Physics 30 Complete Course Study Booklet

Organized Notes Compiled by Anthony Demong from Mr. Christoffel’s Physics 20 Class and Resources

NOTE: This guide does not contain absolutely everything learned in the course, though it covers the most important material. It is designed to accompany questions from the textbooks provided by the teacher.

Kinematics Unit – Study Guide Compiled by Anthony Demong

Definitions and Important Notes

Kinematics: Kinematics is the branch of physics that deals with the description of the motion of objects without reference to the forces or agents that cause said motion. The branch that does deal with this is called dynamics.

Displacement: Displacement describes both the magnitude of a change in position and the direction of that change. If a dog walked 20.0 metres east of its original starting position, it would have a displacement of 20.0 m [E] (20.0 metres is the magnitude of the change in position, and the [E] represents the direction east).

Delta (Δ): Delta is a Greek letter that represents “change of”, “increase of”, “decrease of”, “difference”, or “interval” (e.g. Δt would represent ‘change in time’).

Vectors / Vector Quantities: A vector quantity is a quantity that has a magnitude, unit and direction. Displacement is a vector.

Vectors can be represented graphically as arrows of varying length, according to a certain scale. The following arrow would represent a dog’s displacement 20.0 m [E]:

Scale: 1 cm = 10.0 m

A vector should also have an arrow on top of its variable to indicate that it is a vector:

∆ d⃗=20.0m [E ]

Remember:

∆ d⃗=d⃗2−d⃗1 means displacement = the final position – the initial position

If the value of the vector is negative, draw the arrow facing the other direction

Scalars: Scalars are quantities that only have a unit and magnitude. Time, mass, and length are scalars.

Uniform Velocity / Uniform Motion: Uniform velocity is a constant velocity that does not accelerate or

decelerate over time. It can be found using v⃗=∆ d⃗∆ t

Average Velocity: Average velocity can be calculated using the formula v⃗av=∆ d⃗ tot

∆ t tot

It can also be found by adding all the velocities, then dividing that sum by the number of different velocities.

Instantaneous Velocity: Instantaneous velocity is the same as average velocity if there is uniform velocity. If there is not uniform velocity, then the instantaneous velocity will differ from the average velocity.

In a displacement vs time graph, instantaneous velocity can be found by making a tangent line to the line of best fit, then creating a triangle from the two ends of the tangent line (so the tangent line is the hypotenuse of the triangle).

Based on the two endpoints of the tangent line, instantaneous velocity can be calculated using the

formula v⃗inst=∆ d⃗∆ t

Acceleration: Acceleration is the rate of change of velocity. Constant acceleration on a time vs displacement graph appears as a sloping line, while constant acceleration on a time vs velocity graph appears as a straight line.

Acceleration can be found using the formula a⃗=∆ v⃗∆ t or a⃗=

v⃗ f− v⃗i∆ t

Average Acceleration: Average acceleration can be calculated with the formula a⃗av=∆ v⃗ tot

∆ t tot

Instantaneous Acceleration: Instantaneous acceleration is the same as average acceleration if there is constant acceleration. If there is not constant acceleration, then the instantaneous acceleration will differ from the average acceleration.

In a velocity vs time graph, instantaneous acceleration can be found by making a tangent line to the line of best fit, then creating a triangle from the two ends of the tangent line (so the tangent line is the hypotenuse of the triangle).

Based on the two endpoints of the tangent line, instantaneous acceleration can be calculated using the

formula a⃗ inst=∆ v⃗∆ t

Other Important Formulas and Their Uses

V f

→

=V i

→

+a→

t can be used to calculate the final velocity based on the initial velocity, time, and

acceleration

d→

=V i

→

t+ 12a→

t 2 can be used to calculate the displacement based on the initial velocity, time,

and acceleration

V f

→2−V i

→2=2 a

→

d→

represents that the difference between the final velocity squared and the initial

velocity squared is equivalent to twice the acceleration multiplied by the displacement

Position over Time Graphs

When graphing, remember the following (specific to Mr. Christoffel’s class):

Darken the x-axis and y-axis of the graph to emphasize them Create a meaningful and relevant title for the graph Label the axes with their proper units (e.g. cm, s, etc.) Use the proper number of significant figures on your axes When labelling plotted points on the graph, use the same number of significant figures for the x

and y values Draw a circle around any points to indicate uncertainty Any line of best fit drawn onto the graph should have roughly the same number of points on

either side of it To find the slope, find two points on the line of best fit and use them to draw a triangle Determine the slope of the triangle’s hypotenuse, remember to take account for the proper

number of significant figures in the calculations

Example:

*Don’t make the same mistake that I did. Always state the formula before making calculations, and do not put Slope= riserun

instead

Kinematic Graphing of Displacement, Velocity and Acceleration

Always try to picture a mental image of an object in motion when working on these graphs to assist you in making sense of the increases and decreases in displacement, velocity and acceleration

Set up the graph just like the position over time graph exemplified above, but split the graph into three rows: one for displacement, one for velocity, and one for acceleration

First, outline the time increments in the problem: For the problem below, those would be 10.00 seconds then an added 3.50 seconds

Draw vertical lines in the graph along those times to split it into several regions; you may want to label those regions as A, B, C, etc.

After that, solve for as much data possible: The initial velocity / displacement The final velocity / displacement

(You may need to stop and think about whether or not the object is moving or resting at the beginning or the end)

The velocity / acceleration at each time increment Graph the data, and shade in the acceleration and velocity’s areas to help with calculations.

Remember to use the proper number of significant figures

Example:

In this graph, a tractor is moving at a velocity of 12.00 metres per second for 10.00 seconds, then decelerates to a velocity of 7.00 metres per second in a time of 3.50 seconds.

River Crossing Problems

A river has a current with a velocity of 4.0 metres per second north. A swimmer can swim at a velocity of 6.0 metres per second in the water east. What is the velocity of the swimmer compared to the ground, and in which direction?

Drawing the Diagram

Before the vector diagram is drawn, a basic diagram of the situation should be drawn to represent the scenario. Below, a swimmer is swimming east across a north-flowing river, and is being pushed north as he swims in a northeast direction:

By drawing this essential scenario diagram, we know the basic directions of each vector, and that he will be travelling northeast when compared to the ground.

Next, the vector diagram can be drawn. Each arrow is drawn in scale to each velocity in the diagram (i.e. the water current vector arrow is drawn to the scale of 4), and in the same direction (the river velocity is drawn pointing north (up), the swimmer velocity is drawn pointing east (left), and the resultant velocity is drawn pointing northeast (up and to the left) :

V s /g

→

=V s/w

→

+V w /g

→  

REMEMBER TO LABEL IT WITH THE VECTOR EQUATION hypotenuse = side + side

Finding the Resultant Vector

To calculate the resultant vector’s magnitude, use Pythagorean Theorem:

V s /g

→

=V s/w

→

+V w /g

→

V s /g2=V s/w

2+V w/ g2

V s /g=√V s/w2+V w/g

2

V s /g=√(6.0m /s)2+(4.0m /s)2

V s /g=√36m /s+16m /sV s /g=√52m /sV s /g=7.5m /s

Remember to remove the vector signs (❑→ ) and replace them with exponents (2)

You may need to rearrange the equation if you are solving for a specific side, and not the hypotenuse

To calculate the resultant vector’s direction, use COS SIN TAN based on the vector diagram drawn above:

tanθ=V w /g

V s/w

θ=tan−1(V w/ g

V s /w)

θ=tan−1( 4.0m /s6.0m/ s )

θ=tan−1 (0.67 )θ=34 ° N of E

Therefore the swimmer is swimming at a velocity of 7.5 metres per second 34° north of east.

Projectile Problems

Whether a projectile is projected horizontally or vertically, it will take the same amount of time to fall (both things will hit the ground at the same time)

Since vertical velocity is accelerated, in the equation V f

→2−V i

→2=2 a

→

d→

replace d→

with h→

(height), and a→

with g→

(g→

=−9.81m /s[down]) to make V f

→2−V i

→2=2 g

→

h→

Since horizontal velocity is uniform, in the equation v⃗=∆ d⃗∆ t

replace ∆ d⃗ with ∆ R⃗ (range)

to make v⃗=∆ R⃗∆ t

Circular Motion Problems

Equations

V C

→

=2πRT

aC

→

=2π V C

→

T

aC

→

=4 π2R

T2

aC

→

=V c

→2

R c=2π R2¿

aC

→

=V 2

→

+(−V 1

→

)∆ t

Variables

V C

→ = Velocity of an object in circular

motion T = Amount of time necessary to make

one complete oscillation around the circle

aC

→ = Acceleration of an object in

circular motion R = Radius of the circle

Notes

Centripetal Acceleration (aC

→) is a force acting an object in circular motion that prevents it from

flying out of the circle. In the diagram below, it would be an arrow pointing into the centre of the circle

Diagram

Adding Vectors that Do Not Form a Perfect Right Triangle

The steps will be narrated below in an attempt to explain the process, because heck, I’m not even sure if the title is the right way to describe it.

Problem

A person walked 8.94 km @ 27.0° north of west, then 11.40 km @ 76.0° north of east. Find his total displacement.

Variable Setup

First, the angles must be converted into standard angle format. This is simple converting the angles based on a system where east becomes 0°, north is 90°, west is 180° and south is 270°.

d1→

=8.94 km

d2→

=11.40km

d R

→

=?θ1=153.0 ° std<¿θ2=36.0 ° std<¿

¿

Diagram Stage

The first process is to draw each vector, connected in order. Then, the x and y values of each vector are drawn, a sum of both the x and y vectors are drawn, and the resultant vector is drawn connecting the tail of one vector to the tip of the other:

Calculation Stage

Then, a reference triangle is drawn (don’t worry about magnitudes or directions) to find whether COS, SIN, or TAN would find help to find the sum of the x and y values:

REMEMBER TO ALWAYS REMOVE THE VECTOR SIGN (❑→ ) WHEN DIVIDING VECTORS

Next, calculate the sum of both the x and y values of each vector like shown below:

∑ dx

→

=d1 x→

+d2 x→

∑ dx

→

=d1→

Cosθ1+d2→

Cosθ2

∑ dx

→

=(8.94 km ) (cos153.0 ° )+(11.40 km) (cos36.0 ° )

∑ dx

→

=−7.97 km+9.22km

∑ dx

→

=1.25km

∑ d y

→

=d1 y→

+d2 y→

∑ d y

→

=d1→

Sinθ1+d2→

Sinθ2

∑ d y

→

=(8.94 km) (sin 153.0 ° )+ (11.40 km ) (sin 36.0 ° )

∑ d y

→

=4.06 km+6.70 km

∑ d y

→

=10.76km

Next, draw the vector diagram of the sums of the x and y values for each vector to find the resultant displacement.

d R2=∑ d x

2+∑ d y

2

d R=√∑ d x2+∑ d y

2

d R=√(1.25 km)2+(10.76 km)2

d R=√1.56km+115.8kmd R=√117.4 kmd R=10.83 km

tanθ=∑ d y

∑ d x

θ=tan−1(∑ d y

∑ dx)

θ=tan−1( 10.76km1.25km )θ=tan−1 (8.61 )θ=83.4 ° N of E

Therefore his total displacement was 10.83 kilometers 83.4° north of east.

Dynamics Unit – Study Guide Compiled by Anthony DemongThe following unit relies heavily on the knowledge gained from the kinematics unit. Ensure that you have a comprehensive understanding of kinematics before attempting to solve dynamics problems.

Definitions and Important Notes

Dynamics: Dynamics is the branch of physics that deals with the description of the motion of objects with reference to the forces or agents that cause said motion.

Unbalanced Force: Any force that does not have an equivalent and opposite force acting upon it at the same time. Balanced forces are known as statics.

Newton’s 3 Laws:

A body in motion remains in motion unless an external unbalanced force acts upon it The relationship between an object’s force (F), mass (m) and acceleration (a) is represented by

the formula F=ma For every action there is an equal and opposite reaction

Acceleration: Acceleration is the rate of change of velocity.

Force: Force is measured in newtons (N). One newton is the force required to accelerate 1 kilogram 1 metre per second. Weight is a measurement of force, since it depends on both mass and gravitational acceleration.

Mass: A measurement of matter. In this class it is most often measured in kilograms (kg).

Net Force: Represented byFnet

→, the net force of an object is determined by finding the sum of the

vectors of the applied forces and force of friction:

Fnet

→

=Fapp

→

+F f

→

Force of Gravity: The force of gravity is the force that an object has in the direction of the Earth’s surface. On Earth, the acceleration of gravity is 9.81 metres per second, so the force of gravity acting upon an object can be determined by multiplying the acceleration of gravity by the mass of said object (using Newton’s second law of (F=ma)).

Angle of Inclination: The angle of inclination is the angle in which a slope is measured. For example, a vertical wall would have an angle of inclination of 90°, while a flat plane would have an angle of inclination of 0°.

Normal Force: The normal force is the force acting upon an object perpendicular to the surface that it is in contact with:

The normal force can be calculated using the formula FN

→

=Fg

→

Cosθincl

Downwards Force: Downwards force, represented byFdn

→ is the force of an object in the direction of its

travel down a slope. It is opposite to the force of friction and parallel to a surface:

The downwards force can be calculated using the formulaFdn

→

=Fg

→

Sinθincl. However, it can also be

calculated using the formulaFdn

→

=madn

→.

Sample Calculation

A force of 120N is applied to a mass of 3.50kg. What is the acceleration?

F = 120N F=ma

m = 3.50kg a= Fm

a =? a= 12.0 N3.50 kg

a=3.43m /s2

∴ The mass had an acceleration of 3.43 metres per second squared.

μ – The Coefficient of Friction

The letter μ can be used to represent the coefficient of friction. The coefficient of friction is a ratio of the force of friction compared to the normal force.

μ=F f

FN

QUICK TIP: Never use vector signs () when dividing vectors. Always remove them from the dividend and divisor when putting them into a formula.

The diagram below represents a mass of 13.0kg on a slope with an angle of inclination of 30.0°.

The normal force is acting upon the mass perpendicular to the surface. The object’s force of friction is acting against the downwards force, both of which are parallel to the surface. The force of gravity is acting upon the object downwards towards the Earth. It is also very important to notice that the right triangle formed between the normal force, force of gravity and downwards force has the same angle as the angle of inclination.

SOH CAH TOA can be used to determine the value of the angle of inclination, downwards force, and force of gravity if needed. However, remember to show all of the steps when determining anything from SOH CAH TOA.

Pushing and Pulling Objects

Whenever an object is pushed or pulled, multiple forces are acting upon said object. When an object is pushed, downwards force is also often exerted onto the object which can increase friction. When an object is pulled, upwards force is also often exerted into the object which can decrease friction.

To begin the solving process for this problem, list all of the knowns and unknowns:

Ay

Ay

A

θapp=24.0°

F A

→

=80.0Nm=11.0 kgμ=0.270

anetx

→

=?

g→

=9.81m /s2

Fg

→

=?

F A y

→

=?

F A x→

=?

Step 1: Solve for Fappx

cosθ A=FA x

F A

F Ax=F ACosθA

F Ax=(80.0N )(0.9135)F Ax=73.1N

Step 2: Solve for Fappy

sin θA=F A y

F A

F A y=F ASinθ A

F A y=(80.0N )(0.4067)F A y=32.5N

Step 3: Solve for Fnety

Fnety

→

=F g

→

+F A y

→

Fnety=F g−FA y Converts to negative because gravity is working against the pulling force. If the object was pushed gravity would be added.

Fnety=mg−FA y

Fnety=(11.0 kg)(9.81m /s2)−32.5NFnety=75.4N

Step 4: Solve for Fnetx

Fnetx

→

=F f

→

+F A x

→

Fnetx=F A x−F f Fnetx=F A x−μ Fnety The force of friction is equivalent to μ multiplied by the normal force, which in this case is Fnety.

Fnetx=73.1 N−(0.270)(75.4N )Fnetx=52.7 N

Step 5: Solve for anetx

Fnetx

→

=manetx

→

anetx

→

=Fnetx

→

m

anetx

→

= 52.7N11.0kg

anetx

→

=4.79m /s2

Remember: The difference between push and pull questions is whether or not the force of gravity is added or subtracted from the applied force in the y direction.

Tension and Pulleys

The diagram below represents three masses on a surface, two of which are hanging from pulleys. The larger mass (m1) is affected greater by the force of gravity and is therefore pulling the weaker hanging mass (m3), as well as the mass on the surface (m2) towards it.

For ease, the scenario can also be represented like above, where the masses are all on a single flat

surface. To calculate the net acceleration in the x direction (anetx

→), the net force in the x direction (Fnetx

→)

must be found (if m1 = 250g, m2 = 500g and m3 = 125g):

Fnetx

→

=Fg1

→

+Fg3

→

Fnetx=Fg1−F g3

mtot anetx=m1g−m3 g

anetx=m1g−m3g

mtot

anetx=m1g−m3gm1+m2+m3

anetx=m1−m3

m1+m2+m3g

anetx=250g−125 g

250 g+500g+125g(9.81m /s2)

anetx=1.40m / s2

The tension for each mass can then be found be rearranging the equation for Fnet (demonstrated with m1):

Fnet1

→

=Fg1

→

+T 1→

Fnet1=Fg1−T 1T 1=Fg1−Fnet1

T 1=m1g−m1anet The net acceleration in the x direction is the same for every mass, so it can be used here

T 1=m1(g−anet)

T 1=0.250 g(9 .81m /s2−1.40m /s2)T 1=0.250 g(8.41m /s2)T 1=2.10NThe Universal Gravitational Constant

The universal gravitational constant (G) is equivalent to 6.67×10−11 Nm2

kg2. It can be used in the

following formula to determine the effects of gravity on other celestial bodies:

Fg

→

=mg→

=Gm1m2

d2

The above formula states that the force of gravity (on another celestial body) is equivalent to the universal gravitational constant multiplied by mass 1 (the central mass, in this case the mass of a celestial body) multiplied by mass 2 (the mass of the object being pulled towards the celestial body) divided by the distance between the object and celestial body squared.

Note: The subject of the formula (mass 1) does not need to be a celestial body.

Momentum and Energy Unit – Study Guide Compiled by Anthony Demong

The following unit relies heavily on the knowledge gained from both the kinematics unit and the dynamics unit. Ensure that you have a comprehensive understanding of kinematics and dynamics before attempting to solve any momentum or energy problems.

Definitions and Important Notes

Momentum: Momentum is the quantity of motion of a moving body. It is the product of the object’s velocity and mass:

p→

=mv→

When working with momentum questions, use vectors. It is measured in kg/ms, or Ns (since kg/ms2 is one newton).

Impulse: Impulse is the change in momentum. Like momentum, it is measured in kg/ms or Ns. It is also the product of force and time:

I→

=∆ p→

=F→

·∆ t

Conservation of Momentum: Momentum is conserved when the mass or velocity of an entity is changed. There are three main types of collisions concerned with the conservation of momentum in our class:

Inelastic, where two objects collide and stick together Elastic collision, where two objects collide and bounce of each other

Perfectly elastic collision, where two objects collude and bounce off each other in such a way that all energy is conserved

The sum of the initial momentums of both colliding objects is equivalent to the sum of their final momentums:

∑ pi

→

=∑ pf

→

Conservation of Energy: The sum off all the initial energies is equivalent to the sum off the final energies. Energy can change form, but it cannot be created or destroyed:

∑ E i=∑ E f

In this class we deal with four main types of energy: Kinetic energy, gravitational potential energy, spring energy, and work energy. They are all measured in joules (J). When working with energy, vectors are not used.

Kinetic Energy: Kinetic energy is the energy of motion. It can be found using the following formula:

EK=12mv2

Gravitational Potential Energy: Gravitational Potential energy is the energy of an object when it is dropped. It can be found using the following formula:

EP=mgh

Spring Energy: Spring energy is the energy stored in an elastic device. It can be found using the following formula:

ES=12k x2

Work Energy: Work energy is when a motor or force (such as the force of friction) adds energy to a system. It can be found using the following formula:

W=F ·d

Variable Key:

m = mass of the object in questionv = velocity of the object in questiong = the acceleration of gravity (9.81 metres per second squared)h = height of the object in question

k = spring constant (unique to each spring, measured in newtons per metre)x = compression or stretch of a springF = Force applied on the object in questiond = distance travelled of the object in question

Important:

Work force is involved only when an object moves (hence the use of distance in its equation). For example, pushing on a wall involves applied force, but the wall does not move, so there is no work force involved

Work force can be graphed by force over distance If an object is allowed to “coast”, then only friction is acting upon it

Determining the Type of Energies involved with a Problem

When presented with a physics problem that involves the conservation of energy, you must determine exactly which types of energy are involved in that problem. The best way to do this is make a table listing all possible energies involved:

Initial

Final

EK

EP

ES

WLook at the question and think about what is happening. For example, let’s say the following problem is in question:

A small dog (with a mass of 20.0kg) is held over a ledge at the top of a 50.0 metre building and then dropped. It lands on a trampoline 10.0 metres above the ground that has a spring constant of 100 N/m, and as the dog hits it the trampoline stretches 50.0cm. The dog performs this stunt regularly for amusement, and estimates that he lands on the trampoline at roughly 100 kilometers per hour. Determine if the dog’s estimate is accurate.

The dog has initial gravitational potential energy, but no initial kinetic, spring, or work energy. After hit hits the trampoline it has spring energy and kinetic energy. These energies can be checked off on the above table:

Initial

Final

EK X ✓

EP ✓ X

ES X ✓

W X X

Next, like any other physics question, state all the knowns and unknowns of the question, and you can draw a sketch of the problem if it helps:

m = 20.0kgk = 100 N/mx = 50.0cm = 0.500mg = 9.81m/s2h = 50.0m - 10.0 m = 40.0m

Next, the equation can be written:

∑ E i=∑ E f

EPi=EKf+ESf

mgh=12mv2+1

2k x2

Next, rearrange the formula to solve for the variable in question (velocity):

v=√ 2mgh−k x2

m

Finally, substitute the variables with the correct values and solve for the velocity.

v=√ 2(20.0 kg)(9.81m /ss)(40.0m)−(100N /m)(0.500m)2

(20.0 kg)

v=√784m2/s2

v=28.0m /s = 101km/h

∴ The dog was travelling at about 28.0 metres per second, or 101 kilometers per hour when it hits the trampoline. The dog’s estimate was surprisingly accurate (for a dog).

Important:

Sometimes instead of initial and final, the table will have column headings 1, 2, 3 . . . etc. if there are multiple steps in a problem. For example, if the problem asked to calculate the velocity of the dog as it travelled back up into the air, a third step would be required

Never drop a dog off of a building, even if they give consent Drawing a diagram and labelling each step of the problem can help a lot when determining

which column of the table should contain which energies

Nuclear Energy, Efficiency, and Electrical Energy Unit – Study Guide Compiled by Anthony DemongThe following unit relies on minimal knowledge of how forces operate. Ensure that you have a comprehensive understanding of how forces work before attempting to solve any nuclear energy or electrical energy problems.

Nuclear Energy

Definitions and Important Notes

Atom: A basic unit of a chemical element, and known as the “building block” of all matter. Each atom is of a specific element.

Element: One of more than a hundred substances that cannot be broken down by chemical means into simpler substances. Each element is defined by the number of protons in its nucleus.

Nucleus: The centre structure of an atom, which contains protons and neutrons.

Electron: A minuscule particle that orbits the nucleus of an atom. It carries an electrical charge of -1 and has a mass of 9.11E-31 kilograms or 0.000000000000000000000000000000911 kilograms.

It can be represented by 0

−1e

Proton: A minuscule particle that is inside the nucleus of an atom. It carries an electrical charge of +1 and has a mass of 1.673E-27 kilograms or 0.000000000000000000000000001673 kilograms.

It can be represented by 11H

(H is used as a symbol because it is the atomic symbol for hydrogen, an element that contains only one proton)

Neutron: A minuscule particle that is inside the nucleus of an atom. It carries a neutral electrical charge (0) and has a mass of 1.675E-27 kilograms or 0.000000000000000000000000001675 kilograms.

It can be represented by 10n

Ion: An ion is an atom with more or fewer electrons than a neutral atom of the same element. It therefore has an electrical charge (more electrons than protons = negative charge, more protons than electrons = positive charge).

Isotope: Atoms of the same element with more or fewer neutrons than the most common atom of that element.

Atomic Number: The atomic number of an atom is the number of protons in that atom. It is also equivalent to the nuclear charge of the atom.

Atomic Weight: The atomic weight of an atom is the sum of protons and neutrons in an atom, measured in amu (atomic mass units – both a proton and a neutron have an amu value of 1).

Radiation: An emission of electromagnetic waves or moving subatomic particles.

Alpha Radiation: An alpha radiation particle is equivalent to the nucleus of a helium atom (42He) and

can be identified by the Greek letter alpha (α).

Alpha particles do not penetrate well, so they do not cause much damage when they hit a target. Most do not penetrate human skin, but can be harmful to humans if swallowed or absorbed into the skin. They cannot penetrate clothing, and have a short range (only a few centimetres).

Beta Radiation: An alpha radiation particle is equivalent to a speeding electron (0

−1e) and can be

identified by the Greek letter beta (β).

Beta particles can travel several metres, and can penetrate human skin. They can cause damage to the skin if they remain on the skin for a long period of time. Clothing provides some protection from beta particles, but not complete protection.

Gamma Radiation: Gamma radiation is a form of electromagnetic radiation. It can be identified by the Greek letter beta (γ).

Gamma radiation can be in the form of UVA and UVB rays:

UVB rays have a wavelength between 280 – 315 nanometres, and cause most skin cancer

UVA rays have a wavelength between 315 – 400 nanometres

Strong Force: The nuclear strong force keeps protons from repelling each other inside the nucleus. It has a range of roughly one femtometre, or 1E

-15 metres.

Coulomb Force: The Coulomb force is an electrical force that holds electrons to the nucleus. It operates over a further distance than the Strong Force, but is weaker.

Nuclear Fission: Nuclear fission, or the splitting of an atom, can occur if a travelling neutron slowly passes through the nucleus of an atom (if it goes too fast it will simply pass through the nucleus). When the neutron is passing through the nucleus, it will push the protons apart to a point where the range of the strong force can no longer influence the protons. This results in the atom splitting, which emits some neutrons in the process. If these neutrons cause other nuclei to split, a chain reaction can occur (this produces a great amount of energy).

Chain Reaction: A nuclear reaction that produces materials that cause further reactions (see above).

Moderator: A moderator slows down neutrons so that they can cause atomic nuclei to split.

The CANDU Reactor

The CANDU reactor (CANada Deuterium Uranium reactor) is a Canadian-designed nuclear fission power plant. It fires neutrons at uranium to begin the fission process, which then heats up heavy water (heavy water uses deuterium, a stable isotope of hydrogen with a mass of roughly twice that of a regular hydrogen atom). This heated heavy water – the heavy water is used as a moderator to slow down the neutrons involved in the fission process – then heats up “normal” water. This water turns into steam which rotates a turbine, the spinning of which can be converted into electrical energy.

Below is a diagram of the CANDU reactor (which should be memorized):

NOTE: Whenever an atom is split, energy is produced.

Artificial Transmutations

Transmutations (the change of one kind of atom to another) occur in two ways:

Naturally, as radioactive atoms decay Artificially, when a nucleus is bombarded with subatomic particles

When a subatomic particle is fired at a target, the target could turn into an unstable atom. Subsequently, the atom could then split into other atoms.

For example, when an alpha particle is fired at a Nitrogen-14 atom, it forms an unstable Fluorine atom:

147

N + 42He

189

F

You can determine if the product of a reaction is unstable if has an atomic mass that is not the “normal”

mass of that atom (the normal atomic mass of Fluorine is 19, so it 189

F would be considered unstable.

From this point, Fluorine-18 would split into Oxygen-17 (which is also unstable) and a proton:

147

N + 42He

189

F 178O +

11H

Notice that the sum of the atomic masses between each arrow is the same (14 + 4 18 17 + 1) and so is the nuclear charge (7 + 2 9 8 + 1). Using this information, you may be asked to fill in the blank in the example below:

14?

N + ?2He

189

F ?8O +

1?H

Remember to check whether or not each atom is stable.

Radioactive Decay

Radioactive substances will always decay. When this happens, they will release alpha / beta particles and gamma radiation:

Polonium-211 decays into Lead-207, an alpha particle and gamma radiation:

21184

Po 20782

Pb + 42He + γ

Neutron Decay (β Decay / Beta Decay)

A neutron can decay and be transformed into a neutron. When this happens, some of its mass is lost as energy – but how?

10n

11P +

0−1

e + γ

Einstein said that there was a relationship between mass and energy, in an equation that can be stated as E=mc2 (although this is a simplified version that does not consider all factors). Since the mass of a neutron, proton and electron are already known (see Definitions and Important Notes above) the mass of gamma (γ) can be determined:

10n

11P +

0−1

e + γ

1.0675×10−27 kg 1.0673×10−27 kg+ 9.11×10−31kg + 1.089×10−30 kg

The equation E=mc2 can then convert this mass to energy:

E=mc2

E=¿E=9.80×10−14 J

This 9.80×10−14 joules of energy is known as the binding energy. This is the energy used to bond subatomic particles together.

Gamma Radiation

The energy produced by a UVB or UVA ray can be determined using the following formula:

E=hcλ

Where:

E=Energyh=Planck ' sConstant (6.626×10−34 Js)c=The speed of light (3.00×108m /s )λ=Thewavelength of theray

Half Life

The half-life of a substance is the amount of time it takes for the mass of the substance to be cut in half to form its daughter products (the resulting products from the process). Half-life problems can be solved using the following formulas:

λT 12

=0.693

N t=NOe− λt

Where:

T 12

=Theamount of time for anelement ¿decay ¿half of itsmass

λ=Thedecay constant ,whichwill be statedt=Theamount of time at a givenmomentN t=Thenumber of atomswhen time=t (canbedetermined using Avogadro' s number)N0=Thenumber of atoms whentime=0(canbe determined using Avogadro ' snumber)e=Anirrational number(2.718)

Other Nuclear Energy Notes: 1 amu (atomic mass unit) is equivalent to 1.66×10−27kg Knowledge of what unstable atoms will decay into in unnecessary for this course, but

the ability to identify when atoms are unstable (indicated by an irregular atomic mass) is necessary

Efficiency (A Quick Break from Nuclear and Electrical Energy)

Efficiency (a ratio of input work to output work) can be determined using the following equation:

Efficiency=W out

W ¿×100%

Where:W out=Useful energyoutput (measured∈ joules)W ¿=Useful energy input (measured∈ joules)

Power (the amount of work energy per unit of time) is often essential to efficiency problems. To solve for power, use the following equation:

P=Wt

Where:P=Power (measured∈watts)W=Work energy(measured∈ joules)

t=Time (measured∈seconds )

Electrical Energy

Definitions and Important Notes

Electrostatics: The study of stationary charges or fields (opposed to currents) is called electrostatics. Basic principles of electrostatics state that:

Opposite charges attract each other Like charges repel each other

Electroscope: An electroscope is a device that can detect and measure electricity. It features two metal leaves that will separate when they have the same charge.

If a positively charged object is placed near the electroscope, the electrons in the leaves will jump to the top ball-like structure of the electroscope and the leaves will become positively charged, and therefore repel each other

If a negatively charged object is placed near the electroscope, the electrons in the leaves will be repelled down into the electroscope’s leaves and they will become negatively charged, and therefore repel each other

Note that an electroscope will indicate the presence of a charge, but not what kind of charge it is (positive or negative).

Charging by Induction: This is the type of charge the leaves of an electroscope feature (see above).

Charging by Contact: If an object with an excess number of electrons touches a neutral object, then the electrons from the object will travel to the neutral object making them both negatively charged.

Similarly, if an object with a reduced number of electrons touches an object with a neutral charge, some of the electrons from the neutral object will transfer to the positively charged object, thus making both objects positively charged.

Ampere: An ampere is a measurement of current in an electrical circuit, measured at a rate of one coulomb per second.

Coulomb: A coulomb is the quantity of electricity conveyed in one second by a current of one ampere.

Quantity of Charge: Essential in electrical energy questions, the quantity of charge (q) is a measurement of coulombs (C).

Electric Field: An electrical field (ε) is a region around a charged particle where a force would be exerted on other charged particles / objects.

Electric Force: Electric force (FE) is a force measured in newtons

Determining Electrical Field Strength and Electric Force

The strength of an electric field can be determined using the following equation:

ϵ=FE

q OR ϵ=

kqd2

Where:

ϵ=Electric field (measured∈N /C )FE=Electric force (measured∈N )q=Quantity of charge(measured∈C)

k=Aconstant with the value 9.0×109 Nm2

C2

d=Thedistancebetweenthe twocharges (measured∈metres)

The electric force between two charges can be determined using the following equation:

FE=k q1q2d2

Where:

FE=Electric force (measured∈N )q1=Thequanitiy of chargeof the first charge(measured∈C )q2=Thequanitiy of chargeof the second charge (measured∈C)d=Thedistancebetweenthe twocharges (measured∈metres)

k=Aconstant with the value 9.0×109 Nm2

C2

It should be noted that the electric field strength between two charged plates is the same no matter where the field is measured between the plates.

Calculating Current

Current (I) can be determined using the formulaI=qt , where q represents the quantity of charge of the

current and t represents the time in seconds.

Calculating Voltage

Voltage (V) (the potential difference of electricity between two points) is measured in joules per

coulomb, and can be found using the formula V=EE

q(where EE represents electric work energy and q

represents the quantity of charge).

NOTE: The charge of a single electron is 1.602×10−19C

Circuit Diagrams

There are two primary types of circuits we work with in this class:

Series circuits, where the electrons in a circuit only have one path to travel Parallel circuits, where electrons have multiple paths

NOTES:

Electrons travel from the negative terminal of a battery to the positive terminal A “black box” (a collection of resistors, see below) could be considered a single resistor In a series circuit, the current remains the same In a parallel circuit, the current can changed based on the resistance electrons experience as

they pass through resistors Parallel circuits are often found within series circuits, such as within the “black boxes”

explained above

Kirchhoff’s Law

Kirchhoff’s law states that in an electric circuit, the sum of the increases in voltage is equivalent to the sum of decreases in voltage.

NOTE: Well the voltage may fluctuate throughout several resistors in a circuit, it will remain the same in a parallel circuit (when the flow of electrons in a circuit is split into multiple pathways).

Ohm’s Law

Resistance can be measured in ohms (Ω), and it can be determined using the formulaR=VI , where R

represents the resistance, V the voltage, and I the current.

NOTE: The total resistance in a series circuit can be found using the following formula:

Rtot=R1+R2+R3 . . .

And the total resistance in a parallel circuit can be found using the following formula:

1R tot

= 1R1

+ 1R2

+ 1R3

. ..

Main Parts of a Circuit:

Sample Circuit Problems

In the series circuit above, the second voltage must be determined.

Thanks to Kirchhoff’s Law, VO (the sum of all voltages in the circuit) = V1 + V2:

V o=V 1+V 2

−V 2=V 1−V O

V 2=V O−V !

V 2=240V−100V

V 2=140V

In the parallel circuit above, the first and second resistances must be determined.

Ohm’s law can solve for the total resistance:

Rtot=V O

IO

Rtot=240V8 A

Rtot=30Ω

It can also solve for the second resistance:

R2=V 2

I 2

R2=240V7.0A

R2=34Ω

Since the voltage remains the same throughout parallel circuits, we also know that the first voltage is 240 V. Therefore the first resistance can also be solved:

R1=V 1

I 1

R1=240V1.0 A

R2=240Ω

The formula 1

R1/2= 1

R1+ 1R2

can be used to solve for the total resistance inside the “black box” (the part

of the circuit that features a parallel circuit):

1R1/2

= 1R1

+ 1R2

1R1/2

= 1240Ω

+ 134Ω

1R1/2

=0.0042Ω+0.029Ω

1R1/2

=0.034Ω

R1 /2=29Ω

This is very close to the total resistance calculated earlier, which verifies that the values of the first and second resistance were correct.

Physics 30 Course Complete.