.74 \times 10^6}} = \sqrt{5.638 \times 10^6} = 2.37 \times 10^3\ \text{m/s} \]

Answer: \(2.37 \times 10^3\ \text{m/s}\)

7.6 Schwarzschild Radius

The Schwarzschild radius represents the gravitational event horizon radius for a compressed mass.

Schwarzschild Radius

\[ r_s = \dfrac{2GM}{c^2} \]

Where: \(r_s\) = Schwarzschild radius

Example. Theoretical astrophysicists observe a collapsed star with a mass exactly three times that of our Sun. Calculate the Schwarzschild radius of this star. Use \(M_{\text{sun}} = 1.99 \times 10^{30}\ \text{kg}\), \(G = 6.674 \times 10^{-11}\ \text{m}^3\text{kg}^{-1}\text{s}^{-2}\), and \(c = 3.00 \times 10^8\ \text{m/s}\).

Solution. Determine total stellar mass and calculate the Schwarzschild event horizon radius.

\[ M = 3(1.99 \times 10^{30}\ \text{kg}) = 5.97 \times 10^{30}\ \text{kg} \]
\[ r_s = \dfrac{2GM}{c^2} \]
\[ r_s = \dfrac{2(6.674 \times 10^{-11})(5.97 \times 10^{30})}{(3.00 \times 10^8)^2} = \dfrac{7.969 \times 10^{20}}{9.00 \times 10^{16}} = 8.85 \times 10^3\ \text{m} \]

Answer: \(8.85 \times 10^3\ \text{m}\)

Formula Summary

Formula Name Formula
Weight \(W = mg\)
Resultant Vector Magnitude \(R = \sqrt{R_x^2 + R_y^2 + R_z^2}\)
Newton’s Second Law \(F = ma\)
Static Friction Force \(F_f = \mu N\)
Force on Inclined Plane \(P = W \sin\theta + \mu W \cos\theta\)
Universal Gravitation \(F = \dfrac{G m_1 m_2}{s^2}\)
Work Done \(W = F s \cos\theta\)
Work-Energy Theorem \(F s = \dfrac{1}{2} m v^2\)
Potential Energy \(PE = mgh\)
Power \(P = \dfrac{W}{t}\)
Mass Density \(\rho = \dfrac{m}{V}\)
Specific Weight \(D = \rho g\)
Fluid Hydrostatic Pressure \(P = \gamma h\)
Archimedes Buoyant Force \(BF = \rho g V_s\)
Floating Body Principle \(\text{sp. gr.}_{\text{body}} \cdot V_{\text{total}} = \text{sp. gr.}_{\text{fluid}} \cdot V_{\text{submerged}}\)
Hooke’s Law \(F = ks\)
Elastic Potential Energy \(E = \dfrac{1}{2} k x^2\)
Wave Speed Equation \(v = f \lambda\)
Snell’s Law of Refraction \(n_1 \sin\theta_1 = n_2 \sin\theta_2\)
Critical Angle \(\sin\theta_c = \dfrac{n_2}{n_1}\)
Thin Lens Equation \(\dfrac{1}{f} = \dfrac{1}{d_o} + \dfrac{1}{d_i}\)
Lorentz Factor \(\gamma = \dfrac{1}{\sqrt{1 - \dfrac{v^2}{c^2}}}\)
Time Dilation \(\Delta t' = \gamma \Delta t\)
Length Contraction \(L = \dfrac{L_0}{\gamma}\)
Relativistic Velocity Addition \(u = \dfrac{v + u'}{1 + \dfrac{vu'}{c^2}}\)
Escape Velocity \(v_e = \sqrt{\dfrac{2GM}{R}}\)
Schwarzschild Radius \(r_s = \dfrac{2GM}{c^2}\)