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The maths you need

1. Basic Mathematical Concepts

Multiplication & Notation

  • If there is no multiplication sign (×)(×) or dot ()(∙), numbers or symbols next to each other imply multiplication.
    • Example: m₁m₂ means m1×m2.m₁ × m₂.
    • Alternative notations: m1m2m₁ ∙ m₂ or m1×m2.m₁ × m₂.

Decimals & Commas

  • Commas in numbers indicate decimal points in scientific notation.
    • Example: 4.5 means 4,5 in some calculators.
  • Avoid confusion when multiplying:
    • 4,5 × 4,5 is 20.25, not 4,5 + 4,5.

Scientific Notation & Exponents

  • Divisions often use exponents:
    • Example: 0.5ms=5×104s.0.5 ms = 5 × 10⁻⁴ s.
  • Avoid common mistakes:
    • 0.5ms0.0005s0.5 ms ≠ 0.0005 s (correct value: 0.0005s=5×104s0.0005 s = 5 × 10⁻⁴ s).

Variables & Symbols

  • A variable represents a changing quantity.
  • Scientific variables use single-letter symbols:
    • vv = velocity, aa = acceleration, FF = force.
  • Subscripts provide extra meaning:
    • v1v₁ (initial velocity), v₂ (final velocity).
  • Capital vs lowercase letters matter:
    • VV = voltage, vv= velocity.
    • R = resistance, PP = pressure.

2. Solving for an Unknown

To determine an unknown value in an equation, isolate the required variable by applying basic mathematical operations:

infoNote

Example 1: Basic Algebra

If John has 5 apples and gives Joanna 2, how many are left?

Equation: 52=x5 - 2 = x

Steps:

  1. Add x to both sides: 5=2+x5 = 2 + x

  2. Subtract 2 from both sides: 52=x5 - 2 = x

  3. Solve: x=3x = 3

Conclusion: John has 3 apples left.


infoNote

Example 2: Ohm's Law in Electricity

Using V=IR V=IR, find the current(I)current (I) if voltage(V) voltage (V) is 12 VV and resistance(R)resistance (R) is 3Ω.3Ω.

Equation: V=IRV=IR

Steps:

  1. Divide both sides by RR to isolate I:I: I=VRI = \frac{V}{R}

  2. Substitute values: I=123I = \frac{12}{3}

  3. Solve: I=4AI = 4A

Conclusion: The circuit has a current of 4A4A.


3. Step-by-Step Problem-Solving

infoNote

Example 3: Equilibrium Constant Calculation

Given: Kc=4.5[SO2]=1.5mol/dm3[O2]=0.5mol/dm3[SO3]=xKc= 4.5[SO2]=1.5 mol/dm3[O₂] = 0.5 { mol/dm}^3[SO₃] = x

Find x.

Steps:

  1. Express the equation: Kc=x2[SO2]2[O2]Kc= \frac{x^2}{[SO₂]^2 [O₂]}

  2. Substitute given values: 4.5=x2(1.5)2(0.5)4.5= \frac{x^2}{(1.5)^2 (0.5)}

  3. Simplify denominator: 4.5=x21.1254.5= \frac{x^2}{1.125}

  4. Multiply both sides by 1.125: x2=4.5×1.125x^2 = 4.5 \times 1.125

x2=5.0625x^2 = 5.0625

  1. Take square root: x=2.25x = 2.25

Conclusion: The concentration of SO3SO₃ is 2.25mol/dm32.25 mol/dm³.


Key Takeaways

✔ Use basic algebra to isolate variables. ✔ Apply scientific notation for very small or large numbers. ✔ Recognise common physics equations and their rearrangements. ✔ Break problems into step-by-step solutions for clarity. ✔ Check units and notation for accuracy.


3. Statistics

Key Definitions

  • Dependent variable: The result or effect of an experiment.
  • Independent variable: The cause or input that affects the outcome.
  • Control variable: A variable that remains constant to ensure a fair test.

Understanding Correlation

  • Correlation means two variables appear related but do not necessarily cause each other.
  • A change in one variable might be due to an external factor.
  • Graphs help us analyse these relationships.

Measures of Central Tendency

  • Mean: The average of a data set.
    • Example: In the series 1,3,5,7,9:1, 3, 5, 7, 9:
    • Mean=(1+3+5+7+9)/5=5Mean = (1 + 3 + 5 + 7 + 9) / 5 = 5
  • Median: The middle value when data is arranged in order.
    • Example: In the series 1,3,5,7,9:1, 3, 5, 7, 9:
    • Median=5Median = 5
  • Mode: The most frequently occurring value in a data set.
    • Example: In the series 1,2,2,3,3,3,4,5:1, 2, 2, 3, 3, 3, 4, 5:
    • Mode=3Mode = 3

Proportionality in Science

  • In science, proportionality helps establish relationships between variables.
  • Example: Momentum(p)=mass(m)×velocity(v)Momentum (p) = mass (m) \times velocity (v)
    • Momentum is directly proportional to both mass and velocity.
    • If either increases, so does momentum.

Constants in Equations

  • Some formulas include constants, which are fixed values used to relate variables.
  • Examples:
    • Gravitational constant (G)(G) in Newton's Law of Universal Gravitation.
    • Proportionality constant (k)(k) in various equations.

Summary

  • Statistics in physics help us understand relationships between variables.
  • Key concepts include correlation, mean, median, mode, and proportionality.
  • Constants are used in equations to standardise relationships.

4. Area of a Triangle

image

The area (AA) of a triangle is calculated as:

A=12×Base×HeightA = \frac{1}{2} \times \text{Base} \times \text{Height}

  • Example: A triangle with base=5cmbase = 5 cm and height=3cmheight = 3 cm has an area of: A=12×5×3=:highlight[7.5 cm2]A = \frac{1}{2} \times 5 \times 3 = :highlight[7.5 \text{ cm}^2]

  • This is useful for graphs of motion and acceleration.

Pythagoras' Theorem – Finding Triangle Sides

In a right-angled triangle, the hypotenuse (cc)(cc) is related to the other two sides:

c2=h2+b2c^2 = h^2 + b^2

where:

  • hh=heighthh = height
  • bb=basebb = base
  • cc=hypotenusecc = hypotenuse
infoNote

Worked Example

For a right-angled triangle with:

  • Base=5cmBase = 5 cm
  • Height=3cmHeight = 3 cm Find the hypotenuse (cc):(cc):

c2=32+52=9+25=34c^2 = 3^2 + 5^2 = 9 + 25 = 34

c=34=:success[5.83 cm]c = \sqrt{34} = :success[5.83 \text{ cm}]

Why This is Important in Physics

  • Vector Addition: When dealing with forces, knowing triangle side lengths helps determine resultant force strength and direction.
  • Graph Interpretation: Used when analysing motion and areas under curves in kinematics.

5. Trigonometry in Physics

  • Trigonometry is used to calculate unknown sides or angles in right-angled triangles.
  • The three main trigonometric ratios:
    • sinθsin θ =opposite/hypotenuse= opposite / hypotenuse
    • cosθ=adjacent/hypotenusecos θ = adjacent / hypotenuse
    • tanθ=opposite/adjacenttan θ = opposite / adjacent
  • A useful mnemonic: SOHCAHTOA
    • SOHSine=Opposite/HypotenuseSOH → Sine = Opposite/Hypotenuse
    • CAHCosine=Adjacent/HypotenuseCAH → Cosine = Adjacent/Hypotenuse
    • TOATangent=Opposite/AdjacentTOA → Tangent = Opposite/Adjacent

Key Definitions

  • Hypotenuse: The longest side in a right-angled triangle, always opposite the right angle.
  • Opposite Side: The side opposite to the given angle.
  • Adjacent Side: The side next to the given angle (but not the hypotenuse).
infoNote

Worked Example: Finding an Unknown Side

Given:

  • A right-angled triangle
  • The opposite side = 3 cm
  • The angle = 30°
  • Find the hypotenuse (H).

Solution:

Using SOHCAHTOA:

  • sinθ=opposite/hypotenusesin θ = opposite/hypotenuse sin30°=3/H→ sin 30° = 3/H H=3/sin30°H = 3 / sin 30° H=:success[6cm]→ H = :success[6 cm] Thus, the hypotenuse is 6 cm.
infoNote

Worked Example: Using Trigonometry for Astronomy

  • Earth's distance from the Sun = 149.6 million km (1 Astronomical Unit, AU).
  • If an object in space is observed at an angle of 88° from one position and 92° from another:
  • The unknown object forms a right-angled triangle with Earth and the Sun.
  • Use trigonometry to find its distance.

Solution:

Using the formula:

  • tanθ=opposite/adjacenttan θ = opposite/adjacent tan88°=H/149.6tan 88° = H / 149.6 millionmillion kmkm H=:success[4.2million]km(or:highlight[26.8AU])→ H = :success[4.2 million] km (or :highlight[26.8 AU]) Thus, the object is 4.2 million km away from the reference point.

Inverse Trigonometric Functions

  • Sometimes, you need to find the angle when two sides are known.
  • Use inverse functions:
    • sin1(opposite/hypotenuse)sin⁻¹ (opposite/hypotenuse)
    • cos1(adjacent/hypotenuse)cos⁻¹ (adjacent/hypotenuse)
    • tan1(opposite/adjacent)tan⁻¹ (opposite/adjacent)
  • Example: If opposite = 4 cm and hypotenuse = 5 cm, then: θ=sin1(4/5)=:highlight[53.1°]θ = sin⁻¹(4/5) = :highlight[53.1°]

Summary of Key Concepts

  • Use SOHCAHTOA to solve right-angled triangles.
  • Inverse trigonometry helps find unknown angles.
  • Angles and distances in space can be calculated using trigonometry.
  • Always check that your calculator is in degree mode when solving.

6. Understanding Graphs in Physics

  • Graphs are essential for interpreting motion, forces, and chemical reactions.
  • Cartesian Coordinates help locate points using:
    • x-coordinate: Left-right position.
    • y-coordinate: Up-down position.

2. Working with Coordinates

  • A point on a graph is written as an ordered pair: (x,y).(x, y).
  • Positive values:
    • Right of the yaxisy-axis (x>0).(x > 0).
    • Above the xaxisx-axis (y>0).(y > 0).
  • Negative values:
    • Left of the yaxisy-axis (x<0).(x < 0).
    • Below the xaxisx-axis (y<0).(y < 0).

3. Identifying Linear Relationships

  • A straight-line graph represents a direct relationship.
  • The general equation of a straight line:

where: y=mx+cy=mx+c

  • m=gradientm = gradient (slope).(slope).
  • c = y-$$intercept.

4. Gradient (Slope) of a Graph

  • Gradient(m)Gradient (m) shows how steep the line is.

  • Formula: m=change in ychange in x=y2y1x2x1m = \frac{\text{change in } y}{\text{change in } x} = \frac{y_2 - y_1}{x_2 - x_1}

  • Interpreting the gradient:

    • Positive gradient: Line slopes upward (y increases as x increases).
    • Negative gradient: Line slopes downward (y decreases as x increases).
    • Zero gradient: Horizontal line (constant value).

5. Common Graphs in Physical Sciences

  • Directly Proportional Graph:

    • Passes through the origin (0,0).(0,0).
    • Equation: , where kk is a constant. y=kxy = kx
  • Inverse Proportion Graph:

    • Equation: . y=kxy = \frac{k}{x}

    • As x increases, y decreases (curved graph).

  • Quadratic Graph:

    • Equation: . y=ax2+bx+cy = ax^2 + bx + c

    • Produces a parabola.

6. Application in Physics

  • Velocity-Time Graph:
    • Gradient = acceleration.
    • Area under the graph = displacement.
  • Position-Time Graph:
    • Gradient = velocity.
  • Force vs. Acceleration Graph:
    • Gradient = mass of the object (from Newton's Second Law: ). F=maF = ma

7. Understanding Circles in Physics

image
  • A circle is a two-dimensional shape with a curved boundary.
  • Important parts of a circle:
    • DiameterDiameter (d)(d): The widest part of a circle, passing through the centre.

    • RadiusRadius (r)(r): Half of the diameter. Formula: . r=d2r = \frac{d}{2}

    • CircumferenceCircumference (C)(C): The perimeter (outer boundary) of the circle. Formula: C=2πrC = 2\pi r

    • AreaArea (A)(A): The space inside a circle. Formula: A=πr2A = \pi r^2

2. Key Concepts for Physics

  • Understanding "circumference":
    • Comes from the Latin word meaning "to carry in a circle".
    • Think of how planets orbit the Sun, or how the Earth moves around its axis.
  • Applications in Physical Sciences:
    • Used in calculations for circular motion (e.g. planetary orbits, wheels, and gears).
    • Important for formulas involving rotational motion and wave properties.
    • Helps in understanding the relationship between distance, speed, and time in circular motion.

3. Using Circle Formulas in Exams

  • You may need to calculate radius, diameter, circumference, or area using the formulas provided.
  • Tip: These formulas are usually given in the exam, so focus on knowing how to apply them rather than memorising.
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