Space, time and motion
 Module 4: Understanding processes
4.2 Space, time and motion   
4.2 a)   Describe and explain:
I)                   The use of vectors to represent displacement, velocity and acceleration
II)                 The trajectory of a body moving under constant acceleration, in one or two dimensions
III)              The independent effect of perpendicular components of a force
IV)              Calculation of work done, including cases where the force is not parallel to the displacement
V)                The principle of conservation of energy
VI)              Power as rate of transfer of energy
VII)           Measurement of displacement, velocity and acceleration
VIII)         Newton’s laws of motion
IX)              The principle of conservation of momentum; Newton’s Third Law as a consequence.
b)   Make appropriate use of:
I)                   The terms: displacement, speed, velocity, acceleration, force, mass, vector, scalar, work, energy, power, momentum, impulse
by sketching and interpreting:
II)                 Graphs of accelerated motion; slope of displacement–time and velocity–time graphs; area underneath the line of a velocity–time graph
III)              Graphical representation of addition of vectors and changes in vector magnitude and direction.
c)    Make calculations and estimates involving:
I)                   The resolution of a vector into two components at right angles to each other
II)                 The addition of two vectors, graphically and algebraically
III)              The kinematic equations for constant acceleration derivable from: [math]a = \frac{v – u}{t}[/math] and average velocity [math]=\frac{v + u}{2}; \quad v = u + at, \quad s = ut + \frac{1}{2} at^2, \quad v^2 = u^2 + 2as[/math]
IV)              Momentum p = mv
V)                The equation [math]F = ma = \frac{∆(mv)}{∆t} [/math]  Where the mass is constant
VI)              The principle of conservation of momentum
VII)            Work done ∆E = F∆t
VIII)          Kinetic energy = [math]\frac{1}{2} mv^2[/math]
IX)            Gravitational potential energy = mgh
X)                Force, energy and power:
Power = [math]\frac{∆E}{t}[/math],power = Fv.
XI)              Modelling changes of displacement and velocity in small discrete time steps, using a computational model or graphical representation of displacement and velocity vectors.
d)   Demonstrate and apply knowledge and understanding of the following practical activities (HSW4):
I)                   Investigating the motion and collisions of objects using trolleys, air-track gliders etc. with data obtained from ticker timers, light gates, data-loggers and video techniques
II)                 Determining the acceleration of free fall, using trapdoor and electromagnet arrangement, light gates or video technique
III)              Investigating terminal velocity with experiments such as dropping a ball-bearing in a viscous liquid or dropping paper cones in air

  • a) Detailed Description and Explanation

  • I. The Use of Vectors to Represent Displacement, Velocity, and Acceleration

  • Displacement:
  • – A vector that represents the shortest distance and direction between two points.
  • Example: Moving 5 m northeast can be represented as a vector with magnitude 5 m and a specific angle.
  • Velocity:
  • – A vector representing the rate of change of displacement with time.
  • Example:
  • [math]v = \frac{∆s}{∆t}[/math]
  • Where s is displacement.
  • Acceleration:
  • – A vector representing the rate of change of velocity with time.
  • Example:
  • [math]v = \frac{∆s}{∆t}[/math]
  • Vectors are added geometrically or resolved into perpendicular components to analyze motion.

  • II. The Trajectory of a Body Moving Under Constant Acceleration

  • 1D Motion:
  • – The equations of motion are:
  • [math]v = u + at, \quad s = ut + \frac{1}{2} at^2, \quad v^2 = u^2 + 2as[/math]
  • Where u is initial velocity, v is final velocity, a is acceleration, t is time, and s is displacement.
  • 2D Motion (Projectile Motion):
  • – Horizontal motion: Constant velocity ([math]a_x = 0[/math]).
  • – Vertical motion: Acceleration due to gravity ([math]a_y = -g[/math]).
  • – Trajectory equation: The path of a projectile is parabolic and can be expressed as:
  • [math]y = x \tan\theta – \frac{g x^2}{2u^2 \cos^2\theta}[/math]
  • Where u is the initial velocity, θ is the angle of projection.
  • Figure 1 Projectile motion

  • III. The Independent Effect of Perpendicular Components of a Force

  • A force can be resolved into perpendicular components (e.g., horizontal and vertical).
  • These components act independently and determine motion in their respective directions.
  • – Example: For a force F at an angle θ:
  • – Horizontal component:[math]F_x = F cos⁡θ[/math]
  • – Vertical component:[math]F_y = F sin⁡θ[/math]

  • IV. Calculation of Work Done

  • Definition:
  • – Work is done when a force causes displacement.
  • [math]W = F.d \\ W = Fd cos⁡θ[/math]
  • Where θ is the angle between the force and displacement.
  • Non-parallel Forces: If force and displacement are not parallel, only the component of force along displacement contributes to work.
  • – Example: Pulling an object with a force F at 30° to the horizontal over a displacement d:
  • [math]W = Fd cos 30^0[/math]

  • V. Principle of Conservation of Energy:

  • Definition:
  • – Energy cannot be created or destroyed; it can only be transformed or transferred.
  • Forms of Energy:
  • – Kinetic, potential, thermal, etc.
  • – Total energy in a system remains constant:
  • [math]\text{Total Energy} = \text{Kinetic Energy} + \text{Potential Energy}[/math]
  • Example: A pendulum exchanges kinetic and potential energy during motion.

  • VI. Power as Rate of Transfer of Energy

  • Definition:
  • – Power (P) is the rate at which work is done or energy is transferred.
  • [math]P = \frac{W}{t} \\ P = Fv[/math]
  • Where v is velocity.
  • Example:
  • – A car engine delivering 1000 J of work in 5s:
  • [math]P = \frac{1000}{5} \\ P = 200 W [/math]

  • VII. Measurement of Displacement, Velocity, and Acceleration

  • Displacement:
  • – Measured using rulers, tape measures, or motion sensors.
  • Velocity:
  • – Calculated from the slope of a displacement-time graph.
  • Acceleration:
  • – Calculated from the slope of a velocity-time graph.

  • VIII. Newton’s Laws of Motion

  • First Law (Inertia):
  • – A body remains at rest or in uniform motion unless acted upon by a net external force.
  • Figure 2 First law of motion
  • Example: A stationary car does not move unless pushed.
  • Second Law (Force):
  • – The net force acting on an object is proportional to the rate of change of its momentum.
  • [math]F = ma[/math]
  • Figure 3 Second Law of motion
  • Third Law (Action-Reaction):
  • – For every action, there is an equal and opposite reaction.
  • Example: A rocket propelling upwards due to the downward thrust of exhaust gases.
  • Figure 4 Third Law

  • IX. Principle of Conservation of Momentum

  • Definition:
  • – In the absence of external forces, the total momentum of a system remains constant.
  • [math]m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2[/math]
  • Where [math]m_1, m_2[/math]​ are masses and [math]v, u[/math] are initial and final velocities.
  • Figure 5 Conservation of momentum
  • Example: Two skaters pushing each other will move in opposite directions with equal and opposite momentum.
  • Newton’s Third Law as a Consequence
  • – The conservation of momentum arises because forces between interacting bodies are equal and opposite, ensuring no net external force.

  • b) Detailed Explanation and Application

  • I. Appropriate Use of Key Terms

  • Displacement:
  • – A vector quantity representing the shortest distance and direction between two points.
  • – Example: Moving 10 m east is a displacement of 10 m in the east direction.
  • Speed:
  • – A scalar quantity representing the rate of motion, defined as:
  • [math]\text{Speed} = \frac{\text{Distance}}{\text{Time}}[/math]
  • Velocity:
  • – A vector quantity representing the rate of change of displacement, defined as:
  • [math]v = \frac{∆s}{∆t}[/math]
  • – Includes both magnitude and direction.
  • Acceleration:
  • – A vector quantity representing the rate of change of velocity with respect to time:
  • [math]a = \frac{∆v}{∆t}[/math]
  • Force:
  • – A vector quantity causing acceleration, defined by Newton’s second law:
  • [math]F = ma[/math]
  • – Where m is the mass and is acceleration.
  • Mass:
  • – A scalar quantity representing the amount of matter in an object.
  • Vector:
  • – A quantity with both magnitude and direction, such as velocity, force, and acceleration.
  • Scalar:
  • – A quantity with only magnitude, such as speed, distance, and time.
  • Work:
  • – A scalar quantity defined as the product of force and displacement in the direction of the force:
  • [math]W = Fd \, cos⁡θ[/math]
  • Energy:
  • – The capacity to do work; exists in forms like kinetic energy [math]KE = \frac{1}{2}mv^2 [/math] and potential energy [math]PE = mgh[/math]
  • Power:
  • – The rate at which work is done or energy is transferred:
  • [math]P = \frac{W}{t}[/math]
  • Momentum:
  • – A vector quantity defined as the product of mass and velocity:
  • [math]p = mv[/math]
  • Impulse:
  • – A vector quantity equal to the change in momentum, defined as:
  • [math]J = F∆t[/math]

  • II. Graphs of Accelerated Motion

  • Displacement-Time Graph:
  • – Slope: Represents velocity.
  • – Shape:
  • Straight line: Constant velocity (no acceleration).
  • Curved line: Changing velocity (acceleration).
  • Figure 6 Displacement-time graph
  • Velocity-Time Graph:
  • – Slope: Represents acceleration.
  • – Area Under the Curve: Represents displacement.
  • – Shape:
  • Straight horizontal line: Constant velocity ([math]a = 0[/math]).
  • Straight sloped line: Constant acceleration.
  • Curved line: Changing acceleration.
  • Figure 7 Velocity-time graph
  • Acceleration-Time Graph:
  • – Represents the rate of change of velocity over time.
  • – Area under the curve gives the change in velocity.

  • III. Graphical Representation of Vectors

  • ⇒ Vector Addition:
  • – Use the parallelogram method or tip-to-tail method to add vectors graphically.
  • Example: Adding two vectors A and B gives a resultant vector R.
  • Figure 8 Vectors addition
  • ⇒ Vector Subtraction:
  • Subtraction is equivalent to adding the negative of a vector.
  • Example:
  • [math]A – B \text{is equivalent to} A + (-B)[/math]
  • Figure 9 Vector subtraction
  • ⇒ Changes in Vector Magnitude and Direction:
  • Graphically show the change in magnitude by adjusting the length of the arrow.
  • Change in direction is represented by rotating the arrow.
  • ⇒ Sketching and Interpreting Examples
  • Displacement-Time Graph:
  • – A steep slope indicates a higher velocity.
  • – A horizontal line indicates no motion (zero velocity).
  • Figure 10 Displacement- Time graph
  • Velocity-Time Graph:
  • – A straight line with a positive slope indicates constant acceleration.
  • – A horizontal line indicates constant velocity.
  • – A negative slope indicates deceleration.
  • Vector Addition:
  • – For two perpendicular vectors B, the resultant R can be calculated using Pythagoras:
  • [math]R = \sqrt{A^2 + B^2}[/math]
  • ⇒ Applications
  • Work Done Example:
  • – Pulling a box with a force at an angle of [math]30^\circ[/math] over a displacement of 10m
  • [math]W = Fd \cos\theta \\
    W = 50 \times 10 \times \cos 30^\circ \\
    W = 433 \text{ J} [/math]
  • Impulse Example:
  • – A force of 10 N applied for 5s
  • [math]J = F∆t \\ J = 10 × 5 \\ J = 50N s[/math]
  • Vector Addition Example:
  • – Two vectors, A=3 N north and B=4 N east:
  • [math]R = \sqrt{A^2 + B^2} \\
    R = \sqrt{(3)^2 + (4)^2} \\
    R = 5 \text{ N}[/math]

  • c) Detailed Explanation and Application

  • I. Resolution of a Vector into Two Components

  • Concept:
  • -A vector can be resolved into two perpendicular components: horizontal (xxx) and vertical (yyy) components.
  • – For a vector A at an angle θ to the horizontal:
  • [math]A_x = A \, cos⁡θ \\ A_y = A \, sin⁡θ[/math]
  • Example:
  • – A force of [math]F = 100N[/math] acts at [math]30^\circ [/math] to the horizontal
  • [math]F_x = 100 \cos 30^\circ \\ F_x = 86.6 \text{ N} \\ F_y = 100 \sin 30^ \circ \\ F_y = 50 \text{ N}[/math]

  • II. Addition of Two Vectors

  • Graphical Method:
  • – Place the tail of the second vector at the tip of the first (tip-to-tail method). The resultant vector is drawn from the tail of the first vector to the tip of the second.
  • Figure 11 addition of two vectors
  • Algebraic Method:
  • – For two vectors A and B with components:
  • [math]R_x = A_x + B_x \\ R_y = A_y + B_y[/math]
  • – Magnitude:
  • [math]R = \sqrt{R_x^2 + R_y^2} [/math]
  • – Direction:
  • [math]\theta = \tan^{-1} \left(\frac{R_y}{R_x}\right)[/math]

  • III. Kinematic Equations for Constant Acceleration

  • Derived from definitions of acceleration and average velocity:
  • Equations:
  • [math]v = u + at \\ s = ut + \frac{1}{2} at^2 \\ v^2 = u^2 + 2as[/math]
  • Example:
  • – A car starts from rest [math]u = 0[/math] and accelerations at [math]a = 2 m/s^2 [/math] for t = 5s
  • – Find the velocity and displacement:
  • [math]v = u + at \\
    v = 0 + 2 \times 5 \\
    v = 10 \text{ m/s} \\
    s = ut + \frac{1}{2} at^2 \\
    s = 0 + \frac{1}{2} \times 2 \times (5)^2 \\
    s = 25 \text{ m}[/math]

  • IV. Momentum

  • Definition:
  • – Momentum is defined as: [math]p = mv[/math]
  • Example:
  • – A ball of mass 2 kg moving at 3 m/s
  • [math]p = mv \\ p = 2 × 3 \\ p = 6 kg m/s[/math]

  • V. Newton’s Second Law of Motion

  • Formula:
  • – Newton’s second law:
  • [math]F = \frac{∆p}{∆t}[/math]
  • – For constant mass:
  • [math]F = ma[/math]
  • Example:
  • – A force F=20 N is applied to a 5 kg object:
  • [math]a = \frac{F}{m} \\
    a = \frac{20}{5} \\
    a = 4 \text{ m/s}^2[/math]

  • VI. Conservation of Momentum

  • Principle:
  • – In a closed system, the total momentum before and after an interaction is constant:
  • [math]m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2[/math]
  • Example:
  • – Two carts of mass 3 kg and 2 kg Before the collision, the 3 kg cart moves at 4 m/s, and the 2 kg cart is stationary.
  • – After collision, they move together:
  • [math]m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \\
    (3)(4) + (2)(0) = (3)v + (2)v \\
    12 + 0 = (3 + 2)v \\
    5v = 12 \\
    v = \frac{12}{5} \\
    v = 2.4 \text{ m/s}[/math]

  • VII. Work Done and Energy

  • Work Done:
  • – Defined as:
  • [math]W = F∆x cos⁡θ[/math]
  • Kinetic Energy:
  • – Energy due to motion:
  • [math]E = \frac{1}{2} mv^2[/math]
  • Gravitational Potential Energy:
  • – Energy due to height:
  • [math]PE = mgh[/math]

  • VIII. Power

  • Power as Energy Transfer:
  • [math]P = \frac{∆E}{t}[/math]
  • Power in Terms of Force:
  • [math]P = Fv[/math]

  • IX. Computational Modelling

  • Discrete Time Steps:
  • – A computational model calculates displacement and velocity in small time intervals ([math]Δt[/math]) using:
  • [math]Δv = aΔt \\ Δx = vΔt[/math]
  • Graphical Representation:
  • – Use arrows to represent displacement and velocity vectors, updating their magnitudes and directions at each time step.
  • Figure 12 Velocity, acceleration, and time graph
  • Example Calculation
  • – A 10 kg box slides down a 5m ramp inclined at [math]30^\circ[/math]. Find work done, velocity at the bottom, and power if the motion takes 2s.
  • Work Done:
  • [math]W = F \Delta x \cos\theta \\
    W = (mg) \Delta x \cos\theta \\
    W = (10)(9.8)(5) \cos 30^\circ \\
    W = 424.3 \text{ J}[/math]
  • Velocity:
  • – Using
  • [math]v^2 = u^2 + 2as [/math]
  • – Calculate acceleration and find:
  • [math]v = 6.26 \text{ m/s}[/math]
  • Power:
  • [math]P = \frac{\Delta E}{t} \\
    P = \frac{424.3}{2} \\
    P = 212.15 \text{ W}[/math]

  • d)   Detailed Explanation of Practical Activities

  • I. Investigating the Motion and Collisions of Objects

  • Purpose:
  • – To study velocity, acceleration, momentum, and energy conservation during motion and collisions.
  • Equipment:
  • – Trolleys, air-track gliders, ticker timers, light gates, data-loggers, video recording equipment.
  • Procedure:
  • – Motion Investigation:
  • Set up an air track to minimize friction.
  • Use light gates to measure the time taken for the trolley to pass between them and calculate velocity and acceleration.
  • Alternatively, use a ticker timer to record dots on a tape as the trolley moves, spacing of dots gives speed and acceleration.
  • Collisions:
  • – Set up two gliders on an air track.
  • – Before the collision, measure velocities using light gates.
  • – After the collision, measure velocities and confirm the conservation of momentum:
  • [math]m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2[/math]
  • Analysis:
  • – Plot velocity–time and acceleration–time graphs.
  • – Compare kinetic energy before and after collisions to identify elastic or inelastic collisions.
  • Applications:
  • – Understanding Newton’s laws, conservation of momentum, and energy dissipation.

  • II. Determining the Acceleration of Free Fall

  • Purpose:
  • – To measure the acceleration due to gravity ([math]g ≈ 9.8 m/s^2[/math]).
  • Figure 13 Free falling object
  • Equipment:
  • – Electromagnet and trapdoor setup, light gates, or high-speed camera/video techniques.
  • Figure 14 Determine acceleration with the help of electromagnet
  • Methods:
  • Trapdoor and Electromagnet:
  • – Suspend a small metal ball using an electromagnet.
  • – When the circuit is broken, the ball is released and falls, triggering a timer as it passes through the trapdoor.
  • – Measure the time taken for the ball to fall a known height (h) and use:
  • [math]g = \frac{2h}{t^2}[/math]
  • Light Gates:
  • – Set up light gates at two points along the fall.
  • – Use the time intervals to calculate the acceleration between the two gates.
  • Video Technique:
  • – Record the fall using a camera.
  • – Use frame-by-frame analysis to measure displacement and time.
  • Analysis:
  • – Plot displacement–time and velocity–time graphs.
  • – Determine g as the slope of the velocity–time graph or using equations of motion:
  • [math]s = ut + \frac{1}{2} g t^2[/math]
  • Sources of Error:
  • – Reaction time (trapdoor setup), air resistance, timing inaccuracies.

  • III. Investigating Terminal Velocity

  • Purpose:
  • – To study the motion of objects reaching terminal velocity in a fluid or air and to understand forces acting on the object.
  • Equipment:
  • – Ball bearing, viscous liquid (e.g., glycerin or oil), measuring cylinder, stopwatch, ruler.
  • – Alternatively, use paper cones or parachutes in air.
  • Procedure:
  • – Ball Bearing in Liquid:
  • Fill a transparent measuring cylinder with a viscous liquid.
  • Figure 15  Investigate Terminal Velocity
  • Drop a ball bearing and use markers on the cylinder to measure distances.
  • Measure the time taken to pass between two points where the ball bearing moves at constant velocity (terminal velocity).
  • Paper Cones in Air:
  • – Drop paper cones of different shapes/sizes and measure the time taken to reach the ground.
  • – Observe when the cone stops accelerating (constant velocity).
  • Analysis:
  • – At terminal velocity, forces acting on the object are balanced:
  • [math]F_{\text{gravity}} = F_{\text{drag}} + F_{\text{buoyancy}}[/math]
  • – Use the velocity-time graph to identify when acceleration becomes zero.
  • Applications:
  • – Understanding drag force, buoyancy, and the relationship between terminal velocity, object size, and fluid viscosity.
  • – Applying Stokes’ Law for small spherical objects in fluids:
  • [math]F_{drag} = 6πηrv[/math]
  • Where η is the fluid viscosity, r is the radius of the object, and v is the terminal velocity.
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