to use this document yourself.
• # Introductory Physics Education

An overview to pedagogical perspectives and approaches for A Level/introductory university physics.

• # 01a Measurements

• Readings and measurements
• Systematic vs random errors
• Accuracy vs precision

• # 02 Kinematics

“science of motion,” 1840, from French cinématique (Ampère, 1834), from Greek kinesis “movement, motion”

The study of motion (of bodies)

• Displacement, velocity, acceleration
• Instantaneous & average qty
• Equations
• General observations
• 2D kinematics
• # 03 Dynamics

1827 in the sense “pertaining to force producing motion” (the opposite of static), from French dynamique introduced by German mathematician Gottfried Leibnitz (1646–1716) in 1691 from Greek dynamikos “powerful,” from dynamis “power”.

The study of how forces cause motion

• Newton’s 2nd Law
• Newton’s 3rd Law
• Conservation of Momentum
• Collisions
• # 04 Forces

• Hydrostatic pressure
• Upthrust
• # 05 Work, Energy, Power

• Conservation of energy
• Work done
• Power
• # 06 Motion In A Circle

• Centripetal force
• Deviation from circular motion
• # 07 Gravitational Field

• Gravitational force between two point masses
• Gravitational field of one or more point masses
• Gravitational potential and potential energy
• Geostationary orbit
• Escape velocity
• # 08 Oscillations

• Cyclical and Simple harmonic motion
• Displacement, velocity, acceleration
• Energy in an oscillation
• Damping
• Resonance

• # 09 Thermal Physics

• Pressure, volume, temperature
• Ideal gas equation
• 1st Law of Thermodynamics
• Energy distribution
• Heat capacity
• Latent heat

• # 13 Current of Electricity

• Conventional current
• Resistivity
• Ohm’s Law
• Emf and potential difference (p.d.)
• Internal resistance and terminal voltage
• # 14 DC Circuits

• Resistors in series and parallel
• Potential dividers
• Complex circuits
• Thermistors and light-dependent resistors

• # 20 Nuclear Physics

• ## Active Learning

Keep students actively engaged, provide rapid feedback

1. Students spend class time actively engaged in doing/thinking/talking
2. Students interact with peers
3. Students receive immediate feedback
4. Instructor facilitates
5. Students take responsibility for their knowledge
• ## Phenomena to theory

Focus on phenomena rather than abstractions

• ## Deal explicitly with alternative conceptions

1. Confront student misconceptions directly.
2. Explore the fact that many predictions were wrong.
3. Consider alternative models.
4. Reiterate.
• ## Teaching techniques

Teach and use explicit problem-solving skills and strategies

1. Teach students the specific skills needed to solve complex problems. These include interpretation skills, pictorial skills, graphical skills, and reasoning skills.
2. Show students how those skills re assembled into a powerful problem-solving strategy and demonstrate their use, in detail, in the example problems we work in class.
3. Make explicit the assumptions, decisions, and reasoning that are part of an expert’s problem-solving strategy but which usually go unsaid.
4. Help students organise their knowledge in a more coherent, hierarchical, easily searched structure.
• ## Problem-solving

Give problems that go beyond symbol manipulation
Engage students in qualitative and conceptual analysis of physical phenomena

Appropriate homework and example problems need to:

1. Balance qualitative and quantitative reasoning.
2. Emphasise reasoning, de-emphasise formulas and equations.
3. Deal directly with phenomena and observations. Derivations and “Show that …” problems have little efficacy for students at this level.
• ## Introduction

Man first started learning to measure space, and then gradually time. What other things could we measure, and what do they represent?

• ## A standardised system of units

All quantities we try to express have to be given units to have meaning.

Scientists have to be very precise when presenting their results, so that they can agree on what an observation means. Thus, they need to use a standardised system of units.

#Base units & homogeneity

Base units are physical quantities agreed to be fundamental enough that other units may be derived from them.

These units are used in physics equations, which must be homogeneous to be valid. This means that units on both sides of the equation must match up.

SI unit prefixes provide a compact, economical way to represent very large or very small numbers.

• ## Estimation

In physics, estimating the right order of magnitude is often more important than getting the quantity exactly right

• ## The phenomenon of uncertainty

All readings and measurements involve a certain degree of uncertainty, regardless of the measuring instrument’s accuracy.

At the quantum level, there is a limit to the precision with which we can measure some property-pairs simultaneously. This natural phenomena means, for instance, that we cannot measure the position and momentum of a subelementary particle with zero uncertainty simultaneously.

• ## The process of measurement

Measurement sounds simple, but the actual process of getting a number can sometimes be quite complicated.

Even after we obtain a number, what does that number mean? Will we get the same number again if we repeat the process?

A ‘perfect’ measurement does not exist, since perfect objects do not exist. The true value of a measurement is unknowable.

• ## Accuracy vs precision

Accuracy refers to how closely a measured value agrees with the true value.

Precision refers to how closely individual measurements agree with each other without reference to true value.

• ## Terms of physical motion

In physics, we often study the motion of various objects and systems. We need words that can help us describe these motions precisely.

• A bouncing ball
• A falling stone
• An oscillating spring
• etc

Thus, we need to define common terms so that when we use these terms, other people will understand what we are trying to say, and vice-versa.

• ## Graphical representations

We use graphs to visualise how two quantities change in relation to each other.

The follow graphs are often encountered in the study of kinematics:

• $s$-$t$ graphs
• $v$-$t$ graphs
• $a$-$t$ graphs
• vertical vs. horizontal displacement ($s_y$-$s_x$) graphs
• ## Instantaneous & average quantities

Instantaneous values are an ideal, often used only in derivation; in real life, we deal with values averaged over a short time window (since it is impossible to measure speed at a single point).

Students should understand this difference well so they understand the limitations of measurements and results obtained in kinematics.

## Definitions

Instantaneous velocity/acceleration refers to $v$/$a$ at a particular point in time, while average velocity/acceleration refers to $v$/$a$ over a time interval.

• ## Equations

$v = u + at\\ s = ut + \frac{1}{2}at^2\\ v^2 = u^2 + 2as$

The kinematic equations only apply if $a$ is constant throughout.

The kinematic equations can only be applied to linear motion.

• ## 2D Kinematics

Because $s$, $v$, and $a$ are vectors, they may be resolved in two or more perpendicular directions.

Any pair of perpendicular directions may be used, but the most useful pair for resolving depends on what is required in the situation.

• Typically this means resolving in directions parallel and normal to acceleration, or to velocity.
• Students with sufficient mastery of vector-resolving will find this much easier to grasp.

Components of $a$ do not affect normal components of $v$, which in turn do not affect normal components of $s$. Only components parallel to each other can affect each other.

Deceleration means “acceleration in a direction opposite to velocity”.
Decreasing acceleration means “decreasing magnitude of acceleration”.

• ## Terms of physical interaction

On top of the kinematics terms, we now need terms to describe how forces interact with bodies to cause motion.

• ## The laws of motion

The three laws of motion were first compiled by Isaac Newton in his Philosophiæ Naturalis Principia Mathematica (Mathematical Principles of Natural Philosophy), first published in 1687. Newton used them to explain and investigate the motion of many physical objects and systems.

Today, these laws form the foundation of many mechanical systems in use today. Although we have discovered that very tiny objects (smaller than an atom) and very fast objects (near the speed of light) deviate from the behaviour described by these laws, under ‘normal’ conditions, Newton’s laws of motion provide a simple and adequate model for everyday purposes.

• ## Linear collision of two bodies

In A level Physics, we only analyse linear collisions, i.e. collisions that occur along a straight line. In such collisions, the initial and final velocities of the colliding bodies are always parallel to each other.

Collisions where the initial and final velocities are not parallel are referred to as oblique collisions.

• ## Conservation of momentum

Earlier, we saw that the total momentum of a system remains the same throughout a collision. We can write the above observation in the following form:

$\sum mu = \sum mv$
or
$m_1u_1 + m_2u_2 + ... = m_1v_1 + m_2v_2 + ...$

In prose form:
The total momentum of a system is conserved throughout a collision, if no external forces act on the system (of colliding bodies).

This is known as the principle of conservation of momentum (CoM), and is the underlying formula behind all of collision analysis.

CoM applies to all collisions, if there are no external forces acting on the system.

• ## Hydrostatic Pressure

$p_{fluid} = mgh$

Difficult to explain pressure difference due to variation in height of liquid below surface.
Hard to imagine how difference in pressure at different height can cause upthrust.

• ## Upthrust

$U = \rho_{liquid}gV_{displaced}$

Upthrust is the integral of pressure around the area of an object.

• ## Conservation of energy, revisited

In the absence of external forces, the total energy in a system is always conserved.

• ## Work done

$W.D. = F\cdot s$

The work done by a force is the product of the magnitude of the force and the displacement of the point of action in the direction of the force.

• ## Power

$P = \frac{dE}{dt}$

Power is the rate of change of work done with respect to time.

For an object travelling at constant velocity, $\frac{ds}{dt}$ is constant, so

$P = \frac{d}{dt}(F \cdot s) = F \cdot \frac{ds}{dt} = F \cdot v$

• ## Centripetal force

$F_c = \frac{mv^2}{r} = mr\omega^2$

Centripetal force is not a kind of force (contact, non-contact). It describes a force that is needed to keep an object moving around the centre of a circle.

Centripetal force always acts toward the centre of circular motion.

• ## Centripetal acceleration

$a_c = \frac{v^2}{r} = r\omega^2$

Centripetal acceleration is provided by the centripetal force and points in the same direction as the force.

This links acceleration to force—an extension of dynamics, and N2L.
The connection between direction of acceleration and force is often needed to solve many problems where one is given but not the other; in fact questions commonly assume knowledge of this fact.

• ## Deviation from circular motion

If the force provided is greater than centripetal force, the object will orbit closer to the centre (with accompanying increase in angular velocity).

If the force provided is smaller than centripetal force, the object will orbit further from the centre (with accompanying decrease in angular velocity).

• In a binary star system, centre of rotation and balance point are not at the same position.

This only happens when both stars are the same mass.

• Students still think GPE on Earth = 0

• Students tend to apply conservation of energy to solve problems involving two different orbits, thinking that change in GPE = change in KE

• ## Cyclical motion

In nature and in life, we often observe systems that repeat the same motion over and over again.

• sunrise
• tides
• commuting
• dribbling a basketball
• waiting icon

Clearly, some of these are more special than the others. Some happen more regularly, some more jerkily.

• ## Simple harmonic motion

But there is a special category of regular motions that behave very beautifully. There are many interesting patterns to spot in this class of regular motions. For instance, the further from the equilibrium position they are, the slower they travel.

These occur very commonly, and are perhaps one of the most common kinds of motions in nature. Understanding this motion, and learning its patterns, gives us a mental pattern and an equation we can apply to many things.

• ## Displacement

The displacement varies sinusoidally with time.

## Velocity

The velocity is the rate of change of displacement.

The velocity is greatest at the equilibrium position, and zero at the ends of the oscillation.

## Acceleration

The acceleration is proportional to the displacement, and in the opposite direction, always toward the equilibrium point.

The acceleration is zero at the equilibrium position, and greatest at the ends of the oscillation.

• ## Energy in an oscillation

In an oscillation, the kinetic energy is greatest at the equilibrium position, where velocity is also greatest.

The potential energy is greatest at the ends, where work has been done against the restoring force to slow the oscillator down.

In the absence of external forces (no damping), total system energy is conserved.

• ## Damping

Damping an oscillation means to decrease the energy of the oscillation. This usually takes the form of a retarding force that acts against the direction of motion of the oscillator.

• ## Resonance

Resonance is a phenomenon whereby

• there is maximum transference of energy from an oscillating driver to the oscillating system.
• this happens when the driving frequency is close to the oscillating system’s natural frequency.
• the amplitude is greatest at this point.

The driving frequency (frequency of oscillation) is always less than or equal to the system’s natural frequency.
The discrepancy between driving frequency and system natural frequency is 0 for free oscillations (undamped), and increases as the amount of damping increases.

• ## Ideal gas

An ideal gas is a gas which has no internal PE; its internal energy, $U$, is always equal to its internal KE.

In real life, gases behave like an ideal gas when its internal PE is high (i.e. close to 0). This happens when the KE of the gas particles is high enough to overcome the gas particles’ PE ($T$ must be high), and when the gas particles are sufficiently far apart ($P$ must be low).

For an ideal gas, the gas laws can be combined to form the ideal gas equation, $pV = nRT$

This equation describes possible combinations of $P$, $V$, $n$, and $T$ for any ideal gas system. Each combination of these variables is referred to as a state of the gas system. If the state of the gas system satisfies the ideal gas equation, that state can exist.

• ## The gas laws

We create a simple model of a gas as a collection of particles inside a box. This box has one wall that can move inwards or outwards, allowing us to change its volume. There are $N$ particles ($n$ moles) inside this box.

If the gas is treated as an ideal gas, the following relationships hold between its pressure P, volume V, and temperature T:

• $P \propto T$
• $V \propto T$
• $P \propto \frac{1}{V}$
• ## 0th Law of Thermodynamics

Two systems in thermal equilibrium have the same temperature.

• ## 1st Law of Thermodynamics

$\Delta U_{sys} = Q_{in} + \Delta W_{on}$

Change in internal energy of a gas system ($\Delta U_{sys}$) is the sum of thermal energy transferred into the system ($Q_{in}$), and work done on the system ($\Delta W_{on}$).

## Work done on system

$\Delta W_{on} = -p \Delta V$

The work done by the system is $p \Delta V$, where $p$ is the pressure of the gas system (exerted on the box walls), and $\Delta V$ is the change in volume of the gas system (positive if expanding, negative if contracting/compressing).

Hence, work done on the system is $-p \Delta V$.

Thus, the 1st Law of Thermodynamics can also be written as

$\Delta U_{sys} = Q_{in} - p \Delta V$

• ## Energy distribution

In a gas, particle velocities are not homogeneous.

• ## Heat capacity

Specific heat capacity tells us how much energy is needed to raise 1 unit mass of substance by 1 K.

• ## Latent heat

Specific latent heat tells us how much energy is needed to change 1 unit mass of substance to a different state.
Questions involving thermal energy change causing liquids to heat up/cool down.
Specific heat of fusion is for melting.
Specific heat of vapourisation is for boiling.

• Clarify semiconductor diode: ideal vs real-world graph of I-V characteristics

• Confusion over I-V graph in experiment to determine internal resistance

• Confusing R = V/I (ratio) with R = dV/dI (gradient)

• Identifying and treating short circuits

• Potential divider: why does the test circuit have no current?

• ### Use the classroom for …

• discussion rather than presentation
• clarification rather than solution
• demonstration rather than description
• (instructor-guided) practice rather than copying
• ### Don’t give topic overviews

1. Lay out expectations and reasons for lesson structure, continually repeat for emphasis and reminding as the term progresses
• Do not repeat factual knowledge, expect students to remember it
• Teach students how to read textbook/notes.
• Get them to keep a notebook?
2. Give a day-to-day schedule and stick to it.
• Students need to know when to start each chapter
3. Don’t relent!
• Give short reading quizes to ensure students have done their reading
• Give vocab/terminology quizzes for quick checking
• ### Doing the physics

1. Focus on the reasoning and principles, not on the mathematics
2. Students need to give explanations that aren’t just restating the problem.
3. Pay extra attention to discussing, justifying, and clarifying explanations
• Give examples, and get students to apply the principles
4. Demonstrate and go through extended examples of qualitative reasoning and explaining
5. Strategies should emphasise qualitative steps, by describing and clarifying assumptions and thought
6. Homework should be given that requires analysis and interpretation of problems.
• ### Concrete to abstract

Important focusing questions: “How do we know …?”, “Why do we believe …?”

Start with observations, then concepts and principles

• Gradually generalise to theories
• Apply the theory to phenomena, see if they explain the observations
• ### Novices vs Experts

Novice students

1. Fail to describe problems adequately
2. Do little prior planning or qualitative description
3. Try to assemble solutions by stringing together miscellaneous math formulae
4. Knowledge consists of loosely connected formulas
5. Recognise pieces of info as single chunks, get overwhelmed by too many chunks

Experts

1. Structure their knowledge
2. Use the structure of their knowledge to craft a strategy
3. Recognise and work with larger patterns (fewer chunks)
4. Draws sketches to clarify understanding of problem before solving
• ### Scaffolding

1. Make thought processes and decision-making explicit
2. Make intuition visible—don’t make it a case of “using the right equation”
3. Teach problem-solving skills
4. Give students opportunities to help students organise knowledge coherently
• ### Homework

“A few well-written problems studied carefully—with feedback and, sometimes, in-class discussion—-are a far better learning experience than many problems practised blindly.”

1. Worksheets should serve as a template for following the steps of the strategy
2. Model the strategy and show all steps
3. Insist that students follow and solve the worksheet problems
4. Students need to practise word-to-symbol translation independently of the calculations
• Practise individual steps separately, then approach full problems near the end of the chapter
• ### SI base units

SI base units are standardised and carefully defined so as to be constant in as many scenarios as possible.

They represent the most fundamental properties/measurables we can observe in experiments.

The SI base units are

• mass (kg)
• time (s)
• length (m)
• electric current (A)
• thermodynamic temperature (K)
• amount of substance (mol)
• luminous intensity (cd)
• ### Checking of homogeneity

The base units are the fundamental units by which we check the consistency of equations.

Any two expressions must have the same base units before they can be added or subtracted.

An equation in which the base units are consistent is said to be homogeneous.

• Back-of-the-envelope calculations and mental feasibility evaluations rely heavily on order-of-magnitude estimations for their effectiveness, and they are a very useful system for quick estimations, where exact values are not so important (and often not really knowable).

• Students ought to know some basic numbers from everyday experience, e.g. size of car, power consumption of appliances.

• ### Error vs uncertainty

In addition to a measurement’s uncertainty, the incorrect or imprecise use of an instrument can contribute error to the measurement.

For instance, our human inability to respond to a signal immediately means there will always be human error in our use of a stopwatch.

• ### Instrument error vs measurement uncertainty

It is often easier to know the error of an instrument rather than the uncertainty of a measurement. This is because we can design instruments to be precise to a certain degree.

• Rulers are designed to be precise to the nearest 1mm, vernier calipers to the nearest 0.01mm, micrometers to the nearest 0.001mm (or smaller!)
• ### Error calculation

For complex expressions involving the addition, subtraction, multiplication, or division of multiple variables, we have standard ways of calculating the error in the final value.

• ### Readings and measurements

In scientific terminology, a distinction is drawn between (raw) readings and measurements.

A reading is a number obtained from using an instrument (the actual number recorded during an experiment).

A measurement is the result of analysing readings, its accuracy depending on the measuring instrument and how it is used.

• ### Visual explanations

A set of measurements with a wide spread is imprecise; a narrow spread is precise.

A set of measurements with an average value that is far from the true value is inaccurate; if it is near to the true value, it is accurate.

• ### Systematic vs random errors

Errors are broadly classified as systematic errors and random errors, not mutually exclusive.

Systematic errors yield a definite pattern, consistently over- or under-estimating the true value.

Random errors do not yield a definite pattern; measurements are randomly
greater or less than
the true value.

• ### Displacement

Displacement is the straight-line distance between two points.

Distance is the actual length of the path taken between two points.

Symbol: $s$

### Velocity

Velocity is the rate of change of displacement.

Symbol: $v = \frac{ds}{dt}$

### Acceleration

Acceleration is the rate of change of velocity.

Symbol: $a = \frac{dv}{dt} = \frac{d^2s}{dt^2}$

• ### Deriving the kinematics graphs

The gradient of the $s$-$t$ graph gives the $v$-$t$ graph,
and the gradient of the $v$-$t$ graph gives the $a$-$t$ graph.

Conversely, the integral of the $a$-$t$ graph gives the change in the $v$-$t$ graph,
and the integral of the $v$-$t$ graph gives the change in the $s$-$t$ graph.

• ### Standard examples

Zero $a$

• Horizontal line on $v$-$t$ graph
• Horizontal line on $v$-$t$ graph
• Straight line on $s$-$t$ graph

Constant $a$

• Horizontal line on $a$-$t$ graph
• Straight line on $v$-$t$ graph
• Quadratic curve on $s$-$t$ graph

Constant $v$ (same as zero $a$)

• Horizontal line at a = 0 on $a$-$t$ graph
• Horizontal line on $v$-$t$ graph
• Straight line on $s$-$t$ graph
• Instantaneous acceleration is $\frac{dv}{dt}$,
while average acceleration is $\frac{\Delta v}{\Delta t}$.

There are situations where instantaneous $a$ is not constant, though it seems to be, and therefore the kinematic equations cannot be applied.

• ### Derivation

The kinematic equation $v^2 = u^2 + 2as$ can be derived from the other two equations algebraically.

All three equations can be derived from the same kinematic graph.

• ### Mathematical results

Some mathematical results from algebra and vectors are not explicitly stated, but are understood to be required knowledge:

• Change of a vector need not be along the same axis it is pointing.
• Vector sum allows vectors to be added regardless of their direction.
• ### General results

Some general results can be observed from applying the kinematics equations.

• Displacement, velocity, and acceleration are vectors, and need not be in alignment all the time.
• Any of the quantities $s$, $v$, $a$ can be zero while the others are not.
• A change in displacement must indicate the presence of velocity, and a change in velocity must indicate the presence of acceleration.
• The presence of velocity implies a change in displacement, and the presence of acceleration implies a change in velocity.
• In Mechanics, a large part of being able to do well lies in being able to describe common everyday situations in terms of physics variables. Students should be given sufficient practice to analyse these situations, and picture/visualise them in the form of diagrams or graphs, before translating them into equations.

• ### Projectile Motion

Projectile motion is one instance of motion with constant acceleration in one direction (vertically) and zero acceleration (constant velocity) in another direction (horizontally).

• ### Momentum

When we apply the same force over the same period of time on two bodies, one with greater mass than the other, we known intuitively that one of them (with smaller mass) will have a higher final velocity than the other. But since the force and time are the same for both, something must have remained the same; this is momentum.

#### Equation

$p = mv$
where $m$ represents mass of the body,
$v$ represents velocity of the body.

Momentum $p$ has base units of kg m$^{-1}$ s$^{-1}$.

#### Definition

The momentum of a body is the product of its mass and velocity.

When two bodies collide, momentum may be transferred. That is why in some collisions, one of the bodies may end up with a higher speed that it had initially.

• ### Impulse

In real life, we seldom have situations where a force is applied on an object over an infinite period of time; that is unrealistic.

Instead, if we plot the resultant force exerted against time, we will see that it forms a closed area once there is no longer any resultant force.

#### Equation

$p = Ft$
where $F$ represents resultant force exerted on the body,
$t$ represents time over which the force was exerted.

Impulse has base units of kg m$^{-1}$ s$^{-1}$ (same as $p$).

#### Definition

The impulse applied to a body is the product of the force exerted on the body and the time over which the force is exerted.

Often, the force applied will not be constant, but will taper to zero at the start and end. In such cases, we cannot apply $p = Ft$ directly, but must calculate or estimate the area under the $F$-$t$ graph instead.

• ### Free Body Diagrams

We will regularly encounter situations in which a body is being exerted on by multiple forces. This is the heart of Newtonian mechanics: to analyse the forces being exerted on a body, figure out how it will behave and why, or to analyse a body behaving in a particular way and infer what forces are being exerted on it.

This can be a hugely complicated business, so we need to acquire some tools to help us do this. Free body diagrams (FBDs) are our most important tool for analysing bodies in motion. Without them, any system with more than a handful of forces becomes too much to handle.

The primary purpose of an FBD is to help us determine all the forces acting on a body. We want to do this to aid us in our next step, solving the force equations (N2L).

• ## Newton’s 2nd Law (N2L)

#### Formula

$\sum F = \frac{dp}{dt}$
where $F$ represents resultant force acting on the body,
$p$ represents momentum of the body

#### Statement

When a force is exerted on a body, it experiences a rate of change of momentum equal to the force, in the direction of the force.

At A levels, $F$ is only considered on rigid bodies, but in effect it can also cause shearing, rotating (moment), twisting, compression etc in continuous bodies.

• ## Newton’s 3rd Law (N3L)

An FBD of a single-body system is quite straightforward, since it does not involve contact forces. But once we start to include multiple contact forces, things start to get messy. What is the direction of the contact force? What is it exerted by, and exerted on which body? Newton’s 3rd Law is another tool that helps us in sketching the FBDs.

#### Formula

$F_{AB} = -F_{BA}$

#### Statement

When body A exerts a force on body B, body B exerts an equal and opposite force on body A.

• ### Application of N2L and N3L in a collision

When two bodies A & B collide, they exert equal and opposite forces on each other (according to N2L). As body A is in contact with body B, body B is also in contact with body A. Hence, the time of contact between the two bodies is the same.

Thus, by N3L, the change in momentum ($\Delta p = \int F dt$) experienced by body A & B should also be the same magnitude but in opposite direction.

• ### Momentum-time graph in a collision

If we draw a graph of A’s momentum ($p_A$) and B’s momentum ($p_B$), we will notice that

• They change by the same amounts, in the opposite direction, in the collisions. In other words, the total change in momentum of the system is 0.
• The total momentum of the system (consisting of bodies A and B) remains the same.
• Assumption of compression in collisions is necessary (to avoid infinite acceleration).

While bodies are compressed, they are in contact. They remain in contact until their velocities are the same (relative speed = 0)

• ### Energy in a collision

The change in momentum of bodies in a collision cannot happen instantaneously (this would imply that acceleration is zero!).

We can think of collisions as happening in 2 phases:

1. Bodies moving toward each other
2. Bodies moving away from each other

The moment when the system goes from Phase 1 to Phase 2 is the point when both bodies have the same velocity.

During Phase 1, the total KE of the system is decreasing. If there is no external force acting on the system, then conservation of energy applies and this KE has to be converted to other forms. In such cases, the KE is usually converted to elastic potential energy as the colliding bodies are compressed in the collision.

During Phase 2, the elastic potential energy is converted back to KE of the colliding bodies. This conversion may not always be 100% efficient; there may be some loss of energy in the collision. This does not violate the principle of conservation of momentum.

The efficiency of conversion from potential energy back to KE determines the type of collision that occurs.

• ### Misconceptions

1. Identifying correct $m$, $u$, $v$ in a collision.
2. Using correct signs for velocity directions.
3. Identifying where it is appropriate to apply CoM and/or CoE (i.e. whether system is isolated, whether there is external force).
4. Correctly identifying type of graph: force-time, momentum-time, change in momentum, etc.
5. Correctly identifying type of collision.
6. Applying incorrect conditions for type of collision.
7. Correctly identifying colliding bodies in a graph (e.g. mistaking $p_A$ and $p_B$ in a completely inelastic collision for the total momentum of the system.).
8. Mixing up formulas for momentum ($mv$) and kinetic energy ($\frac{1}{2}mv^2$).

Students will find it difficult to understand where energy has gone to, how it is lost. Difficult to go into depth except with high-ability students.

• ### Types of collisions

• Use taupok as an analogy: would you rather be at the top or at the bottom? Why?

• Difficult to explain the origin of pressure, as stated above. It is a contact force exerted betwen molecules in a liquid, and between liquid molecule other objects. It has a vertical gradient only when an external force (typically gravitational force) is present.

• What happens when gravity is not present? Use video of balloon in a bottle of water as example. What happens to upthrust and water pressure in free-fall?

• Students have difficulty remembering and applying the fact that CoE not only takes place at one point, it takes place between any two chosen points, and in fact throughout the entire process. They might need many examples for scaffolding.

• centrum “”centre”” + petere “”to seek, to go after”” [e.g. petition]). Not to be confused with centrifugal (centrum “”centre”” + fugere “”to flee”” [e.g. fugitive]

• This is a direct result of N2L: since object undergoes change in velocity, an acceleration must be present, which can only be provided by a force.

Understanding centripetal accel. and force will help cement students’ understanding of Newton’s Laws of Motion in non-linear contexts.

• Centripetal force in elliptical motion is not constant throughout. Circular motion is a special case of elliptical motion.

• Students will constantly forget that an object undergoing circular motion is not at rest, and in fact experiences an acceleration. It will take some time for them to get used to this.

Students tend to confuse centripetal force for centrifugal (outwards) force. There needs to be constant reminding that without centripetal force, the object will continue with linear motion per N1L.

• Once gravitation is taught, students might form the impression that where circular motion around Earth is involved, the centripetal force always points towards the centre of Earth – this is only true if circular motion is centred at Earth’s centre.

• Centripetal acceleration does indeed always point toward the centre of motion even in elliptical orbits/motion. Hence, the centripetal acceleration indicates the direction of motion, rather than vice-versa.

• An object need not necessarily always experience the same centripetal force; this is an assumption we make when we observe an object undergoing uniform circular motion.

• Students may fall under the misconception that minimum velocity (around a bend, at the top of a vertical loop) is 0. They should be encouraged to consider the consequences of zero velocity at such points.

• Examples include car going over hump, ball rolling over a curved surface, rollercoaster loop, etc.

Students often are unable to translate problem descriptions to physical equations. For instance, normal force is zero when an object loses contact with the surface. This may need to be explicitly taught.

• 3 situations:
1) Provided force > required force
2) Provided force = required force
3) Provided force < required force

What happens in each of the situations? Useful as a summary of the topic.
(1) and (3) occur at a point of force imbalance; the result is deviation from circular motion. (2) occurs when forces are balanced again and circular motion is restored.

• Velocity is the greatest at the equilibrium position, and zero at the ends. Hence, the graph of velocity vs displacement forms an ellipse.

[graph]

Acceleration is proportional to the displacement and always pointing towards the equilibrium position (opposite direction from displacement), hence the graph of acceleration vs displacement forms a straight line with negative gradient that passes through the origin.

[graph]

• The displacement is $\frac{\pi}{2}$ out of phase with the velocity, which is in turn $\frac{\pi}{2}$ out of phase with the acceleration.

[graph of displacement, velocity, acceleration]

• The graph of kinetic energy vs displacement is a quadratic curve with the maximum in the middle.

The graph of potential energy is a quadratic curve with the minimum in the middle (if total system is conserved).

• #Light Damping

Under light damping, an object oscillates about the equilibrium position a few times as the amplitude gets smaller, before it finally comes to a stop.

The amplitude usually decreases exponentially.

#Critical damping

As the damping on an object increases, the time taken for the oscillator to come to rest decreases, until a point.

This point is known as the critical damping point. At this point, the oscillator comes to rest in the shortest possible time.

If the oscillator is damped any further, the time for it to come to a stop increases instead; it undergoes heavy damping.

Critical damping is the threshold between light damping and heavy damping.

#Heavy Damping

Under heavy damping, the oscillator experiences a retarding force great enough that it does not overshoot the equilibrium point.

• When there is no damping in the system, at resonance the total energy of the oscillating system can grow to infinity.

When there is critical or heavy damping, there can be no oscillation since there is no overshoot in displacement.

• Finite forced oscillations can only occur with light damping.

At the resonance point, the driving frequency is always less than or equal to the natural (undamped) frequency of the oscillating system, since damping introduces a delay in the period of oscillation.

• ### Example

In H2 Chemistry, you are taught that 1 mol of any gas occupies a volume of 24 dm3 at room temperature. Does this agree with the ideal gas equation?

P = 1E5 Pa
V = ? m3
n = 1 mol
R = 8.31 J K–1
T = 298 K (25 deg C)

• The equation can be modelled using first principles; from a minimal-energy state, work done on system causes pressure and volume to increase; also causes temperature to increase.

• Equalities are easy to misapply. Students can easily confuse variables with change in variables (V vs ΔV).

Also, NKT vs nRT, and p being pressure exerted by gas, not on gas.

• There are higher-order terms to model non-ideal gases.

• ### Pressure

When the gas particles collide with the walls of the box, they exert a force on the walls of the box. The gas particles hence exert a pressure, P, on the walls of the box.

• This pressure always acts outwards, and is taken to be normal to the surface of the box.
• Since the direction is always taken to be outwards, when we consider the pressure of the gas, it is considered to be scalar and is always positive.
• However, when we consider the pressure of the gas on a single wall, pressure is treated as a vector, since it is no longer acting outwards but normal to the wall. Hence, the gas can still do work on the wall if the movement of the wall is along the orientation of the pressure exerted on it.
• ### Volume

The box has a volume, V, and when the movable wall shifts, the box’s volume changes by ΔV. The gas particles will travel throughout the entire volume of the box, hence we take the volume of the gas to be the volume of the box.

• When the movable wall shifts outwards, the box expands in volume and ΔV is positive (ΔV > 0).
• When the movable wall shifts inwards, the box is compressed and ΔV is negative (ΔV < 0)
• ### Internal potential energy

If the gas particles exert no forces on each other, we take their potential energy to be 0.

If the gas particles are not too near to each other, they may exert attractive forces on each other, as the positively charged nuclei of the particles are attracted to the negatively charged electron clouds of other particles.

• Since the forces are attractive in nature, the potential energy of the gas particles are negative.

If the gas particles come very close to each other, their electron clouds exert a large repelling force against each other, allowing them to “collide” and bounce off each other.

• Since the forces are repulsive in nature, the potential energy of these gas particles in close proximity is positive.

Overall, we would expect the potential energy of these gas particles to be negative.

Since these potential energies of the particles arise from internal forces (within the system of gas particles), they constitute the internal potential energy of the system. We differentiate this from (external) potential energy arising from weight and other external forces.

• ### Internal kinetic energy

If the box is moving, it will have a kinetic energy of $\frac{1}{2}Mv^2$, where $M$ is the total mass of the gas and box, and $v$ is the velocity of the box.

However, if the box is not moving, this does not mean that the gas particles have 0 KE. Each gas particle is still moving, but since the direction of the gas particles is essentially random, in effect the vector sum of their velocities is close to 0.

This implies that the gas particles have internal kinetic energy even when the box is at rest. This internal energy is $\sum\limits_i^N \frac{1}{2}m_i v_i^2$, the sum of the KE of each gas particle.

### Temperature

The thermodynamic temperature is a measure of this internal KE of the gas system. This thermodynamic temperature is measured in units of Kelvin (K). At 0 K, the gas has 0 internal KE. At temperature $T$, the gas has internal KE $U = \frac{3}{2}nRT$ or $U = \frac{3}{2}NKT$.

• Thermal equilibrium does not mean no thermal energy is being transferred; it means thermal energy transfer in both directions occurs at the same rate.

This may not be intuitive for differently-sized systems, where one has clearly more total energy than the other (but the same average KE per particle for both)

This law enables us to determine the direction of heat transfer, from a logical extension of energy conservation and the formulaic definition of temperature.

• Students are likely to consider total quantity of energy in the system, rather than “density” of energy/average energy per particle.

• ### Heat transfer

Sign/direction of ΔQ and ΔW.

Various types of energy-change processes: isobaric, isochoric, isothermal, adiabatic.

• Conservation of energy is a subject-wide concept, and its specific consequences in each topic cannot be under-emphasised. It is the underlying assumption of many concepts in Physics.

• Entropy and its related equations.

• ### Internal energy

Internal energy can be analysed as PE and KE.

PE arises from particle interactions.
KE arises from particle movement.
Together, they describe all of the substance’s internal energy.

It is a simple but powerful basis for understanding energy: all energy is ultimately potential or kinetic in nature.

• ### Work done

Doing work on the system involves a change in volume
$\Delta W = ‒p\Delta V$

• ### Boltzmann distribution

Particle velocity follows Boltzmann distribution.
Derivation of Boltzmann distribution; likelihood of each energy state being occupied, etc.

Thinking of particle velocities as a distribution is necessary to explain some phenomena, e.g. evaporation rate on windy day.

• ### Root-mean-square velocity

Velocities of particles are semi-randomly distributed.

ΣKE = KE_1 + KE_2 + …
= 0.5m(v_12 + v_22 + …)
= 0.5mv_r^2
v_r = (v_12 + v_22 + …)^0.5

v_r is the velocity all particles must have for the substance to have the same KE.

• This enables us to calculate energy requirements for various daily tasks/processes, and is important for calculating energy efficiency.
A standard amount of energy is transferred for any process, and specific heat capacity is one way for us to calculate this amount of energy for the process of heating/cooling.

• Specific heat capacity involves particle KE; particle PE is not affected (or its effects are not considered).

• Specific heat capacity does not change with mass.
The same substance in different states has different heat capacities as well; students often mix them up.
Two substances of different heat capacities, when fused, have an equivalent heat capacity that depends on their mass proportions.

• This enables us to calculate energy requirements for various daily tasks/processes, and is important for calculating energy efficiency.
A standard amount of energy is transferred for any process, and specific heat capacity is one way for us to calculate this amount of energy for these processes (e.g. melting and boiling).

• Specific latent heat involves particle PE; particle KE is not affected (or its effects are not considered).

• Specific latent heat does not change with mass.
Students frequently mix up specific latent heat of fusion and vapourisation.

• Specific heat of vapourisation is greater than specific heat of fusion.

Boiling/evaporation involves more energy than freezing/melting; an understanding of this is important to understand the cooling effects of water (as used daily or in industry), which is also mentioned in other sciences (i.e. biology, chemistry).

• ### Types of variables and expressions

• Physical constants represent properties of objects/systems, and may either have units or be dimensionless.

• Numerical coefficients are considered to have no unit and can be ignored in homogeneity checks.

• Logarithmic variables sometimes have units, but they are always dimensionless.

• e.g. decibels (dB)
• Trigonometric ratios are always dimensionless.

• ### Rules of homogeneity-checking

• Only SI base units may be used. All other units must be converted to SI base units using known standard formulas.
• Two expressions cannot be added unless they have the same base units.
• When two expressions are multiplied, their units must likewise be multiplied.
• The notation ‘[ ]’ means ‘units of’. It should not be used with units, only with variables.
• Units on the LHS and RHS of an equation must match for it to be homogeneous.
• An ‘order of magnitude’ refers to a change by a factor of about 10 (1 decimal place).
Broadly speaking, within the same order of magnitude, changing most quantities do not result in a significant change in phenomenon and observations.

• Order-of-magnitude considerations are particularly important for scientists and engineers when it comes to dimensionality.

• Car
• Laptop
• Refrigerator
• Aircon
• Smartphone
• ### Absolute vs fractional error

Absolute error is the numerical value of the instrument’s precision, with an associated unit.

• e.g. ±1mm, ±0.1cm

Fractional error is the proportion of absolute error to the reading/measurement. It is given in %, or as a number.

• e.g. 10%, 0.1

Fractional error increases when the reading/measurement decreases, and decreases when the reading/measurement increases. For this reason, we try to measure large values so as to reduce fractional error.

• ### Significant figures (s.f.)

In calculation of variables with values of different s.f., use the lower s.f.

Reasoning:
ds/s = 0.01 for 2 s.f., 0.001 for 3 s.f., etc
When 2 variables a (2s.f.) and b (3 s.f.) are multiplied/divided, error = da/a + db/b = 0.01 + 0.001 = 0.011 ~= 0.01 (2 s.f.)
Therefore use the lower s.f.
In calculation of variables with values of different s.f., use the lower s.f.

Reasoning:
ds/s = 0.01 for 2 s.f., 0.001 for 3 s.f., etc
When 2 variables a (2s.f.) and b (3 s.f.) are multiplied/divided, error = da/a + db/b = 0.01 + 0.001 = 0.011 ~= 0.01 (2 s.f.)
Therefore use the lower s.f.

• ### Graphical methods

1. Draw a tangent to the curve at the point
2. Mark out two points, calculate the vertical and horizontal displacement
3. Calculate the gradient of the tangent.

Area

1. Estimate the number of 5×5 squares, by counting.
2. Calculate the quantity represented by one 5×5 square (a unit square).
3. Multiply the number of squares by the unit-square quantity.
• ### Numerical methods

The gradient of a graph can be taken at one point only of the graph is smooth and continuous (and ideally if its function is known).

If the data points are discrete (non-continuous), we are unable to take the gradient at points of discontinuity. We will have to take the average gradient over short intervals.

• ### Misconceptions

• When presented with an $a$-$t$ graph showing increasing/decreasing acceleration, students treat it mentally as increasing/decreasing velocity.
• A misleading idea is the impression that if $a$ increases, $v$ must increase as well, and vice-versa implied.
• Broad “truths”, e.g. “when $v$ = 0, $a$ = 0” (true only when $v$ is 0 and constant).
• Probe students and ask them to explain. Often these misconceptions dispel themselves once they come to light.
• ### Situations without constant $a$

• Free-fall in the presence of air resistance

### Assumptions/results that no longer hold without air resistance

• ground-to-peak time = peak-to-ground time
• ### Graphical derivation

Image: Standard $v$-$t$ graph

1. $a = \frac{v - u}{t}$:
Seen from the form for gradient of line
2. $s = ut + \frac{1}{2}at^2$:
Split the area under graph into a bottom rectanglar section and an upper triangular section.
Area of rectangle = $ut$, area of triangle = $\frac{1}{2}(v-u)(t) = \frac{1}{2}at^2$ (subbed from 1).
3. $v^2 = u^2 - 2as$:
“Average” the area under graph by flattening it.
Height of flattened section: $\frac{1}{2}(v+u)$
Width of flattened section: $t$, or $\frac{v-u}{a}$ (subbed from 1 to eliminate $t$)
Area under $v$-$t$ graph is displacement, represented by $s$:
$s = \frac{1}{2}(v+u)\times\frac{v-u}{a} = \frac{v^2 - u^2}{2a}$
• ### Misconceptions

• A common impression is that each equation fits a different scenario (Equation 1 for this kind of problem, Equation 2 for another kind, …), and one only needs to identify which scenario is for which equation.

• All three equations essentially refer to the same kinematic picture.
• If you use the ‘wrong’ equation, you get a different result.

• The three equations are internally consistent (if the assumptions hold), and any of them can be used, though some will be more useful in some situations than others.
• The graphical method and the equation method are completely different methods of solving, and cannot be used together.

• The equations and graphs refer to the same kinematic picture, and mean the same thing. They are different ways of looking at this kinematic picture, with emphasis on different things.
• The equations are “magical formulas”; if you use them in a different form, they won’t work.

• The equations are algebraic representations of the graphical forms, and can be manipulated just like other algebraic expressions. However, students have to be clear about the use of variables.
• Students might have calculation shortcuts that they might mistake for concepts.
• ### General results

• If energy is conserved, an object always has the same speed at the same height.

• Speed of an object in free-fall is not affected by distance travelled (is not path-dependent) if there is no external force.

• An object starting from one point will always have the same speed at another point, no matter how it gets there.
• ### Graphical scaffolding

Start with the $a$-$t$ graph, then multiply the y-axis by $m$.

• What does the y-axis represent now?
• What does the gradient of the graph represent?
• What does the area under the graph represent?

This exercise can help to link $F = \frac{dp}{dt}$ with the $F$-$t$ graph.

• ### Framework of analysis:

1. Draw in all non-contact forces (typically $W$)
2. Observe all objects in contact with body, by going around the perimeter of the object. Label their forces where relevant.
• ### Thinking exercises

Taupok

1. Would you rather be at the top or at the bottom?
2. Why?
3. Explain using FBDs.
• ### Demonstrations

1. Get students to predict and explain mass distribution of slinky held vertically by the top end.
2. Why is it more stretched out near the top than bottom?
3. Explain using FBD.

1. Get students to predict behaviour of falling slinky.
• Bottom springs up first
• Top springs down first
• Both fall at the same time
2. Drop slinky, get students to observe.
3. Explain using FBD.

Observation: Quite a number of students expect both top and bottom to fall together.

• ### Single-body systems

$F = ma$ is a subset of this general equation that can only be applied to single-body examples. In continuous flow examples (e.g. water flowing through a pipe, sand on a conveyor belt), there is no single body ($m$) under consideration, and it has to be analysed as a rate of change of $p$.

• ### Pitfalls

1. An insufficient understanding and intuition of momentum may hamper understanding of force as dp/dt.

2. Intuiting the effects of $a$ vs $v$ at this point is probably still not natural to students yet. Might want to emphasise that the $a$ experienced by the body does not translate directly to $v$.

• ### Force pairs

The first thing that N3L implies is that forces never act alone; they always act in pairs. And these force pairs have the following properties:

1. They have the same magnitude.
2. They act in opposite directions.
3. They act on a pair of bodies.
A exerts a force on B, B exerts a force back on A. The forces in a force pair must never be drawn acting on the same body.
4. The forces must be of the same type.
If A exerts a frictional force on B, B exerts a frictional force back on A.

### Force pairs in an FBD

We only draw one of the force pairs in an FBD because of condition 3. If you have both forces in a pair drawn in the same FBD, something is terribly wrong!

• ## FBDs for systems with more than one body

FBDs are usually drawn with only one bod, but sometimes we combine multiple bodies into a system and analyse that system with a FBD. We can do this when all the bodies in the system behave identically, i.e. when these bodies always have the same velocity and acceleration (for example if they are tied or glued together).

When we do this, we can ignore internal forces in the system. These internal forces are force pairs exerted between bodies within the system. We are able to do this because internal forces do not affect the behaviour of the system. This allows us to simplify the system FBD without missing any crucial information.

• #Superelastic collisions

Superelastic collisions are collisions where the total KE of the system increases as a result of energy converted/transferred from other sources.

Typical examples:

• Exploding bodies
• #Elastic collisions

Elastic collisions are collisions where the total KE of the system is conserved before and after an elastic collision, if no external forces act on the system.

This means that all the PE stored during the collision is fully restored as KE after the collision.

• #Inelastic collisions

Inelastic collisions are collisions where the total KE of the system is not conserved. The total KE of the system before the collision is less than the total KE of the system after the collision.

This means that some of the PE stored during the collision is lost to the environment, and not all of it is restored as KE after the collision.

• #Completely inelastic collisions

Completely inelastic collisions are inelastic collisions where the colliding bodies stick together (coalesce) after the impact, and move off with a common velocity.

Since the bodies do not separate after the collision, none of the Pe stored during the collision is restored as KE, and all of it is lost to the environment or converted to other forms of energy.

• Students typically have some intuitive understanding of this principle, but its applications are so wide-ranging that they might have difficulty applying it in various situations. Variety is key.

• Begin without centripetal force: object is initially in linear motion.

What is needed for uniform circular motion? Since speed is constant, acceleration must be perpendicular to velocity. In circular motion, acceleration would thus also point towards the centre (since velocity is tangential)

How much acceleration is required? $\frac{mv^2}{r}$. This is centripetal acceleration, the required acceleration for uniform circular motion of radius r and velocity v.

What provides centripetal acceleration? A resultant force, which may be provided by gravitational force, tension, normal reaction force, etc, or a combination of these. This resultant force causes centripetal acceleration and is thus known as centripetal force.

• The word “critical” in physics describes thresholds in behaviour, e.g. critical angle in refraction of light, or critical mass in nuclear chain reaction. This change in behaviour is usually drastic and happens over a very short threshold.

• Pseudo-real-life examples probably pose the friendliest learning curve. Get students to imagine the following scenarios:
1) They are trying to improve the performance of their bike suspension by using different kinds of oils in the spring-loaded tube. Which thickness/consistency of oil provides the fastest return to equilibrium position?
2) You have a set of swinging doors in your kitchen/office/room/wherever. You don’t like it when thy keep swinging for a long time after someone walks through, nor do you want them to take too long to return to rest. How tight should the hinges be, or how much frictional force should they exert?

• Use Algodoo simulation to show critical damping and how it relates to light, heavy damping?

• Students will tend to associate T with U, and hence with PE and KE both. Will need scaffolding to bring forth the idea of ideal gas having 0 PE.

Very easy to confuse heat (energy) with temperature.

• Adding heat might not raise temperature (e.g. during phase change)
• root-mean-square concept is used again in AC electricity, where r.m.s. voltage/current is used to calculate average AC power.

• T also indicates KE in other degrees of freedom, e.g. rotational, vibrational.

• Get students to explain, if they think so, why two systems at the same temperature might still exchange energy.
Simulations might be useful to illustrate that this occurs for any two system of colliding particles.

• Questions linking other concepts, e.g. describing changes to one system (from 1st Law), and then asking if it is in thermal equilibrium with another system.

• This topic is new to students, might take more care in explanation.
Might need to make assumptions explicit, e.g. when are ΔU, ΔQ and ΔW equal to 0, and why.

• There are more detailed breakdowns for PE and KE.
In abstract form, PE varies depending on how we define particle interactions. These interactions are defined as potentials.

Potential energy (“that which is yet to be”) is a very abstract concept, can’t be easily visualised or seen in motion. Many analogies may be needed.

Students might pull in other forms of energy they don’t perceive as being either PE or KE.

Might have to be clear about what PE, KE refer to.
PE → Any interaction (gravitational, electrostatic, etc)
KE → Movement of individual particles.
KE of gas → sum of particle KEs.

• Work done on/by the system involves the same underlying concept as work done on/by a moving object: applying force over a distance.

• There is a standard definition of work done on/by a gas in physics, and it is important to differentiate this from other forms of energy change (e.g. isobaric change). This is important in understanding which processes involve mechanical input and output, critical for engineering.

• Pressure is assumed to be constant. If it is not, students might not know how to calculate an equivalent pressure.
Change in volume is assumed to take place at constant pressure; if this cannot be assumed, the formula cannot/should not be used.

• What is v_r for? Students will most likely ask.
Might need worked examples. From KE = 3/2KT maybe? Or avg. v vs vr in relation to T.

The point/purpose of v_r will probably be vague or ambiguous until worked examples are given.
Should also show more examples of avg. v being inadequate. Graphs of v vs KE, for instance.

This particle distribution concept is used in chemistry as well, to explain why endothermic reaction rate increases with temperature.

• Students have learned this at O levels; A levels mainly adds measurement of s.h.c. by electrical methods. Do not need to go into too much concept detail.
Focus on applications that involve other topic concepts, e.g. conversion of energy, efficiency calculations, etc.

• Students have learned this at O levels; A levels mainly adds measurement of s.l.h. by electrical methods. Do not need to go into too much concept detail.
Focus on applications that involve other topic concepts, e.g. conversion of energy, efficiency calculations, etc.

• The difference between s.h.v. and s.h.f. can be (naïvely) calculated for any substance using the 1st Law of thermodynamics.

• Both specific heats involve a change in phase, and students may perceive them as being the same process, failing to see that vapourisation involves a much greater change in volume and particle PE.

• Students might not be aware that specific heat of vapourisation involves work done against atmospheric pressure; drawing link to formula could be useful if properly scaffolded.

• Dimensionless quantities might be a bit harder to teach
E.g. angular displacement, counts/frequency, fixed quantities such as mol

Might be difficult getting students to accept the idea that though they don’t have SI units, there is still something being counted/measured

• Students have probably been drilled beforehand into overemphasising the exact answer (to 3 s.f.), and are uncomfortable with reasoned estimations.

Students most likely turn to the calculator straight away when they encounter an arithmetic problem.

• Two main types of completely inelastic collisions tend to appear in exams:

• Losses due to friction (parcel falling into trolley)
• Losses due to mechanical compression/deformation (i.e. pin-and-cork trolleys).
• Go through changes of phase again.
As heat is applied, what happens re. PE, KE? How does temp. change?
Students should be able to describe a transfer of heat to/from the substance, and how this manifests in physical properties (i.e. temperature, state of substance).

• Concept-based questions describing changes to temperature or to KE.
Calculation-based questions giving sample velocities, getting students to calculate r.m.s. velocity.
MCQ questions describing possible relationships between v_fms, v_avg, T, and KE.

• Introduce various examples of internal energy changes in systems, get students to describe explicitly what each change involves. Is thermal energy being transferred? Is work being done?
Once each example is described, calculate ΔU, ΔQ, ΔW for each example. Give sample numbers where necessary.

• Calculation questions involving calculation of one variable’s value given other variables’ values.
MCQ involving direction of ΔQ or ΔW, e.g. a) Work is done on the system, b) Work is done by the system, c) Thermal energy is transferred to the system, d) thermal energy is transferred away from the system.

• Draw analogies between gas PE and other PE, e.g. GPE.

• Objects have yet to move together, but are able to do so due to interactions between objects.

Ask students to surface types of energy they are unable to categorise, help them to categorise.

This process should help students see similarities between PE and KE as defined in the kinetic model, and in other topics e.g. gravitation.

• Multiple-choice questions:
1) PE increases
2) KE increases
3) U increases
Give various examples, get students to pick from combinations of the above.
Likely problems: Neither PE nor KE increases though U increases.
U is 0 but one of the other two increases.

• Work done is the definite integral of pressure vs volume; we usually assume constant pressure (or processes with minimal change in pressure), or take the average pressure as indicative of the equivalent pressure.

• Show how a uniform velocity distribution ends up after some collisions.
Explain aspects of distribution

• few high-KE particles
• mostly mid-KE
• Concept-based questions asking about velocities of particles. E.g.
As water is heated, a) the number of particles above x speed increases, b) the number of particles below y speed increases, c) the speed of each particle increases proportionally, …

• Recap O level content. Get students to think about what is happening to internal energy; how are PE, KE being affected?

• Questions based on calorimetry, s.h.c. measurement.
Questions involving thermal energy change causing liquids to heat up/cool down.

• Recap O level content. Get students to think about what is happening to internal energy; how are PE, KE being affected?

• Questions based on calorimetry, s.l.h. measurement.
Questions involving thermal energy change causing liquids to change state.

• Make students aware of the mechanical processes involved in melting and boiling.
What is happening to the particles? What prevents them from expanding naturally without any thermal energy input? How do they overcome these “energy obstacles”?

• Structured, explanation-type questions asking students to explain related phenomena.

• Draw explicit links to this earlier equation. Taking W = F.s on a per-unit-area average gives us the formula.
Similar to 1st Law: introduce various examples, calculate ΔW for each case. Throw in some cases where pressure is non-constant, or volume change does not take place under constant pressure.

• MCQ involving direction of ΔW, e.g. a) Work is done on the system, b) Work is done by the system.
Structured questions asking for equation assumptions/conditions, or to explain the equation (why minus sign).

{"cards":[{"_id":"46f04e0d4ad99b488e0000de","treeId":"45f4536692911004a000003d","seq":2954208,"position":0.5,"parentId":null,"content":"# Introductory Physics Education\n\nAn overview to pedagogical perspectives and approaches for A Level/introductory university physics."},{"_id":"46f04faf4ad99b488e0000df","treeId":"45f4536692911004a000003d","seq":2954236,"position":1,"parentId":"46f04e0d4ad99b488e0000de","content":"## Active Learning\n\nKeep students actively engaged, provide rapid feedback\n\n1. Students spend class time actively engaged in doing/thinking/talking\n2. Students interact with peers\n3. Students receive immediate feedback\n4. Instructor facilitates\n5. Students take responsibility for their knowledge"},{"_id":"46f0546a4ad99b488e0000e4","treeId":"45f4536692911004a000003d","seq":2954296,"position":1,"parentId":"46f04faf4ad99b488e0000df","content":"### Use the classroom for ...\n\n+ discussion rather than presentation\n+ clarification rather than solution\n+ demonstration rather than description\n+ (instructor-guided) practice rather than copying"},{"_id":"46f059244ad99b488e0000e5","treeId":"45f4536692911004a000003d","seq":2954298,"position":2,"parentId":"46f04faf4ad99b488e0000df","content":"### Don't give topic overviews\n\n1. Lay out expectations and reasons for lesson structure, continually repeat for emphasis and reminding as the term progresses\n - Do not repeat factual knowledge, expect students to remember it\n - Teach students how to read textbook/notes.\n - Get them to keep a notebook?\n3. Give a day-to-day schedule and stick to it.\n - Students need to know when to start each chapter\n4. Don't relent!\n - Give short reading quizes to ensure students have done their reading\n - Give vocab/terminology quizzes for quick checking"},{"_id":"46f069744ad99b488e0000e6","treeId":"45f4536692911004a000003d","seq":2954303,"position":3,"parentId":"46f04faf4ad99b488e0000df","content":"### Doing the physics\n\n1. Focus on the reasoning and principles, not on the mathematics\n2. Students need to give explanations that aren't just restating the problem.\n3. Pay extra attention to discussing, justifying, and clarifying explanations\n - Give examples, and get students to apply the principles\n4. Demonstrate and go through extended examples of qualitative reasoning and explaining\n5. Strategies should emphasise qualitative steps, by describing and clarifying assumptions and thought\n6. Homework should be given that requires analysis and interpretation of problems."},{"_id":"46f051594ad99b488e0000e0","treeId":"45f4536692911004a000003d","seq":2954237,"position":2,"parentId":"46f04e0d4ad99b488e0000de","content":"## Phenomena to theory\n\nFocus on phenomena rather than abstractions"},{"_id":"46f06fd14ad99b488e0000e7","treeId":"45f4536692911004a000003d","seq":2954305,"position":1,"parentId":"46f051594ad99b488e0000e0","content":"### Concrete to abstract\n\nImportant focusing questions: “How do we know ...?”, “Why do we believe ...?”\n\nStart with observations, then concepts and principles\n- Gradually generalise to theories\n- Apply the theory to phenomena, see if they explain the observations"},{"_id":"46f051e04ad99b488e0000e1","treeId":"45f4536692911004a000003d","seq":7782837,"position":3,"parentId":"46f04e0d4ad99b488e0000de","content":"## Deal explicitly with alternative conceptions\n\n1. Confront student misconceptions directly.\n2. Explore the fact that many predictions were wrong.\n3. Consider alternative models.\n4. Reiterate.\n\n"},{"_id":"46f0523b4ad99b488e0000e2","treeId":"45f4536692911004a000003d","seq":7782836,"position":4,"parentId":"46f04e0d4ad99b488e0000de","content":"## Teaching techniques\n\nTeach and use explicit problem-solving skills and strategies\n\n1. Teach students the specific skills needed to solve complex problems. These include interpretation skills, pictorial skills, graphical skills, and reasoning skills.\n2. Show students how those skills re assembled into a powerful problem-solving strategy and demonstrate their use, in detail, in the example problems we work in class.\n3. Make explicit the assumptions, decisions, and reasoning that are part of an expert's problem-solving strategy but which usually go unsaid.\n4. Help students organise their knowledge in a more coherent, hierarchical, easily searched structure."},{"_id":"46f07a1b4ad99b488e0000ea","treeId":"45f4536692911004a000003d","seq":2954371,"position":1,"parentId":"46f0523b4ad99b488e0000e2","content":"### Novices vs Experts\n\n**Novice students**\n1. Fail to describe problems adequately\n2. Do little prior planning or qualitative description\n3. Try to assemble solutions by stringing together miscellaneous math formulae\n4. Knowledge consists of loosely connected formulas\n5. Recognise pieces of info as single chunks, get overwhelmed by too many chunks\n\n\n**Experts**\n1. Structure their knowledge\n2. Use the structure of their knowledge to craft a strategy\n3. Recognise and work with larger patterns (fewer chunks)\n4. Draws sketches to clarify understanding of problem before solving"},{"_id":"46f0830a4ad99b488e0000eb","treeId":"45f4536692911004a000003d","seq":2954372,"position":2,"parentId":"46f0523b4ad99b488e0000e2","content":"### Scaffolding\n\n1. Make thought processes and decision-making explicit\n2. Make intuition visible—don't make it a case of “using the right equation”\n3. Teach problem-solving skills\n4. Give students opportunities to help students organise knowledge coherently"},{"_id":"46f052f74ad99b488e0000e3","treeId":"45f4536692911004a000003d","seq":7782852,"position":5,"parentId":"46f04e0d4ad99b488e0000de","content":"## Problem-solving\n\nGive problems that go beyond symbol manipulation\nEngage students in qualitative and conceptual analysis of physical phenomena\n\nAppropriate homework and example problems need to:\n1. Balance qualitative and quantitative reasoning.\n2. Emphasise reasoning, de-emphasise formulas and equations.\n3. Deal directly with phenomena and observations. Derivations and \"Show that ...\" problems have little efficacy for students at this level."},{"_id":"46f073cd4ad99b488e0000e9","treeId":"45f4536692911004a000003d","seq":2954377,"position":1,"parentId":"46f052f74ad99b488e0000e3","content":"### Homework\n\n*“A few well-written problems studied carefully—with feedback and, sometimes, in-class discussion—-are a far better learning experience than many problems practised blindly.”*\n\n1. Worksheets should serve as a template for following the steps of the strategy\n2. Model the strategy and show all steps\n3. Insist that students follow and solve the worksheet problems\n4. Students need to practise word-to-symbol translation independently of the calculations\n - Practise individual steps separately, then approach full problems near the end of the chapter"},{"_id":"45f4543592911004a0000040","treeId":"45f4536692911004a000003d","seq":2954209,"position":1,"parentId":null,"content":"# 01a Measurements\n\n* Readings and measurements\n* Systematic vs random errors\n* Accuracy vs precision"},{"_id":"4b24248895e5163d2d000103","treeId":"45f4536692911004a000003d","seq":2954393,"position":0.5,"parentId":"45f4543592911004a0000040","content":"## Introduction\n\nMan first started learning to measure space, and then gradually time. What other things could we measure, and what do they represent?"},{"_id":"4602a2fd4b4bc7be43000048","treeId":"45f4536692911004a000003d","seq":2954395,"position":1,"parentId":"45f4543592911004a0000040","content":"## A standardised system of units\n\nAll *quantities* we try to express have to be given *units* to have meaning.\n\nScientists have to be very precise when presenting their results, so that they can agree on what an observation means. Thus, they need to use a *standardised system of units*.\n\n#Base units & homogeneity\n\nBase units are **physical quantities** agreed to be fundamental enough that other units may be derived from them.\n\nThese units are used in physics equations, which must be **homogeneous** to be valid. This means that units on both sides of the equation must match up.\n\nSI unit prefixes provide a compact, economical way to represent very large or very small numbers."},{"_id":"4602ab5f4b4bc7be43000052","treeId":"45f4536692911004a000003d","seq":2954403,"position":2,"parentId":"4602a2fd4b4bc7be43000048","content":"### SI base units\n\n**SI base units** are *standardised* and carefully defined so as to be *constant* in as many scenarios as possible.\n\nThey represent the most *fundamental properties/measurables* we can observe in experiments.\n\nThe **SI base units** are\n* mass (kg)\n* time (s)\n* length (m)\n* electric current (A)\n* thermodynamic temperature (K)\n* amount of substance (mol)\n* luminous intensity (cd)"},{"_id":"4b079342f8f3ec95df000103","treeId":"45f4536692911004a000003d","seq":2954405,"position":3,"parentId":"4602a2fd4b4bc7be43000048","content":"### Checking of homogeneity\n\nThe **base units** are the fundamental units by which we check the consistency of equations.\n\nAny two expressions must have the same **base units** before they can be added or subtracted.\n\nAn equation in which the **base units** are consistent is said to be **homogeneous**."},{"_id":"503c37d4a4b1c36f8800011a","treeId":"45f4536692911004a000003d","seq":2954407,"position":0.5,"parentId":"4b079342f8f3ec95df000103","content":"### Types of variables and expressions\n\n* Physical **constants** represent properties of objects/systems, and may either have units or be **dimensionless**.\n\n* Numerical **coefficients** are considered to have *no unit* and can be ignored in homogeneity checks.\n\n* Logarithmic variables sometimes have units, but they are always **dimensionless**.\n * e.g. decibels (dB)\n\n* Trigonometric ratios are always **dimensionless**.\n"},{"_id":"503c4e34a4b1c36f8800011b","treeId":"45f4536692911004a000003d","seq":1673767,"position":1,"parentId":"503c37d4a4b1c36f8800011a","content":"**Dimensionless** quantities might be a bit harder to teach\nE.g. angular displacement, counts/frequency, fixed quantities such as mol\n\nMight be difficult getting students to accept the idea that though they don't have SI units, there is still *something* being counted/measured"},{"_id":"503c2fa9a4b1c36f88000119","treeId":"45f4536692911004a000003d","seq":2954408,"position":1,"parentId":"4b079342f8f3ec95df000103","content":"### Rules of homogeneity-checking\n\n* Only **SI base units** may be used. All other units must be converted to **SI base units** using known standard formulas.\n* Two expressions cannot be added unless they have the same **base units**.\n* When two expressions are multiplied, their units must likewise be multiplied.\n* The notation ‘[ ]’ means ‘units of’. It should not be used with units, only with variables.\n* Units on the LHS and RHS of an equation must match for it to be **homogeneous**."},{"_id":"4602a37e4b4bc7be4300004a","treeId":"45f4536692911004a000003d","seq":2954396,"position":3,"parentId":"45f4543592911004a0000040","content":"## Estimation\n\nIn physics, estimating the right order of magnitude is often more important than getting the quantity exactly right"},{"_id":"4602af504b4bc7be43000057","treeId":"45f4536692911004a000003d","seq":1673557,"position":2,"parentId":"4602a37e4b4bc7be4300004a","content":"Back-of-the-envelope calculations and mental feasibility evaluations rely heavily on **order-of-magnitude** estimations for their effectiveness, and they are a very useful system for quick estimations, where exact values are not so important (and often not really knowable)."},{"_id":"4602b0454b4bc7be43000058","treeId":"45f4536692911004a000003d","seq":894511,"position":1,"parentId":"4602af504b4bc7be43000057","content":"An ‘order of magnitude’ refers to a change by a factor of about 10 (1 decimal place).\nBroadly speaking, within the same order of magnitude, changing most quantities do not result in a significant change in phenomenon and observations."},{"_id":"4602b1274b4bc7be43000059","treeId":"45f4536692911004a000003d","seq":305371,"position":2,"parentId":"4602af504b4bc7be43000057","content":"Order-of-magnitude considerations are particularly important for scientists and engineers when it comes to dimensionality."},{"_id":"4602b1954b4bc7be4300005a","treeId":"45f4536692911004a000003d","seq":305373,"position":1,"parentId":"4602b1274b4bc7be43000059","content":"Students have probably been drilled beforehand into overemphasising the exact answer (to 3 s.f.), and are uncomfortable with reasoned estimations.\n\nStudents most likely turn to the calculator straight away when they encounter an arithmetic problem."},{"_id":"503bd0291cae506498000109","treeId":"45f4536692911004a000003d","seq":1673559,"position":3,"parentId":"4602a37e4b4bc7be4300004a","content":"Students ought to know some basic numbers from everyday experience, e.g. size of car, power consumption of appliances."},{"_id":"503bd2201cae50649800010a","treeId":"45f4536692911004a000003d","seq":1673560,"position":1,"parentId":"503bd0291cae506498000109","content":"* Car\n* Laptop\n* Refrigerator\n* Aircon\n* Smartphone"},{"_id":"503be415a4b1c36f8800010e","treeId":"45f4536692911004a000003d","seq":2954398,"position":4,"parentId":"45f4543592911004a0000040","content":"## The phenomenon of uncertainty\n\nAll readings and measurements involve a certain degree of **uncertainty**, regardless of the measuring instrument’s accuracy.\n\nAt the quantum level, there is a limit to the precision with which we can measure some property-pairs simultaneously. This natural phenomena means, for instance, that we cannot measure the position and momentum of a subelementary particle with zero uncertainty **simultaneously**."},{"_id":"503bee64a4b1c36f8800010f","treeId":"45f4536692911004a000003d","seq":2954429,"position":1,"parentId":"503be415a4b1c36f8800010e","content":"### Error vs uncertainty\n\nIn addition to a measurement’s **uncertainty**, the incorrect or imprecise use of an instrument can contribute **error** to the measurement.\n\nFor instance, our human inability to respond to a signal immediately means there will always be **human error** in our use of a stopwatch."},{"_id":"503bf38aa4b1c36f88000110","treeId":"45f4536692911004a000003d","seq":2954430,"position":2,"parentId":"503be415a4b1c36f8800010e","content":"### Instrument error vs measurement uncertainty\n\nIt is often easier to know the *error of an instrument* rather than the *uncertainty of a measurement*. This is because we can design instruments to be **precise** to a certain degree.\n* Rulers are designed to be **precise** to the nearest 1mm, vernier calipers to the nearest 0.01mm, micrometers to the nearest 0.001mm (or smaller!)"},{"_id":"503bfa16a4b1c36f88000111","treeId":"45f4536692911004a000003d","seq":2954431,"position":3,"parentId":"503be415a4b1c36f8800010e","content":"### Error calculation\n\nFor complex expressions involving the addition, subtraction, multiplication, or division of multiple variables, we have standard ways of calculating the error in the final value."},{"_id":"503c0cb2a4b1c36f88000113","treeId":"45f4536692911004a000003d","seq":2954432,"position":2.25,"parentId":"503bfa16a4b1c36f88000111","content":"### Absolute vs fractional error\n\n**Absolute error** is the *numerical value* of the instrument’s precision, with an associated unit.\n* e.g. ±1mm, ±0.1cm\n\n**Fractional error** is the *proportion* of absolute error to the reading/measurement. It is given in %, or as a number.\n* e.g. 10%, 0.1\n\n**Fractional error** increases when the reading/measurement decreases, and decreases when the reading/measurement increases. For this reason, we try to measure large values so as to reduce **fractional error**."},{"_id":"503c1ef2a4b1c36f88000118","treeId":"45f4536692911004a000003d","seq":2954436,"position":4,"parentId":"503bfa16a4b1c36f88000111","content":"### Significant figures (s.f.)\n\nIn calculation of variables with values of different s.f., use the lower s.f.\n\nReasoning:\nds/s = 0.01 for 2 s.f., 0.001 for 3 s.f., etc\nWhen 2 variables a (2s.f.) and b (3 s.f.) are multiplied/divided, error = da/a + db/b = 0.01 + 0.001 = 0.011 ~= 0.01 (2 s.f.)\nTherefore use the lower s.f.\nIn calculation of variables with values of different s.f., use the lower s.f.\n\nReasoning:\nds/s = 0.01 for 2 s.f., 0.001 for 3 s.f., etc\nWhen 2 variables a (2s.f.) and b (3 s.f.) are multiplied/divided, error = da/a + db/b = 0.01 + 0.001 = 0.011 ~= 0.01 (2 s.f.)\nTherefore use the lower s.f.\n"},{"_id":"4602a4624b4bc7be4300004c","treeId":"45f4536692911004a000003d","seq":2954399,"position":5,"parentId":"45f4543592911004a0000040","content":"## The process of measurement\n\nMeasurement sounds simple, but the actual process of getting a number can sometimes be quite complicated.\n\nEven after we obtain a number, what does that number mean? Will we get the same number again if we repeat the process?\n\nA ‘perfect’ measurement does not exist, since perfect objects do not exist. The **true value** of a measurement is unknowable."},{"_id":"503c9aaba4b1c36f8800011d","treeId":"45f4536692911004a000003d","seq":2954444,"position":1.5,"parentId":"4602a4624b4bc7be4300004c","content":"### Readings and measurements\n\nIn scientific terminology, a distinction is drawn between (raw) **readings** and **measurements**.\n\nA **reading** is a number obtained from using an instrument (the actual number recorded during an experiment).\n\nA **measurement** is the result of analysing readings, its accuracy depending on the measuring instrument and how it is used."},{"_id":"4602a51f4b4bc7be4300004f","treeId":"45f4536692911004a000003d","seq":2954401,"position":6,"parentId":"45f4543592911004a0000040","content":"## Accuracy vs precision\n\n**Accuracy** refers to how closely a measured value agrees with the true value.\n\n**Precision** refers to how closely individual measurements agree with each other without reference to true value."},{"_id":"4602bcc6772e023bdc00002c","treeId":"45f4536692911004a000003d","seq":2954445,"position":1,"parentId":"4602a51f4b4bc7be4300004f","content":"### Visual explanations\n\nA set of measurements with a *wide spread* is **imprecise**; a *narrow spread* is **precise**.\n\nA set of measurements with an **average value** that is *far from the true value* is **inaccurate**; if it is *near to the true value*, it is **accurate**."},{"_id":"4602a4c24b4bc7be4300004e","treeId":"45f4536692911004a000003d","seq":2954446,"position":3,"parentId":"4602a51f4b4bc7be4300004f","content":"### Systematic vs random errors\n\nErrors are broadly classified as **systematic errors** and **random errors**, not mutually exclusive.\n\n**Systematic errors** yield a definite pattern, **consistently over- or under-estimating** the **true value**.\n\n**Random errors** do not yield a definite pattern; measurements are **randomly\n greater or less than** the **true value**."},{"_id":"45f4549092911004a0000041","treeId":"45f4536692911004a000003d","seq":2954211,"position":2,"parentId":null,"content":"# 01b Vectors"},{"_id":"45f454c892911004a0000042","treeId":"45f4536692911004a000003d","seq":2954213,"position":3,"parentId":null,"content":"# 02 Kinematics\n\n> “science of motion,” 1840, from French *cinématique* (Ampère, 1834), from Greek *kinesis* “movement, motion”\n\n**The study of motion (of bodies)**\n\n* Displacement, velocity, acceleration\n* Instantaneous & average qty\n* Equations\n* General observations\n* 2D kinematics\n"},{"_id":"4602c075772e023bdc00002f","treeId":"45f4536692911004a000003d","seq":2954449,"position":1,"parentId":"45f454c892911004a0000042","content":"## Terms of physical motion\n\nIn physics, we often study the motion of various objects and systems. We need words that can help us describe these motions precisely.\n* A bouncing ball\n* A falling stone\n* An oscillating spring\n* etc\n\nThus, we need to define common terms so that when we use these terms, other people will understand what we are trying to say, and vice-versa."},{"_id":"4602c696772e023bdc00003a","treeId":"45f4536692911004a000003d","seq":2954457,"position":1,"parentId":"4602c075772e023bdc00002f","content":"### Displacement\n\n**Displacement** is the straight-line distance between two points.\n\n**Distance** is the actual length of the path taken between two points.\n\nSymbol: $s$\n\n\n\n### Velocity\n\n**Velocity** is the *rate of change* of **displacement**.\n\nSymbol: $v = \\frac{ds}{dt}$\n\n\n\n### Acceleration\n\n**Acceleration** is the *rate of change* of **velocity**.\n\nSymbol: $a = \\frac{dv}{dt} = \\frac{d^2s}{dt^2}$"},{"_id":"503da2d90d8d29f95800021a","treeId":"45f4536692911004a000003d","seq":2954452,"position":1.625,"parentId":"45f454c892911004a0000042","content":"## Graphical representations\n\nWe use graphs to visualise how two quantities change in relation to each other.\n\nThe follow graphs are often encountered in the study of kinematics:\n\n* $s$-$t$ graphs\n* $v$-$t$ graphs\n* $a$-$t$ graphs\n* vertical vs. horizontal displacement ($s_y$-$s_x$) graphs"},{"_id":"4602c833772e023bdc00003c","treeId":"45f4536692911004a000003d","seq":2954458,"position":1,"parentId":"503da2d90d8d29f95800021a","content":"### Deriving the kinematics graphs\n\nThe **gradient** of the $s$-$t$ graph gives the $v$-$t$ graph,\nand the **gradient** of the $v$-$t$ graph gives the $a$-$t$ graph.\n\nConversely, the **integral** of the $a$-$t$ graph gives the **change in** the $v$-$t$ graph,\nand the **integral** of the $v$-$t$ graph gives the **change in** the $s$-$t$ graph."},{"_id":"4602c8c1772e023bdc00003d","treeId":"45f4536692911004a000003d","seq":1674347,"position":1,"parentId":"4602c833772e023bdc00003c","content":"###Graphical methods\n\n**Gradient**\n1. Draw a tangent to the curve at the point\n2. Mark out two points, calculate the vertical and horizontal displacement\n3. Calculate the gradient of the tangent.\n\n**Area**\n1. Estimate the number of 5×5 squares, by counting.\n2. Calculate the quantity represented by one 5×5 square (a unit square).\n3. Multiply the number of squares by the unit-square quantity."},{"_id":"4602c916772e023bdc00003e","treeId":"45f4536692911004a000003d","seq":1696690,"position":2,"parentId":"4602c833772e023bdc00003c","content":"###Numerical methods\n\nThe **gradient** of a graph can be taken at *one point* only of the graph is smooth and continuous (and ideally if its function is known).\n\nIf the data points are discrete (non-continuous), we are unable to take the gradient at points of discontinuity. We will have to take the *average* **gradient** over short intervals."},{"_id":"503dc5dd0d8d29f95800021b","treeId":"45f4536692911004a000003d","seq":2954461,"position":3,"parentId":"503da2d90d8d29f95800021a","content":"### Standard examples\n\n**Zero $a$**\n* *Horizontal* line on $v$-$t$ graph\n* *Horizontal* line on $v$-$t$ graph\n* *Straight* line on $s$-$t$ graph\n\n**Constant $a$**\n* *Horizontal* line on $a$-$t$ graph\n* *Straight* line on $v$-$t$ graph\n* Quadratic curve on $s$-$t$ graph\n\n**Constant $v$** (same as zero $a$)\n* *Horizontal* line at **a = 0** on $a$-$t$ graph\n* *Horizontal* line on $v$-$t$ graph\n* *Straight* line on $s$-$t$ graph"},{"_id":"4602c9c5772e023bdc000040","treeId":"45f4536692911004a000003d","seq":2954496,"position":0.5,"parentId":"503dc5dd0d8d29f95800021b","content":"### Misconceptions\n\n* When presented with an $a$-$t$ graph showing *increasing/decreasing* **acceleration**, students treat it mentally as *increasing/decreasing* **velocity**.\n* A misleading idea is the impression that if $a$ increases, $v$ must increase as well, and vice-versa implied.\n* Broad “truths”, e.g. “when $v$ = 0, $a$ = 0” (true only when $v$ is 0 and constant).\n * Probe students and ask them to explain. Often these misconceptions dispel themselves once they come to light."},{"_id":"4602c2c9772e023bdc000034","treeId":"45f4536692911004a000003d","seq":2954499,"position":2.25,"parentId":"45f454c892911004a0000042","content":"## Instantaneous & average quantities\n\n**Instantaneous** values are an ideal, often used only in derivation; in real life, we deal with values **averaged** over a short time window (since it is impossible to measure speed at a single point).\n\nStudents should understand this difference well so they understand the limitations of measurements and results obtained in kinematics.\n\n## Definitions\n\n**Instantaneous velocity**/**acceleration** refers to $v$/$a$ *at a particular point in time*, while **average velocity**/**acceleration** refers to $v$/$a$ *over a time interval*."},{"_id":"4602d2b7772e023bdc00004d","treeId":"45f4536692911004a000003d","seq":1697174,"position":1,"parentId":"4602c2c9772e023bdc000034","content":"*Instantaneous* **acceleration** is $\\frac{dv}{dt}$,\nwhile *average* **acceleration** is $\\frac{\\Delta v}{\\Delta t}$.\n\nThere are situations where instantaneous $a$ is **not constant**, though it seems to be, and therefore the kinematic equations cannot be applied."},{"_id":"4602d805772e023bdc000053","treeId":"45f4536692911004a000003d","seq":2954503,"position":2,"parentId":"4602d2b7772e023bdc00004d","content":"### Situations without constant $a$\n\n* Free-fall in the presence of air resistance\n\n### Assumptions/results that no longer hold without air resistance\n\n* ground-to-peak time = peak-to-ground time"},{"_id":"4602c347772e023bdc000035","treeId":"45f4536692911004a000003d","seq":2954506,"position":2.5,"parentId":"45f454c892911004a0000042","content":"## Equations\n\n$v = u + at\\\\\ns = ut + \\frac{1}{2}at^2\\\\\nv^2 = u^2 + 2as$\n\nThe kinematic equations only apply if **$a$ is constant throughout**.\n\nThe kinematic equations can only be applied to *linear motion*."},{"_id":"4602c4ad772e023bdc000036","treeId":"45f4536692911004a000003d","seq":2954508,"position":1,"parentId":"4602c347772e023bdc000035","content":"### Derivation\n\nThe kinematic equation $v^2 = u^2 + 2as$ can be derived from the other two equations algebraically.\n\nAll three equations can be derived from the same kinematic graph."},{"_id":"4602dbb6772e023bdc00005c","treeId":"45f4536692911004a000003d","seq":2954510,"position":0.25,"parentId":"4602c4ad772e023bdc000036","content":"### Graphical derivation\n\nImage: Standard $v$-$t$ graph\n\n1. $a = \\frac{v - u}{t}$:\nSeen from the form for gradient of line\n2. $s = ut + \\frac{1}{2}at^2$:\nSplit the area under graph into a bottom rectanglar section and an upper triangular section.\nArea of rectangle = $ut$, area of triangle = $\\frac{1}{2}(v-u)(t) = \\frac{1}{2}at^2$ (subbed from 1).\n3. $v^2 = u^2 - 2as$:\n“Average” the area under graph by flattening it.\nHeight of flattened section: $\\frac{1}{2}(v+u)$\nWidth of flattened section: $t$, or $\\frac{v-u}{a}$ (subbed from 1 to eliminate $t$)\nArea under $v$-$t$ graph is displacement, represented by $s$:\n$s = \\frac{1}{2}(v+u)\\times\\frac{v-u}{a} = \\frac{v^2 - u^2}{2a}$"},{"_id":"4602d9ae772e023bdc000057","treeId":"45f4536692911004a000003d","seq":2954511,"position":0.5,"parentId":"4602c4ad772e023bdc000036","content":"### Misconceptions\n\n* A common impression is that *each equation fits a different scenario* (Equation 1 for this kind of problem, Equation 2 for another kind, ...), and one only needs to identify which scenario is for which equation.\n * All three equations essentially refer to the same kinematic picture.\n\n* *If you use the ‘wrong’ equation, you get a different result.*\n * The three equations are internally consistent (if the assumptions hold), and any of them can be used, though some will be more useful in some situations than others.\n\n* *The graphical method and the equation method are completely different methods of solving, and cannot be used together.*\n * The equations and graphs refer to the same kinematic picture, and mean the same thing. They are different ways of *looking* at this kinematic picture, with emphasis on different things.\n\n* *The equations are “magical formulas”; if you use them in a different form, they won’t work.*\n * The equations are algebraic representations of the graphical forms, and can be manipulated just like other algebraic expressions. However, students have to be clear about the use of variables.\n* *Students might have calculation shortcuts that they might mistake for concepts.*"},{"_id":"4602c189772e023bdc000031","treeId":"45f4536692911004a000003d","seq":2954514,"position":3,"parentId":"45f454c892911004a0000042","content":"## Applications\n\n"},{"_id":"50713be38fdebc9158000108","treeId":"45f4536692911004a000003d","seq":2954516,"position":0.25,"parentId":"4602c189772e023bdc000031","content":"### Mathematical results\n\nSome mathematical results from algebra and vectors are not explicitly stated, but are understood to be required knowledge:\n\n* Change of a vector need not be along the same axis it is pointing.\n* Vector sum allows vectors to be added regardless of their direction.\n"},{"_id":"507137708fdebc9158000106","treeId":"45f4536692911004a000003d","seq":2954519,"position":0.5,"parentId":"4602c189772e023bdc000031","content":"### General results\n\nSome general results can be observed from applying the kinematics equations.\n\n* **Displacement**, **velocity**, and **acceleration** are vectors, and need not be in alignment all the time.\n* Any of the quantities $s$, $v$, $a$ can be zero while the others are not.\n* A *change in* **displacement** must indicate the presence of **velocity**, and a *change in* **velocity** must indicate the presence of **acceleration**.\n* The presence of **velocity** implies a *change in* **displacement**, and the presence of **acceleration** implies a *change in* **velocity**."},{"_id":"4603ba47772e023bdc0000e7","treeId":"45f4536692911004a000003d","seq":2954528,"position":3.5,"parentId":"45f454c892911004a0000042","content":"## 2D Kinematics\n\nBecause $s$, $v$, and $a$ are vectors, they may be resolved in two or more perpendicular directions.\n\nAny pair of perpendicular directions may be used, but the most useful pair for resolving depends on what is required in the situation.\n* Typically this means resolving in directions *parallel* and *normal* to **acceleration**, or to **velocity**.\n* Students with sufficient mastery of vector-resolving will find this much easier to grasp.\n\nComponents of $a$ do not affect *normal* components of $v$, which in turn do not affect *normal* components of $s$. Only components parallel to each other can affect each other.\n\n**Deceleration** means “acceleration in a direction opposite to velocity”.\n**Decreasing acceleration** means “decreasing magnitude of acceleration”."},{"_id":"4602d26b772e023bdc00004c","treeId":"45f4536692911004a000003d","seq":1697353,"position":4,"parentId":"4603ba47772e023bdc0000e7","content":"In Mechanics, a large part of being able to do well lies in being able to describe common everyday situations in terms of physics variables. Students should be given sufficient practice to analyse these situations, and picture/visualise them in the form of diagrams or graphs, before translating them into equations."},{"_id":"4602c597772e023bdc000038","treeId":"45f4536692911004a000003d","seq":2954529,"position":5,"parentId":"4603ba47772e023bdc0000e7","content":"### Projectile Motion\n\nProjectile motion is one instance of motion with constant acceleration in one direction (vertically) and zero acceleration (constant velocity) in another direction (horizontally)."},{"_id":"4602c5e0772e023bdc000039","treeId":"45f4536692911004a000003d","seq":2954531,"position":1,"parentId":"4602c597772e023bdc000038","content":"### General results\n\n* If energy is conserved, an object always has the same speed at the same height.\n\n* Speed of an object in free-fall is not affected by distance travelled (is not path-dependent) *if there is no external force*.\n * An object starting from one point will always have the same speed at another point, no matter how it gets there."},{"_id":"45f454ea92911004a0000043","treeId":"45f4536692911004a000003d","seq":2954215,"position":4,"parentId":null,"content":"# 03 Dynamics\n\n> 1827 in the sense “pertaining to force producing motion” (the opposite of static), from French *dynamique* introduced by German mathematician Gottfried Leibnitz (1646–1716) in 1691 from Greek *dynamikos* “powerful,” from *dynamis* “power”.\n\n**The study of how forces cause motion**\n\n* Newton’s 2nd Law\n* Newton’s 3rd Law\n* Conservation of Momentum\n* Collisions"},{"_id":"5072b5988fdebc915800010c","treeId":"45f4536692911004a000003d","seq":2954532,"position":0.25,"parentId":"45f454ea92911004a0000043","content":"## Terms of physical interaction\n\nOn top of the kinematics terms, we now need terms to describe how forces interact with bodies to cause motion."},{"_id":"507293468fdebc915800010a","treeId":"45f4536692911004a000003d","seq":2954603,"position":1,"parentId":"5072b5988fdebc915800010c","content":"### Momentum\n\nWhen we apply the *same force over the same period of time* on two bodies, one with greater mass than the other, we known intuitively that one of them (with smaller mass) will have a higher final velocity than the other. But since the force and time are the same for both, *something* must have remained the same; this is momentum.\n\n#### Equation\n$p = mv$\nwhere $m$ represents **mass** of the body,\n$v$ represents **velocity** of the body.\n\n**Momentum** $p$ has base units of kg m$^{-1}$ s$^{-1}$.\n\n#### Definition\n> The **momentum** of a body is the product of its mass and velocity.\n\nWhen two bodies collide, momentum may be transferred. That is why in some collisions, one of the bodies may end up with a higher speed that it had initially."},{"_id":"50729e0a8fdebc915800010b","treeId":"45f4536692911004a000003d","seq":2954607,"position":2,"parentId":"5072b5988fdebc915800010c","content":"### Impulse\n\nIn real life, we seldom have situations where a force is applied on an object over an infinite period of time; that is unrealistic.\n\nInstead, if we plot the **resultant force** exerted against time, we will see that it forms a closed area once there is no longer any resultant force.\n\n#### Equation\n$p = Ft$\nwhere $F$ represents **resultant force** exerted on the body,\n$t$ represents **time** over which the force was exerted.\n\n**Impulse** has base units of kg m$^{-1}$ s$^{-1}$ (same as $p$).\n\n#### Definition\n> The **impulse** applied to a body is the product of the force exerted on the body and the time over which the force is exerted.\n\nOften, the force applied will not be constant, but will taper to zero at the start and end. In such cases, we cannot apply $p = Ft$ directly, but must calculate or estimate the area under the $F$-$t$ graph instead."},{"_id":"5072cc838fdebc915800010e","treeId":"45f4536692911004a000003d","seq":2954614,"position":1,"parentId":"50729e0a8fdebc915800010b","content":"### Graphical scaffolding\n\nStart with the $a$-$t$ graph, then multiply the y-axis by $m$.\n* What does the y-axis represent now?\n* What does the gradient of the graph represent?\n* What does the area under the graph represent?\n\nThis exercise can help to link $F = \\frac{dp}{dt}$ with the $F$-$t$ graph."},{"_id":"503d8d170d8d29f958000219","treeId":"45f4536692911004a000003d","seq":2954534,"position":0.5,"parentId":"45f454ea92911004a0000043","content":"## The laws of motion\n\nThe three laws of motion were first compiled by Isaac Newton in his Philosophiæ Naturalis Principia Mathematica (Mathematical Principles of Natural Philosophy), first published in 1687. Newton used them to *explain and investigate the motion of many physical objects and systems*.\n\nToday, these laws form the foundation of many mechanical systems in use today. Although we have discovered that very tiny objects (smaller than an atom) and very fast objects (near the speed of light) deviate from the behaviour described by these laws, under ‘normal’ conditions, Newton’s laws of motion provide a simple and adequate model for everyday purposes."},{"_id":"4602ee04772e023bdc00006b","treeId":"45f4536692911004a000003d","seq":2954616,"position":0.5,"parentId":"503d8d170d8d29f958000219","content":"### Free Body Diagrams\n\nWe will regularly encounter situations in which a body is being exerted on by multiple forces. This is the heart of Newtonian mechanics: to analyse the forces being exerted on a body, figure out how it will behave and why, or to analyse a body behaving in a particular way and infer what forces are being exerted on it.\n\nThis can be a hugely complicated business, so we need to acquire some tools to help us do this. *Free body diagrams (FBDs)* are our most important tool for analysing bodies in motion. Without them, any system with more than a handful of forces becomes too much to handle.\n\nThe primary purpose of an FBD is to help us **determine all the forces acting on a body**. We want to do this to aid us in our next step, solving the force equations (N2L)."},{"_id":"4602ef76772e023bdc00006e","treeId":"45f4536692911004a000003d","seq":2954617,"position":3,"parentId":"4602ee04772e023bdc00006b","content":"### Framework of analysis:\n1. Draw in all non-contact forces (typically $W$)\n2. Observe all objects in contact with body, by going around the perimeter of the object. Label their forces where relevant."},{"_id":"4602f0d8772e023bdc000071","treeId":"45f4536692911004a000003d","seq":2954618,"position":4,"parentId":"4602ee04772e023bdc00006b","content":"### Thinking exercises\n\n**Taupok**\n1. Would you rather be at the top or at the bottom?\n2. Why?\n3. Explain using FBDs."},{"_id":"4602f048772e023bdc00006f","treeId":"45f4536692911004a000003d","seq":2954619,"position":5,"parentId":"4602ee04772e023bdc00006b","content":"### Demonstrations\n\n**Slinky demonstration 1**\n1. Get students to predict and explain mass distribution of slinky held vertically by the top end.\n2. Why is it more stretched out near the top than bottom?\n3. Explain using FBD.\n\n**Slinky demonstration 2**\n1. Get students to predict behaviour of falling slinky.\n * Bottom springs up first\n * Top springs down first\n * Both fall at the same time\n2. Drop slinky, get students to observe.\n3. Explain using FBD.\n\nObservation: Quite a number of students expect both top and bottom to fall together."},{"_id":"4602e691772e023bdc000061","treeId":"45f4536692911004a000003d","seq":2954638,"position":1,"parentId":"503d8d170d8d29f958000219","content":"## Newton’s 2nd Law (N2L)\n\n#### Formula\n$\\sum F = \\frac{dp}{dt}$\nwhere $F$ represents **resultant force** acting on the body,\n$p$ represents **momentum** of the body\n\n#### Statement\nWhen a force is exerted on a body, it experiences a **rate of change** of **momentum** equal to the force, **in the direction of** the force.\n\nAt A levels, $F$ is only considered on rigid bodies, but in effect it can also cause shearing, rotating (moment), twisting, compression etc in continuous bodies."},{"_id":"5072c0298fdebc915800010d","treeId":"45f4536692911004a000003d","seq":2954630,"position":0.25,"parentId":"4602e691772e023bdc000061","content":"### Single-body systems\n\n$F = ma$ is a subset of this general equation that can only be applied to single-body examples. In continuous flow examples (e.g. water flowing through a pipe, sand on a conveyor belt), there is no single body ($m$) under consideration, and it has to be analysed as a rate of change of $p$.\n"},{"_id":"4602e9ac772e023bdc000063","treeId":"45f4536692911004a000003d","seq":2954633,"position":1,"parentId":"4602e691772e023bdc000061","content":"### Pitfalls\n\n1. An insufficient understanding and intuition of momentum may hamper understanding of force as dp/dt.\n\n2. Intuiting the effects of $a$ vs $v$ at this point is probably still not natural to students yet. Might want to emphasise that the $a$ experienced by the body does not translate directly to $v$."},{"_id":"4602e76c772e023bdc000062","treeId":"45f4536692911004a000003d","seq":2954626,"position":2,"parentId":"503d8d170d8d29f958000219","content":"## Newton’s 3rd Law (N3L)\n\nAn FBD of a single-body system is quite straightforward, since it does not involve contact forces. But once we start to include multiple contact forces, things start to get messy. What is the direction of the contact force? What is it exerted by, and exerted on which body? Newton’s 3rd Law is another tool that helps us in sketching the FBDs.\n\n#### Formula\n$F_{AB} = -F_{BA}$\n\n#### Statement\nWhen body A exerts a force on body B, body B exerts an equal and opposite force on body A."},{"_id":"50748a698fdebc915800010f","treeId":"45f4536692911004a000003d","seq":2954639,"position":0.5,"parentId":"4602e76c772e023bdc000062","content":"### Force pairs\n\nThe first thing that N3L implies is that **forces never act alone; they always act in pairs**. And these force pairs have the following properties:\n\n1. **They have the same magnitude.**\n2. **They act in opposite directions.**\n3. **They act on a pair of bodies.**\nA exerts a force on B, B exerts a force back on A. The forces in a force pair must never be drawn acting on the same body.\n4. **The forces must be of the same type.**\nIf A exerts a frictional force on B, B exerts a frictional force back on A.\n\n### Force pairs in an FBD\n\nWe only draw one of the force pairs in an FBD because of condition 3. If you have both forces in a pair drawn in the same FBD, something is terribly wrong!"},{"_id":"507499338fdebc9158000110","treeId":"45f4536692911004a000003d","seq":1698425,"position":2,"parentId":"4602e76c772e023bdc000062","content":"##FBDs for systems with more than one body\n\nFBDs are usually drawn with only one bod, but sometimes we combine multiple bodies into a **system** and analyse that system with a FBD. We can do this when **all the bodies in the system behave identically**, i.e. when these bodies always have the same velocity and acceleration (for example if they are tied or glued together).\n\nWhen we do this, we can ignore **internal forces** in the system. These internal forces are force pairs exerted *between bodies within the system*. We are able to do this because internal forces *do not affect the behaviour* of the system. This allows us to simplify the system FBD without missing any crucial information."},{"_id":"4602f41a772e023bdc000075","treeId":"45f4536692911004a000003d","seq":2954642,"position":2.125,"parentId":"45f454ea92911004a0000043","content":"## Linear collision of two bodies\n\nIn A level Physics, we only analyse **linear collisions**, i.e. collisions that occur along a straight line. In such collisions, the initial and final velocities of the colliding bodies are always parallel to each other.\n\nCollisions where the initial and final velocities are not parallel are referred to as **oblique collisions**."},{"_id":"4a029e9c6f14bce92f000102","treeId":"45f4536692911004a000003d","seq":2954645,"position":0.25,"parentId":"4602f41a772e023bdc000075","content":"### Application of N2L and N3L in a collision\n\nWhen two bodies A & B collide, they exert **equal and opposite forces** on each other (according to **N2L**). As body A is in contact with body B, body B is also in contact with body A. Hence, the **time of contact** between the two bodies is the same.\n\nThus, by **N3L**, the **change in momentum** ($\\Delta p = \\int F dt$) experienced by body A & B should also be the same magnitude but in opposite direction."},{"_id":"4a02a0106f14bce92f000103","treeId":"45f4536692911004a000003d","seq":2954646,"position":0.375,"parentId":"4602f41a772e023bdc000075","content":"### Momentum-time graph in a collision\n\nIf we draw a graph of A's momentum ($p_A$) and B's momentum ($p_B$), we will notice that\n* They change by the same amounts, in the opposite direction, in the collisions. In other words, the **total change in momentum of the system** is 0.\n* The **total momentum of the system** (consisting of bodies A and B) remains the same."},{"_id":"4603174c772e023bdc000081","treeId":"45f4536692911004a000003d","seq":306616,"position":0.53125,"parentId":"4602f41a772e023bdc000075","content":"Assumption of compression in collisions is necessary (to avoid infinite acceleration).\n\nWhile bodies are compressed, they are in contact. They remain in contact until their velocities are the same (relative speed = 0)"},{"_id":"4a03457c6f14bce92f000105","treeId":"45f4536692911004a000003d","seq":2954651,"position":0.765625,"parentId":"4602f41a772e023bdc000075","content":"### Energy in a collision\n\nThe change in momentum of bodies in a collision cannot happen instantaneously (this would imply that acceleration is zero!).\n\nWe can think of collisions as happening in 2 phases:\n1. Bodies moving toward each other\n2. Bodies moving away from each other\n\nThe moment when the system goes from Phase 1 to Phase 2 is the point when both bodies have the same velocity.\n\nDuring Phase 1, the **total KE of the system** is decreasing. If there is no external force acting on the system, then *conservation of energy* applies and this KE has to be converted to other forms. In such cases, the KE is usually converted to **elastic potential energy** as the colliding bodies are compressed in the collision.\n\nDuring Phase 2, the elastic potential energy is converted back to KE of the colliding bodies. This conversion may not always be 100% efficient; there may be some loss of energy in the collision. This does not violate the **principle of conservation of momentum**.\n\nThe efficiency of conversion from potential energy back to KE determines the type of collision that occurs."},{"_id":"4602f2f0772e023bdc000074","treeId":"45f4536692911004a000003d","seq":2954654,"position":1,"parentId":"4602f41a772e023bdc000075","content":"### Misconceptions\n\n1. Identifying correct $m$, $u$, $v$ in a collision.\n2. Using correct signs for velocity directions.\n3. Identifying where it is appropriate to apply **CoM** and/or **CoE** (i.e. whether system is isolated, whether there is external force).\n4. Correctly identifying type of graph: force-time, momentum-time, change in momentum, etc.\n5. Correctly identifying type of collision.\n6. Applying incorrect conditions for type of collision.\n7. Correctly identifying colliding bodies in a graph (e.g. mistaking $p_A$ and $p_B$ in a completely inelastic collision for the total momentum of the system.).\n8. Mixing up formulas for momentum ($mv$) and kinetic energy ($\\frac{1}{2}mv^2$).\n\nStudents will find it difficult to understand where energy has gone to, how it is lost. Difficult to go into depth except with high-ability students."},{"_id":"4a02cbe36f14bce92f000104","treeId":"45f4536692911004a000003d","seq":2954656,"position":3,"parentId":"4602f41a772e023bdc000075","content":"### Types of collisions\n\n"},{"_id":"46030a33772e023bdc00007f","treeId":"45f4536692911004a000003d","seq":783841,"position":1,"parentId":"4a02cbe36f14bce92f000104","content":"#Superelastic collisions\n\n**Superelastic collisions** are collisions where the **total KE of the system** increases as a result of energy converted/transferred from other sources.\n\nTypical examples:\n* Exploding bodies"},{"_id":"460308a3772e023bdc00007c","treeId":"45f4536692911004a000003d","seq":783825,"position":2,"parentId":"4a02cbe36f14bce92f000104","content":"#Elastic collisions\n\n**Elastic collisions** are collisions where the **total KE of the system** is conserved before and after an elastic collision, if no external forces act on the system.\n\nThis means that all the PE stored during the collision is fully restored as KE after the collision."},{"_id":"46030997772e023bdc00007d","treeId":"45f4536692911004a000003d","seq":783828,"position":3,"parentId":"4a02cbe36f14bce92f000104","content":"#Inelastic collisions\n\n**Inelastic collisions** are collisions where the **total KE of the system** is not conserved. The **total KE of the system** *before* the collision is less than the **total KE of the system** *after* the collision.\n\nThis means that some of the PE stored during the collision is lost to the environment, and not all of it is restored as KE after the collision."},{"_id":"4a010f9d6f14bce92f0000ff","treeId":"45f4536692911004a000003d","seq":783830,"position":4,"parentId":"4a02cbe36f14bce92f000104","content":"#Completely inelastic collisions\n\n**Completely inelastic collisions** are inelastic collisions where the colliding bodies **stick together** (coalesce) after the impact, and move off with a **common velocity**.\n\nSince the bodies do not separate after the collision, none of the Pe stored during the collision is restored as KE, and all of it is lost to the environment or converted to other forms of energy."},{"_id":"46031ae0772e023bdc000084","treeId":"45f4536692911004a000003d","seq":783833,"position":1,"parentId":"4a010f9d6f14bce92f0000ff","content":"Two main types of completely inelastic collisions tend to appear in exams:\n* Losses due to friction (parcel falling into trolley)\n* Losses due to mechanical compression/deformation (i.e. pin-and-cork trolleys)."},{"_id":"4602f188772e023bdc000072","treeId":"45f4536692911004a000003d","seq":2954658,"position":2.25,"parentId":"45f454ea92911004a0000043","content":"## Conservation of momentum\n\nEarlier, we saw that the total momentum of a system remains the same throughout a collision. We can write the above observation in the following form:\n\n$\\sum mu = \\sum mv$\nor\n$m_1u_1 + m_2u_2 + ... = m_1v_1 + m_2v_2 + ...$\n\nIn prose form:\nThe **total momentum of a system** is **conserved throughout a collision**, if **no external forces** act on the system (of colliding bodies).\n\nThis is known as the **principle of conservation of momentum** (**CoM**), and is the underlying formula behind all of collision analysis.\n\n**CoM** applies to **all collisions**, if there are no external forces acting on the system."},{"_id":"4a0298f06f14bce92f000100","treeId":"45f4536692911004a000003d","seq":2954659,"position":3.125,"parentId":"45f454ea92911004a0000043","content":"## Conservation of kinetic energy\n\n"},{"_id":"45f4550c92911004a0000044","treeId":"45f4536692911004a000003d","seq":2954216,"position":5,"parentId":null,"content":"# 04 Forces\n\n* Hydrostatic pressure\n* Upthrust"},{"_id":"46032122772e023bdc000085","treeId":"45f4536692911004a000003d","seq":2954660,"position":1,"parentId":"45f4550c92911004a0000044","content":"## Hydrostatic Pressure\n\n$p_{fluid} = mgh$\n\nDifficult to explain pressure difference due to variation in height of liquid below surface.\nHard to imagine how difference in pressure at different height can cause upthrust."},{"_id":"46032227772e023bdc000086","treeId":"45f4536692911004a000003d","seq":306633,"position":1,"parentId":"46032122772e023bdc000085","content":"Use taupok as an analogy: would you rather be at the top or at the bottom? Why?"},{"_id":"4603224d772e023bdc000087","treeId":"45f4536692911004a000003d","seq":2954661,"position":2,"parentId":"45f4550c92911004a0000044","content":"## Upthrust\n\n$U = \\rho_{liquid}gV_{displaced}$\n\nUpthrust is the integral of pressure around the area of an object."},{"_id":"460322e3772e023bdc000088","treeId":"45f4536692911004a000003d","seq":306635,"position":1,"parentId":"4603224d772e023bdc000087","content":"Difficult to explain the origin of pressure, as stated above. It is a contact force exerted betwen molecules in a liquid, and between liquid molecule other objects. It has a vertical gradient only when an external force (typically gravitational force) is present."},{"_id":"46032320772e023bdc000089","treeId":"45f4536692911004a000003d","seq":306637,"position":2,"parentId":"4603224d772e023bdc000087","content":"What happens when gravity is not present? Use video of balloon in a bottle of water as example. What happens to upthrust and water pressure in free-fall?"},{"_id":"45f4552f92911004a0000045","treeId":"45f4536692911004a000003d","seq":2954217,"position":6,"parentId":null,"content":"# 05 Work, Energy, Power\n\n* Conservation of energy\n* Work done\n* Power"},{"_id":"46032dba772e023bdc00008a","treeId":"45f4536692911004a000003d","seq":2954663,"position":1,"parentId":"45f4552f92911004a0000045","content":"## Conservation of energy, revisited\n\nIn the absence of external forces, the total energy in a system is always conserved."},{"_id":"46032ead772e023bdc00008b","treeId":"45f4536692911004a000003d","seq":906815,"position":1,"parentId":"46032dba772e023bdc00008a","content":"Students have difficulty remembering and applying the fact that CoE not only takes place at one point, it takes place between any two chosen points, and in fact throughout the entire process. They might need many examples for scaffolding."},{"_id":"46032f22772e023bdc00008c","treeId":"45f4536692911004a000003d","seq":306697,"position":1,"parentId":"46032ead772e023bdc00008b","content":"Students typically have some intuitive understanding of this principle, but its applications are so wide-ranging that they might have difficulty applying it in various situations. Variety is key."},{"_id":"4604c014e5fcc1253b0000e4","treeId":"45f4536692911004a000003d","seq":2954664,"position":1.5,"parentId":"45f4552f92911004a0000045","content":"## Work done\n\n$W.D. = F\\cdot s$\n\nThe **work done** by a force is the product of the **magnitude of the force** and the **displacement** of the point of action in the direction of the force."},{"_id":"4604b772e5fcc1253b0000e3","treeId":"45f4536692911004a000003d","seq":2954665,"position":2,"parentId":"45f4552f92911004a0000045","content":"## Power\n\n$P = \\frac{dE}{dt}$\n\n**Power** is the *rate of change* of **work done** with respect to time.\n\nFor an object travelling at constant velocity, $\\frac{ds}{dt}$ is constant, so\n\n$P = \\frac{d}{dt}(F \\cdot s) = F \\cdot \\frac{ds}{dt} = F \\cdot v$"},{"_id":"45f4557292911004a0000046","treeId":"45f4536692911004a000003d","seq":2954218,"position":7,"parentId":null,"content":"# 06 Motion In A Circle\n\n* Centripetal force\n* Deviation from circular motion"},{"_id":"46032f8d772e023bdc00008d","treeId":"45f4536692911004a000003d","seq":2954697,"position":1,"parentId":"45f4557292911004a0000046","content":"## Centripetal force\n\n$F_c = \\frac{mv^2}{r} = mr\\omega^2$\n\nCentripetal force is not a kind of force (contact, non-contact). It describes a force that is needed to keep an object moving around the centre of a circle.\n\nCentripetal force always acts toward the centre of circular motion."},{"_id":"46033119772e023bdc00008e","treeId":"45f4536692911004a000003d","seq":306700,"position":1,"parentId":"46032f8d772e023bdc00008d","content":"centrum \"\"centre\"\" + petere \"\"to seek, to go after\"\" [e.g. petition]). Not to be confused with centrifugal (centrum \"\"centre\"\" + fugere \"\"to flee\"\" [e.g. fugitive]"},{"_id":"46033384772e023bdc00008f","treeId":"45f4536692911004a000003d","seq":306708,"position":2,"parentId":"46032f8d772e023bdc00008d","content":"This is a direct result of N2L: since object undergoes change in velocity, an acceleration must be present, which can only be provided by a force.\n\nUnderstanding centripetal accel. and force will help cement students' understanding of Newton's Laws of Motion in non-linear contexts.\n"},{"_id":"46033430772e023bdc000090","treeId":"45f4536692911004a000003d","seq":306709,"position":3,"parentId":"46032f8d772e023bdc00008d","content":"Centripetal force in elliptical motion is not constant throughout. Circular motion is a special case of elliptical motion."},{"_id":"4603346e772e023bdc000091","treeId":"45f4536692911004a000003d","seq":306711,"position":4,"parentId":"46032f8d772e023bdc00008d","content":"Students will constantly forget that an object undergoing circular motion is not at rest, and in fact experiences an acceleration. It will take some time for them to get used to this.\n\nStudents tend to confuse centripetal force for centrifugal (outwards) force. There needs to be constant reminding that without centripetal force, the object will continue with linear motion per N1L."},{"_id":"460335bb772e023bdc000093","treeId":"45f4536692911004a000003d","seq":306727,"position":1,"parentId":"4603346e772e023bdc000091","content":"Begin without centripetal force: object is initially in linear motion.\n\nWhat is needed for uniform circular motion? Since speed is constant, acceleration must be perpendicular to velocity. In circular motion, acceleration would thus also point towards the centre (since velocity is tangential)\n\nHow much acceleration is required? $\\frac{mv^2}{r}$. This is centripetal acceleration, the required acceleration for uniform circular motion of radius r and velocity v.\n\nWhat provides centripetal acceleration? A resultant force, which may be provided by gravitational force, tension, normal reaction force, etc, or a combination of these. This resultant force causes centripetal acceleration and is thus known as centripetal force."},{"_id":"46033572772e023bdc000092","treeId":"45f4536692911004a000003d","seq":306721,"position":5,"parentId":"46032f8d772e023bdc00008d","content":"Once gravitation is taught, students might form the impression that where circular motion around Earth is involved, the centripetal force always points towards the centre of Earth – this is only true if circular motion is centred at Earth's centre."},{"_id":"4603377e772e023bdc000094","treeId":"45f4536692911004a000003d","seq":2954698,"position":2,"parentId":"45f4557292911004a0000046","content":"## Centripetal acceleration\n\n$a_c = \\frac{v^2}{r} = r\\omega^2$\n\nCentripetal acceleration is provided by the centripetal force and points in the same direction as the force.\n\nThis links acceleration to force—an extension of dynamics, and N2L.\nThe connection between direction of acceleration and force is often needed to solve many problems where one is given but not the other; in fact questions commonly assume knowledge of this fact."},{"_id":"460339c9772e023bdc000096","treeId":"45f4536692911004a000003d","seq":306730,"position":0.5,"parentId":"4603377e772e023bdc000094","content":"Centripetal acceleration does indeed always point toward the centre of motion even in elliptical orbits/motion. Hence, the centripetal acceleration indicates the direction of motion, rather than vice-versa."},{"_id":"46033896772e023bdc000095","treeId":"45f4536692911004a000003d","seq":306729,"position":1,"parentId":"4603377e772e023bdc000094","content":"An object need not necessarily always experience the same centripetal force; this is an assumption we make when we observe an object undergoing uniform circular motion."},{"_id":"46033a1e772e023bdc000097","treeId":"45f4536692911004a000003d","seq":2954700,"position":3,"parentId":"45f4557292911004a0000046","content":"## Deviation from circular motion\n\nIf the force provided is greater than centripetal force, the object will orbit closer to the centre (with accompanying increase in angular velocity).\n\nIf the force provided is smaller than centripetal force, the object will orbit further from the centre (with accompanying decrease in angular velocity)."},{"_id":"46033c8c772e023bdc000098","treeId":"45f4536692911004a000003d","seq":306738,"position":1,"parentId":"46033a1e772e023bdc000097","content":"Students may fall under the misconception that minimum velocity (around a bend, at the top of a vertical loop) is 0. They should be encouraged to consider the consequences of zero velocity at such points."},{"_id":"46033cca772e023bdc000099","treeId":"45f4536692911004a000003d","seq":306739,"position":2,"parentId":"46033a1e772e023bdc000097","content":"Examples include car going over hump, ball rolling over a curved surface, rollercoaster loop, etc.\n\nStudents often are unable to translate problem descriptions to physical equations. For instance, normal force is zero when an object loses contact with the surface. This may need to be explicitly taught."},{"_id":"46033d0e772e023bdc00009a","treeId":"45f4536692911004a000003d","seq":306740,"position":3,"parentId":"46033a1e772e023bdc000097","content":"3 situations:\n1) Provided force > required force\n2) Provided force = required force\n3) Provided force < required force\n\nWhat happens in each of the situations? Useful as a summary of the topic.\n(1) and (3) occur at a point of force imbalance; the result is deviation from circular motion. (2) occurs when forces are balanced again and circular motion is restored."},{"_id":"45f455ba92911004a0000047","treeId":"45f4536692911004a000003d","seq":2954220,"position":8,"parentId":null,"content":"# 07 Gravitational Field\n\n* Gravitational force between two point masses\n* Gravitational field of one or more point masses\n* Gravitational potential and potential energy\n* Geostationary orbit\n* Escape velocity"},{"_id":"46033d9d772e023bdc00009b","treeId":"45f4536692911004a000003d","seq":459116,"position":1,"parentId":"45f455ba92911004a0000047","content":"In a binary star system, centre of rotation and balance point are not at the same position.\n\nThis only happens when both stars are the same mass."},{"_id":"5a1673af35656afeab0000eb","treeId":"45f4536692911004a000003d","seq":3190664,"position":2,"parentId":"45f455ba92911004a0000047","content":"Students still think GPE on Earth = 0"},{"_id":"6db4609203854927ab0000f2","treeId":"45f4536692911004a000003d","seq":8283797,"position":3,"parentId":"45f455ba92911004a0000047","content":"Students tend to apply conservation of energy to solve problems involving two different orbits, thinking that change in GPE = change in KE"},{"_id":"45f4562b92911004a0000048","treeId":"45f4536692911004a000003d","seq":2954221,"position":9,"parentId":null,"content":"# 08 Oscillations\n\n* Cyclical and Simple harmonic motion\n* Displacement, velocity, acceleration\n* Energy in an oscillation\n* Damping\n* Resonance"},{"_id":"4bde96abdc2e24928a000104","treeId":"45f4536692911004a000003d","seq":8283798,"position":0.5,"parentId":"45f4562b92911004a0000048","content":"## Cyclical motion\n\nIn nature and in life, we often observe systems that repeat the same motion over and over again.\n\n* sunrise\n* tides\n* commuting\n* dribbling a basketball\n* waiting icon\n\nClearly, some of these are more special than the others. Some happen more regularly, some more jerkily."},{"_id":"4bdebcdcdc2e24928a000105","treeId":"45f4536692911004a000003d","seq":2954703,"position":0.75,"parentId":"45f4562b92911004a0000048","content":"## Simple harmonic motion\n\nBut there is a special category of regular motions that behave very beautifully. There are many interesting patterns to spot in this class of regular motions. For instance, the further from the equilibrium position they are, the slower they travel.\n\nThese occur very commonly, and are perhaps one of the most common kinds of motions in nature. Understanding this motion, and learning its patterns, gives us a mental pattern and an equation we can apply to many things."},{"_id":"460357c4772e023bdc00009d","treeId":"45f4536692911004a000003d","seq":3051196,"position":1,"parentId":"45f4562b92911004a0000048","content":"## Displacement\n\nThe displacement varies sinusoidally with time.\n$$x = x_0 \\sin\\omega t$$\n\n## Velocity\nThe velocity is the rate of change of displacement.\n\n$$v = \\frac{dx}{dt} = \\omega(x_0\\cos\\omega t)$$\n\nThe velocity is greatest at the equilibrium position, and zero at the ends of the oscillation.\n\n## Acceleration\nThe acceleration is proportional to the displacement, and in the opposite direction, always toward the equilibrium point.\n\n$$a = \\frac{dv}{dt} = -\\omega^2(x_0\\sin\\omega t) = -\\omega^2x$$\n\nThe acceleration is zero at the equilibrium position, and greatest at the ends of the oscillation."},{"_id":"461db93442de20aefe0000dd","treeId":"45f4536692911004a000003d","seq":998226,"position":0.75,"parentId":"460357c4772e023bdc00009d","content":"Velocity is the greatest at the equilibrium position, and zero at the ends. Hence, the graph of **velocity vs displacement forms an ellipse**.\n\n[graph]\n\nAcceleration is proportional to the displacement and always pointing towards the equilibrium position (opposite direction from displacement), hence the graph of **acceleration vs displacement forms a straight line with negative gradient that passes through the origin**.\n\n[graph]"},{"_id":"461d8a3742de20aefe0000db","treeId":"45f4536692911004a000003d","seq":321046,"position":1,"parentId":"460357c4772e023bdc00009d","content":"The displacement is $\\frac{\\pi}{2}$ out of phase with the velocity, which is in turn $\\frac{\\pi}{2}$ out of phase with the acceleration.\n\n[graph of displacement, velocity, acceleration]"},{"_id":"4603b26d772e023bdc0000e5","treeId":"45f4536692911004a000003d","seq":2954707,"position":1.25,"parentId":"45f4562b92911004a0000048","content":"## Energy in an oscillation\n\nIn an oscillation, the kinetic energy is greatest at the equilibrium position, where velocity is also greatest.\n\nThe potential energy is greatest at the ends, where work has been done against the restoring force to slow the oscillator down.\n\nIn the absence of external forces (no damping), total system energy is conserved."},{"_id":"461dcc9342de20aefe0000de","treeId":"45f4536692911004a000003d","seq":321247,"position":1,"parentId":"4603b26d772e023bdc0000e5","content":"The graph of kinetic energy vs displacement is a quadratic curve with the maximum in the middle.\n\nThe graph of potential energy is a quadratic curve with the minimum in the middle (if total system is conserved)."},{"_id":"4603ce05772e023bdc0000e8","treeId":"45f4536692911004a000003d","seq":2954709,"position":1.375,"parentId":"45f4562b92911004a0000048","content":"## Damping\n\n**Damping** an oscillation means to decrease the energy of the oscillation. This usually takes the form of a retarding force that acts against the direction of motion of the oscillator."},{"_id":"460358c4772e023bdc00009f","treeId":"45f4536692911004a000003d","seq":321319,"position":1,"parentId":"4603ce05772e023bdc0000e8","content":"#Light Damping\n\nUnder light damping, an object oscillates about the equilibrium position a few times as the amplitude gets smaller, before it finally comes to a stop.\n\nThe amplitude usually decreases exponentially.\n\n#Critical damping\n\nAs the damping on an object increases, the time taken for the oscillator to come to rest decreases, until a point.\n\nThis point is known as the **critical damping point**. At this point, the oscillator **comes to rest in the shortest possible time**.\n\nIf the oscillator is damped any further, the time for it to come to a stop *increases* instead; it undergoes heavy damping.\n\nCritical damping is the threshold between light damping and heavy damping.\n\n#Heavy Damping\n\nUnder heavy damping, the oscillator experiences a retarding force great enough that it does not overshoot the equilibrium point."},{"_id":"46035908772e023bdc0000a0","treeId":"45f4536692911004a000003d","seq":306974,"position":1,"parentId":"460358c4772e023bdc00009f","content":"The word \"critical\" in physics describes thresholds in behaviour, e.g. critical angle in refraction of light, or critical mass in nuclear chain reaction. This change in behaviour is usually drastic and happens over a very short threshold."},{"_id":"46035966772e023bdc0000a1","treeId":"45f4536692911004a000003d","seq":306976,"position":2,"parentId":"460358c4772e023bdc00009f","content":"Pseudo-real-life examples probably pose the friendliest learning curve. Get students to imagine the following scenarios:\n1) They are trying to improve the performance of their bike suspension by using different kinds of oils in the spring-loaded tube. Which thickness/consistency of oil provides the fastest return to equilibrium position?\n2) You have a set of swinging doors in your kitchen/office/room/wherever. You don't like it when thy keep swinging for a long time after someone walks through, nor do you want them to take too long to return to rest. How tight should the hinges be, or how much frictional force should they exert?"},{"_id":"4bdef9fddc2e24928a000106","treeId":"45f4536692911004a000003d","seq":998235,"position":3,"parentId":"460358c4772e023bdc00009f","content":"Use Algodoo simulation to show critical damping and how it relates to light, heavy damping?"},{"_id":"4603b3f9772e023bdc0000e6","treeId":"45f4536692911004a000003d","seq":2954711,"position":3,"parentId":"45f4562b92911004a0000048","content":"## Resonance\n\n**Resonance** is a phenomenon whereby\n- there is maximum transference of energy from an oscillating driver to the oscillating system.\n- this happens when the **driving frequency** is close to the oscillating **system’s natural frequency**.\n- the amplitude is greatest at this point.\n\nThe driving frequency (frequency of oscillation) is always less than or equal to the system's natural frequency.\nThe discrepancy between driving frequency and system natural frequency is 0 for **free oscillations** (undamped), and increases as the amount of damping increases."},{"_id":"477211f1a5f488e6f00000eb","treeId":"45f4536692911004a000003d","seq":458643,"position":1,"parentId":"4603b3f9772e023bdc0000e6","content":"When there is no damping in the system, at resonance the total energy of the oscillating system can grow to infinity.\n\nWhen there is critical or heavy damping, there can be no oscillation since there is no overshoot in displacement."},{"_id":"47721451a5f488e6f00000ed","treeId":"45f4536692911004a000003d","seq":458641,"position":3,"parentId":"4603b3f9772e023bdc0000e6","content":"Finite forced oscillations can only occur with light damping.\n\nAt the resonance point, the driving frequency is always less than or equal to the natural (undamped) frequency of the oscillating system, since damping introduces a delay in the period of oscillation."},{"_id":"49648e752c36b00d340000ef","treeId":"45f4536692911004a000003d","seq":2954222,"position":11,"parentId":null,"content":"# 10 Wave Motion\n\n"},{"_id":"460359ed772e023bdc0000a2","treeId":"45f4536692911004a000003d","seq":2954224,"position":11.5,"parentId":null,"content":"# 09 Thermal Physics\n\n* Pressure, volume, temperature\n* Ideal gas equation\n* 1st Law of Thermodynamics\n* Energy distribution\n* Heat capacity\n* Latent heat"},{"_id":"46035be5772e023bdc0000a5","treeId":"45f4536692911004a000003d","seq":2954738,"position":2,"parentId":"460359ed772e023bdc0000a2","content":"## Ideal gas\nAn ideal gas is a gas which has no **internal PE**; its internal energy, $U$, is always equal to its **internal KE**.\n\nIn real life, gases behave like an **ideal gas** when its **internal PE** is high (i.e. close to 0). This happens when the KE of the gas particles is high enough to overcome the gas particles' PE ($T$ must be high), and when the gas particles are sufficiently far apart ($P$ must be low).\n\nFor an ideal gas, the gas laws can be combined to form the **ideal gas equation**, $pV = nRT$\n\nThis equation describes possible combinations of $P$, $V$, $n$, and $T$ for any ideal gas system. Each combination of these variables is referred to as a **state** of the gas system. If the state of the gas system satisfies the ideal gas equation, that state can exist."},{"_id":"4603740b772e023bdc0000b2","treeId":"45f4536692911004a000003d","seq":2954740,"position":0.5,"parentId":"46035be5772e023bdc0000a5","content":"### Example\nIn H2 Chemistry, you are taught that 1 mol of any gas occupies a volume of 24 dm<sup>3</sup> at room temperature. Does this agree with the ideal gas equation?\n\nP = 1E5 Pa\nV = ? m<sup>3</sup>\nn = 1 mol\nR = 8.31 J K<sup>–1</sup>\nT = 298 K (25 deg C)"},{"_id":"46037522772e023bdc0000b4","treeId":"45f4536692911004a000003d","seq":307143,"position":0.625,"parentId":"46035be5772e023bdc0000a5","content":"The equation can be modelled using first principles; from a minimal-energy state, work done on system causes pressure and volume to increase; also causes temperature to increase."},{"_id":"460374b4772e023bdc0000b3","treeId":"45f4536692911004a000003d","seq":732974,"position":0.75,"parentId":"46035be5772e023bdc0000a5","content":"Equalities are easy to misapply. Students can easily confuse variables with change in variables (*V* vs Δ*V*).\n\nAlso, NKT vs nRT, and p being pressure exerted by gas, not on gas."},{"_id":"46037271772e023bdc0000b1","treeId":"45f4536692911004a000003d","seq":307125,"position":1,"parentId":"46035be5772e023bdc0000a5","content":"There are higher-order terms to model non-ideal gases."},{"_id":"496495a32c36b00d340000fb","treeId":"45f4536692911004a000003d","seq":7475116,"position":3,"parentId":"460359ed772e023bdc0000a2","content":"## The gas laws\n\nWe create a simple model of a gas as a collection of particles inside a box. This box has one wall that can move inwards or outwards, allowing us to change its volume. There are $N$ particles ($n$ moles) inside this box.\n\nIf the gas is treated as an ideal gas, the following relationships hold between its pressure P, volume V, and temperature T:\n* $P \\propto T$\n* $V \\propto T$\n* $P \\propto \\frac{1}{V}$"},{"_id":"4964b81d2c36b00d340000fc","treeId":"45f4536692911004a000003d","seq":2954715,"position":1,"parentId":"496495a32c36b00d340000fb","content":"### Pressure\nWhen the gas particles collide with the walls of the box, they exert a force on the walls of the box. The gas particles hence exert a **pressure, P**, on the walls of the box.\n* This pressure always acts outwards, and is taken to be normal to the surface of the box.\n* Since the direction is always taken to be outwards, when we consider the pressure of the gas, it is considered to be scalar and is always positive.\n* However, when we consider the pressure of the gas *on a single wall*, pressure is treated as a vector, since it is no longer acting outwards but normal to the wall. Hence, the gas can still do work on the wall if the movement of the wall is along the orientation of the pressure exerted on it."},{"_id":"4964b9d92c36b00d340000fd","treeId":"45f4536692911004a000003d","seq":2954717,"position":2,"parentId":"496495a32c36b00d340000fb","content":"### Volume\nThe box has a **volume, V**, and when the movable wall shifts, the box's volume changes by ΔV. The gas particles will travel throughout the entire volume of the box, hence we take the volume of the gas to be the volume of the box.\n* When the movable wall shifts outwards, the box expands in volume and ΔV is positive (ΔV > 0).\n* When the movable wall shifts inwards, the box is compressed and ΔV is negative (ΔV < 0)"},{"_id":"4964ba1b2c36b00d340000fe","treeId":"45f4536692911004a000003d","seq":2954718,"position":3,"parentId":"496495a32c36b00d340000fb","content":"### Internal potential energy\nIf the gas particles exert no forces on each other, we take their potential energy to be 0.\n\nIf the gas particles are not too near to each other, they may exert attractive forces on each other, as the positively charged nuclei of the particles are attracted to the negatively charged electron clouds of other particles.\n* Since the forces are attractive in nature, the potential energy of the gas particles are negative.\n\nIf the gas particles come very close to each other, their electron clouds exert a large repelling force against each other, allowing them to \"collide\" and bounce off each other.\n* Since the forces are repulsive in nature, the potential energy of these gas particles in close proximity is positive.\n\nOverall, we would expect the potential energy of these gas particles to be negative.\n\nSince these potential energies of the particles arise from internal forces (within the system of gas particles), they constitute the **internal potential energy** of the system. We differentiate this from (external) *potential energy* arising from weight and other external forces."},{"_id":"4964d6bf2c36b00d34000100","treeId":"45f4536692911004a000003d","seq":2954735,"position":4,"parentId":"496495a32c36b00d340000fb","content":"### Internal kinetic energy\nIf the box is moving, it will have a kinetic energy of $\\frac{1}{2}Mv^2$, where $M$ is the total mass of the gas and box, and $v$ is the velocity of the box.\n\nHowever, if the box is not moving, this does not mean that the gas particles have 0 KE. Each gas particle is still moving, but since the direction of the gas particles is essentially random, in effect the vector sum of their velocities is close to 0.\n\nThis implies that the gas particles have **internal kinetic energy** even when the box is at rest. This internal energy is $\\sum\\limits_i^N \\frac{1}{2}m_i v_i^2$, the sum of the KE of each gas particle.\n\n### Temperature\nThe **thermodynamic temperature** is a measure of this internal KE of the gas system. This thermodynamic temperature is measured in units of Kelvin (K). At 0 K, the gas has 0 internal KE. At temperature $T$, the gas has internal KE $U = \\frac{3}{2}nRT$ or $U = \\frac{3}{2}NKT$."},{"_id":"460381c8772e023bdc0000b8","treeId":"45f4536692911004a000003d","seq":2954725,"position":1,"parentId":"4964d6bf2c36b00d34000100","content":"Students will tend to associate T with U, and hence with PE and KE both. Will need scaffolding to bring forth the idea of ideal gas having 0 PE.\n\nVery easy to confuse heat (energy) with temperature.\n* Adding heat might not raise temperature (e.g. during phase change)\n\n"},{"_id":"460382fb772e023bdc0000bb","treeId":"45f4536692911004a000003d","seq":307184,"position":1,"parentId":"460381c8772e023bdc0000b8","content":"Go through changes of phase again.\nAs heat is applied, what happens re. PE, KE? How does temp. change?\nStudents should be able to describe a transfer of heat to/from the substance, and how this manifests in physical properties (i.e. temperature, state of substance)."},{"_id":"46038343772e023bdc0000bc","treeId":"45f4536692911004a000003d","seq":307187,"position":2,"parentId":"460381c8772e023bdc0000b8","content":"Concept-based questions describing changes to temperature or to KE.\nCalculation-based questions giving sample velocities, getting students to calculate r.m.s. velocity.\nMCQ questions describing possible relationships between v_fms, v_avg, T, and KE."},{"_id":"460382c8772e023bdc0000ba","treeId":"45f4536692911004a000003d","seq":2954727,"position":2,"parentId":"4964d6bf2c36b00d34000100","content":"root-mean-square concept is used again in AC electricity, where r.m.s. voltage/current is used to calculate average AC power."},{"_id":"46038199772e023bdc0000b7","treeId":"45f4536692911004a000003d","seq":2954728,"position":3,"parentId":"4964d6bf2c36b00d34000100","content":"T also indicates KE in other degrees of freedom, e.g. rotational, vibrational."},{"_id":"46035c68772e023bdc0000a7","treeId":"45f4536692911004a000003d","seq":2954741,"position":4,"parentId":"460359ed772e023bdc0000a2","content":"## 0th Law of Thermodynamics\n\nTwo systems in thermal equilibrium have the same temperature.\n"},{"_id":"46038818772e023bdc0000bd","treeId":"45f4536692911004a000003d","seq":307203,"position":1,"parentId":"46035c68772e023bdc0000a7","content":"Thermal equilibrium does not mean no thermal energy is being transferred; it means thermal energy transfer in both directions occurs at the same rate.\n\nThis may not be intuitive for differently-sized systems, where one has clearly more total energy than the other (but the same average KE per particle for both)\n\nThis law enables us to determine the direction of heat transfer, from a logical extension of energy conservation and the formulaic definition of temperature."},{"_id":"460388dd772e023bdc0000be","treeId":"45f4536692911004a000003d","seq":307204,"position":2,"parentId":"46035c68772e023bdc0000a7","content":"Students are likely to consider total quantity of energy in the system, rather than “density” of energy/average energy per particle."},{"_id":"46038922772e023bdc0000bf","treeId":"45f4536692911004a000003d","seq":307205,"position":1,"parentId":"460388dd772e023bdc0000be","content":"Get students to explain, if they think so, why two systems at the same temperature might still exchange energy.\nSimulations might be useful to illustrate that this occurs for any two system of colliding particles."},{"_id":"46038968772e023bdc0000c0","treeId":"45f4536692911004a000003d","seq":307206,"position":2,"parentId":"460388dd772e023bdc0000be","content":"Questions linking other concepts, e.g. describing changes to one system (from 1st Law), and then asking if it is in thermal equilibrium with another system."},{"_id":"46035d25772e023bdc0000ac","treeId":"45f4536692911004a000003d","seq":2954743,"position":4.5,"parentId":"460359ed772e023bdc0000a2","content":"## 1st Law of Thermodynamics\n\n$\\Delta U_{sys} = Q_{in} + \\Delta W_{on}$\n\n**Change in internal energy** of a gas system ($\\Delta U_{sys}$) is the sum of **thermal energy transferred into** the system ($Q_{in}$), and **work done on** the system ($\\Delta W_{on}$).\n\n## Work done on system\n\n$\\Delta W_{on} = -p \\Delta V$\n\nThe **work done *by* the system** is $p \\Delta V$, where $p$ is the pressure of the gas system (exerted on the box walls), and $\\Delta V$ is the change in volume of the gas system (positive if expanding, negative if contracting/compressing).\n\nHence, **work done *on* the system** is $-p \\Delta V$.\n\nThus, the **1st Law of Thermodynamics** can also be written as\n\n$\\Delta U_{sys} = Q_{in} - p \\Delta V$"},{"_id":"46039712772e023bdc0000db","treeId":"45f4536692911004a000003d","seq":2954744,"position":0.5,"parentId":"46035d25772e023bdc0000ac","content":"### Heat transfer\n\nSign/direction of ΔQ and ΔW.\n\nVarious types of energy-change processes: isobaric, isochoric, isothermal, adiabatic."},{"_id":"4603978d772e023bdc0000dc","treeId":"45f4536692911004a000003d","seq":307288,"position":1,"parentId":"46039712772e023bdc0000db","content":"This topic is new to students, might take more care in explanation.\nMight need to make assumptions explicit, e.g. when are ΔU, ΔQ and ΔW equal to 0, and why."},{"_id":"460398dc772e023bdc0000dd","treeId":"45f4536692911004a000003d","seq":307301,"position":1,"parentId":"4603978d772e023bdc0000dc","content":"Introduce various examples of internal energy changes in systems, get students to describe explicitly what each change involves. Is thermal energy being transferred? Is work being done?\nOnce each example is described, calculate ΔU, ΔQ, ΔW for each example. Give sample numbers where necessary."},{"_id":"46039917772e023bdc0000de","treeId":"45f4536692911004a000003d","seq":307305,"position":2,"parentId":"4603978d772e023bdc0000dc","content":"Calculation questions involving calculation of one variable’s value given other variables’ values.\nMCQ involving direction of ΔQ or ΔW, e.g. a) Work is done on the system, b) Work is done by the system, c) Thermal energy is transferred to the system, d) thermal energy is transferred away from the system."},{"_id":"4603966a772e023bdc0000d8","treeId":"45f4536692911004a000003d","seq":307280,"position":1,"parentId":"46035d25772e023bdc0000ac","content":"Conservation of energy is a subject-wide concept, and its specific consequences in each topic cannot be under-emphasised. It is the underlying assumption of many concepts in Physics."},{"_id":"460396cb772e023bdc0000d9","treeId":"45f4536692911004a000003d","seq":307281,"position":2,"parentId":"46035d25772e023bdc0000ac","content":"Entropy and its related equations."},{"_id":"46035b10772e023bdc0000a4","treeId":"45f4536692911004a000003d","seq":2954747,"position":2.5,"parentId":"46035d25772e023bdc0000ac","content":"### Internal energy\n\nInternal energy can be analysed as PE and KE.\n\nPE arises from particle interactions.\nKE arises from particle movement.\nTogether, they describe all of the substance’s internal energy.\n\nIt is a simple but powerful basis for understanding energy: all energy is ultimately potential or kinetic in nature."},{"_id":"46036dc1772e023bdc0000ae","treeId":"45f4536692911004a000003d","seq":307108,"position":1,"parentId":"46035b10772e023bdc0000a4","content":"There are more detailed breakdowns for PE and KE.\nIn abstract form, PE varies depending on how we define particle interactions. These interactions are defined as potentials.\n\nPotential energy (“that which is yet to be”) is a very abstract concept, can’t be easily visualised or seen in motion. Many analogies may be needed.\n\nStudents might pull in other forms of energy they don’t perceive as being either PE or KE.\n\nMight have to be clear about what PE, KE refer to.\nPE → Any interaction (gravitational, electrostatic, etc)\nKE → Movement of individual particles.\nKE of gas → sum of particle KEs."},{"_id":"46036f08772e023bdc0000af","treeId":"45f4536692911004a000003d","seq":307112,"position":1,"parentId":"46036dc1772e023bdc0000ae","content":"Draw analogies between gas PE and other PE, e.g. GPE.\n- Objects have yet to move together, but are able to do so due to interactions between objects.\n\nAsk students to surface types of energy they are unable to categorise, help them to categorise.\n\nThis process should help students see similarities between PE and KE as defined in the kinetic model, and in other topics e.g. gravitation."},{"_id":"46036faf772e023bdc0000b0","treeId":"45f4536692911004a000003d","seq":307114,"position":2,"parentId":"46036dc1772e023bdc0000ae","content":"Multiple-choice questions:\n1) PE increases\n2) KE increases\n3) U increases\nGive various examples, get students to pick from combinations of the above.\nLikely problems: Neither PE nor KE increases though U increases.\nU is 0 but one of the other two increases."},{"_id":"46035d51772e023bdc0000ad","treeId":"45f4536692911004a000003d","seq":2954748,"position":3,"parentId":"46035d25772e023bdc0000ac","content":"### Work done\n\nDoing work on the system involves a change in volume\n$\\Delta W = ‒p\\Delta V$"},{"_id":"46039a8b772e023bdc0000df","treeId":"45f4536692911004a000003d","seq":307312,"position":1,"parentId":"46035d51772e023bdc0000ad","content":"Work done on/by the system involves the same underlying concept as work done on/by a moving object: applying force over a distance."},{"_id":"46039ad4772e023bdc0000e0","treeId":"45f4536692911004a000003d","seq":307313,"position":2,"parentId":"46035d51772e023bdc0000ad","content":"There is a standard definition of work done on/by a gas in physics, and it is important to differentiate this from other forms of energy change (e.g. isobaric change). This is important in understanding which processes involve mechanical input and output, critical for engineering."},{"_id":"46039afd772e023bdc0000e1","treeId":"45f4536692911004a000003d","seq":307326,"position":1,"parentId":"46039ad4772e023bdc0000e0","content":"Work done is the definite integral of pressure vs volume; we usually assume constant pressure (or processes with minimal change in pressure), or take the average pressure as indicative of the equivalent pressure."},{"_id":"46039bf8772e023bdc0000e3","treeId":"45f4536692911004a000003d","seq":307323,"position":1,"parentId":"46039afd772e023bdc0000e1","content":"Draw explicit links to this earlier equation. Taking W = F.s on a per-unit-area average gives us the formula.\nSimilar to 1st Law: introduce various examples, calculate ΔW for each case. Throw in some cases where pressure is non-constant, or volume change does not take place under constant pressure."},{"_id":"46039c35772e023bdc0000e4","treeId":"45f4536692911004a000003d","seq":307325,"position":2,"parentId":"46039afd772e023bdc0000e1","content":"MCQ involving direction of ΔW, e.g. a) Work is done on the system, b) Work is done by the system.\nStructured questions asking for equation assumptions/conditions, or to explain the equation (why minus sign)."},{"_id":"46039b30772e023bdc0000e2","treeId":"45f4536692911004a000003d","seq":307319,"position":4,"parentId":"46035d51772e023bdc0000ad","content":"Pressure is assumed to be constant. If it is not, students might not know how to calculate an equivalent pressure.\nChange in volume is assumed to take place at constant pressure; if this cannot be assumed, the formula cannot/should not be used."},{"_id":"46035c93772e023bdc0000a8","treeId":"45f4536692911004a000003d","seq":2954749,"position":5,"parentId":"460359ed772e023bdc0000a2","content":"## Energy distribution\n\nIn a gas, particle velocities are not homogeneous."},{"_id":"46038cc2772e023bdc0000c2","treeId":"45f4536692911004a000003d","seq":2954750,"position":0.5,"parentId":"46035c93772e023bdc0000a8","content":"### Boltzmann distribution\n\nParticle velocity follows Boltzmann distribution.\nDerivation of Boltzmann distribution; likelihood of each energy state being occupied, etc.\n\nThinking of particle velocities as a distribution is necessary to explain some phenomena, e.g. evaporation rate on windy day."},{"_id":"46038b32772e023bdc0000c1","treeId":"45f4536692911004a000003d","seq":2954751,"position":1,"parentId":"46035c93772e023bdc0000a8","content":"### Root-mean-square velocity\n\nVelocities of particles are semi-randomly distributed.\n\nΣKE = KE_1 + KE_2 + …\n= 0.5m(v_12 + v_22 + …)\n= 0.5mv_r^2\nv_r = (v_12 + v_22 + …)^0.5\n\nv_r is the velocity all particles must have for the substance to have the same KE."},{"_id":"46038dc2772e023bdc0000c3","treeId":"45f4536692911004a000003d","seq":307223,"position":1,"parentId":"46038b32772e023bdc0000c1","content":"What is v_r for? Students will most likely ask.\nMight need worked examples. From KE = 3/2KT maybe? Or avg. v vs vr in relation to T.\n\nThe point/purpose of v_r will probably be vague or ambiguous until worked examples are given.\nShould also show more examples of avg. v being inadequate. Graphs of v vs KE, for instance.\n\nThis particle distribution concept is used in chemistry as well, to explain why endothermic reaction rate increases with temperature."},{"_id":"46038e91772e023bdc0000c4","treeId":"45f4536692911004a000003d","seq":307226,"position":1,"parentId":"46038dc2772e023bdc0000c3","content":"Show how a uniform velocity distribution ends up after some collisions.\nExplain aspects of distribution\n- few high-KE particles\n- mostly mid-KE"},{"_id":"46038ed4772e023bdc0000c5","treeId":"45f4536692911004a000003d","seq":307227,"position":2,"parentId":"46038dc2772e023bdc0000c3","content":"Concept-based questions asking about velocities of particles. E.g.\nAs water is heated, a) the number of particles above x speed increases, b) the number of particles below y speed increases, c) the speed of each particle increases proportionally, ..."},{"_id":"46035cb7772e023bdc0000a9","treeId":"45f4536692911004a000003d","seq":2954752,"position":6,"parentId":"460359ed772e023bdc0000a2","content":"## Heat capacity\n\nSpecific heat capacity tells us how much energy is needed to raise 1 unit mass of substance by 1 K.\n"},{"_id":"46038f98772e023bdc0000c6","treeId":"45f4536692911004a000003d","seq":307231,"position":1,"parentId":"46035cb7772e023bdc0000a9","content":"This enables us to calculate energy requirements for various daily tasks/processes, and is important for calculating energy efficiency.\nA standard amount of energy is transferred for any process, and specific heat capacity is one way for us to calculate this amount of energy for the process of heating/cooling."},{"_id":"46038fe5772e023bdc0000c7","treeId":"45f4536692911004a000003d","seq":307232,"position":2,"parentId":"46035cb7772e023bdc0000a9","content":"Specific heat capacity involves particle KE; particle PE is not affected (or its effects are not considered)."},{"_id":"46039013772e023bdc0000c8","treeId":"45f4536692911004a000003d","seq":307235,"position":3,"parentId":"46035cb7772e023bdc0000a9","content":"Specific heat capacity does not change with mass.\nThe same substance in different states has different heat capacities as well; students often mix them up.\nTwo substances of different heat capacities, when fused, have an equivalent heat capacity that depends on their mass proportions."},{"_id":"4603904b772e023bdc0000c9","treeId":"45f4536692911004a000003d","seq":307236,"position":1,"parentId":"46039013772e023bdc0000c8","content":"Students have learned this at O levels; A levels mainly adds measurement of s.h.c. by electrical methods. Do not need to go into too much concept detail.\nFocus on applications that involve other topic concepts, e.g. conversion of energy, efficiency calculations, etc."},{"_id":"460390aa772e023bdc0000cb","treeId":"45f4536692911004a000003d","seq":307237,"position":1,"parentId":"4603904b772e023bdc0000c9","content":"Recap O level content. Get students to think about what is happening to internal energy; how are PE, KE being affected?"},{"_id":"460390c9772e023bdc0000cc","treeId":"45f4536692911004a000003d","seq":307239,"position":2,"parentId":"4603904b772e023bdc0000c9","content":"Questions based on calorimetry, s.h.c. measurement.\nQuestions involving thermal energy change causing liquids to heat up/cool down."},{"_id":"46035ce0772e023bdc0000aa","treeId":"45f4536692911004a000003d","seq":2954753,"position":7,"parentId":"460359ed772e023bdc0000a2","content":"## Latent heat\n\nSpecific latent heat tells us how much energy is needed to change 1 unit mass of substance to a different state.\nQuestions involving thermal energy change causing liquids to heat up/cool down.\nSpecific heat of fusion is for melting.\nSpecific heat of vapourisation is for boiling.\n"},{"_id":"460391dd772e023bdc0000cd","treeId":"45f4536692911004a000003d","seq":307243,"position":1,"parentId":"46035ce0772e023bdc0000aa","content":"This enables us to calculate energy requirements for various daily tasks/processes, and is important for calculating energy efficiency.\nA standard amount of energy is transferred for any process, and specific heat capacity is one way for us to calculate this amount of energy for these processes (e.g. melting and boiling)."},{"_id":"460392bf772e023bdc0000ce","treeId":"45f4536692911004a000003d","seq":307247,"position":1.5,"parentId":"46035ce0772e023bdc0000aa","content":"Specific latent heat involves particle PE; particle KE is not affected (or its effects are not considered)."},{"_id":"460392f0772e023bdc0000cf","treeId":"45f4536692911004a000003d","seq":307248,"position":1.75,"parentId":"46035ce0772e023bdc0000aa","content":"Specific latent heat does not change with mass.\nStudents frequently mix up specific latent heat of fusion and vapourisation."},{"_id":"46039362772e023bdc0000d0","treeId":"45f4536692911004a000003d","seq":307250,"position":1,"parentId":"460392f0772e023bdc0000cf","content":"Students have learned this at O levels; A levels mainly adds measurement of s.l.h. by electrical methods. Do not need to go into too much concept detail.\nFocus on applications that involve other topic concepts, e.g. conversion of energy, efficiency calculations, etc."},{"_id":"460393ac772e023bdc0000d1","treeId":"45f4536692911004a000003d","seq":307251,"position":1,"parentId":"46039362772e023bdc0000d0","content":"Recap O level content. Get students to think about what is happening to internal energy; how are PE, KE being affected?"},{"_id":"460393e8772e023bdc0000d2","treeId":"45f4536692911004a000003d","seq":307255,"position":2,"parentId":"46039362772e023bdc0000d0","content":"Questions based on calorimetry, s.l.h. measurement.\nQuestions involving thermal energy change causing liquids to change state."},{"_id":"46035d09772e023bdc0000ab","treeId":"45f4536692911004a000003d","seq":307261,"position":2,"parentId":"46035ce0772e023bdc0000aa","content":"Specific heat of vapourisation is greater than specific heat of fusion.\n\nBoiling/evaporation involves more energy than freezing/melting; an understanding of this is important to understand the cooling effects of water (as used daily or in industry), which is also mentioned in other sciences (i.e. biology, chemistry)."},{"_id":"46039515772e023bdc0000d3","treeId":"45f4536692911004a000003d","seq":307263,"position":1,"parentId":"46035d09772e023bdc0000ab","content":"The difference between s.h.v. and s.h.f. can be (naïvely) calculated for any substance using the 1st Law of thermodynamics."},{"_id":"46039540772e023bdc0000d4","treeId":"45f4536692911004a000003d","seq":307268,"position":2,"parentId":"46035d09772e023bdc0000ab","content":"Both specific heats involve a change in phase, and students may perceive them as being the same process, failing to see that vapourisation involves a much greater change in volume and particle PE."},{"_id":"46039566772e023bdc0000d5","treeId":"45f4536692911004a000003d","seq":307269,"position":3,"parentId":"46035d09772e023bdc0000ab","content":"Students might not be aware that specific heat of vapourisation involves work done against atmospheric pressure; drawing link to formula could be useful if properly scaffolded."},{"_id":"46039586772e023bdc0000d6","treeId":"45f4536692911004a000003d","seq":307272,"position":1,"parentId":"46039566772e023bdc0000d5","content":"Make students aware of the mechanical processes involved in melting and boiling.\nWhat is happening to the particles? What prevents them from expanding naturally without any thermal energy input? How do they overcome these “energy obstacles”?"},{"_id":"460395b6772e023bdc0000d7","treeId":"45f4536692911004a000003d","seq":307275,"position":2,"parentId":"46039566772e023bdc0000d5","content":"Structured, explanation-type questions asking students to explain related phenomena."},{"_id":"4964905c2c36b00d340000f1","treeId":"45f4536692911004a000003d","seq":2954225,"position":12,"parentId":null,"content":"# 11 Superposition\n\n"},{"_id":"4964909b2c36b00d340000f2","treeId":"45f4536692911004a000003d","seq":2954226,"position":13,"parentId":null,"content":"# 12 Electric Field\n\n"},{"_id":"496491292c36b00d340000f3","treeId":"45f4536692911004a000003d","seq":2954228,"position":14,"parentId":null,"content":"# 13 Current of Electricity\n\n* Conventional current\n* Resistivity\n* Ohm's Law\n* Emf and potential difference (p.d.)\n* Internal resistance and terminal voltage"},{"_id":"63c01cfe8ec8a3a3240000ed","treeId":"45f4536692911004a000003d","seq":6214235,"position":1,"parentId":"496491292c36b00d340000f3","content":"Clarify semiconductor diode: ideal vs real-world graph of I-V characteristics"},{"_id":"63c19672f6d92542eb0000ef","treeId":"45f4536692911004a000003d","seq":6214492,"position":2,"parentId":"496491292c36b00d340000f3","content":"Confusion over I-V graph in experiment to determine internal resistance"},{"_id":"651012eb44b223b9e9000107","treeId":"45f4536692911004a000003d","seq":6464309,"position":3,"parentId":"496491292c36b00d340000f3","content":"Confusing R = V/I (ratio) with R = dV/dI (gradient)"},{"_id":"651013fc44b223b9e9000108","treeId":"45f4536692911004a000003d","seq":6464310,"position":4,"parentId":"496491292c36b00d340000f3","content":"Identifying and treating short circuits"},{"_id":"6510149544b223b9e9000109","treeId":"45f4536692911004a000003d","seq":6464311,"position":5,"parentId":"496491292c36b00d340000f3","content":"Potential divider: why does the test circuit have no current?"},{"_id":"496491a52c36b00d340000f4","treeId":"45f4536692911004a000003d","seq":2954229,"position":15,"parentId":null,"content":"# 14 DC Circuits\n\n* Resistors in series and parallel\n* Potential dividers\n* Complex circuits\n* Thermistors and light-dependent resistors"},{"_id":"496491e42c36b00d340000f5","treeId":"45f4536692911004a000003d","seq":2954230,"position":16,"parentId":null,"content":"# 15 Electromagnetism"},{"_id":"496493012c36b00d340000f6","treeId":"45f4536692911004a000003d","seq":6464312,"position":17,"parentId":null,"content":"# 16 EM Induction"},{"_id":"4964933d2c36b00d340000f7","treeId":"45f4536692911004a000003d","seq":6464313,"position":18,"parentId":null,"content":"# 17 AC Circuits"},{"_id":"496493b62c36b00d340000f8","treeId":"45f4536692911004a000003d","seq":6464314,"position":19,"parentId":null,"content":"# 18 Quantum Physics"},{"_id":"496493f42c36b00d340000f9","treeId":"45f4536692911004a000003d","seq":6464315,"position":20,"parentId":null,"content":"# 19 Lasers & Semiconductors"},{"_id":"4964947d2c36b00d340000fa","treeId":"45f4536692911004a000003d","seq":6464316,"position":21,"parentId":null,"content":"# 20 Nuclear Physics"}],"tree":{"_id":"45f4536692911004a000003d","name":"9646 H2 Physics","publicUrl":"9646-h2-physics","latex":true}}