NCEA Physics 2.4 Mechanics Study Guide

Kinematic Equations of Motion

• To calculate instantaneous quantities for distances and velocities of an accelerating object, specific mathematical relationships known as the kinematic equations of motion are used.

• The five fundamental variables in kinematic motion are:

  • $v_f$: Final velocity ($\text{m\,s}^{-1}$)

  • $v_i$: Initial velocity ($\text{m\,s}^{-1}$)

  • $a$: Acceleration ($\text{m\,s}^{-2}$)

  • $t$: Time ($\text{s}$)

  • $d$: Displacement or distance ($\text{m}$)

• The four standard kinematic equations of motion are:

  • $v_f = v_i + a t$

  • $d = \left(\frac{v_i + v_f}{2}\right) t$

  • $v_f^2 = v_i^2 + 2 a d$

  • $d = v_i t + \frac{1}{2} a t^2$

• Each kinematic equation omits exactly one of the five kinematic variables. Problem-solving requires identifying the variable that is neither given nor required, and choosing the equation where that variable is absent.

• Four-Step Problem Solving Procedure:

  1. List all given data values including units and explicitly state the unknown variable.

  2. Select the kinematic formula that omits the absent 5th variable.

  3. Substitute the given values into the chosen formula.

  4. Algebraically rearrange and solve for the unknown value.

• Step-by-Step Kinematic Examples and Applications:

  • Deceleration of a Braking Car:

  • Scenario: A car travelling along a road at $25\,\text{m\,s}^{-1}$ comes to a complete stop ($0\,\text{m\,s}^{-1}$) over a distance of $30\,\text{m}$.

  • List of Data: $v_i = 25\,\text{m\,s}^{-1}$, $v_f = 0\,\text{m\,s}^{-1}$, $d = 30\,\text{m}$, $a = ?$

  • Absent Variable: Time ($t$

  • Formula Selection: $v_f^2 = v_i^2 + 2 a d$

  • Substitution and Solution:       $0^2 = 25^2 + (2 \times a \times 30)$       $0 = 625 + 60 a$       $60 a = -625$       $a = -10.4\,\text{m\,s}^{-2}$

  • Tennis Ball Collision and Shot History:

  • Part A (Collision Acceleration): A tennis ball is travelling at $45.0\,\text{m\,s}^{-1}$ when it collides with a tennis racket. After the collision, the ball moves in the opposite direction at $-55.0\,\text{m\,s}^{-1}$. The collision takes $0.60\,\text{s}$.

    • List of Data: $v_i = 45.0\,\text{m\,s}^{-1}$, $v_f = -55.0\,\text{m\,s}^{-1}$, $t = 0.60\,\text{s}$, $a = ?$

    • Formula Selection: $a = \frac{v_f - v_i}{t}$

    • Substitution and Solution:         $a = \frac{-55.0 - 45.0}{0.60} = \frac{-100}{0.60} = -166.66\dots\,\text{m\,s}^{-2} = -167\,\text{m\,s}^{-2}\text{ (3 s.f.)}$

  • Part B (Original Shot Velocity): Before hitting the racket, the incoming ball slowed down at a rate of $1.25\,\text{m\,s}^{-2}$ as it travelled across the court over a distance of $18.0\,\text{m}$ to reach its $45.0\,\text{m\,s}^{-1}$ velocity. Calculate its velocity when hit $18.0\,\text{m}$ earlier.

    • List of Data: $v_f = 45.0\,\text{m\,s}^{-1}$, $a = -1.25\,\text{m\,s}^{-2}$, $d = 18.0\,\text{m}$, $v_i = ?$

    • Formula Selection: $v_f^2 = v_i^2 + 2 a d \implies v_i = \sqrt{v_f^2 - 2 a d}$

    • Substitution and Solution:         $v_i = \sqrt{45.0^2 - (2 \times -1.25 \times 18.0)} = \sqrt{2025 - (-45)} = \sqrt{2070} = 45.5\,\text{m\,s}^{-1}$

  • Car Accelerating Over Distance:

  • Scenario: A car travels at $4.0\,\text{m\,s}^{-1}$ and accelerates at $10.0\,\text{m\,s}^{-2}$ over a distance of $49.0\,\text{m}$. Calculate final velocity.

  • List of Data: $v_i = 4.0\,\text{m\,s}^{-1}$, $a = 10.0\,\text{m\,s}^{-2}$, $d = 49.0\,\text{m}$, $v_f = ?$

  • Formula Selection: $v_f = \sqrt{v_i^2 + 2 a d}$

  • Substitution and Solution:       $v_f = \sqrt{4.0^2 + (2 \times 10.0 \times 49.0)} = \sqrt{16 + 980} = \sqrt{996} = 31.559\dots\,\text{m\,s}^{-1} = 32\,\text{m\,s}^{-1}$

  • Cyclist Uniform Acceleration:

  • Scenario: A cyclist travelling at $3.0\,\text{m\,s}^{-1}$ accelerates uniformly to $8.0\,\text{m\,s}^{-1}$ over $5.0\,\text{s}$. Calculate distance travelled.

  • List of Data: $v_i = 3.0\,\text{m\,s}^{-1}$, $v_f = 8.0\,\text{m\,s}^{-1}$, $t = 5.0\,\text{s}$, $d = ?$

  • Formula Selection: $d = \left(\frac{v_i + v_f}{2}\right) t$

  • Substitution and Solution:       $d = \left(\frac{3.0 + 8.0}{2}\right) \times 5.0 = 5.5 \times 5.0 = 27.5\,\text{m} = 28\,\text{m}\text{ (2 s.f.)}$

  • Car Accelerating from Rest:

  • Scenario: A car at rest ($v_i = 0\,\text{m\,s}^{-1}$) accelerates uniformly at $1.5\,\text{m\,s}^{-2}$.

  • Distance in $20\,\text{s}$:       $d = v_i t + \frac{1}{2} a t^2 = (0 \times 20) + \left(\frac{1}{2} \times 1.5 \times 20^2\right) = 0.75 \times 400 = 300\,\text{m}$

  • Speed after $100\,\text{m}$:       $v_f = \sqrt{v_i^2 + 2 a d} = \sqrt{0^2 + (2 \times 1.5 \times 100)} = \sqrt{300} = 17.32\dots\,\text{m\,s}^{-1} = 17\,\text{m\,s}^{-1}$

  • Boat Race Comparison:

  • Scenario: Boat A and Boat B race over a distance of $800\,\text{m}$. Boat A starts from rest ($v_i = 0\,\text{m\,s}^{-1}$) and accelerates at a constant rate of $4\,\text{m\,s}^{-2}$. Boat B has an initial velocity of $12\,\text{m\,s}^{-1}$ and accelerates uniformly until crossing the finish line at $52\,\text{m\,s}^{-1}$.

  • Race Winner Calculation:

    • Boat A Time: $d = \frac{1}{2} a t^2 \implies t = \sqrt{\frac{2 d}{a}} = \sqrt{\frac{2 \times 800}{4}} = \sqrt{400} = 20\,\text{s}$

    • Boat B Time: $d = \left(\frac{v_i + v_f}{2}\right) t \implies t = \frac{2 d}{v_i + v_f} = \frac{2 \times 800}{12 + 52} = \frac{1600}{64} = 25\,\text{s}$

    • Result: Boat A wins the race ($20\,\text{s} < 25\,\text{s}$).

  • Acceleration of Boat B:       $a = \frac{v_f - v_i}{t} = \frac{52 - 12}{25} = \frac{40}{25} = 1.6\,\text{m\,s}^{-2}$

  • Point of Equal Velocity:

    • Set $v_{fA} = v_{fB}$ using $v_f = v_i + a t$

    • $v_{iA} + a_A t = v_{iB} + a_B t \implies 0 + 4 t = 12 + 1.6 t$

    • $4 t - 1.6 t = 12 \implies 2.4 t = 12 \implies t = \frac{12}{2.4} = 5\,\text{s}$

    • Common Velocity: $v = 0 + (4 \times 5) = 20\,\text{m\,s}^{-1}$

Projectile Motion

• Horizontal projectile motion occurs when an object is projected into the air at an angle relative to the Earth's surface and moves along a curved path called a trajectory due to the force of gravity alone.

• Fundamental Principles of Trajectory:

  • Assuming air resistance and friction are negligible, horizontal velocity ($v_{\text{horiz}}$) remains constant throughout the flight because no horizontal forces act on the object.

  • Vertical velocity ($v_{\text{vert}}$) changes continuously due to downward acceleration caused by gravity ($a = +9.8\,\text{m\,s}^{-2}$ downwards).

  • As the projectile ascends, its vertical velocity decreases in magnitude.

  • At the peak/top of the trajectory, the vertical instantaneous velocity is exactly zero ($v_{\text{vert}} = 0\,\text{m\,s}^{-1}$), but the object still possesses its constant horizontal velocity.

  • As the projectile descends, its vertical velocity increases in magnitude downward.

• Component Rule for Formulas:

  • Standard kinematic equations must NEVER use the total angled initial velocity vector directly.

  • Velocity must always be resolved into separate horizontal and vertical independent components before applying kinematic equations.

• Vector Decomposition Equations:

  • Horizontal Velocity Component: $v_{\text{horiz}} = v_{\text{object}} \cos(\theta)$

  • Vertical Velocity Component: $v_{\text{vert}} = v_{\text{object}} \sin(\theta)$

• Component Calculation Examples:

  • Golf Ball hit at $40.0\,\text{m\,s}^{-1}$ at an angle of $20^\circ$ from the ground:

  • $v_{\text{horiz}} = 40 \cos(20^\circ) = 37.6\,\text{m\,s}^{-1}$

  • $v_{\text{vert}} = 40 \sin(20^\circ) = 13.7\,\text{m\,s}^{-1}$

  • Projectile with initial velocity of $25.0\,\text{m\,s}^{-1}$ at $70.0^\circ$ from horizontal:

  • $v_{\text{vert}} = 25 \sin(70^\circ) = 23.492\dots\,\text{m\,s}^{-1} = 23.5\,\text{m\,s}^{-1}$

  • $v_{\text{horiz}} = 25 \cos(70^\circ) = 8.5505\dots\,\text{m\,s}^{-1} = 8.55\,\text{m\,s}^{-1}$

  • Fishing Sinker cast at $28.0\,\text{m\,s}^{-1}$ upwards at $32.0^\circ$ and to the left:

  • $v_{\text{vert}} = 28 \sin(32^\circ) = 14.837\dots\,\text{m\,s}^{-1} = 14.8\,\text{m\,s}^{-1}$

  • $v_{\text{horiz}} = 28 \cos(32^\circ) = 23.745\dots\,\text{m\,s}^{-1} = 23.7\,\text{m\,s}^{-1}$

• Complete Trajectory Analysis (Stone Thrown Horizontally from Cliff):

  • Scenario: A stone is thrown horizontally at $12\,\text{m\,s}^{-1}$ from a cliff top with a height of $30\,\text{m}$.

  • Time to reach the ground:

  • Vertical parameters: $v_{i,\text{vert}} = 0\,\text{m\,s}^{-1}$, $a_{\text{vert}} = +9.8\,\text{m\,s}^{-2}$, $d_{\text{vert}} = 30\,\text{m}$

  • Formula: $d = v_i t + \frac{1}{2} a t^2 \implies d = \frac{1}{2} a t^2 \implies t = \sqrt{\frac{2 d}{a}}$

  • Calculation: $t = \sqrt{\frac{2 \times 30}{9.8}} = \sqrt{6.1224\dots} = 2.47433\dots\,\text{s} = 2.5\,\text{s}\text{ (2 s.f.)}$

  • Range (Horizontal Distance Travelled):

  • Horizontal parameters: $v_{\text{horiz}} = 12\,\text{m\,s}^{-1}$, $t = 2.4743\dots\,\text{s}$

  • Formula: $\Delta d = v_{\text{horiz}} \Delta t$

  • Calculation: $\Delta d = 12 \times 2.47433\dots = 29.692\dots\,\text{m} = 30\,\text{m}$

  • Comparison of Initial vs. Final Impact Velocity:

  • Final vertical velocity component just before impact:       $v_{f,\text{vert}} = \sqrt{v_{i,\text{vert}}^2 + 2 a d} = \sqrt{0^2 + (2 \times 9.8 \times 30)} = \sqrt{588} = 24.248\dots\,\text{m\,s}^{-1} = 24\,\text{m\,s}^{-1}$

  • Magnitude of total velocity at impact:       $v_{\text{total}} = \sqrt{v_{\text{vert}}^2 + v_{\text{horiz}}^2} = \sqrt{(24.248\dots)^2 + 12^2} = \sqrt{588 + 144} = \sqrt{732} = 27.055\dots\,\text{m\,s}^{-1} = 27\,\text{m\,s}^{-1}$

  • Angle of impact relative to horizontal:       $\theta = \tan^{-1}\left(\frac{v_{\text{vert}}}{v_{\text{horiz}}}\right) = \tan^{-1}\left(\frac{24.248\dots}{12}\right) = \tan^{-1}(2.0207\dots) = 63.669^\circ = 64^\circ$

Forces, Mass, and Weight

• Force Definitions and Terminologies:

  • Force ($F$): A push or pull acting on an object, measured in newtons ($\text{N}$). Forces are vector quantities possesses both magnitude and direction.

  • Thrust: A push force that propels an object in a given direction, particularly when generated by the object itself (such as by an engine or rocket).

  • Weight: A downward force exerted on an object due to its mass and local gravitational acceleration ($F_g = m g$). Its direction is always downward toward the Earth's centre.

  • Friction: A force opposing relative motion. Forms include static friction, kinetic friction, and drag (fluid/air friction). Drag always acts in the exact opposite direction to motion (e.g., if a vehicle travels left, air drag acts to the right).

  • Gravity: Acceleration arising from gravitational attraction between the mass of an object and the mass of the Earth, Moon, or other planetary body.

  • Support Force: An upward reaction force holding an object in place against gravity. For an object resting on a horizontal surface, support force acts directly upward, opposing weight.

  • Net Force ($F_{\text{net}}$): The resultant vector sum of all forces acting on an object. Determines the object's change in motion (acceleration or deceleration). When forces cancel out (equilibrium), net force is zero ($0\,\text{N}$).

  • Fundamental Equation: $F = m a$, relating mass and acceleration under a net force.

• Distinguishing Mass and Weight:

  • Mass: The quantity of matter comprising an object. It is a scalar quantity (no direction), constant regardless of location, measured in kilograms ($\text{kg}$), and is not a force.

  • Weight: The gravitational force pulling on a mass. It is a vector quantity pointing toward the celestial body's centre, measured in newtons ($\text{N}$), and varies based on local gravitational acceleration ($g$).

  • Planetary Comparison for a $10\,\text{kg}$ Mass:

  • On Earth ($g = 9.8\,\text{m\,s}^{-2}$): Mass = $10\,\text{kg}$, Weight = $98\,\text{N}$

  • On the Moon ($g = 1.6\,\text{m\,s}^{-2}$): Mass = $10\,\text{kg}$, Weight = $16\,\text{N}$

  • On Mars ($g = 3.7\,\text{m\,s}^{-2}$): Mass = $10\,\text{kg}$, Weight = $37\,\text{N}$

Hooke's Law and Spring Mechanics

• Hooke's Law governs the mechanical behaviour of springs when subject to stretching or compressing loads.

• Hooke's Law Equation: $F = -k x$

  • $F$: Restoring force generated by the spring ($\text{N}$)

  • $k$: Spring constant, measuring the stiffness of the spring ($\text{N\,m}^{-1}$)

  • $x$: Extension or compression displacement from equilibrium ($\text{m}$)

• Physical Interpretation of Variables and Signs:

  • Spring Constant ($k$): A larger spring constant corresponds to a stiffer spring that requires greater force to compress or stretch (e.g., automobile suspension springs).

  • Negative Sign: Indicates that the restoring force produced by the spring acts in the direction opposite to the displacement (extension or compression). The restoring force continuously tries to restore the spring to its original equilibrium position.

  • Omission of Negative Sign in Mass Experiments: When hanging a mass vertically on a spring at equilibrium ($m g = k x$), the negative sign can be ignored because downward gravity force ($m g$) and upward spring force ($k x$) are equal in magnitude and opposite in direction.

• Worked Calculations for Spring Systems:

  • Calculating Spring Constant from Stretch:

  • Scenario: A $10.0\,\text{N}$ brass mass stretches a spring by $8.00\,\text{cm}$ ($0.0800\,\text{m}$).

  • Calculation: $k = \frac{F}{x} = \frac{10.0}{0.0800} = 125\,\text{N\,m}^{-1}$

  • Effect of Compression: If compressed by $8.0\,\text{cm}$ instead of stretched, the force magnitude remains identical ($10\,\text{N}$), but acts in the opposite direction.

  • Compression Mass Calculation:

  • Scenario: Calculate mass required to compress a spring ($k = 2.5\,\text{N\,m}^{-1}$) by $0.015\,\text{m}$.

  • Formula: $m g = k x \implies m = \frac{k x}{g}$

  • Calculation: $m = \frac{2.5 \times 0.015}{9.8} = 3.826\dots \times 10^{-3}\,\text{kg} = 3.8\,\text{g}$

  • Two-Mass Spring Constant Determination:

  • Scenario: A mass of $0.120\,\text{kg}$ hung on a spring results in a ground distance of $1.10\,\text{m}$. Replacing it with a mass of $0.150\,\text{kg}$ results in a ground distance of $0.980\,\text{m}$.

  • Added Mass: $\Delta m = 0.150 - 0.120 = 0.030\,\text{kg}$

  • Added Weight Force: $F = \Delta m \times g = 0.030 \times 9.8 = 0.294\,\text{N}$

  • Extension Produced: $\Delta x = 1.10 - 0.98 = 0.12\,\text{m}$

  • Spring Constant: $k = \frac{F}{\Delta x} = \frac{0.294}{0.12} = 2.45\,\text{N\,m}^{-1}$

Circular Motion and Centripetal Acceleration

• Kinematics of Circular Motion:

  • An object in uniform circular motion travels at constant speed while constantly changing its direction of motion.

  • Because velocity is defined by both speed and direction, continuous directional change requires continuous velocity change, meaning the object accelerates constantly even at constant speed.

  • Tangential Velocity: At any given instant, the object's velocity vector points along a tangent to the circular path.

• Centripetal Acceleration ($a_c$):

  • Acceleration directed toward the centre of the circular path ("centripetal" means "towards the centre").

  • Equation: $a_c = \frac{v^2}{r}$

  • $v$: Tangential velocity ($\text{m\,s}^{-1}$)

  • $r$: Radius of circular orbit or path ($\text{m}$)

• Centripetal Force ($F_c$):

  • Derived by substituting centripetal acceleration into Newton's second law ($F_{\text{net}} = m a$):

  • Equation: $F_c = \frac{m v^2}{r}$

  • $F_c$ represents the net force maintaining circular motion, always directed toward the centre of the circle.

• Vector Subtraction Analysis:

  • Change in velocity is determined via vector subtraction: $\Delta v = v_f - v_i = v_f + (-v_i)$.

  • Vector Construction: Draw $v_f$ first, draw $-v_i$ backwards from the tip of $v_f$, and complete the triangle with resultant vector $\Delta v$.

  • The resultant $\Delta v$ vector points directly toward the centre of the circle, confirming that acceleration and net force point centrally.

  • Interval Effect: As distance between measured velocity points increases, $\Delta v$ increases; however, because time interval $\Delta t$ increases proportionally, centripetal acceleration magnitude remains constant for constant speed circular motion.

• Worked Example (Hammer Throw Sport):

  • Scenario: A $7.26\,\text{kg}$ hammer is spun in a circle of radius $2.1\,\text{m}$ at a speed of $18\,\text{m\,s}^{-1}$.

  • Acceleration Calculation:     $a_c = \frac{v^2}{r} = \frac{18^2}{2.1} = \frac{324}{2.1} = 154.285\dots\,\text{m\,s}^{-2} = 150\,\text{m\,s}^{-2}$

  • Name and Meaning: Centripetal acceleration, meaning acceleration directed towards the centre of the circle.

  • Rotating Force Calculation:     $F_c = m a_c = 7.26 \times 154.285\dots = 1120.114\dots\,\text{N} = 1100\,\text{N}\text{ directed to the centre of the circle}$

Work, Power, and Energy Transfer

• Work ($W$):

  • Energy is defined as the capability to do work. Work measures energy transfer.

  • Equation: $W = F d$

  • $W$: Work done ($\text{J}$)

  • $F$: Force applied ($\text{N}$)

  • $d$: Distance travelled in direction of force ($\text{m}$)

  • Alignment Requirement: Force and displacement MUST be parallel or have a parallel component. Moving an object perpendicular to an applied force results in zero work done on that object by that force.

  • Historical Context: Named in connection with energy transfer relationships formulated around the era of James Prescott Joule (24 December 1816 – October 1889).

• Power ($P$):

  • Power is defined as the rate at which work is done, or the quantity of energy transformed per second.

  • Equation: $P = \frac{W}{t}$

  • $P$: Power output ($\text{W}$)

  • $W$: Work completed ($\text{J}$)

  • $t$: Time taken ($\text{s}$)

  • Unit Equivalence: One Watt equals one Joule of energy transformed per second ($1\,\text{W} = 1\,\text{J\,s}^{-1}$).

• Detailed Work and Power Calculations:

  • Student Lifting and Carrying Bag:

  • Part A (Vertical Lift): Student lifts a school bag upwards $1.5\,\text{m}$ using a $30\,\text{N}$ upward force.       $W = F d = 30 \times 1.5 = 45\,\text{J}$

  • Part B (Horizontal Walk): Student walks horizontally $5\,\text{m}$ while applying the same $30\,\text{N}$ upward force.       Work done on the bag = $0\,\text{J}$. The upward force is perpendicular to horizontal motion. The bag maintains constant gravitational potential energy, so no energy is transferred to the bag. Muscular effort felt by the student does not equate to energy transferred to the object.

  • Crane Operations on Heavy Load:

  • Vertical Lift: Crane lifts a $10000\,\text{kg}$ load $12\,\text{m}$ vertically at constant speed.       $W = F d = m g d = 10000 \times 9.8 \times 12 = 1176000\,\text{J} = 1200000\,\text{J}\text{ (}1.2 \times 10^6\,\text{J)}$

  • Held Stationary: Crane holds load stationary at height of $12\,\text{m}$ for several minutes.       Work done = $0\,\text{J}$ because displacement is zero ($d = 0\,\text{m}$) and no energy transformation occurs.

  • Horizontal Transport and Lowering:

    • Horizontal move of $10\,\text{m}$: Work done = $0\,\text{J}$ because initial and final kinetic/potential energies are identical (height and speed unchanged).

    • Lowering $5\,\text{m}$ vertically at constant speed: Gravitational potential energy lost by load:         $W = m g d = 10000 \times 9.8 \times 5 = 490000\,\text{J}\text{ of GPE lost}$

  • Bench Press Weightlifting:

  • Scenario: A weightlifter lifts $130\,\text{kg}$ at constant speed over $0.7\,\text{m}$ in $2.0\,\text{s}$.

  • Force Required: $F = m g = 130 \times 9.8 = 1274\,\text{N}$

  • Work Done: $W = F d = 1274 \times 0.7 = 891.8\,\text{J}$

  • Power Output: $P = \frac{W}{t} = \frac{891.8}{2.0} = 445.9\,\text{W}$

  • Outboard Motor Calculations:

  • Scenario: A $15\,\text{hp}$ outboard motor has a power output of $11000\,\text{W}$.

  • Work in $2\,\text{minutes}$ ($120\,\text{s}$):       $W = P t = 11000 \times 120 = 1320000\,\text{J} = 1300000\,\text{J}\text{ (}1.3 \times 10^6\,\text{J)}$

  • Time Required for $1\,\text{MJ}$ ($1 \times 10^6\,\text{J}$) of Work:       $t = \frac{W}{P} = \frac{1 \times 10^6}{11000} = 90.909\dots\,\text{s} = 91\,\text{s}$

  • Chin-Up Competition Comparison:

  • Dean: Mass = $100\,\text{kg}$, Lift distance = $0.70\,\text{m}$, Time = $2.0\,\text{s}$.

    • Work: $W = m g d = 100 \times 9.8 \times 0.70 = 686\,\text{J}$

    • Power: $P = \frac{W}{t} = \frac{686}{2.0} = 343\,\text{W}$

  • Russell: Mass = $85\,\text{kg}$, Lift distance = $0.80\,\text{m}$, Time = $1.8\,\text{s}$.

    • Work: $W = m g d = 85 \times 9.8 \times 0.80 = 666.4\,\text{J}$

    • Power: $P = \frac{W}{t} = \frac{666.4}{1.8} = 370.22\dots\,\text{W} = 370\,\text{W}$

  • Conclusion: Russell generates greater power output ($370\,\text{W} > 343\,\text{W}$).