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IGCSE Physics: Units, Forces, Weight, and Energy

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Section 1

IGCSE Physics: Units, Forces, Weight, and Energy

STUDY GUIDE

๐ŸŽ“ IGCSE Physics (CAIE & Edexcel) - Study Guide

๐Ÿ“– Chapter 1: Units, Prefixes, and Forces

๐Ÿ”‘ Essential Concepts & Formulas

Concept/FormulaDefinition/EquationWhen to Use
Unit ConversionMultiply or divide by conversion factorConverting between units (km to m)
Standard Formaร—10na \times 10^n, where 1โ‰คa<101 \leq a < 10Representing very large/small numbers
Resultant ForceVector sum of all forcesFinding net force on an object
Balanced ForcesResultant force = 0Object at constant velocity

๐Ÿ› ๏ธ Problem Types

Type A: Unit Conversion

Setup: "When you see a measurement in one unit and need it in another."

Method: Identify conversion factor, multiply or divide accordingly.

Example: Convert 5 km to meters. 5 km * 1000 m/km = 5000 m

Type B: Resultant Force Calculation

Setup: "If given multiple forces acting on an object."

Method: Add forces as vectors, accounting for direction.

Example: 5 N right, 2 N left. Resultant = 3 N right.

๐Ÿงฎ Solved Example

Problem: Convert 0.0000025 seconds to microseconds. Steps:

  1. Identify the conversion factor: 1 microsecond = 10โˆ’610^{-6} seconds.
  2. Divide: 0.0000025/10โˆ’6=2.50.0000025 / 10^{-6} = 2.5 microseconds.
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Answer: 2.5 microseconds
Section 2

๐Ÿ“– Chapter 2: Weight, Work Done, and Hooke's Law

๐Ÿ”‘ Essential Concepts & Formulas

Concept/FormulaDefinition/EquationWhen to Use
WeightW=mgW = mgCalculating force due to gravity
Work DoneW=FdW = FdCalculating energy transferred by a force
GPEGPE=mghGPE = mghCalculating gravitational potential energy
Hooke's LawF=keF = keCalculating force in a spring
Elastic Potential EnergyE=12ke2E = \frac{1}{2}ke^2Calculating energy stored in a spring

๐Ÿ› ๏ธ Problem Types

Type A: Weight Calculation

Setup: "Given mass and gravitational field strength."

Method: Use W=mgW = mg.

Example: Mass = 5 kg, g = 9.8 N/kg. W = 5 kg * 9.8 N/kg = 49 N

Type B: Work Done Calculation

Setup: "Given force and distance."

Method: Use W=FdW = Fd.

Example: Force = 10 N, distance = 2 m. W = 10 N * 2 m = 20 J

๐Ÿงฎ Solved Example

Problem: A spring with spring constant 200 N/m is extended by 0.1 m. Calculate the elastic potential energy stored. Steps:

  1. Use the formula: E=12ke2E = \frac{1}{2}ke^2
  2. Substitute: E=12ร—200ร—(0.1)2=1JE = \frac{1}{2} \times 200 \times (0.1)^2 = 1 J
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Answer: 1 J

๐Ÿ“– Chapter 3: Moments, Speed, Velocity, and Acceleration

๐Ÿ”‘ Essential Concepts & Formulas

Concept/FormulaDefinition/EquationWhen to Use
MomentM=FdM = FdCalculating turning force
Speedv=dtv = \frac{d}{t}Calculating rate of distance covered
Accelerationa=vโˆ’uta = \frac{v-u}{t}Calculating rate of change of velocity
SUVAT Equation 1v=u+atv = u + atFinding final velocity
SUVAT Equation 2s=ut+12at2s = ut + \frac{1}{2}at^2Finding displacement

๐Ÿ› ๏ธ Problem Types

Type A: Moment Calculation

Setup: "Given force and perpendicular distance to pivot."

Method: Use M=FdM = Fd.

Example: Force = 5 N, distance = 0.2 m. M = 5 N * 0.2 m = 1 Nm

Type B: Acceleration Calculation

Setup: "Given initial velocity, final velocity, and time."

Method: Use a=vโˆ’uta = \frac{v-u}{t}.

Example: u = 2 m/s, v = 8 m/s, t = 3 s. a = (8-2)/3 = 2 m/s2^2

๐Ÿงฎ Solved Example

Problem: A car accelerates from rest to 20 m/s in 5 seconds. Calculate the acceleration. Steps:

  1. Use the formula: a=vโˆ’uta = \frac{v-u}{t}
  2. Substitute: a=20โˆ’05=4m/s2a = \frac{20-0}{5} = 4 m/s^2
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Answer: 4 m/s2^2

๐Ÿ“– Chapter 4: Newton's Laws of Motion and Momentum

๐Ÿ”‘ Essential Concepts & Formulas

Concept/FormulaDefinition/EquationWhen to Use
Newton's 2nd LawF=maF = maCalculating force, mass, or acceleration
Momentump=mvp = mvCalculating momentum of an object
Conservation of Momentumm1u1+m2u2=m1v1+m2v2m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2Analyzing collisions
Force (Momentum Change)F=ฮ”pฮ”tF = \frac{\Delta p}{\Delta t}Calculating force from momentum change

๐Ÿ› ๏ธ Problem Types

Type A: Force Calculation (Newton's 2nd Law)

Setup: "Given mass and acceleration."

Method: Use F=maF = ma.

Example: Mass = 2 kg, acceleration = 3 m/s2^2. F = 2 kg * 3 m/s2^2 = 6 N

Type B: Momentum Calculation

Setup: "Given mass and velocity."

Method: Use p=mvp = mv.

Example: Mass = 3 kg, velocity = 4 m/s. p = 3 kg * 4 m/s = 12 kg m/s

๐Ÿงฎ Solved Example

Problem: A 2 kg object moving at 5 m/s collides with a stationary 1 kg object. After the collision, the 2 kg object moves at 2 m/s. What is the velocity of the 1 kg object? Steps:

  1. Use conservation of momentum: m1u1+m2u2=m1v1+m2v2m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2
  2. Substitute: (2ร—5)+(1ร—0)=(2ร—2)+(1ร—v2)(2 \times 5) + (1 \times 0) = (2 \times 2) + (1 \times v_2)
  3. Solve: 10=4+v210 = 4 + v_2, so v2=6m/sv_2 = 6 m/s
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Answer: 6 m/s

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