Cambridge IGCSE Physics Motion Forces and Energy Revision Guide

Nature of Science, Physical Quantities, and Forces

• Cambridge Core Definitions

  • Scalar Quantity: A quantity that possesses magnitude (size) only. Examples include:
    • Mass
    • Distance
    • Speed
    • Time
    • Energy
  • Vector Quantity: A quantity that possesses both magnitude and direction. Examples include:
    • Displacement
    • Velocity
    • Acceleration
    • Force
    • Weight
    • Momentum

• Scientific Laws vs. Theories

  • Scientific Law: A concise mathematical statement describing what happens under specific physical conditions. An example is Newton's Law of Universal Gravitation:
    • $F = \frac{G \times m_1 \times m_2}{r^2}$
  • Scientific Theory: A comprehensive, empirically verified explanation of why phenomena occur. An example is Einstein's Theory of General Relativity.

• Classification of Forces

  • Forces are pushes or pulls measured in Newtons ($N$) using a spring balance or vector quantity rules.
  • Contact Forces:
    • Friction ($F_f$): The force that opposes motion between two surfaces sliding past each other.
    • Air Resistance / Drag ($F_{air}$): Friction exerted by fluid particles on a moving body.
    • Normal Contact Force ($R$ or $N$): The perpendicular reaction force exerted by a solid surface.
    • Tension ($T$): The pulling force transmitted through a stretched string, cable, or spring.
  • Non-Contact Forces:
    • Gravitational Force / Weight ($W$): The downward force of gravity acting on a mass, defined by the formula:
      • $W = m \times g$
    • Electrostatic Force: The attraction or repulsion between stationary electric charges.
    • Magnetic Force: The attraction or repulsion between magnetic poles or moving charges.

• Four Primary Effects of Forces

  • An unbalanced force acting on an object can cause a change in:
    1. Speed: Speeding up (acceleration) or slowing down (deceleration/retardation).
    2. Direction: Changing the path of motion, such as circular or curved motion.
    3. Shape / Size: Causing elastic or plastic deformation. This is described by Hooke's Law:
       • $F = k \times x$

Resultant Force and Friction Reduction

• Resultant (Net) Force Calculation ($F_{net} = \Sigma F$)

  • Parallel / Same Direction: $F_{net} = F_1 + F_2$
  • Opposite Directions: $F_{net} = F_{large} - F_{small}$ (the resultant force acts in the direction of the larger force).
  • Perpendicular Vectors: $F_{net} = \sqrt{F_1^2 + F_2^2}$
    • The angle is calculated as $\theta = \tan^{-1}(\frac{F_2}{F_1})$

• Methods of Reducing Friction

  • Lubrication: Interposing oil, grease, or fluid film between sliding solid surfaces.
  • Rolling Bearings: Utilizing ball or roller bearings to convert sliding friction into rolling resistance.
  • Streamlining: Designing teardrop or aerodynamic profiles to reduce fluid air resistance/drag.
  • Polishing: Reducing microscopic surface asperities.

• Terminal Velocity Step-by-Step Sequence

  • For an object falling in a uniform gravitational field with air resistance:
    1. At Release ($t = 0$): Velocity $v = 0$, and Drag $F_{air} = 0$. Therefore, Net force $F_{net} = Weight (W) = m \times g$. Downward acceleration is at its maximum ($a = g = 9.8\,m/s^2$ or $10\,m/s^2$).
    2. Accelerating Phase: As velocity ($v$) increases, air resistance increases ($F_{air} \propto v^2$). As a result, the net force decreases ($F_{net} = W - F_{air}$), which causes the downward acceleration ($a$) to decrease.
    3. Terminal Velocity Achieved: Eventually, $F_{air} = W$. Thus, $F_{net} = 0\,N$, acceleration drops to $a = 0\,m/s^2$, and velocity remains constant.

Kinematics and Graph Analysis

• Speed, Velocity, and Acceleration Formulas

  • $v = \frac{d}{t}$
  • $v_{avg} = \frac{\Delta x}{\Delta t}$
  • $a = \frac{v - u}{t}$
  • Variables:
    • $v$ = final speed/velocity ($m/s$)
    • $u$ = initial velocity ($m/s$)
    • $d, \Delta x$ = distance/displacement ($m$)
    • $t$ = time taken ($s$)
    • $a$ = acceleration ($m/s^2$)

• Average vs. Instantaneous Speed

  • Average Speed: The total distance covered divided by the total time taken ($v_{avg} = \frac{d_{total}}{t_{total}}$).
  • Instantaneous Speed: The speed at a specific moment in time, given by a speedometer or the gradient of a tangent on a graph.

• Interpretation of Motion Graphs

  • Distance–Time Graph:
    • Slope / Gradient: Represents Speed ($v = \frac{\Delta d}{\Delta t}$).
    • Area under graph: No physical meaning.
  • Speed–Time Graph:
    • Slope / Gradient: Represents Acceleration ($a = \frac{\Delta v}{\Delta t}$).
    • Area under graph: Represents Total Distance Traveled ($d = Area$).

• Cambridge Exam Tips for Speed-Time Graphs

  • Horizontal line = constant speed.
  • Constant straight slope = uniform acceleration.
  • Curving line upwards = increasing acceleration.
  • The area under the speed-time curve equals the distance.

Newtonian Dynamics and Bridge Engineering

• Newton's Laws of Motion

  1. 1st Law: An object remains at rest or moves with constant velocity unless acted upon by a resultant force ($F_{net} \neq 0$).
  2. 2nd Law: $F_{net} = m \times a$. Acceleration is directly proportional to resultant force and inversely proportional to mass.
  3. 3rd Law: When body A exerts a force on body B, body B exerts an equal and opposite force on body A.

• Structural Mechanics in Bridge Engineering

  • Static Equilibrium: For structural stability, two criteria must be satisfied:
    1. No resultant force: $\Sigma F = 0$
    2. No resultant moment (Principle of Moments): $\Sigma M = 0$ (Clockwise Moments = Anticlockwise Moments).
  • Suspension Bridges: Main cables are subjected to extreme tension forces, which transfer load to vertical towers that are under compression.
  • Arch Bridges: Designed to push outwards at the abutments, keeping structural stones or steel under direct compression.
  • Truss Bridges: Triangular arrangements resolve external loads into axial tension and compression forces.

Energy, Work, and Mechanical Power

• Law of Conservation of Energy: Energy cannot be created or destroyed, only transferred from one store to another (e.g., potential, kinetic, thermal, electrical, radiation).

• Gravitational Potential Energy

  • $\Delta E_p = m \times g \times \Delta h$
  • $\Delta E_p$ = change in potential energy ($J$)
  • $m$ = mass ($kg$)
  • $g$ = gravitational field strength ($9.8$ or $10\,N/kg$)
  • $\Delta h$ = height change ($m$)

• Kinetic Energy

  • $E_k = \frac{1}{2} \times m \times v^2$
  • $E_k$ = kinetic energy ($J$)
  • $v$ = velocity ($m/s$)

• Work Done and Mechanical Power

  • $W = F \times d$
  • $P = \frac{W}{t} = \frac{\Delta E}{t}$
  • $W$ = Work done (Joules, $J$ or $N \cdot m$)
  • $F$ = Force ($N$)
  • $d$ = distance moved in the direction of the force ($m$)
  • $P$ = Power (Watts, $W$ or $J/s$)

• Conservation Mechanics (Free-Fall Drop)

  • Assuming negligible air drag resistance during a free-fall drop:
    • $E_p(top) = E_k(bottom)$
    • $m \times g \times h = \frac{1}{2} \times m \times v^2$
    • $v = \sqrt{2 \times g \times h}$

Worked Cambridge Exam-Style Question

• Worked Example 1: Kinematics and Energy Conservation
  • Question: A roller coaster car of mass $m = 500\,kg$ starts from rest at height $h = 20\,m$. Taking $g = 9.8\,m/s^2$ and ignoring friction:
    • (a) Calculate the gravitational potential energy of the car at the top.
    • (b) Determine the maximum speed of the car at the bottom of the slope.
    • (c) If the car is brought to rest at the station in $t = 4.0\,s$ over a distance of $40\,m$, calculate the braking force required.
  • Solution (a): Potential Energy Calculation
    • $E_p = m \times g \times h = 500\,kg \times 9.8\,m/s^2 \times 20\,m = 98,000\,J = 98\,kJ$
  • Solution (b): Velocity at Bottom
    • By Conservation of Energy: $E_k = E_p = 98,000\,J$
    • $\frac{1}{2} \times m \times v^2 = 98,000$
    • $\frac{1}{2} \times (500) \times v^2 = 98,000$
    • $250 \times v^2 = 98,000$
    • $v^2 = 392$
    • $v = \sqrt{392} \approx 19.80\,m/s$
  • Solution (c): Braking Force and Power
    • Work done by brakes = Initial Kinetic Energy = $98,000\,J$
    • $W = F \times d$
    • $98,000\,J = F \times 40\,m$
    • $F = \frac{98,000}{40} = 2,450\,N$
    • Braking Power: $P = \frac{W}{t} = \frac{98,000\,J}{4.0\,s} = 24,500\,W = 24.5\,kW$