Comprehensive Physics Study Guide: Forces, Energy Conservation, and Momentum

Fundamentals of Dynamics: Newton's Laws and Types of Forces

• Definition and Properties of Force:

  • A force is defined as a push or a pull acting on an object.

  • Every force is exerted by an identifiable agent and acts directly on a recipient object.

  • Force is a vector quantity, possessing both magnitude and direction.

  • Forces are categorized by interaction type into contact forces (requiring direct physical contact between surfaces) and long-range forces (acting across space without physical contact, such as gravity).

  • The standard SI unit of force is the Newton ($\text{N}$), defined as the compound unit:     $1\,\text{N} = 1\,\text{kg}\cdot\text{m/s}^2$

  • English equivalent units include the pound ($\text{lb}$) and slug for mass, where on Earth's surface a mass of $1\,\text{kg}$ corresponds to approximately $2.2\,\text{lbs}$.

• Fundamental Forces of Nature:

  • All known forces in the universe reduce to four fundamental interactions:

  • Gravitational Force: An attractive force between masses, responsible for weight.

  • Electromagnetic Force: Responsible for all macroscopic contact forces, including normal force, tension, friction, and spring forces, as well as chemical bonding.

  • Weak Nuclear Force: Governs subatomic interactions and radioactive decay processes.

  • Strong Nuclear Force: Binds quarks together inside protons and neutrons, holding atomic nuclei together against electromagnetic repulsion.

• Newton's Laws of Motion:

  • Newton's First Law (Law of Inertia): An object in motion remains in motion at a constant velocity, and an object at rest remains at rest, unless acted upon by a net external force. When the net force is zero ($\Sigma \mathbf{F} = 0$), the acceleration is zero ($\mathbf{a} = 0$), maintaining constant velocity.

  • Newton's Second Law: The net force acting on an object of mass $m$ is directly proportional to its acceleration:     $\Sigma \mathbf{F} = m\mathbf{a}$

  • Mass ($m$) is an inherent scalar property measuring an object's inertia (resistance to acceleration), with standard SI units of kilograms ($\text{kg}$, where $1000\,\text{g} = 1\,\text{kg}$).

  • Broken down into orthogonal Cartesian components:       $\Sigma F_x = m a_x$       $\Sigma F_y = m a_y$

  • Newton's Third Law (Action-Reaction): For every force exerted on an object by an agent, there is an equal in magnitude and opposite in direction force exerted by the object back on the agent:     $\mathbf{F}_{1 \text{ on } 2} = -\mathbf{F}_{2 \text{ on } 1}$

  • Pushing against an object with $100\,\text{N}$ causes the object to exert $100\,\text{N}$ back on the pusher.

  • When the Earth exerts a downward gravitational force of $10\,\text{N}$ on an object, the object exerts an equal upward gravitational force of $10\,\text{N}$ on the Earth.

• Standard Force Classifications in Mechanics:

  • Weight ($F_g$ or $mg$): The downward gravitational force exerted on mass $m$ by the Earth:     $F_g = mg$

  • On Earth's surface, the standard acceleration due to gravity is $g = 9.81\,\text{m/s}^2$.

  • Normal Force ($N$): The supporting contact force exerted perpendicular to the interface between two contacting surfaces. The normal force must always be solved for using equilibrium equations; it is never automatically assumed equal to $mg$.

  • Tension ($T$): The pulling force transmitted through an ideal string, rope, or cable. Tension always pulls away from the object along the line of the rope. An ideal light string running over an ideal massless, frictionless pulley maintains the same tension throughout its length.

  • Spring Force ($F_s$): Governed by Hooke's Law, representing the restoring force exerted by a deformed spring:     $F_s = kx$

  • $k$ is the spring constant (stiffness) measured in $\text{N/m}$, and $x$ is the displacement from equilibrium.

  • Drag Force ($D$): Fluid resistance (air or water) opposing velocity:     $D = b v^2$

  • $b$ represents an aerodynamic drag factor.

  • Thrust: Driving force generated internally (e.g., rocket or jet exhaust).

Force Problem-Solving Methodology and Incline Plane Analysis

• Standardized Systematic Procedure:

  • Step 1: Read the problem carefully and identify the physical system.

  • Step 2: Draw a physical diagram showing all interactions.

  • Step 3: Draw an isolated Free Body Diagram (FBD) for each object, replacing the object with a dot and drawing all external forces as vectors pointing outward from the dot.

  • Step 4: Establish a Cartesian coordinate system. Identify the direction of acceleration and align axes to maximize the number of forces lying on the axes.

  • Step 5: Resolve non-aligned force vectors into components:     $F_x = F \cos(\theta)$     $F_y = F \sin(\theta)$

  • In general, if angle $\phi$ is measured relative to the vertical axis:       $F_x = F \sin(\phi)$       $F_y = F \cos(\phi)$

  • Step 6: Write Newton's Second Law component equations:     $\Sigma F_x = m a_x$     $\Sigma F_y = m a_y$

  • Step 7: List all knowns and unknowns.

  • Step 8: Solve the algebraic system symbolically for the requested variables before substituting numerical values.

• Conditions Where $\Sigma F = 0$:

  • Static Equilibrium: The object is not moving in the specified coordinate direction ($v = 0$, $a = 0$).

  • Dynamic Equilibrium: The object is moving with constant velocity in that coordinate direction ($v = \text{constant}$, $a = 0$).

• Apparent Weight and Elevator Dynamics:

  • Apparent weight is defined as the magnitude of the supporting normal force ($N$) felt by or measured on an object.

  • Stationary or Constant Velocity Elevator ($a_y = 0$):     $\Sigma F_y = N - mg = 0 \implies N = mg$

  • For a $75.0\,\text{kg}$ person:       $N = 75.0(9.81) = 735.75\,\text{N}$

  • Upward Acceleration ($a_y = +2.00\,\text{m/s}^2$):     $\Sigma F_y = N - mg = m a_y \implies N = m(g + a_y)$     $N = 75.0(2.00) + 75.0(9.81) = 150 + 735.75 = 885.75\,\text{N} \approx 886\,\text{N}$

  • The scale reads higher than the resting weight.

  • Downward Acceleration ($a_y = -2.00\,\text{m/s}^2$):     $\Sigma F_y = N - mg = -m a \implies N = m(g - a)$     $N = -75.0(2.00) + 75.0(9.81) = -150 + 735.75 = 585.75\,\text{N} \approx 586\,\text{N}$

  • The scale reads lower than the resting weight.

  • Free Fall ($a_y = -g = -9.81\,\text{m/s}^2$):     $N = m(g - g) = 0\,\text{N}$

  • Complete apparent weightlessness occurs because both the person and the elevator floor accelerate downward at $g$.

• Inclined Planes (Ramps):

• Standard axes are tilted such that the $x'$-axis is parallel to the incline surface and the $y'$-axis is perpendicular to the surface.

• This alignment ensures that acceleration occurs purely along the $x'$-axis ($a_x = a$), while acceleration perpendicular to the ramp is zero ($a_y = 0$).

• Gravity ($mg$) acts straight down, making an angle $\theta$ with the negative $y'$-axis (where $\theta$ is the ramp angle with the horizontal):   $(mg)_x = mg \sin(\theta)$   $(mg)_y = mg \cos(\theta)$

• Equilibrium Perpendicular to Ramp:   $\Sigma F_y = N - mg \cos(\theta) = 0 \implies N = mg \cos(\theta)$

• Acceleration Down a Frictionless Incline:   $\Sigma F_x = mg \sin(\theta) = ma \implies a = g \sin(\theta)$

Friction Mechanics: Static vs. Kinetic Friction

• Physical Origin and Properties:

  • Friction is a resistive electromagnetic contact force arising from microscopic surface irregularities (asperities) interlocking between two contacting bodies.

  • Direction rules:

  • Friction always acts parallel to the contacting surface.

  • Friction always acts in the direction directly opposing relative motion or attempted relative motion.

• Kinetic Friction ($f_k$):

  • Operates when an object is actively sliding relative to the contact surface.

  • The magnitude of kinetic friction depends solely on the coefficient of kinetic friction ($\mu_k$) and the normal force ($N$):     $f_k = \mu_k N$

  • $\mu_k$ is a dimensionless scalar determined by the specific material composition of both contacting surfaces.

  • Representative material coefficients:

  • Steel on steel: $\mu_k = 0.57$

  • Aluminum on steel: $\mu_k = 0.47$

  • Rubber on concrete: $\mu_k = 0.8$

  • Ice on ice: $\mu_k = 0.03$

  • Approximations of zero friction ($f_k \approx 0$):

  • Objects on wheels or low-friction bearings.

  • Smooth ice surfaces.

  • Explicitly stated "frictionless" conditions.

• Static Friction ($f_s$):

  • Operates when external forces attempt to move an object, but no relative sliding occurs ($v = 0$, $a = 0$).

  • Static friction matches whatever applied force is necessary to prevent motion up to a threshold:     $f_s \le \mu_s N$

  • Maximum Static Friction (Breakaway Threshold):     $f_{s,\text{max}} = \mu_s N$

  • The coefficient of static friction is almost universally larger than the coefficient of kinetic friction ($\mu_s > \mu_k$).

  • Steel on steel: $\mu_s = 0.74$

  • Aluminum on steel: $\mu_s = 0.61$

  • Rubber on concrete: $\mu_s = 1.0$

  • Ice on ice: $\mu_s = 0.1$

  • Two-Question Protocol for Static Friction Problems:

  • Question 1: Will the object move?

    1. Calculate normal force $N$ from $\Sigma F_y = 0$.

    2. Calculate the threshold $f_{s,\text{max}} = \mu_s N$.

    3. Calculate the required holding friction $f_s$ assuming static equilibrium ($\Sigma F_x = 0$).

    4. If $f_s \le \mu_s N$, the object remains stationary ($a = 0$), and the actual friction force is $f_s$.

    5. If $f_s > \mu_s N$, static friction fails, the object accelerates, and friction immediately transitions to kinetic friction ($f_k = \mu_k N$).

  • Question 2: What is the minimum force required to initiate motion?

    • Set applied force equal to $f_{s,\text{max}} = \mu_s N$.

• Comprehensive Example: Horizontal Block with Friction:

  • A block of mass $m = 1.00\,\text{kg}$ with initial velocity $v_0 = 10.0\,\text{m/s}$ moves across a surface with $\mu_k = 0.250$.     $\Sigma F_y = N - mg = 0 \implies N = mg = 1.00(9.81) = 9.81\,\text{N}$     $\Sigma F_x = -f_k = -\mu_k N = -m a \implies a = \mu_k g = 0.250(9.81) = 2.45\,\text{m/s}^2 \text{ to the left}$

  • A block of mass $m = 15.0\,\text{kg}$ sits at rest with $\mu_s = 0.450$ and $\mu_k = 0.250$. An external horizontal force $F = 150\,\text{N}$ is applied.     $N = mg = 15.0(9.81) = 147.15\,\text{N}$     $f_{s,\text{max}} = \mu_s N = 0.450(147.15) = 66.2\,\text{N}$

  • Since $F = 150\,\text{N} > 66.2\,\text{N}$, the box moves.

  • The friction force becomes kinetic:       $f_k = \mu_k N = 0.250(147.15) = 36.79\,\text{N}$

  • Acceleration:       $\Sigma F_x = F - f_k = ma \implies a = \frac{150 - 36.79}{15.0} = 7.55\,\text{m/s}^2$

  • The minimum force required to keep this $15.0\,\text{kg}$ block moving without accelerating ($a=0$) once in motion is:     $F = f_k = \mu_k mg = 0.250(15.0)(9.81) = 36.8\,\text{N}$

  • The maximum force that can be applied to this block while remaining stationary is:     $F_{\text{max}} = \mu_s mg = 0.450(15.0)(9.81) = 66.2\,\text{N}$

• Comprehensive Example: Box Pulled at an Angle with Friction:

• A $25.0\,\text{kg}$ box is dragged across a horizontal floor where $\mu_k = 0.300$ by a rope tension $T = 105\,\text{N}$.

• Case A: Rope pulled horizontally ($\theta = 0^\circ$):     $\Sigma F_y = N - mg = 0 \implies N = mg = 25.0(9.81) = 245.25\,\text{N}$     $f_k = \mu_k N = 0.300(245.25) = 73.575\,\text{N}$     $\Sigma F_x = T - f_k = ma \implies a = \frac{105 - 73.575}{25.0} = 1.257\,\text{m/s}^2 \approx 1.26\,\text{m/s}^2$

• Case B: Rope pulled at $\theta = 25.0^\circ$ above horizontal:     $\Sigma F_y = N - mg + T \sin(\theta) = 0 \implies N = mg - T \sin(\theta)$     $N = 25.0(9.81) - 105 \sin(25.0^\circ) = 245.25 - 44.375 = 200.875\,\text{N}$     $f_k = \mu_k N = 0.300(200.875) = 60.26\,\text{N}$     $\Sigma F_x = T \cos(\theta) - f_k = ma$     $a = \frac{105 \cos(25.0^\circ) - 60.26}{25.0} = \frac{95.163 - 60.26}{25.0} = 1.396\,\text{m/s}^2 \approx 1.40\,\text{m/s}^2$

  • Pulling upward at an angle decreases the normal force, reducing the kinetic friction and increasing the overall forward acceleration.

  • Comprehensive Example: Sliding Block on Incline with Friction:

• A $2.00\,\text{kg}$ wooden block sits on an incline set to $\theta = 40.0^\circ$ with $\mu_k = 0.300$.     $\Sigma F_y = N - mg \cos(\theta) = 0 \implies N = mg \cos(\theta)$     $\Sigma F_x = mg \sin(\theta) - f_k = ma$     $mg \sin(\theta) - \mu_k mg \cos(\theta) = ma$     $a = g(\sin(\theta) - \mu_k \cos(\theta))$     $a = 9.81(\sin(40.0^\circ) - 0.300 \cos(40.0^\circ)) = 9.81(0.6428 - 0.2298) = 4.05\,\text{m/s}^2$

Coupled Multi-Body Systems and Kinematics Integration

• Principles of Connected Systems:

  • When bodies are connected by an inextensible taut string over an ideal pulley:

  • All attached bodies share the exact same magnitude of acceleration ($a_1 = a_2 = a$).

  • The tension in the continuous string is uniform throughout ($T_1 = T_2 = T$).

• Half-Atwood Machine (Cart on Table Connected to Hanging Mass):

• Consider a frictionless cart ($m_1 = 2.30\,\text{kg}$) on a horizontal table attached via a light string to a hanging mass ($m_2 = 1.50\,\text{kg}$), held at rest at height $h = 0.860\,\text{m}$ above the floor.

• Cart Equations ($m_1$):     $\Sigma F_y = N - m_1 g = 0 \implies N = m_1 g$     $\Sigma F_x = T = m_1 a$

• Hanging Mass Equations ($m_2$):     $\Sigma F_y = T - m_2 g = -m_2 a \implies m_2 g - T = m_2 a$

• Solving by Substitution:     $m_2 g - (m_1 a) = m_2 a$     $m_2 g = (m_1 + m_2) a$     $a = \frac{m_2 g}{m_1 + m_2} = \frac{1.50(9.81)}{2.30 + 1.50} = \frac{14.715}{3.80} = 3.872\,\text{m/s}^2 \approx 3.87\,\text{m/s}^2$

• String Tension:     $T = m_1 a = 2.30(3.872) = 8.91\,\text{N}$

• Kinematics Integration for Impact Velocity:

  • The hanging mass falls a distance $d = 0.860\,\text{m}$ from rest ($v_0 = 0$):       $v^2 - v_0^2 = 2 a d$       $v = \sqrt{2 a d} = \sqrt{2(3.872)(0.860)} = \sqrt{6.660} = 2.58\,\text{m/s}$

  • Since both objects are physically coupled, the velocity of the cart when the mass hits the floor is also identically $2.58\,\text{m/s}$.

  • Kinematic Linkages between Force and Motion:

• Constant acceleration kinematic equations:

  1. $x = x_0 + v_0 t + \frac{1}{2} a t^2$

  2. $v = v_0 + a t$

  3. $v^2 - v_0^2 = 2 a (x - x_0)$

• Linkage workflows:

  • Given forces $\implies \Sigma F = ma \implies a \implies$ use kinematics to find $x$, $v$, or $t$.

  • Given motion ($x$, $v$, $t$) $\implies$ kinematics yields $a \implies \Sigma F = ma \implies$ determine unknown forces.

• Example: A force $F = 100\,\text{N}$ acts horizontally on a $10.0\,\text{kg}$ cart initially at rest across a distance $d = 5.00\,\text{m}$ on a frictionless surface.     $\Sigma F_x = F = ma \implies a = \frac{100}{10.0} = 10.0\,\text{m/s}^2$     $v = \sqrt{v_0^2 + 2 a (x - x_0)} = \sqrt{0 + 2(10.0)(5.00)} = \sqrt{100} = 10.0\,\text{m/s}$

Work and Energy Conservation Principles

• Energy Fundamentals:

  • Energy ($E$) is a scalar quantity representing the capacity to do work or cause physical change.

  • Units: Joules ($\text{J}$, where $1\,\text{J} = 1\,\text{N}\cdot\text{m} = 1\,\text{kg}\cdot\text{m}^2/\text{s}^2$).

  • Energy can be positive, negative, or zero, but carries no vector direction.

• Work ($W$):

  • Work is the quantitative transfer of energy across the boundary of a system produced when an external force acts across a displacement:     $W = F d \cos(\theta)$

  • $F$ is the applied force magnitude.

  • $d$ is the displacement magnitude over which the force acts.

  • $\theta$ is the angle between the force vector and the displacement vector.

  • Sign Conventions of Work:

  • Positive Work ($W > 0$, or $W_{in}$): Occurs when $\theta < 90^\circ$ (force has a component in the direction of motion). Kinetic energy increases.

    • When $\theta = 0^\circ$, $W = +Fd$.

  • Negative Work ($W < 0$, or $W_{out}$): Occurs when $\theta > 90^\circ$ (force opposes motion). Kinetic energy decreases.

    • When $\theta = 180^\circ$, $W = -Fd$.

  • Zero Work ($W = 0$): Occurs when $\theta = 90^\circ$ (force is strictly perpendicular to displacement, e.g., normal force on a horizontal surface or gravity acting on horizontal motion).

• Work-Energy Theorem:

  • The net work done on an object by all acting forces equals the net change in its kinetic energy:     $W_{net} = \Delta KE = KE_f - KE_i$

  • Kinetic energy is defined as:       $KE = \frac{1}{2} m v^2$

• Total Energy Conservation Equation:

  • In an isolated or general system, total energy is strictly conserved:     $E_i = E_f$     $KE_i + PE_i + W_{in} = KE_f + PE_f + E_{th} + |W_{out}|$

  • Alternatively expressed in terms of non-conservative work ($W_{nc}$):       $W_{nc} = \Delta KE + \Delta PE$

  • Limitations of Energy Conservation methods:

  • Vector direction information is lost because kinetic energy depends on speed squared ($v^2$), not directional velocity.

  • Elapsed time ($t$) is not an explicit parameter in standard energy equations.

• Comprehensive Example: Truck Braking Distance:

  • A $4000\,\text{kg}$ truck travels at $v_0 = 30.0\,\text{m/s}$. The brakes exert a stopping force of $50,000\,\text{N}$.

  • (a) Kinetic Energy:     $KE = \frac{1}{2} m v^2 = \frac{1}{2}(4000)(30.0)^2 = 1,800,000\,\text{J} = 1.80 \times 10^6\,\text{J}$

  • (b) Work Required to Stop:     $W_{net} = KE_f - KE_i = 0 - 1,800,000 = -1,800,000\,\text{J} = -1.80 \times 10^6\,\text{J}$

  • (c) Stopping Distance:     $W = F d \cos(180^\circ) = -F d$     $-1,800,000 = -50,000 d \implies d = \frac{1,800,000}{50,000} = 36.0\,\text{m}$

Gravitational and Elastic Potential Energy Systems

• Conservative vs. Non-Conservative Forces:

  • Conservative Forces (Gravity, Spring Force):

  • The work done depends only on initial and final positions, independent of the path taken.

  • The net work done around any closed loop is identically zero ($W_{loop} = 0$).

  • Energy expended against a conservative force is stored reversibly as Potential Energy ($PE$).

  • Non-Conservative Forces (Friction, Drag, Applied Forces, Tension):

  • Path-dependent.

  • Mechanical energy is dissipated irreversibly into thermal energy ($E_{th}$) or transferred in/out as external work ($W_{in}$, $W_{out}$).

• Gravitational Potential Energy ($PE_g$):

  • Represents stored energy due to vertical elevation against gravity:     $PE_g = mgh$

  • The choice of zero datum ($h = 0$) is arbitrary; physical behaviors depend solely on the potential energy difference ($\Delta PE_g$):

  • A $5.00\,\text{kg}$ block elevated $2.00\,\text{m}$ above a table that is $3.00\,\text{m}$ above a floor:

    • Relative to the table ($h=0$): at the top $PE = 5(9.81)(2) = 98.1\,\text{J}$; at the table $PE = 0\,\text{J}$; at the floor ($h = -3\,\text{m}$) $PE = 5(9.81)(-3) = -147.15\,\text{J}$.

    • Relative to the floor ($h=0$): at the top $PE = 5(9.81)(5) = 245.25\,\text{J}$; at the table $PE = 5(9.81)(3) = 147.15\,\text{J}$.

    • In both frameworks, $\Delta PE$ between top and table is identically $98.1\,\text{J}$, and between table and floor is identically $147.15\,\text{J}$.

• Free-Fall and Trajectory Independence:

  • Two balls of mass $m = 0.0500\,\text{kg}$ are released from height $h = 3.00\,\text{m}$. Ball 1 travels straight down at $4.00\,\text{m/s}$; Ball 2 is projected horizontally at $4.00\,\text{m/s}$.     $KE_i = \frac{1}{2}(0.0500)(4.00)^2 = 0.400\,\text{J}$     $PE_i = mgh = 0.0500(9.81)(3.00) = 1.4715\,\text{J}$     $E_{\text{total}} = 0.400 + 1.4715 = 1.8715\,\text{J}$

  • At ground level ($PE_f = 0$):     $KE_f = E_{\text{total}} = 1.8715\,\text{J}$     $v = \sqrt{\frac{2 KE_f}{m}} = \sqrt{\frac{2(1.8715)}{0.0500}} = 8.65\,\text{m/s}$

  • Both balls impact the ground with the identical speed ($8.65\,\text{m/s}$) regardless of launch direction, demonstrating that potential energy depends strictly on elevation change, not trajectory shape.

• Elastic (Spring) Potential Energy ($PE_s$):

  • Stored by compressing or stretching an elastic medium from its relaxed equilibrium position ($x = 0$):     $PE_s = \frac{1}{2} k x^2$

  • Dart Gun Example:

  • A spring gun compresses $x = 0.0500\,\text{m}$ ($5.00\,\text{cm}$) loaded with a $m = 0.0100\,\text{kg}$ ($10.0\,\text{g}$) dart. Upon horizontal release, it fires at $v = 3.00\,\text{m/s}$.

  • Spring constant:       $\frac{1}{2} k x^2 = \frac{1}{2} m v^2 \implies k = \frac{m v^2}{x^2} = \frac{0.0100(3.00^2)}{0.0500^2} = 36.0\,\text{N/m}$

  • If held at height $h = 1.40\,\text{m}$ above the floor and discharged horizontally:       $E_i = mgh + \frac{1}{2} m v_0^2 = KE_f = \frac{1}{2} m v^2$       $v = \sqrt{v_0^2 + 2gh} = \sqrt{(3.00)^2 + 2(9.81)(1.40)} = \sqrt{9 + 27.468} = 6.04\,\text{m/s}$

• Vertical Spring Oscillations with Coupled Potential Energies:

  • A $1.50\,\text{kg}$ box hangs vertically from a spring ($k = 34.0\,\text{N/m}$).

  • Static Equilibrium Extension ($a = 0$):     $\Sigma F_y = k x_0 - mg = 0 \implies x_0 = \frac{mg}{k} = \frac{1.50(9.81)}{34.0} = 0.4328\,\text{m} \approx 0.433\,\text{m}$

  • Dynamic Release:

  • The box is pulled down an additional $d = 0.500\,\text{m}$ below equilibrium ($x_{\text{total}} = 0.4328 + 0.500 = 0.9328\,\text{m}$ total extension) and released from rest.

  • Define the low release position as $h = 0$, so the equilibrium height is $h = 0.500\,\text{m}$.

  • Initial energy at release point ($v_i = 0$):       $E_i = \frac{1}{2} k x_{\text{total}}^2 + 0 + 0 = \frac{1}{2}(34.0)(0.9328)^2 = 14.792\,\text{J}$

  • Final energy as it passes through equilibrium:       $E_f = \frac{1}{2} k x_0^2 + mgh + \frac{1}{2} m v^2$       $E_f = \frac{1}{2}(34.0)(0.4328)^2 + 1.50(9.81)(0.500) + \frac{1}{2}(1.50) v^2$       $14.792 = 3.184 + 7.3575 + 0.75 v^2$       $14.792 - 10.542 = 0.75 v^2 \implies 4.250 = 0.75 v^2$       $v = \sqrt{\frac{4.250}{0.75}} = 2.38\,\text{m/s}$

Non-Conservative Forces, Thermal Dissipation, and Power

• Thermal Energy Generated by Friction:

• Kinetic friction removes mechanical energy from a system, transforming it into internal thermal energy ($E_{th}$):     $W_{f_k} = f_k d \cos(180^\circ) = -f_k d = -\mu_k N d$     $E_{th} = |W_{f_k}| = \mu_k N d$

• On a horizontal surface where $N = mg$:     $E_{th} = \mu_k mg d$

• On an inclined plane where $N = mg \cos(\theta)$, with displacement $d$ along the ramp:     $E_{th} = \mu_k (mg \cos(\theta)) d$

  • Comprehensive Example: Sled on Frictionless Hill with Rough Run-Out:

• A child and sled ($m = 38.0\,\text{kg}$) start from rest atop a frictionless hill of height $h = 2.30\,\text{m}$. At the bottom, the sled encounters horizontal snow with $\mu_k = 0.100$.

• (a) Initial Potential Energy:     $PE_i = mgh = 38.0(9.81)(2.30) = 857.394\,\text{J} \approx 857\,\text{J}$

• (b) Speed at the Foot of the Hill:     $mgh = \frac{1}{2} m v^2 \implies v = \sqrt{2gh} = \sqrt{2(9.81)(2.30)} = 6.716\,\text{m/s} \approx 6.72\,\text{m/s}$

• (c) Stopping Distance across Rough Snow:

  • The total initial mechanical energy converts entirely into thermal dissipation:       $PE_i = E_{th} = \mu_k mg d$       $mgh = \mu_k mg d \implies d = \frac{h}{\mu_k} = \frac{2.30}{0.100} = 23.0\,\text{m}$

  • Power ($P$):

• Power is defined as the time rate of doing work or transferring energy:     $P = \frac{W}{\Delta t} = \frac{\Delta E}{\Delta t}$

• SI Unit: Watt ($\text{W}$), where $1\,\text{W} = 1\,\text{J/s}$.

• Imperial Unit: Horsepower ($\text{hp}$):     $1\,\text{hp} = 746\,\text{W}$

• Power Delivered at Constant Velocity:

  • When an applied force $F$ moves an object at constant speed $v$ against resistance:       $P = F v$

  • Example: An object of mass $m = 5.00\,\text{kg}$ moved at $v = 5.00\,\text{m/s}$ where $\mu_k = 0.300$:       $F = f_k = \mu_k mg = 0.300(5.00)(9.81) = 14.715\,\text{N}$       $P = F v = 14.715(5.00) = 73.6\,\text{W}$

• Automobile Power Calculation:

  • A $1500\,\text{kg}$ automobile accelerates from $0$ to $26.8\,\text{m/s}$ ($60\,\text{mph}$) in $5.30\,\text{s}$.

  • Work required:       $W = \Delta KE = \frac{1}{2}(1500)(26.8)^2 = 538,680\,\text{J}$

  • Average power output:       $P = \frac{538,680}{5.30} = 101,638\,\text{W} \approx 102\,\text{kW}$

  • In horsepower:       $P = \frac{101,638}{746} = 136.2\,\text{hp} \approx 136\,\text{hp}$

• Commercial Energy Billing (Kilowatt-Hour):

  • The kilowatt-hour ($\text{kWh}$) is a unit of energy, not power:       $1\,\text{kWh} = (1000\,\text{W})(3600\,\text{s}) = 3.60 \times 10^6\,\text{J} = 3,600,000\,\text{J}$

Momentum, Impulse, and Conservation in Collisions

• Linear Momentum (Vector $\mathbf{p}$):

  • Defined as the product of mass and velocity:     $\mathbf{p} = m\mathbf{v}$

  • SI Units: $\text{kg}\cdot\text{m/s}$ or Newton-seconds ($\text{N}\cdot\text{s}$).

  • Direction strictly follows the velocity vector. Momentum is directional, so opposite directions require opposing algebraic signs ($+\mathbf{v}$ vs $-\mathbf{v}$).

  • Mathematical link to kinetic energy:     $KE = \frac{p^2}{2m}$

• Newton's Second Law in Terms of Momentum:

  • Net force equals the instantaneous rate of change of linear momentum:     $\Sigma \mathbf{F} = \frac{\Delta \mathbf{p}}{\Delta t}$

• Impulse ($\mathbf{J}$ or $\mathbf{I}$):

  • Impulse is the integrated effect of a force over an elapsed time interval:     $\mathbf{J} = \mathbf{F}_{avg} \Delta t = \Delta \mathbf{p} = m\mathbf{v}_f - m\mathbf{v}_i$

• Baseball Impact Example:

  • A $0.100\,\text{kg}$ baseball approaches home plate horizontally at $v_i = -20.0\,\text{m/s}$ and is struck by a bat, rebounding at $v_f = +40.0\,\text{m/s}$. The impact duration is $\Delta t = 0.100\,\text{s}$.

  • Initial and final momentum:       $p_i = m v_i = 0.100(-20.0) = -2.00\,\text{kg}\cdot\text{m/s}$       $p_f = m v_f = 0.100(40.0) = +4.00\,\text{kg}\cdot\text{m/s}$

  • Impulse delivered:       $\Delta p = p_f - p_i = 4.00 - (-2.00) = +6.00\,\text{kg}\cdot\text{m/s}$

  • Average contact force:       $F_{avg} = \frac{\Delta p}{\Delta t} = \frac{6.00}{0.100} = 60.0\,\text{N}$

  • Law of Conservation of Linear Momentum:

• In an isolated system experiencing zero net external force ($\Sigma \mathbf{F}_{ext} = 0$):     $\mathbf{P}_i = \mathbf{P}_f$

• Internal interaction forces between colliding bodies form Newton's third law action-reaction pairs ($\mathbf{F}_{12} = -\mathbf{F}_{21}$). The internal impulses cancel identically ($\Delta \mathbf{p}_1 + \Delta \mathbf{p}_2 = 0$), conserving total momentum.

  • Taxonomy of Collision Types:

• Completely (Perfectly) Inelastic Collisions:

  • Colliding bodies stick together, moving with a single shared final velocity ($v_3 = v_4 = v_f$).

  • Momentum is fully conserved; mechanical kinetic energy is maximally lost to deformation and thermal energy.

  • Governing equation:       $m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_f$

• Inelastic Collisions (General):

  • Colliding bodies rebound independently.

  • Momentum is conserved, but kinetic energy is not conserved ($KE_f < KE_i$).

  • Governing equation:       $m_1 v_1 + m_2 v_2 = m_1 v_3 + m_2 v_4$

• Perfectly Elastic Collisions:

  • Colliding bodies rebound without permanent deformation or thermal dissipation.

  • Both total momentum and total kinetic energy are conserved simultaneously:       $m_1 v_1 + m_2 v_2 = m_1 v_3 + m_2 v_4$       $\frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 = \frac{1}{2} m_1 v_3^2 + \frac{1}{2} m_2 v_4^2$

  • Relative Velocity Theorem (1D Elastic Collision):

    • Combining the momentum and kinetic energy conservation laws yields the relative velocity condition:         $v_1 - v_2 = -(v_3 - v_4) = v_4 - v_3$

    • The relative speed of approach equals the relative speed of separation.

• Explosions:

  • A unified body at initial velocity $v_i$ separates into multiple fragments:       $(m_1 + m_2) v_i = m_1 v_3 + m_2 v_4$

  • Kinetic energy increases dramatically ($KE_f > KE_i$), converted from internal chemical or mechanical potential energy.

One-Dimensional and Two-Dimensional Collision Dynamics

• One-Dimensional Inelastic Collision Examples:

  • Two Carts Sticking Together:

  • A $2.30\,\text{kg}$ cart moving at $4.50\,\text{m/s}$ collides with a stationary $3.50\,\text{kg}$ cart on a frictionless track:       $m_1 v_1 + m_2(0) = (m_1 + m_2) v_f$       $v_f = \frac{2.30(4.50)}{2.30 + 3.50} = \frac{10.35}{5.80} = 1.78\,\text{m/s}$

  • Opposing Motion (Mouse Jumping onto Cart):

  • A $0.980\,\text{kg}$ mouse runs at $+1.20\,\text{m/s}$ toward an oncoming $2.30\,\text{kg}$ cart traveling at $-2.50\,\text{m/s}$. The mouse jumps onto the cart:       $v_f = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} = \frac{0.980(1.20) + 2.30(-2.50)}{0.980 + 2.30} = \frac{1.176 - 5.75}{3.28} = -1.39\,\text{m/s}$

  • The combined unit moves at $1.39\,\text{m/s}$ in the cart's initial direction of travel.

• One-Dimensional Elastic Collisions between Equal Masses ($m_1 = m_2 = 1.30\,\text{kg}$, $v_1 = 3.40\,\text{m/s}$):

  • Case A: Cart 2 initially at rest ($v_2 = 0$):

  • Using $v_4 = v_1 - v_2 + v_3$ into momentum equation:       $v_3 = \frac{(m_1 - m_2) v_1 + 2 m_2 v_2}{m_1 + m_2} = \frac{0 + 0}{2.60} = 0\,\text{m/s}$       $v_4 = v_1 + v_3 = 3.40 + 0 = +3.40\,\text{m/s}$

  • Identical masses completely swap velocities upon elastic collision; Cart 1 stops dead.

  • Case B: Cart 2 initially moving oppositely ($v_2 = -3.40\,\text{m/s}$):     $v_3 = \frac{(1.30 - 1.30)(3.40) + 2(1.30)(-3.40)}{2.60} = -3.40\,\text{m/s}$     $v_4 = v_1 - v_2 + v_3 = 3.40 - (-3.40) + (-3.40) = +3.40\,\text{m/s}$

  • The velocities reflect symmetrically.

  • Case C: Cart 2 initially moving oppositely at higher speed ($v_2 = -6.80\,\text{m/s}$):     $v_3 = \frac{0 + 2(1.30)(-6.80)}{2.60} = -6.80\,\text{m/s}$     $v_4 = 3.40 - (-6.80) + (-6.80) = +3.40\,\text{m/s}$

• Bullet-Block Interactions ($m_1 = 0.0220\,\text{kg}$, $v_1 = 240\,\text{m/s}$, $m_2 = 2.00\,\text{kg}$, $v_2 = 0$):

  • Embedded Bullet (Inelastic):     $v_f = \frac{m_1 v_1}{m_1 + m_2} = \frac{0.0220(240)}{2.00 + 0.0220} = \frac{5.28}{2.022} = 2.61\,\text{m/s}$

  • Perforating Bullet (Exits at $v_3 = 140\,\text{m/s}$):     $m_1 v_1 = m_1 v_3 + m_2 v_4$     $v_4 = \frac{m_1(v_1 - v_3)}{m_2} = \frac{0.0220(240 - 140)}{2.00} = \frac{0.0220(100)}{2.00} = 1.10\,\text{m/s}$

  • Elastic Rebound (Theoretical Vibranium Block):     $v_3 = \frac{m_1 - m_2}{m_1 + m_2} v_1 = \frac{0.0220 - 2.00}{2.022}(240) = -234.78\,\text{m/s} \approx -235\,\text{m/s}$     $v_4 = v_1 + v_3 = 240 + (-234.78) = 5.22\,\text{m/s}$

• Explosion Recoil Examples:

  • Hunter on Ice: A $99.0\,\text{kg}$ hunter standing motionless ($v_0 = 0$) on frictionless ice fires a $0.0220\,\text{kg}$ bullet at $305\,\text{m/s}$.     $0 = m_1 v_1 + m_2 v_2 \implies v_1 = -\frac{m_2 v_2}{m_1} = -\frac{0.0220(305)}{99.0} = -0.0678\,\text{m/s}$

  • Ice Skater Throwing a Weight: A $60.0\,\text{kg}$ skater moving at $5.00\,\text{m/s}$ on ice throws a $0.500\,\text{kg}$ snowball forward at $25.0\,\text{m/s}$.     $(m_1 + m_2) v_i = m_1 v_3 + m_2 v_4$     $v_3 = \frac{(60.0 + 0.500)(5.00) - 0.500(25.0)}{60.0} = \frac{302.5 - 12.5}{60.0} = 4.83\,\text{m/s}$

• Two-Dimensional Collisions:

  • Total linear momentum is independently conserved along each orthogonal axis:     $\Sigma P_{xi} = \Sigma P_{xf} \implies m_1 v_{1x} + m_2 v_{2x} = m_1 v_{3x} + m_2 v_{4x}$     $\Sigma P_{yi} = \Sigma P_{yf} \implies m_1 v_{1y} + m_2 v_{2y} = m_1 v_{3y} + m_2 v_{4y}$

  • Vector Reconstruction:     $v = \sqrt{v_x^2 + v_y^2}$     $\theta = \tan^{-1}\left(\frac{v_y}{v_x}\right)$

  • If $v_x < 0$, add $180^\circ$ to the principal inverse tangent result to locate the correct quadrant.

  • 2D Inelastic Clay Collision:

  • Lump 1 ($m_1 = 2.00\,\text{kg}$) moves at $v_1 = 3.00\,\text{m/s}$ at $30.0^\circ$ above the positive $x$-axis.

  • Lump 2 ($m_2 = 3.00\,\text{kg}$) moves at $v_2 = 2.00\,\text{m/s}$ along the negative $x$-axis ($\theta_2 = 180^\circ$).

  • Resolving initial components:       $v_{1x} = 3.00 \cos(30.0^\circ) = 2.598\,\text{m/s}$       $v_{1y} = 3.00 \sin(30.0^\circ) = 1.500\,\text{m/s}$       $v_{2x} = -2.00\,\text{m/s}$       $v_{2y} = 0$

  • Total Initial Momentum Components:       $P_{ix} = 2.00(2.598) + 3.00(-2.00) = 5.196 - 6.00 = -0.804\,\text{kg}\cdot\text{m/s}$       $P_{iy} = 2.00(1.500) + 3.00(0) = +3.00\,\text{kg}\cdot\text{m/s}$

  • Final velocity components after sticking ($m_{\text{total}} = 5.00\,\text{kg}$):       $v_{fx} = \frac{P_{ix}}{m_{\text{total}}} = \frac{-0.804}{5.00} = -0.1608\,\text{m/s} \approx -0.161\,\text{m/s}$       $v_{fy} = \frac{P_{iy}}{m_{\text{total}}} = \frac{3.00}{5.00} = +0.600\,\text{m/s}$

  • Resultant Speed:       $v_f = \sqrt{(-0.1608)^2 + (0.600)^2} = \sqrt{0.02586 + 0.360} = 0.621\,\text{m/s}$

  • Direction:       $\theta = \tan^{-1}\left(\frac{0.600}{-0.1608}\right) = -74.98^\circ + 180^\circ = 105.02^\circ \text{ from the } +x\text{-axis}$       (or $74.98^\circ$ above the negative $x$-axis).