Combined Physics: Motion & Forces

Prerequisite Mathematical Skills for Physics

• Rearranging Algebraic Equations:

  • To isolate an unknown variable on one side of an equation, apply the inverse (opposite) mathematical operation to both sides of the equation.
  • Multiplication and division are inverse operations; addition and subtraction are inverse operations; squaring and taking the square root are inverse operations.
  • Worked Example 1 (Direct Multiplication):
    • Equation: $a = b \times c$
    • Given values: $a = 2$, $c = 4$
    • Substitution: $2 = b \times 4$
    • Rearrangement: Divide both sides by $4$ to make $b$ the subject:             $b = \frac{a}{c} = \frac{2}{4} = 0.5$
  • Worked Example 2 (Power Terms):
    • Equation: $a = b^2 \times c$
    • Given values: $a = 16$, $c = 4$
    • Substitution: $16 = b^2 \times 4$
    • Rearrangement: Divide both sides by $4$:             $b^2 = \frac{16}{4} = 4$
    • Square root both sides to solve for $b$:             $b = \sqrt{4} = 2$
  • Worked Example 3 (Fractions/Division):
    • Equation: $a = \frac{b}{c}$
    • Given values: $a = 2$, $c = 4$
    • Rearrangement: Multiply both sides by $c$:             $b = a \times c = 2 \times 4 = 8$
  • Worked Example 4 (Three Variables):
    • Equation: $z = \frac{x}{y}$
    • Given values: $z = 5$, $x = 15$
    • Rearrangement: Multiply both sides by $y$ ($z \times y = x$), then divide both sides by $z$:             $y = \frac{x}{z} = \frac{15}{5} = 3$
  • Worked Example 5 (Complex Expression):
    • Equation: $a = \frac{b \times c}{d}$
    • Given values: $a = 2$, $c = 4$, $d = 4$
    • Rearrangement: Multiply both sides by $d$ and divide by $c$:             $b = \frac{a \times d}{c} = \frac{2 \times 4}{4} = 2$

• Operations with Positive and Negative Numbers:

  • Two identical adjacent signs result in addition: two negative signs make a positive ($-(-a) = +a$). Example: $5 - (-3) = 5 + 3 = 8$.
  • Two differing adjacent signs result in subtraction: a positive and a negative sign make a negative ($+(-a) = -a$). Example: $7 + (-3) = 7 - 3 = 4$.
  • Calculation with signed numbers:
    • $-3 + (-3) = -3 - 3 = -6$
    • $3 - (-3) = 3 + 3 = 6$
    • $-2 + (-1) = -2 - 1 = -3$
    • $-5 - (-3) = -5 + 3 = -2$

• Rules for Significant Figures ($\text{s.f.}$):

  • In scientific calculations, values are rounded to a specific number of significant figures rather than decimal places.
  • Rule 1: All non-zero digits are always significant.
    • $123$ contains $3\,\text{s.f.}$
    • $1.78$ contains $3\,\text{s.f.}$
  • Rule 2: Any zeros located between two non-zero significant digits are significant.
    • $108$ contains $3\,\text{s.f.}$
    • $10003$ contains $5\,\text{s.f.}$
    • $1.006$ contains $4\,\text{s.f.}$
    • $10000001$ contains $8\,\text{s.f.}$
  • Rule 3: Leading zeros (zeros preceding non-zero digits) are not significant; they serve only as placeholders.
    • $0.183$ contains $3\,\text{s.f.}$ (digits $1, 8, 3$)
    • $0.012$ contains $2\,\text{s.f.}$ (digits $1, 2$)
  • Rule 4: Trailing zeros in a whole number without a decimal point are not significant.
    • $1390$ contains $3\,\text{s.f.}$ (digits $1, 3, 9$)
    • $23120$ contains $4\,\text{s.f.}$ (digits $2, 3, 1, 2$)
  • Rule 5: Trailing zeros in the decimal portion (after the decimal point) are significant.
    • $1.40$ contains $3\,\text{s.f.}$ (digits $1, 4, 0$)
    • $1.420$ contains $4\,\text{s.f.}$ (digits $1, 4, 2, 0$)
  • Rounding Examples:
    • $186.21$ rounded to $2\,\text{s.f.}$ is $190$
    • $0.0291444$ rounded to $3\,\text{s.f.}$ is $0.0291$
    • $0.0400156$ rounded to $4\,\text{s.f.}$ is $0.04002$

• Standard Form (Scientific Notation):

  • Numbers written in standard form take the structure: $a \times 10^n$
  • Condition 1: The coefficient $a$ must satisfy $1 \le a < 10$.
  • Condition 2: The exponent $n$ is a positive integer ($n > 0$) for numbers greater than or equal to $10$, representing how many times $a$ is multiplied by $10$.
    • $3 \times 10^8 = 300\,000\,000$ ($3$ multiplied by $10$ eight times)
  • Condition 3: The exponent $n$ is a negative integer ($n < 0$) for numbers strictly between $0$ and $1$, representing how many times $a$ is divided by $10$.
    • $2 \times 10^{-5} = 0.00002$ ($2$ divided by $10$ five times)
  • Rounding in Standard Form:
    • When rounding a value in standard form, only the coefficient $a$ is rounded; the factor $10^n$ does not affect significant figure count.
    • $5.18 \times 10^6$ rounded to $2\,\text{s.f.}$ is $5.2 \times 10^6$
    • $143\,000\,000$ in standard form to $2\,\text{s.f.}$ is $1.4 \times 10^8$
    • $0.00000005366$ in standard form to $3\,\text{s.f.}$ is $5.37 \times 10^{-8}$
    • $736\,000$ in standard form to $2\,\text{s.f.}$ is $7.4 \times 10^5$
    • $0.00000767$ in standard form is $7.67 \times 10^{-6}$
    • $0.000000555$ in standard form to $1\,\text{s.f.}$ is $6 \times 10^{-7}$
    • $3.445 \times 10^9$ written as an ordinary number is $3\,445\,000\,000$ and contains $4\,\text{s.f.}$

Scalar and Vector Quantities

• Definitions:

  • Magnitude: The numerical size, amount, or extent of a physical quantity.
  • Scalar Quantity: A physical quantity completely described by magnitude alone, lacking any directional component.
    • Examples: Mass, distance, distance travelled, speed, energy, time, temperature, density.
  • Vector Quantity: A physical quantity completely described by both a magnitude and a specific direction.
    • Examples: Displacement, velocity, acceleration, force, weight, momentum.

• Distance versus Displacement:

  • Distance / Distance Travelled: A scalar quantity describing the total path length traversed by an object between two positions, regardless of directional changes.
  • Displacement: A vector quantity defined as the straight-line distance moved in a specified direction from an initial starting point to a final endpoint.
  • Track Runner Example:
    • A runner completes exactly one full lap around a $400\,m$ running track and returns to the initial starting line.
    • $\text{Distance travelled} = 400\,m$
    • $\text{Displacement} = 0\,m$ (since the final position coincides with the initial starting position).
  • Mountain Road Geographical Scenario:
    • The linear distance from Southvale to Northwold across an intervening mountain is $3\,\text{miles}$ North (displacement).
    • The winding road around the mountain between both towns measures $7\,\text{miles}$ long (distance travelled).
    • A car driver navigating the road requires knowledge of the distance travelled ($7\,\text{miles}$) to estimate fuel consumption and journey duration.
    • A civil engineer planning a straight road tunnel through the mountain requires knowledge of the displacement ($3\,\text{miles}$ North) to direct excavation.

• Graphical Representation of Vectors:

  • Vectors are drawn as straight arrows:
    • The length of the arrow is scaled proportionally to represent the vector's magnitude.
    • The orientation and arrowhead indicate the vector's direction of action.
  • Directional Sign Conventions on a Number Line:
    • Vectors directed horizontally to the right are conventionally assigned a positive value ($+$, e.g., $+4$).
    • Vectors directed horizontally to the left are assigned a negative value ($-$, e.g., $-3$).
    • Vectors directed vertically upwards are assigned a positive value ($+$).
    • Vectors directed vertically downwards are assigned a negative value ($-$).

• Resultant Vectors in One Dimension (Collinear Vectors):
  • A resultant vector is a single vector quantity whose magnitude and direction equal the combined effect of two or more component vectors acting simultaneously at a single point.
  • When vectors act along the same parallel line (collinear):
    • Vectors pointing in identical directions are added algebraically.
    • Vectors pointing in opposite directions are subtracted algebraically.
  • Collinear Vector Calculation Examples:
    • Two horizontal vectors acting rightwards: $(+3) + (+7) = +10$
    • Two opposing horizontal vectors: $(+10) + (-3) = +7$
    • Two upward vertical vectors: $(+4) + (+4) = +8$
    • Opposing vectors: $(-6) + (+3) = -3$
    • Opposing vectors: $(+3) + (-1) = +2$
    • Two vectors of value $+6$ acting in the same direction yield a resultant vector of $+12$.

Vector Diagrams and Resolution of Forces (Higher Tier)

• Tip-to-Tail Vector Diagrams:

  • A graphical method used to combine two or more vectors acting in non-parallel directions into a single resultant vector.
  • The tail of the second vector arrow is positioned directly at the tip (arrowhead) of the first vector arrow. The resultant vector is drawn from the tail of the first vector to the tip of the final vector.

• Perpendicular Vectors and Pythagoras' Theorem:

  • When two vectors act at right angles ($90^\circ$) to one another, they form the adjacent and opposite sides of a right-angled triangle.
  • The resultant vector represents the hypotenuse ($c$) of the triangle.
  • Pythagoras' Theorem is formulated as:         $c^2 = a^2 + b^2$$c = \sqrt{a^2 + b^2}$

• Perpendicular Displacement Examples:

  • An object moves from position A by travelling $4\,m$ East (right), followed by $3\,m$ North (up) to reach position B:         $\text{Displacement} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5\,m$
  • Two perpendicular components of $6\,m$ and $8\,m$:         $c = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\,m$
  • Two perpendicular components of $5\,m$ and $2\,m$:         $c = \sqrt{5^2 + 2^2} = \sqrt{25 + 4} = \sqrt{29} \approx 5.39\,m$
  • Two perpendicular sides of $9\,m$ and $4\,m$:         $c = \sqrt{9^2 + 4^2} = \sqrt{81 + 16} = \sqrt{97} \approx 9.85\,m$
  • Two perpendicular forces of $11\,N$ and $14\,N$:         $\text{Resultant} = \sqrt{11^2 + 14^2} = \sqrt{121 + 196} = \sqrt{317} \approx 17.80\,N$

• Representing Vectors with Specified Angles:

  • Vectors acting at an angle to the horizontal require both a defined angle (measured using a protractor) and a proportional arrow length.
  • Tennis Ball Example: Two tennis balls are thrown at an angle of $45^\circ$ above the horizontal; Ball 1 travels at a velocity of $5\,m/s$, and Ball 2 travels at $10\,m/s$. Both arrows point in the identical angular direction ($45^\circ$), but the vector arrow for Ball 2 is precisely twice as long as Ball 1.

• Scaled Vector Drawings for Non-Perpendicular Vectors:
  • When vectors act at non-right angles, their resultant cannot be determined by simple right-triangle trigonometry and must be found using scale drawings.
  • Step-by-Step Procedure:
    • Step 1: Establish and state an appropriate scale (e.g., $1\,cm = 1\,kN$ or $1\,cm = 10\,m$).
    • Step 2: Connect the component vectors head-to-tail maintaining their exact angular bearings.
    • Step 3: Draw the resultant vector connecting the start of the first vector to the end of the last.
    • Step 4: Measure the resultant line with a ruler to find its physical length.
    • Step 5: Convert the ruler measurement back into the real physical units using the scale factor.
    • Step 6: Measure the angle of the resultant vector using a protractor (referencing North for navigation bearings or the horizontal axis for forces).

• Scale Conversion Exercises:

  • Scale $2\,cm = 5\,km$:
    • $4\,cm = 10\,km$
    • $12\,cm = 30\,km$
    • $5\,cm = 12.5\,km$
  • Scale $3\,cm = 3\,km$ ($1\,cm = 1\,km$):
    • $6\,cm = 6\,km$
    • $15\,cm = 15\,km$
    • $4.5\,cm = 4.5\,km$
  • Scale $2.5\,cm = 5\,km$ ($1\,cm = 2\,km$):
    • $7.5\,cm = 15\,km$

• Worked Case Study: Rhinos Pushing a Vehicle:

  • Two rhinos push against a vehicle at right angles: Rhino 1 exerts a forward force of $7.0\,kN$; Rhino 2 exerts a perpendicular lateral force of $5.0\,kN$.
  • Scale chosen: $1\,cm = 1.0\,kN$.
  • Vector 1 is drawn $7.0\,cm$ horizontally; Vector 2 is drawn $5.0\,cm$ vertically from the tip of Vector 1.
  • The resultant arrow connecting the origin to the tip measures $8.6\,cm$.
  • Resultant force: $8.6\,kN$

• Parallelogram of Forces Method:
  • When two forces act outward from a common point at non-right angles, construct a parallelogram:
    • Draw both force arrows originating from the common origin at their correct relative angles and lengths.
    • Construct parallel projection lines from the head of each vector parallel to the opposite vector.
    • Draw the diagonal of the parallelogram starting from the origin to the opposite intersecting vertex; this diagonal represents the resultant force.

• Worked Aircraft Case Study:
  • An aeroplane moving along a runway experiences an engine thrust of $50\,kN$ forward and rightward, while experiencing a crosswind force of $15\,kN$ backward and leftward.
  • Scale: $1\,\text{grid square} = 10\,kN$.
  • A parallelogram constructed using the $50\,kN$ vector and $15\,kN$ vector produces a horizontal resultant forward force aligned directly along the runway centreline.

![Resultant force on an aeroplane using parallelogram method](https://assets.knowt.com/pdf-flow-prod/199786b6-8bc3-4cb