AceGamsat - The Physics Bible

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The Physics Bible Everything you need to know for GAMSAT Physics www.AceGAMSAT.com

The information contained in this guide is for informational purposes only.

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Publication of such Third Party Material is simply a recommendation and an expression of our own opinion of that material. No part of this publication shall be reproduced, transmitted, or sold in whole or in part in any form, without the prior written consent of the author. All trademarks and registered trademarks appearing in this guide are the property of their respective owners. Users of this guide are advised to do their own due diligence when it comes to making decisions and all information, products, services that have been provided should be independently verified by your own qualified professionals. By utilising this guide, you agree that the company AceGAMSAT is not responsible for the success or failure relating to any information presented in this guide. ©2018 AceGAMSAT. All Rights Reserved. AceGAMSAT is not affiliated with ACER in any way.

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About GAMSAT

The GAMSAT (Graduate Australia Medical School Admissions Test) has been developed by the Australian Council for Educational Research (ACER). This test selects for students that have the greatest capacity to advance their studies in medicine, dentistry, and pharmacy. The GAMSAT was developed for Australian universities and its component of use for admissions has spread to institutions in the UK and Ireland.

GAMSAT evaluates the nature and extent of abilities and skills gained through prior experience and learning, including the mastery and use of concepts in basic science as well as the acquisition of more general skills in problem solving, critical thinking and writing.

Candidates whose first degree is in a non-scientific field of study can still sit GAMSAT and succeed in an application for admission to one of the graduate-entry programs. A science degree is not always a prerequisite and institutions encourage applications from candidates who have achieved academic excellence in the humanities and social sciences. However, it must be stressed that success in GAMSAT is unlikely without knowledge and ability in the biological and physical sciences.

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Table of Contents TRANSLATIONAL MOTION

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SCALARS AND VECTORS

ADDING AND SUBTRACTING VECTORS

RESOLUTION OF VECTORS AND TRIGONOMETRIC FUNCTIONS DISTANCE AND DISPLACEMENT SPEED AND VELOCITY ACCELERATION

UNIFORMLY ACCELERATED MOTION

FORCE, MOTION, AND GRAVITATION

1 2 2 5 6 7 8

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MASS, WEIGHT, CENTRE OF MASS

11 12 13 14 15 16 16 17 18 19 20 21 22 23

NEWTON’S FIRST LAW

NEWTON’S SECOND LAW NEWTON’S THIRD LAW GRAVITATION

SPECIAL RIGHT TRIANGLES FREE FALL MOTION

PROJECTILE MOTION FRICTION

INCLINED PLANES

CIRCULAR MOTION AND CENTRIPETAL FORCE HOOKE’S LAW TENSION

PULLEY SYSTEMS

EQUILIBRIUM

24 24

TRANSLATIONAL AND ROTATIONAL MOTION

MOMENTUM AND COLLISIONS

27

MOMENTUM ELASTIC COLLISIONS INELASTIC COLLISIONS

27 27 28

IV

ENERGY AND WORK

29 29 32 33 33

ENERGY WORK

ENERGY CONSERVATION POWER

FLUIDS AND SOLIDS

35

DENSITY

SPECIFIC GRAVITY PRESSURE

BUOYANCY AND ARCHIMEDES’ PRINCIPLE FLUIDS IN MOTION

FLUID VISCOSITY AND DETERMINATION OF FLOW SURFACE TENSION SOLIDS

WAVE CHARACTERISTICS AND PERIODIC MOTION TRANSVERSE AND LONGITUDINAL MECHANICAL WAVES MATHEMATICAL REPRESENTATION OF A WAVE

SUPERPOSITION OF WAVES, PHASE, INTERFACE RESONANCE

STANDING WAVES HARMONICS

PERIODIC MOTION

SOUND

35 37 38 39 41 42 42 43

44 44 45 46 46 47 47 48

49 49 49 50 50

PRODUCTION OF SOUND INTENSITY AND PITCH BEATS

DOPPLER EFFECT

ELECTROSTATICS

52

CHARGE, CONDUCTORS, INSULATORS

52 53 54 55 56

COULOMB’S LAW, ELECTRIC FORCE

ELECTRIC FIELD, ELECTRIC FIELD LINES POTENTIAL ENERGY

EQUIPOTENTIAL LINES AND ELECTRIC DIPOLES

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ELECTROMAGNETISM

57 57 58

ELECTROMAGNETISM

ELECTROMAGNETIC SPECTRUM

ELECTRICITY: ELECTRIC CIRCUITS

60 60 62 63 63 64 65 66 67

CURRENT

OHM’S LAW

RESISTANCE AND RESISTIVITY RESISTANCE IN CIRCUITS

BATTERIES AND ELECTROMOTIVE FORCE (EMF) KIRCHHOFF’S LAWS

CAPACITORS AND DIELECTRICS POWER

LIGHT AND OPTICS

69 69 70 71 74 75 76

VISUAL SPECTRUM AND POLARISATION IMAGES

MIRRORS

REFRACTION AND DISPERSION SNELL’S LAW LENSES

ATOMIC AND NUCLEAR STRUCTURE

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PROTONS, NEUTRONS, ELECTRONS

78 79 80 80 81 83

ISOTOPES, ATOMIC NUMBER, ATOMIC WEIGHT FISSION AND FUSION RADIOACTIVITY

NUCLEAR REACTIONS AND RADIOACTIVE DECAY HALF-LIFE

APPENDIX

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Translational Motion Video: Introduction to Vectors and Scalars Video: Adding Vectors Video: Acceleration

Video: Deriving Displacement as a Function of Time, Acceleration, and Initial Velocity Video: Plotting Displacement, Acceleration, and Velocity Video: Projectile Height Vs Time

Video: Deriving max projectile displacement given time Video: Impact velocity from given Height

The following video shows one-dimensional projectile motion. This is useful to watch to become familiar with the manipulation of the kinematic equations.

Video: One-Dimensional Motion Scalars and Vectors Knowing the difference between vectors and scalars will be advantageous when solving physics problems in the GAMSAT. A vector is a physical quantity that has both direction and magnitude; a scalar is a physical quantity that has magnitude but no direction. Arrows can be used to reveal the direction of the vector. The length of the arrow will reveal the magnitude of the vector. The following table shows examples of vectors and scalars:

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Table 1. – Examples of scalar and vector quantities.

Adding and Subtracting Vectors

When adding vectors, the head of the first vector is placed towards the tail of the second vector. An arrow is drawn from the tail of the first vector to the head of the second one. The resulting arrow represents the vector sum of the two vectors. For the subtraction of vectors, place the heads of the two vectors together and draw an arrow from the tail of the first vector to the tail of the second one. The resulting arrow represents the vector difference between the two vectors.

Resolution of Vectors and Trigonometric Functions

A vector can be divided into two components that are perpendicular to each other. A vector is resolved into its scalar components. The lengths of the component vectors are found through Pythagoras’s Theorem and trigonometric ratios. This skill is required in projectile motion problems in the GAMSAT. It is essential that students are able to understand and apply Pythagoras’s Theorem and trigonometric ratios.

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Table 2. – Pythagoras’s theorem.

Sample question: Calculate the vertical and horizontal components of a 50 N force, which is acting 40 degrees to the horizontal.

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Solution: The question requires students to determine the values of the vertical and horizontal components. The first step involves splitting the vector into its vertical and horizontal components, as seen in the diagram below:

Now that both components have been identified, the next thing task is to label the components. Label the component that is adjacent to the given angle as ‘F × cos θ’, where F is the force and θ is the given angle. Then find out which component is opposite to the given angle and label it ‘F × sin θ’.

In these questions, the component that is adjacent to the given angle is the horizontal component (horizontal axis), which is labelled 50 × cos 40°. The component that is opposite the given angle is the vertical component (vertical axis) so it is labelled 50 × sin 40°.

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The vertical and horizontal components are: Vertical component = 50 × sin 40 = 32.14 N

Horizontal component = 50 × cos 40 = 38.30 N

Distance and Displacement Distance refers to how much ground an object has covered, and displacement is the object’s overall change in position. Distance and displacement are scalar and vector components, respectively. However, displacement is distance with an added dimension— direction. To test an understanding of this distinction, consider the motion depicted in the diagram below. A lady walks 4 m east, 2 m south, 4 m west, and finally 2 m north.

Fig 1. - Distance and displacement example.

Even though the lady has walked a total distance of 12 m, her displacement is 0 m. During the course of her motion, she covered 12 meters of ground (distance = 12 m). Yet when she is finished walking, she is not out of place. In other words, there is no displacement in her motion (displacement = 0 m). Displacement, being a vector quantity, must relate to direction. www.acegamsat.com

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The 4 m east cancels the 4 m west, and the 2 m south cancels the 2 m north. Vector quantities, such as displacement, are direction aware. Scalar quantities, such as distance, are ignorant of direction. In determining the overall distance travelled by the lady, the various directions of motion can be ignored.

Speed and Velocity Speed and velocity are scalar and vector components, respectively. Velocity is speed with the added dimension of direction. The units for velocity are expressed in length divided by time (e.g., m/s, ft/s, and mi/h). The following two equations must be memorised for the GAMSAT: Speed = distance/time

Velocity = displacement/time

Sample question:

A man walks 8 km east in 2 hours and then 7 km west in 1 hour.

a) What is the man's average speed for the whole journey?

b) What is the man's average velocity for the whole journey? distance 8 km + 7 km 15 km = = = 5 km/h time 2 hours + 1 hour 3 hours b) displacement 8 km - 7 km 1 km average velocity = = = = 0.333 km/h time 2 hours + 1 hour 3 hours average speed =

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Acceleration Acceleration is the rate of change in velocity. Acceleration is a vector, meaning that it has a direction and a magnitude. An object is accelerating if it is changing its velocity. The units for acceleration are expressed as velocity divided by time (m/s2). When an object experiences negative acceleration, it is termed deceleration. The formula for acceleration must be memorised for the GAMSAT: Acceleration = rate of change of velocity/time

Sample question:

Use the acceleration equation to determine the acceleration for the following motion.

To find the change in velocity, subtract the initial velocity from the final velocity. First use the formula a = (vf - vi) / t and then pick out any two points. a = (8 m/s - 0 m/s) / (4 s) a = (8 m/s) / (4 s) a = 2 m/s2

Answer: a = 2 m/s2

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Uniformly Accelerated Motion Uniformly accelerated motion is motion that involves constant acceleration. Constant acceleration means that both the magnitude and direction of acceleration must remain constant. A particle that exhibits uniform accelerated motion will accelerate at a constant rate. When this particle is accelerating on a linear path, there are four variables that can describe its motion: Time (t) - scalar

Displacement (x) - vector Velocity (v) - vector

Acceleration (a) – vector

Three basic equations can be used to derive the values of these variables. These equations are usually provided on the GAMSAT stimulus material, but they should be memorised for speed and efficiency. Manipulation of these equations will become second nature through repetition and practice. The linear motion equations are as follows:

1. x = xo + vot + 1/2at2 2. v = vo + at 3. v2 = vo2 + 2ax

When deciding which equation to use, choose the equation where the value of all variables, except one, are known. Remember: constant acceleration and linear motion are required in order to use these equations.

Sample questions:

A truck starts from rest and accelerates in a straight line at 2 m/s2 for 6 seconds. 1. What is the final velocity of the truck?

2. How far has the truck travelled in this time?

3. If the brakes are applied and the truck travels a further 30m before stopping, what is the deceleration? www.acegamsat.com

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Firstly, write down what is known: Initial velocity: vo = 0 m/s Acceleration: a = 2 m/s2 Time: t = 6 s

Final velocity: v = ? Question 1. Find the final velocity (v):

(use equation 2) v = vo + at

v=0+2x6 = 12 m/s

Question 2. Determine how far the truck has travelled (x): (use equation 1)

x = xo + vot + 1/2at2

x = 0 + 0 + 1/2 x 2 x 62 = 36 m

Question 3. Firstly, write down what is known: Initial velocity: vo = 12 m/s

Final velocity: v = 0 m/s (truck coming to a stop)

Distance: x = 30 m Acceleration: a = ? (use equation 3)

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v2 = vo2 + 2ax

0 = 122 + 2 x a x 30

0 = 144 + 60a 60a = - 144

a = - 144/60

a = - 2.4 m/s2 Another important concept in the GAMSAT is average velocity. When an object experiences uniformly accelerated motion, the average velocity can be determined using the following formula: Vavg = 1/2 (v + vo)

Sample question:

A car is travelling with an initial velocity of 16m/s and it reaches the final destination at a velocity of 40m/s. Calculate the average velocity of the car.

Solution:

Given initial velocity: vo = 16 m/s, Final velocity: v = 40 m/s.

Average velocity: Vavg = 1/2 (v + vo)

= 1/2 (16 m/s + 40 m/s) = 1/2 (56) m/s = 28 m/s

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Force, Motion and Gravitation Video: Newton’s first law of motion

Video: Newton’s first law of motion concepts Video: Newton’s second law of motion Video: Newton’s third law of motion Mass, Weight, and Centre of Mass The mass of an object is a fundamental property of the object. Mass is a measure of the amount of matter in the object or a numerical measure of its inertia. The symbol for mass is m and usually measured in kilograms (kg) in the GAMSAT. Mass does not change—it will be the same on earth and in space. However, weight will be virtually zero in space.

The weight of an object is defined as the force of gravity on the object and may be calculated as the mass times the acceleration of gravity (w = mg). Since the weight is a force, its SI unit is the Newton (N). For an object in free fall, so that gravity (gravity at the surface of the earth = 9.8m/s2) is the only force acting on it, the expression for weight comes from Newton’s second law.

The centre of mass is the point where all of the mass of an object is concentrated. When an object is supported at its centre of mass, it will remain in static equilibrium. An easy way to determine the location of the centre of mass of a rigid pole is to support the pole horizontally on one finger from each hand. Gently slide your fingers together. When your fingers meet, you will be at the centre of mass at which time you can easily hold up the pole with only one finger as long as it can withstand the entire weight of the pole. If the object is uniform, a meter stick for example, the centre of mass will be at the exact geometric centre; if the object is irregular in shape, such as a hammer, the centre of mass will be closer to the heavier end.

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Fig. 2 - Centre of mass of a hammer.

In the above figure, the hammer is supported by a rope from above. We can see that the hammer is well balanced and therefore it is hanging from its centre of mass (closer to the heavier end).

The centre of gravity is a geometric property of any object. The centre of gravity is also the average location of the weight of an object. The motion of any object through space can be described by the translation of the centre of gravity of the object from one place to another and the rotation of the object about its centre of gravity if it is free to rotate.

Newton’s First Law The first law is termed the law of inertia. It states that an object in a state of motion or a state of rest will remain in that state unless a net force acts upon it.

Let’s consider the movement of a bus. When a bus moves from a standstill, the passengers sitting or standing in the bus tend to fall backward. This is due to inertia of rest. So what is actually happening here?

When the bus takes off, the lower part of the body of the passenger, which is in contact with the bus, moves along with the bus, while the upper part of the body tends to retain its state of rest due to inertia. Consequently, the passenger will fall backward.

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Fig. 3 - Demonstration of Newton’s first law.

Newton’s Second Law The second law states that when an object is acted upon by a net force, the change in that object’s state of motion will be inversely proportional to the mass (m) of the object and directly proportional to the net force (F) acting upon the object. The formula for Newton’s second law is:

Fig. 4 - Demonstration of Newton’s second law.

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Newton’s Third Law The third law states that for every action, there is an equal and opposite reaction. This means that when object A applies a force to object B, object A will experience a force of equal magnitude to B but in the opposite direction. Let’s consider the motion of a flying eagle. An eagle uses its wings to fly. The eagle’s wings push air downwards. Since forces act from mutual interactions, the air must also be pushing the eagle upwards.

The size of the force on the air is equal to the size of the force on the eagle. The direction of the force on the air (downwards) is opposite the direction of the force on the eagle (upwards). For every action, there is an equal (in size) and opposite (in direction) reaction. This actionreaction pair of forces makes it possible for the eagle to fly. Another example we can look at is the action of hitting a nail with a hammer. When hitting a nail with a hammer, the force of the nail on the hammer is equal to the force of the hammer on the nail. These forces are acting in the opposite direction of each other.

Fig. 5 - Demonstration of Newton’s third law.

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Gravitation This law states that there is a force of attraction existing between any two bodies of masses in the universe. The force is proportional to both of the masses m1 and m2 and inversely proportional to the square of the distance, r, between their centre of mass. Note: r is measured as the distance from the centre of one mass to the centre of the other (not from their surfaces). The formula used is: F = G(m1,m2/r2)

Where G is 6.673 x 10-11 m3 kg-1 s-2.

This formula gives the magnitude of the force without direction. The direction is from the centre of mass of one object to the centre of mass of the other object. Taking Newton’s third law into consideration, it is assumed that both masses experience a force of the same magnitude.

Students should memorise that one N is the force needed to accelerate one kg of mass at the rate of one m/s2.

Sample question

Determine the force of gravitational attraction between the earth (m = 5.98 x 1024 kg) and a 70-kg physics student if the student is standing at sea level at a distance of 6.38 x 106 m from the earth's centre. The solution to this problem involves substituting known values of G (6.673 x 10-11 N m1(5.98 x 1024 kg) and m2 (70 kg) and r (6.38 x 106 m) into the universal gravitation equation and solving for Fgrav. The solution is as follows:

m2/kg2),

Fgrav = (6.673 x 10-11 N m2/kg2) x (5.98 x 1024 kg) x (70 kg) Fgrav = 686 N

(6.38 x 106 m)2

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Special Right Triangles

Since calculators are prohibited in the GAMSAT, it is crucial that students memorise the following special right triangles. These triangles contain the standard values of the trigonometric functions:

Fig. 6 - Special right triangles and their trigonometric values.

Free Fall Motion Free fall motion is the vertical motion of an object with reference towards earth. Air resistance is taken to be zero, and the motion is always uniformly accelerated. Acceleration is taken as g. Kinematic equations are used to solve for free fall motion. Note that a = g, since the only acceleration is a result of gravity.

x = xo + vot + 1/2at2 v = vo + at

v2 = vo2 + 2ax The following video works through a simple example of free fall motion: www.acegamsat.com

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Video: Free Fall Motion Problem Projectile Motion Projectile motion is the motion of an object launched at a certain angle (θ) from the horizontal. The motion defines a parabola, so the linear motion equations do not apply. Separate the projectile motion into two components (x and y) so it can be analysed as two distinct linear motion problems. The linear motion equations now come into practice:

x = xo + vot + 1/2at2 v = vo + at

v2 = vo2 + 2ax Linear component y: vertical component

- Acceleration is constant due to gravity (9.8m/s2).

Linear component x: horizontal component

- In the GAMSAT, projectile motion questions assume no air resistance; the horizontal acceleration is thus a constant zero.

Fig. 7 - Projectile motion.

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Projectile motion is a difficult concept to grasp for beginners. Projectile motion is easier learned through visual practice problems. It is crucial that students understand the following videos to excel in the projectile motion problems in the GAMSAT.

Video: Visualising vectors in 2 dimensions Video: Projectile at an angle

Video: Different ways to determine time in air

Video: Horizontally launched projectile problems Friction Friction is a force that provides resistance to the relative motion of two solid objects. The resistance is proportional to the force that presses the surfaces together, as well as the roughness of the surfaces (coefficient of friction is different for different surfaces).

Fig. 8 - The force of friction.

The maximum frictional force has the following expression:

fmax = μN μ: coefficient of friction N: normal force to the surface on which the object rests

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Two things that students must remember for the GAMSAT:

1. A frictional force is always parallel to the surface in contact with the object.

2. The normal force is always perpendicular to the surface in contact with the object.

Inclined Planes

Inclined plane problems generally arise in the GAMSAT exam. When an object is placed on an inclined surface, it will generally slide down the surface. Objects will slide down this tilted surface at different rates, depending on how tilted the surface is. It is common sense that the greater the tilt of the surface, the faster the rate at which the object will slide down.

An unbalanced force is responsible for an object sliding down an inclined plane. To understand this type of motion, it is important to analyse the forces that act upon the object on the inclined plane. It is a general rule in the GAMSAT that problems of inclined planes assume no friction between surfaces unless specified. There are at least two forces acting upon any object on an inclined plane:

Force of gravity – acts in the downward direction. It is also known as weight.

Normal force – acts in the direction perpendicular to the surface of the object.

The force of gravity of the object is resolved into two component forces – one perpendicular to the inclined surface and the other parallel to the inclined surface.

The following figure shows that the force of gravity can be resolved into two components. Together, these two components can replace the effect of the force of gravity.

Fig. 9 - Inclined plane.

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The following equations are used to determine the magnitude of the two components of the force of gravity (FI and FII). Perpendicular component: FI = m.g.cosθ Parallel component FII = m.g.sinθ

When no friction is present, the acceleration of the object on an incline is the value of the parallel component (m.g.sinθ) divided by the mass (m).

So if the parallel component is m.g.sinθ, we will need to divide it by ‘m’ to find the acceleration. As a result, we are left with m.g.sinθ / m

A = g.sinθ

Video: Inclined plane force components Video: Ice accelerating down an incline

Circular Motion and Centripetal Force

When a particle is moving in a circular motion at constant speed, the velocity vector continuously changes. However, the magnitude of the vector remains the same. The constantly changing direction of the velocity (v) causes a radially inward acceleration called centripetal acceleration (ac).

Fig. 10 - Circular motion and centripetal force.

The magnitude of this acceleration ac can be determined by v2/r. Thus, to calculate centripetal force:

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Fc = mac Since ac = v2/r

Then Fc = mv2/r.

Fc = centripetal force, m = mass of particle, v = velocity of particle, r = radius of circle Video: Constant speed around curve

Video: Centripetal force and acceleration intuition

Video: Visually understanding the centripetal acceleration formula Hooke’s Law

Video: Introduction to springs and Hooke’s Law Hooke’s law comes into practice in the GAMSAT when a problem involves force due to a stretched or compressed object. The force is directly proportional to the change in position (x) of an object being stretched or compressed.

Fig. 11 - Visual representation of Hooke’s law. www.acegamsat.com

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Hooke’s law is given by the following equation:

F = kx

k = a constant unique to a given object – also known as the spring constant. Sample question:

What is the force required to stretch a spring whose constant value is 80 N/m by an amount of 0.40 m?

Solution:

We will use to formula F = kx to solve this question F = force (N)

k = force constant (N/m) x = change in position F = (80)(0.4) F = 32 N

Tension

Tension is a force acting on a flexible object with no mass such as a rope or string. If there is no friction on the rope, it is assumed that tension is equal throughout the rope. Tension requires an equal force at both ends of a rope, although the tension of the rope is equal to only one of the forces.

Video: The force of Tension

Video: Introduction to Tension

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Pulley Systems A pulley allows a force to act over a greater distance and therefore do the same amount of work with less force. The rope in a frictionless pulley is assumed to be weightless, and therefore the tension throughout the rope is constant. In the figure below, we can see that a 25 N force applied to a particular pulley system will allow for a 100 N weight to be lifted.

Fig. 12 - Forces in a pulley system.

The concept of pulley systems is better learned visually. The following videos will cover everything required for solving pulley-related problems in the GAMSAT:

Video: Pulley systems 1 (hard way) Video: Pulley systems 2 (easy way)

Video: Two masses hanging from a pulley www.acegamsat.com

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Equilibrium For the GAMSAT, students should understand that equilibrium exists when mass moves with constant velocity or is at rest. Students should clearly understand the following:

• • •

The net forces acting on a system in equilibrium are zero. The sum of the magnitudes of the downward forces equals the sum of the magnitude of the upward forces. The sum of the magnitudes of the leftward forces equals the sum of the magnitude of the rightward forces.

Translational and Rotational Motion Students should be familiar with two types of motion - translational motion (straight) and rotational motion (rotating), as these commonly arise in the GAMSAT.

When an object experiences a force, it will either undergo translational motion, rotational motion, or a combination of both. Rotational motion deals with the rotation of a body about its axis. Perpendicular forces cause the rotation of the object around the axis. The force that causes the rotation around the axis is termed torque (τ).

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Consider the following to be a bird’s eye view of a door:

Fig. 13 - Important aspects in rotational motion.

Pivot point: the axis in which the door rotates (hinge).

Force acting perpendicular to the lever arm: the torque force for the rotation of the door. The torque is defined as the force applied multiplied by the perpendicular distance from the axis (lever arm). The following equation should be memorised:

τ=FxL

τ: Torque

F: Force

L: Lever arm

Torque can cause both clockwise and anticlockwise motion.

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Remember: clocks are NEGATIVE. Clockwise is taken as negative (-) and anticlockwise is taken as positive (+). The clockwise and anticlockwise torques are added together to determine the net torque. In terms of the GAMSAT, the torque problems will commonly be static. The following video shows examples of torque and static equilibrium. It is very important to understand the concepts in these videos.

Video: Centre of mass

Video: Introduction to torque Video: Moments

Video: Moments (part 2)

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Momentum and Collisions Video: Introduction to momentum Momentum Momentum problems commonly arise in the GAMSAT exam.

Momentum is a vector quantity and a measure of a moving object’s tendency to continue along its present path. The momentum of an object is calculated by multiplying its mass by its velocity. If an object’s mass or velocity is increased, its momentum is increased. A higher momentum of an object corresponds to an increased difficulty in changing its path.

M = mV

M: momentum m: mass

V: velocity

The units for momentum are kg m/s. When approaching momentum problems, students should remember that the momentum of an isolated system is always conserved.

Knowledge about collisions is crucial, as problems involving collisions have shown up in almost every previous GAMSAT exam. A collision occurs when two bodies collide, in which each exerts a force upon the other, causing the exchange of energy or momentum. For the GAMSAT, students should be aware of the two types of collisions and their characteristics – elastic and non-elastic collisions.

Elastic collisions -

Momentum and kinetic energy are conserved in elastic collisions. Objects bounce off each other (e.g., collision of rubber balls and collision of gas molecules).

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Inelastic collisions -

Momentum is conserved. Kinetic energy is lost as heat or sound in inelastic collisions. Objects stick together upon collision (e.g., collision between truck and car becoming stuck together).

When two isolated objects collide, momentum is always conserved. Use the following equation in problems that involve the collision of isolated objects:

m1v1i + m2v2i = m1v1f + m2v2f m1 = mass of object 1 m2 = mass of object 2

v1i = initial velocity of object 1 v2i = initial velocity of object 2 v1f = final velocity of object 1 v2f = final velocity of object 2

Sample question:

A 10g bullet is stopped in a block of wood with a mass of 5kg. The speed of the bulletwood combination immediately after the collision is 0.6m/s. What was the original speed of the bullet?

Solution:

This is a perfectly inelastic collision. The two objects stuck together after the collision and moved with a common velocity of vf. Assuming the motion is along the x-direction:

m1v1 + m2v2 = (m1 + m2) vf

m1 is the mass of the bullet, v1 its initial velocity, m2 is the mass of the block, and v2 is its initial velocity. Initially, the block is at rest (v2 = 0). Therefore, 0.01kg x v1 = 5.01kg x 0.6 m/s v1 = 300.6 m/s

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Energy and Work Video: Introduction to work and energy Video: Work and energy (part 2)

Video: Work as the transfer of energy Video: Work example problems Video: Conservation of energy Video: Power Energy

Energy is the capacity for doing work. There are many types of energy that students should be aware of for the GAMSAT: kinetic energy, potential energy, and gravitational potential energy. Kinetic energy (KE) is energy possessed by a moving object due to its motion. The equation for kinetic energy is:

KE = 1/2 mv2

Potential energy (Ep) is the stored energy of position possessed by an object. In terms of the GAMSAT, there are three important types of potential energy: gravitational potential energy, elastic potential energy, and electric potential energy. 1. Gravitational potential energy is the energy derived from the force of gravity. It is the energy stored in an object as a result of its height or vertical position.

Equation: Ep = mgh

Sample question: Determine the gravitational potential energy of a skydiver with a mass of 100 kg about to jump out of a plane at an altitude of 6000 m.

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Solution: Ep = mgh Ep = 100 x 9.8 x 6000 (acceleration due to gravity is 9.8 m/s2) Ep = 5 880 000 J Ep = 5880 kJ

m = mass of object, h = height of object, g = acceleration due to gravity (9.8N/kg)

Fig. 14 - Visual representation of gravitational potential energy.

2. Elastic potential energy is the energy stored in elastic materials as a result of their

compressing or stretching. It can be stored in rubber bands and compression springs.

Equation:

Ep = kx2/2 k = spring constant x = displacement of spring (amount of compression)

Sample question:

If the force to stretch a spring is given by F = (80 N/m)x, then what is the potential energy of the spring if it is stretched 4 meters from rest? www.acegamsat.com

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Solution: Here, k = 80 N/m and x = 4 m.

Therefore: Ep = (1/2)kx2 = (1/2)(80N/m)(4 m)2 = 640 joules 3. Electric potential energy is the energy that results from Coulomb forces associated with point charges within a defined system.

Equation:

Ep = kq1q2/r r = distance between point charges q1 and q2 (m) Ep = force (N) q = charge (C) k = 8.99 x 109 Nm2/C2

Sample question:

One charge of 2.0 C is 1.5m away from a –2.0 C charge. Determine the force they exert on each other.

Solution:

Ep = kq1q2/r Ep = (8.99 x 109 x 2 x -2)/2 Ep = 8.99 x 109 Ep = -1.798 x 1010

The negative sign indicates that one charge is positive and the other is negative, so there is an attractive force between them.

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Work Work is defined as the transfer of energy by a force. Students should remember that all work is energy transfer. Work is measured in units of energy (joules or J).

To determine the work produced by a force (F), multiply the distance travelled by an object by its force in the direction of displacement.

W = F.d.cosθ W = work

F = force on system

d = displacement of system θ = angle between F and d

Note: Both energy and work are measured in joules (1 J = 1N x 1m).

Sample question:

A boy pushes a lawn mower across the yard. The force he is applying to the handle of the lawn mower is angled down at 60.0° from the horizontal plane. This force has a magnitude of 500 N. If the boy pushes the lawn mower 25.0 m, how much work has been done to move the mower?

Solution:

So the force is at an angle of 60.0° with respect to the movement. The work can be found using the formula: W = F.d.cosθ

W = F.dcos60° W = F.d(0.5)

W = (500 N)(25.0 m)(0.5)

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W = 6250 N∙m

W = 6250 J (remember 1 N.m of energy = 1 J in energy)

Therefore, the work done while moving the lawn mower the given distance was 6250 J.

Energy Conservation The sum of the kinetic energy (Ek) and potential energy (Ep) of a system is defined as the mechanical energy (ET). In an isolated system, the sum of the kinetic and potential energy will remain the same, although the kinetic energy and potential energy may vary.

ET = Ep + Ek

Power Power is the rate at which work is done. Power is calculated by dividing the work by time necessary to do the work. The unit for power is the watt (W), which is equivalent to joules/second (J/S).

P = W/t P = power (watt)

W = work (joules) t = time (second) Sample question: A gymnast lifts his partner, who weighs 500 N, across 1 m in 2 seconds. How much power is required for the gymnast to do this?

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Solution: We will break this down into steps to make it as simple as possible. 1. We will first write the formula needed to calculate power. P=W/t 2. We will also write the formula used to calculate work (W). W=Fxd

3. We will now substitute the given values into the formulas. W = 500 N x 1 m = 500 J P = 500 J / 2 sec

4. We will now do the math. Remember that power is measured in Watts and a Watt is defined as a J/sec. P = 250 J / sec P = 250 Watts

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Fluids and Solids Video: Specific gravity

Video: Pressure and Pascal’s principle 1 Video: Pressure and Pascal’s principle 2 Video: Pressure at a depth in a fluid

Video: Archimedes’ principle and buoyant force Video: Buoyant force example problems

Video: Volume flow rate and equation of continuity Video: Surface tension and adhesion

A fluid is either a liquid or a gas. The molecules in a fluid are not arranged in any order or structure and thus move about in random directions relative to each other. The molecules of a fluid bond weakly, spin, and move past each other.

The molecules in solids are held in place by permanent molecular bonds. The molecules bond strongly and vibrate in a fixed position.

Density Density (ρ) can be defined as the amount of mass (m) a fluid contains in a specified volume (V). It is defined as the ratio of its mass to its volume.

ρ = m/V

ρ = density m= mass

V = volume

The units for density are g/cm3 or kg/m3.

Students should know that density increases from gas to liquid to solid.

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Fig. 15 - Demonstration of objects with different densities.

Sample question:

A rock has a volume of 20 cm3 and a mass of 80 g. What is the density of the rock?

Solution:

Realise that density is equal to mass divided by volume.

ρ = m/V

ρ = 80 g / 20 cm3 ρ = 4 g / cm3

Density: solid > liquid > gas Gases are the least dense and can be compressed unlike solids and liquids, which are treated in the GAMSAT as incompressible.

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Specific Gravity Specific gravity is the ratio of the density of a substance to the density of water.

S.G. = ρsubstance/ρwater S.G. = specific gravity

ρsubstance = density of a substance ρwater = density of water

Note that we can use the weight or mass of a substance to find the specific gravity if

the density is not known. The specific gravity will still be the same.

Sample question:

What is the specific gravity of a substance that weighs 80 g and occupies a volume of 160 mL?

Solution: SG = weight of substance / weight of an equal volume of water SG = 80 / 160 g SG = 0.5

The following 4 points must be remembered for the GAMSAT: 1. The density of water is 1g/ml = 1g/cm3. 1g/cm3 is the same as 103kg/m3. 2. A specific gravity of less than one = a substance lighter than water.

3. A specific gravity of exactly one = a substance equally as heavy as water. 4. A specific gravity of more than one = a substance heavier than water.

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Use the specific gravity of a substance to gain an intuitive understanding of the relative

weight of the substance to water.

For example, mercury has a specific gravity of 13.63, so lifting a container full of

mercury will be equivalent to lifting 13.63 of the same containers filled with water.

Pressure

Problems about pressure arise in almost every GAMSAT exam. Pressure (P) can be defined as the force (F) per unit area (A): P = F/A

F = perpendicular force to area in Newtons (N)

P = pressure in Pascal (Pa). 1 Pa = 1 N/m2 A = area in square meters (m2)

Pressure can also be calculated as potential energy per unit volume. The following equation can be used to calculate different pressures in liquids at different depths. Pressure is calculated by multiplying the density of the fluid (ρ) with the acceleration due to gravity (g) and the depth below the surface of the fluid (h): P = ρgh P = ρgh

P = pressure (Pa)

ρ = density of a gas or fluid (kg/m3)

g = acceleration due to gravity (9.80 m/s2)

h = the height of a column of gas or fluid (m)

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Sample question:

The wreckage of the ship, the Titanic, is 4000 m under the Atlantic Ocean. If the density of the saltwater above the ship is 1050 kg / m3, what is the pressure that that depth?

Solution: To find the pressure the following formula is used: P = ρgh

P = (1050 kg/m3)(9.80 m/s2)(4000 m) P = 41 160 000 kg/m.s2 P = 41 160 000 Pa P = 41.16 MPa

It is important to understand the following characteristics of force and pressure of liquid

fluids when solving these problems in the GAMSAT: -

At a certain depth, the pressure of the fluid is the same in all directions.

The shape or surface area of the container does not affect fluid pressure.

Fluids exert forces that are always 90 degrees (perpendicular) to the surface of the container.

The pressure of a fluid is directly proportional to the density of the fluid and to its depth.

Buoyancy and Archimedes’ Principle

When an object is submerged in water (object replaces water), an upward force acts on the submerged object. This is termed the buoyant force. Archimedes’ principle states that the buoyant force (Fb) is an upward force that acts on a submerged object and equal to the weight of the fluid that is displaced by the submerged object. www.acegamsat.com

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The equation for the buoyant force is:

Fb = ρfluidVg

Fb = buoyant force

ρfluid = density of fluid

V = volume of fluid displaced

g = acceleration due to gravity

Fig. 16 - Components of Archimedes’ principle.

In previous years of the GAMSAT, there have been questions that involve using the specific gravity of a fluid to determine the height of an object below or above the surface of the water.

The specific gravity of the fluid is equivalent to the height of an object partially submerged in the fluid. If the specific gravity of the fluid is 0.5, then 50% of the height of the object will be immersed in water and 50% of the height will be above water. If the specific gravity is 0.6, then 60% of the height will be immersed and 40% of the height will be above water. www.acegamsat.com

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Fluids in Motion

Molecules of a moving fluid can be described as having two types of motion: laminar flow and turbulent flow.

Laminar flow is the fluid motion in which all the particles in the fluid are moving in a straight line. The particles within a layer are moving at the same rate and move in the same direction. Turbulent flow is an irregular flow of particles. Unlike the linear motion of laminar flow, the particles of turbulent flow move in a state of chaos, with some particles opposing the direction of others and causing collisions.

Fig. 17 - Turbulent versus laminar flow.

The rate of laminar flow through a pipe can be determined by the following: To determine the volume (V) past a point, multiply the cross-sectional area (A) by length/distance (d).

Volume = (cross sectional area) x (distance)

Distance = (velocity) x (time) d = vt

Therefore,

Volume = Avt V = Avt

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Now that the volume is calculated, determine the rate (R):

Rate (R) is given by dividing the volume past a point by time.

R = (volume past a point) / time

R = Avt/t R = Av The continuity equation, which is used for a fluid in an enclosed tube, can be written from the above equation:

A1V1 = A2V2

The subscripts 1 and 2 represent different points in the line of flow of a fluid.

Fluid Viscosity and Determination of Flow

Fluid velocity is the resistance of fluid layers to flow past each other. The higher the viscosity of a fluid, the slower the fluid will flow. So if a fluid has a high viscosity coefficient (honey), it will flow slower than a fluid with a low viscosity constant (water). Reynolds number (R) can be used to determine if the flow of a fluid is laminar or turbulent. The following equation is used:

R = vdρ/η

fluid,

R = Reynolds number, v = velocity of flow, d = diameter of tube, ρ = density of η = viscosity coefficient

When Reynolds number is less than 2,000, flow in a pipe is generally laminar; when greater than 2,000, flow is generally turbulent.

Surface Tension Surface tension can be defined as the intensity of intermolecular forces per unit length.

The surface tension of liquid results from an imbalance of the cohesive forces between molecules: www.acegamsat.com

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1. A molecule in the bulk liquid experiences cohesive forces with other molecules in all directions. 2. A molecule at the surface of a liquid experiences only net inward cohesive forces.

These cohesive forces between water molecules can allow some insects, that are usually denser than water, to float and stride on the water surface.

Solids

Solids are elastic to some extent. They can change their dimensions by compressing or stretching while maintaining their bonds. Stress and strain are important concepts to understand when examining the elasticity of solids. 1. Stress is defined as the ratio of force applied to an object on the area over which the force is applied. The units for stress are N/m2.

Stress = force/area

2. Strain is defined as the fractional change in dimension of an object caused by stress. Strain has no units.

Stress and strain are proportional to each other, and this proportionality can be given as a ratio known as the modulus of elasticity (ME). To determine ME, simply divide the stress by the strain.

ME = stress/strain

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Wave Characteristics and Periodic Motion Video: Introduction to waves

Video: Amplitude, period, frequency and wavelength of periodic waves Transverse and Longitudinal Mechanical Waves

A wave is the transfer of energy and momentum from one point to another. During this transfer, there is a disturbance in a medium where each particle in the medium vibrates in simple harmonic motion respectively to an equilibrium point. There are two types of waves that students must be aware of for the GAMSAT— mechanical and electromagnetic waves.

Mechanical waves are made up of two types of waves: transverse and longitudinal waves.

Transverse wave: a wave that causes the medium to be displaced perpendicularly to the direction the wave is traveling. For example, the waves on a guitar string are transverse waves. Longitudinal wave: a wave that causes the medium to be displaced parallel to the direction the wave is traveling. For example, the waves from sound are longitudinal waves.

Fig. 18 - Transverse and longitudinal waves. www.acegamsat.com

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Mathematical Representation of a Wave

It is crucial that students understand all the following terms that relate to the mathematical representation of a wave. These include wavelength, frequency, amplitude, intensity, and period.

Wavelength (λ): measured from any point in the wave to the point where the wave begins to repeat itself. It is easiest to measure the distance from crest to crest or trough to trough of a transverse wave. Wavelength is measured in meters. Frequency (f): the number of wavelengths that pass a single fixed point in one second. In other words, it is the number of cycles per second. It is measured in hertz (Hz). By multiplying wavelength by frequency, the velocity of the wave can be determined:

v = fλ

Amplitude (A): the maximum displacement of a particle from its equilibrium point. Amplitude is always positive. Intensity (I): the square of the amplitude.

Period (T): the reciprocal of frequency. It is the number of seconds it takes for one wavelength to pass a fixed point.

T = 1/f

Fig. 19 - Components of a wave.

x axis (horizontal) = distance from the source y axis (vertical) = displacement/elevation

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Superposition of Waves, Phase, and Interference

If two waves begin at the same point and have the same wavelength, they are said to be in phase with each other.

When two transverse waves share the same space, their displacements will add together at each point along to wave. The superposition of these waves will create a new wave in a process termed interference. There are two types of interference that students must know for the GAMSAT:

1. Constructive interference: waves add together to create a wave with a larger displacement than either original wave. 2. Destructive interference: waves add together to create a wave with a smaller displacement than either original wave.

Fig. 20 - Constructive versus destructive interference.

Resonance

When a series of waves hit an object, they cause vibrations within the object. If the series of waves cause an object to vibrate at one of its natural frequencies, this phenomenon is termed resonance. Resonance only occurs when the first object is vibrating at the natural frequency of the second object.

Video: Wave resonance

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Standing Waves Standing waves are produced from the repeated interference of two waves of identical frequency moving in opposite directions along the same medium. All standing waves consist of nodes and antinodes.

Fig. 22 - Standing waves – nodes and antinodes.

Nodes result from the destructive interference of the two waves and are therefore points of no displacement. Antinodes result from the constructive interference of the two waves and therefore undergo maximum displacement from the rest position.

Video: Standing waves Harmonics

The following videos show a visual representation of different harmonics –

Video: Harmonics

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For a given situation, one can list all of the wavelengths from largest to smallest of the possible standing waves. This is called a harmonic series. Harmonics are numbered from the longest to shortest wavelength. The first harmonic has fewest numbers of nodes, which is two. The second harmonic has one more node than the first harmonic. Each successive harmonic in the series has an extra node.

Fig. 23 - Harmonic series.

Periodic Motion Motion that is regular and repeating is referred to as a periodic motion. Most objects that vibrate do so in a regular and repeated fashion, and thus their vibrations are periodic. www.acegamsat.com

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Sound Video: Production of sound Video: Sound properties Video: Decibel Scale

Production of Sound Sound is a longitudinal, mechanical wave. Sound can travel through any medium, with the exception of travel through a vacuum.

A region of increased pressure on a sound wave is called a compression, and a region of decreased pressure is called a rarefaction. An increase in temperature will increase the velocity of sound, while a decrease in temperature will decrease the velocity.

Students should know that the velocity of sound is fastest in solids, then liquids, and slowest in gases.

Intensity and Pitch Sound intensity (I) is the rate of energy movement through space.

I = f 2A 2

I = intensity, f = frequency, A = amplitude

The decibel (dB) is used to measure sound level. Questions in previous GAMSAT exams have required students to manipulate the following equation:

dB = 10log10(I/I0)

These questions commonly require students to be able to understand and apply the rules of logarithms. www.acegamsat.com

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Sample question:

Calculate the level of sound (dB) when a dog generates a sound wave 1 million times (1,000,000) more intense than I0? Assume that I/I0 = 1,000,000

When substituting this into the formula: dB = 10log101000000 dB = 10log10106

dB = 6 x 10log1010 (rule of logarithm) dB = 60 log1010 (rule of logarithm)

dB = 60

Beats

When two sound waves of different frequency are heard together, the alternating constructive and destructive interference causes the sound to be alternatively soft and loud. This phenomenon is termed producing beats.

The beat frequency is equal to the absolute value of the difference in frequency of the two waves that produce the beats. Therefore, if two sound waves with frequencies of 262 Hz and 252 Hz are played simultaneously, a beat frequency of 10 Hz will be detected

Doppler Effect

The relative motion of the observer (receiver of waves) and the source (production of waves) is responsible for the Doppler effect. This phenomenon causes a shift in the frequency of sounds received by the observer.

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If the distance between the observer and source is decreasing, there is a shift to shorter wavelengths and higher frequencies.

If the distance between the observer and source is increasing, there is a shift to longer wavelengths and lower frequencies.

Fig. 24 - The Doppler effect.

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Electrostatics Video: Triboelectric effect and charge Video: Coulomb’s Law

Video: Conductors and Insulators Video: Electric field Charge

Charge forms the identity of subatomic particles. There are two kinds of charge – positive (+) and negative (-). Like charges repel and opposite charges attract. Positive charge comes from having more protons than electrons, and negative charge comes from having more electrons than protons. Charge is given in units of coulombs (C).

The law of conservation of charge states that the net charge of an isolated system remains constant. Net charge cannot be created, but that charge can be transferred from one object to another. Students should remember that charge can be created and destroyed, but only in positive-negative pairs.

Example

Let’s take the concept of rubbing a balloon on human hair as an example. The frictional charging process results in a transfer of electrons between the two objects that are rubbed together. Rubber has a greater attraction for electrons than human hair. As a result, the atoms of rubber pull electrons from the atoms of human hair, leaving both objects with an imbalance of charge. The rubber balloon has an excess of electrons and the hair has a shortage of electrons. Having an excess of electrons, the rubber balloon is charged negatively. Similarly, the shortage of electrons on the hair leaves it with a positive charge. The two objects have become charged with opposite types of charges as a result of the transfer of electrons from the least electron-loving material to the most electron-loving material.

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Conductors and Insulators Conductors are materials that allow electrons to flow through. They allow for the transfer of charge via the free movement of electrons. In contrast to this, insulators are materials that resist the free flow of electrons and thus resist charge

Coulomb’s Law and Electric Force Coulomb’s law states that the electrical forces between two charged objects are directly proportional to the product of the quantity of charge on the objects and inversely proportional to the square of the distance separating the two charged objects. The force (F) between any two charges, q1 and q2, can be determined using Coulomb’s Law.

F = k(q1q2)/r2 F = force

k = Coulomb’s constant = 8.99 x 109 Nm2/C2

r = distance between the centres of charge

Fig. 25 - Coulomb’s law.

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Electric Field and Electric Field Lines

A charge will generate an electric field in its surrounding space. An electric field is an electric force per unit charge. The electric field of an object will exert a force on a second object that is in the proximity of its field. This force exerted on the second object depends upon the charge of that object and the charge generating the field. The following equation is used to determine the electric field (electrostatic force per unit charge):

E = F/q = kQ/r2 E = electric field, F = force, q = charge placed in field,

k = Coulomb’s constant = 8.99 x 109 Nm2/C2, Q = charge creating the field Electric field lines are a man-made concept that illustrates the direction of force. The arrowheads are pointed away from positive charges and toward negative charges. The phenomenon of electric fields should be understood as it generally arises in the GAMSAT exam.

Fig. 26 - Electric field lines between opposing charges.

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If a beam of electrons is moving perpendicular to an electric field, created between two horizontal plates, they are accelerated vertically (toward the positive plate). Varying the potential applied to the places can alter this angle of vertical acceleration. The velocity of the electrons is not affected.

Fig. 27 - Electrons moving perpendicular to an electric field between plates.

The electron beam is negative. It will be attracted by the positively charged upper plate and repelled by the negatively charged lower plate.

Potential Energy

A charge will exert a force on any other charge and potential energy arises from any collection of charges. For instance, if a positive charge Q is fixed at some point in space, any other positive charge close to it will experience a repulsive force and will thus have potential energy. The following video will provide an in-depth explanation of electrical potential energy

Video: Electric potential energy

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Equipotential Lines and Electric Dipoles

Perpendicular movement to an electric field does not result in a change in potential. The perpendicular movement follows lines called equipotential lines/surfaces. All points along the equipotential surface will have the same voltage. Therefore, when a charge is moved along an equipotential line, no work is done.

Fig. 28 - Equipotential lines and field lines.

An electric dipole is created by two opposite charges that have equal magnitude. To determine the magnitude of an electric dipole moment, multiply the charge (q) on one of the charges with the distance between the charges.

p = qd

p = electric dipole moment

q = charge on one of the charges d = distance

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Electromagnetism Video: Magnetic field created by a current carrying a wire Electromagnetism A moving charge (current) creates a changing electric field, which in turn creates a magnetic field. Questions commonly arise in the GAMSAT where students are required to predict the direction of the magnetic field, given the direction of current through a wire.

The following method known as the right-hand rule determines the direction of magnetic field when a current is moving through a wire: 1. Use your right hand and place your thumb in the direction of the current. 2. Grab the wire.

3. The direction in which your fingers wrap around the wire is the direction of the magnetic field.

Fig. 29 - Predicting the direction of the magnetic field.

I = current

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When a charge is moving through a magnetic field, it experiences a force. The following equation is used to determine the force acting on a charge that is moving with a certain velocity through a magnetic field: F = qvBsinθ F = force

q = charge

v = velocity

B = magnetic field

θ = angle formed between the velocity of the charge and the magnetic field

Electromagnetic Spectrum

Previous GAMSAT exams have directly asked students to choose the colour of visible light with the highest wavelength or lowest frequency. Some other questions have required students to choose which colour of light has the highest energy. The following equation was developed by Max Planck: E = hf E = energy

h = Planck’s constant f = frequency

From this equation, the visible light with the highest frequency has the highest energy. Students should know that a shorter wavelength has a higher frequency. Overall, a shorter wavelength = higher frequency = high energy. Colours of visible light from longest to shortest wavelength are:

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ROY G BIV

Red, orange, yellow, green, blue, indigo, and violet.

For example, red has the longest wavelength, lowest frequency, and thus lowest energy.

The following shows the electromagnetic spectrum:

Fig. 30 - Electromagnetic spectrum.

By being familiar with the relative positions of each type of radiation, students can identify the relative wavelengths and thus frequencies with other forms of radiation when asked in the GAMSAT.

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Electricity: Electric Circuits An electric circuit is a closed path that allows the flow of electrons.

Video: Introduction to circuits and Ohm’s law Video: Resistors in series

Video: Resistors in parallel

Video: Complex resistor circuit

Video: Resistor circuit with two batteries Video: Resistivity and conductivity Video: Voltmeter and ammeters

Video: Capacitors and capacitance Video: Dielectrics in capacitors Current Current (I) is the flow of charge (Q) through an electrical circuit. It can be determined by measuring the amount of charge that flows past a point in a certain amount of time (t). Current is measured in amperes (A) and is expressed by the following equation: I = Q/t

I = current

Q = amount of change t = time

In an electric circuit, electrons move toward to the positive connection. The electrons are removed at the positive connection, while more electrons are entering the circuit at the negative connection. The movement of electrons occurs because there are two points on the circuit that have a significant potential difference.

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Fig. 31 - Direction of the flow of charge. I (current); V (voltage); R (resistance).

Sample questions: 1. Calculate the amount of charge, Q, obtained when a current of 60 amps runs for 3 minutes. I = 60 A t = 3 minutes = 180 seconds Q=?C I = Q/t

Rearrange the equation to find Q: Q=Ixt Q = 60 x 180 = 10800 C

2. Calculate the current required to provide 12 000 C of energy in 10 minutes. Q = 12000 C t = 10 minutes = 600 seconds I=?A I = Q/t I = 12000/600 I = 20 A

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Ohm’s Law

Ohm’s law is very important to learn, as questions involving its manipulation commonly arise in the GAMSAT.

Ohm’s law states that the voltage can be determined by multiplying the current by resistance.

Fig. 32 - Ohm’s law.

V = IR V = voltage, I = current, R = resistance.

A 12 V battery supplies power to a camera with a resistance of 16 ohms. How much current is flowing through the camera? V = IR

Rearrange the equation to find I: I = V/R

I = 12/16

I = 3/4 = 0.75 A www.acegamsat.com

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Resistance and Resistivity When a charge flows through a circuit, it will encounter hindrance to its flow. Resistance (R) is the measure of the hindrance of electron flow. Resistance is measured in ohms (Ω).

Resistivity (ρ) is the measure of a material’s tendency to resist electron flow. Students should remember that resistivity increases with an increase in temperature. The resistance can be calculated using the following formula:

R = ρl/A

R = resistance (Ω - ohms)

ρ = resistivity (Ω⋅m)

l = length of wire (m)

A = cross section area of wire (m2).

Resistance in Circuits As previously mentioned, a circuit is an enclosed pathway for moving charge. The components of a circuit are either in series or parallel.

Series circuit: components are lined up in a row along the wire. There is only one path for electrons to flow. The voltage is the same over each component in a series circuit. When resistors are in series, their total resistance is the sum of their resistances:

RT = R1 + R2 + R3 + R4 + . . .

Parallel circuit: components are in alternate paths. There is more than one path for electrons to flow. The total current is the sum of currents for each path. The voltage is the same throughout the circuit for all the paths in parallel. The total resistance in a parallel circuit can be determined using the following formula:

1/RT = 1/R1 + 1/R2 + 1/R3 + . . .

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Fig. 33 - Resistance in series and parallel circuits.

Batteries and Electromotive Force (EMF) A battery adds energy to a circuit by raising the voltage from one point to another. This energy added by a battery can be referred to as an electromotive force (emf – units are volts). The battery (emf source) replaces the energy lost from electrons as they flow through the circuit. The following is a symbol for an emf source/battery:

Fig. 34 – Symbol for an emf source (battery).

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In the past, Benjamin Franklin envisioned positive charges as the carriers of charge. As such, an early convention for the direction of an electric current was established to be in the direction that positive charges would move. The convention has stuck and is still used today. The direction of an electric current is, by convention, the direction in which a positive charge would move. Therefore, the current in the external circuit is directed away from the positive terminal and toward the negative terminal of the battery. Electrons would actually move through the wires in the opposite direction. Knowing that the actual charge carriers in wires are negatively charged electrons may make this convention seem a bit odd and outdated. However, it is the convention that is used worldwide and one that a student must use in the GAMSAT exam. If two batteries are connected in opposition (positive end to positive end), energy will be lost from the charge when passing through the second battery. The following phenomenon is crucial to learn as it commonly arises in the GAMSAT.

If there is more than one battery (emf source) in a circuit, the following applies: If the emf sources are in opposition, the total sum of the individual sources in opposition will be subtracted from the total sum of the emf sources not in opposition.

Kirchhoff’s Laws

There are two Kirchhoff laws that students must know for the GAMSAT.

Kirchhoff’s first law: This first law states that the amount of current flowing into any node must be the same amount that flows out. The intersection of wires is termed a node. All current arriving at a node is defined as positive and all current leaving is negative. Kirchhoff’s second law: This second law states that the sum of voltages around any path in a circuit must equal to zero.

Video: Kirchhoff’s Laws

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Capacitors and Dielectrics The function of a capacitor is to temporarily store energy in a circuit. Energy is stored in the form of separated charge. Dielectrics are a type of material that increases capacitance.

Capacitance is the ability to store charge per unit volume. It can be calculated using the following formula:

C = Q/V C = capacitance

Q = charge (coulomb) V = voltage (volts)

Capacitance is the number of coulombs that are needed to raise the potential of a conductor by one volt. Questions relating to parallel plate capacitors commonly arise in the GAMSAT. There are two conductive plates, which are separated by a small space. In a charged parallel plate capacitor, one plate holds a positive charge, and the other holds a negative charge of the same magnitude. This separation of charge creates a constant electric field. As the area of the plates increases, so does the capacitance. However, an increase in the distance between the two plates will cause a decrease in capacitance.

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Fig. 35 - Parallel plate capacitors.

Another important equation to solve common problems is:

V = Ed

V = voltage

E = electric field strength

d = distance between the plates of the capacitor The total capacitance can be calculated by looking at both series and parallel circuits. The following formulas apply:

Series circuit: 1/CT = 1/C1 + 1/C2 + 1/C3 + 1/C4 . . . Parallel circuit: CT = C1 + C2 + C3 + C4 . . .

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Power The electric power (in watts) associated with an electric circuit represents the rate at which energy is converted from the electrical energy of the moving charges into some other form of energy (mechanical energy, heat or energy stored in electric fields). The following three equations can be used to solve problems relating to power:

P = IV

P = I2R P = V2/R

Fig. 36 - Equations used to calculate power.

Video: Power practice problems www.acegamsat.com

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Light and Optics Video: Electromagnetic waves and the electromagnetic spectrum The following link provides a list of videos that should be studied. These videos are important in providing intuition and knowledge of geometrical optics, a topic that is commonly examined in the GAMSAT.

Video: Geometric optics series Visual Spectrum and Polarisation

The visual spectrum of light takes up only a small section of the entire electromagnetic spectrum. The visible spectrum ranges from wavelengths of 390nm – 700nm.

Fig. 37 - Electromagnetic spectrum and the visible spectrum.

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Light is made up of photons, and each photon represents an electromagnetic wave. Light acts like both a wave and a particle.

The energy transformation of light can be described by the particle theory, and the propagation of light can be described by the wave theory. The wave theory is used when analysing diffraction. Polarised light waves are waves in which the vibrations occur in a single plane.

Unpolarised light can be transformed into polarised light by a process termed polarisation.

Images Lenses refract light and mirrors reflect light in order to produce images. Students must know the difference between the two types of images formed: Real image: light rays pass through image and can be projected onto a screen.

Virtual image: no light passes through image and cannot be projected onto a screen.

Video: Types of images

Understanding Virtual Images:

An object placed in front of a mirror will have a corresponding image located behind the mirror. The distance from the image to the mirror is always identical to the distance from the object to the mirror. If a person stands 5 m in front of a mirror, then the image will be located an identical 5 m behind the mirror. This image is termed a virtual image.

When viewing this virtual image in the mirror, it would appear as though the light is coming from a location of 5 m behind the mirror. However, there would be nothing physically there when moving behind the mirror and looking at this so-called virtual image location.

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Mirrors There are two types of mirrors: plane (flat) and spherical. Spherical mirrors can be further divided into convex and concave mirrors.

Plane Mirrors:

The behaviour of light can be predicted when observed approaching and reflecting off a flat mirror. The mirror below visualises the law of reflection:

Fig. 38 - The law of reflection.

The light ray approaching the mirror is termed the incident ray. The ray that leaves the mirror is termed the reflected ray. When a light ray strikes the mirror, a line can be drawn perpendicular to the surface of the mirror. This line is termed the normal, which divides the angle of the incident and reflected ray into two equal angles. The angle formed between the incident ray and the normal is termed the angle of incidence. The angle formed between the reflected ray and the normal is termed the angle of reflection. For a plane (flat) mirror, all images seen are virtual, left-right reversed, and erect.

For a non-plane mirror (spherical), the surfaces may have a reflective surface that converges or diverges light. A convex surface will diverge light and a concave surface will converge light. www.acegamsat.com

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Spherical Mirrors: Concave and Convex Analyse the following diagram to become familiar with the terminology used with spherical mirrors:

Fig. 39 - Concave and convex mirrors.

Notes: Focus (in above figure) is another name for focal point. www.acegamsat.com

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The focal length (f) is the length from F to the centre of the mirror itself. The focal point is separated from the mirror by the focal length. The focal length is always half of the distance of the radius. The object distance (o) is the distance from the centre of the mirror (C) to the object along the axis.

Concave Mirrors: With concave mirrors, light is converged toward the axis (shown above). The following conditions affects the image produced by a concave (converging) lens: If object distance is less than focal length = virtual and erect image

If object distance is more than focal length = real and inverted image If object distance is the same as the focal length = no image formed AND

If object distance is less than the radius = size of image is increased

If object distance is more than the radius = size of image is reduced

If object distance is the same as the radius = the size of the image will be the same

Questions in the GAMSAT have required students to understand that incident rays that pass through F (when extended) will reflect at an angle parallel to the axis.

Convex Mirrors:

It is easier when dealing with convex mirror problems as the image formed by a convex mirror is always virtual, erect, and smaller than the object.

Students should be familiar with manipulating the following formulas and be able to memorise the formula through practicing relevant problems, although these formulas are usually provided in the stimulus material. These following equations are relevant for both convex and concave mirrors.

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Mirror Equation: The mirror equation expresses the quantitative relationship between the object distance (o), the image distance (i), and the focal length (f). The equation is stated as follows:

1/i = 1/o = 1/f

Magnification Equation: The magnification equation is used to determine the magnification (M) between the object (o) and the image formed (i).

M = -i/o

It is very important that the following three steps of convention are memorised for the GAMSAT:

1. If M is more than 1, the image is larger than the object. If M is less than 1, the image is smaller than the object. If M is positive, the image is erect. If M is negative, the image is inverted. 2. For i and o, positive values = real image. Negative values = virtual image. 3. For r and f, positive values = converging. Negative values = diverging.

Refraction and Dispersion Refraction is the bending of a light when it enters a medium that changes its speed. The refraction of light when it passes from a fast medium to a slow medium bends the light ray toward the normal to the boundary between the two media. The amount of bending depends on the indices of refraction of the two media. The index of refraction (n) can be determined by dividing the speed of light in vacuum by the speed of light in the medium. A more optically dense medium will have a higher index of refraction (n), than one that has a less optically dense medium.

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Snell’s Law The following figure can be to visually understand Snell’s law:

Fig. 40 - Visual representation of Snell’s law.

θ1 is the angle of the incident ray to the normal.

θ2 is the angle of the transmitted (refracted) ray.

n1 and n2 represent the index of refraction of two different media.

Light is bending toward the normal in the transmitted ray, thus n2 is a slow medium and is thus more optically dense than n1.

Formula for Snell’s law:

sinθ1/sinθ2 = n1/n2 = v1/v2 = λ1/ λ2 v = velocity

λ = wavelength

n = index of refraction Remember:

1. A smaller wavelength of the incident light will cause a smaller angle between the transmitted ray and the normal. 2. The angle θ will be smaller in a medium that is more optically dense, and thus larger in a material that is less optically dense. www.acegamsat.com

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Video: Snell’s law and refraction Video: Snell’s law example 1 Video: Snell’s law example 2 Lenses A lens is a piece of transparent material that refracts light rays to form an image. There are two types of lenses: diverging (concave) and converging (convex).

A converging lens converges light rays that are traveling parallel to its principal axis. These lenses are wider across their middle and thinner at the ends. A diverging lens diverges light rays that are traveling parallel to its principal axis. These lenses are thinner across their middle and thicker at the ends.

Fig. 41 – Convex lens.

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Fig. 42 – Concave lens.

The focal point is the point at which rays of light initially parallel converge towards after emerging from a converging lens or diverge from after emerging from a diverging lens. The backward extensions of the diverging light rays will also form a focal point.

Diverging (concave) lenses form images that are always virtual, erect, and reduced in size.

Converging (convex) lenses form different types of images depending on relative focal length and object distance. If object distance is less than focal length = virtual and erect image If object distance is more than focal length = real and inverted image If object distance is the same as the focal length = no image formed

If object distance is less than the radius = size of image is increased If object distance is more than the radius = size of image is reduced If object distance is the same as the radius = the size of the image will be the same Note: Radius = 2 x focal length

The equation previously covered for mirrors is also used in calculations involving converging and diverging lenses:

1/i = 1/o = 1/f i = image distance, o = object distance, f = focal length

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Atomic and Nuclear Structure Video: Protons, neutrons, electrons, and isotopes Video: Fission and fusion

Protons, Neutrons, and Electrons The nucleus of an atom is composed of protons (+) and neutrons (neutral). Electrons (-) are found in clouds surrounding the nucleus.

The negative charge of the electrons is the same magnitude as the positive charge of the protons.

Fig. 43 - Components of an atom.

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Isotopes, Atomic Number, and Atomic Weight An element is a chemical substance containing atoms that have the same number of protons in their atomic nuclei. Elements are displayed as follows:

dfsdfsdfsdf

dddsfdfs

Fig. 44 - The display of an element.

A is the mass number of the atom. This is the number of protons plus neutrons in the nucleus. Z is the atomic number. This is the number of protons in the nucleus.

An element is characterised by the number of protons it has. Any atom can only have one number of protons. However, the number of neutrons or electrons can change and the atom will still retain its identity.

If there are two atoms of the same element that have a different mass number (different number of neutrons), these are said to be isotopes. Three isotopes of carbon include: 12C, 13C

and 14C.

Each isotope of carbon contains 6 protons (gives the element its identity as carbon). 12C contains 6 neutrons, 13C contains 7 neutrons, and 14C contains 8 neutrons. www.acegamsat.com

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There are also isotopes of hydrogen, which are shown in the figure below:

Fig. 45 - Different isotopes of hydrogen.

The atomic weight or molar mass (M) of an atom is given in atomic mass units (amu). It is the average atomic weight of naturally occurring isotopes of an element.

Fission and Fusion

It can be helpful in the GAMSAT to understand the difference between fission and fusion. Fission occurs when a nucleus splits into smaller nuclei, and fusion occurs when smaller nuclei join together to form a larger nucleus.

Radioactivity

Radioactivity refers to the particles that are emitted from nuclei as a result of nuclear instability. There are many unstable nuclear isotopes that emit some kind of radiation. The most common types of radiation are alpha, beta, gamma particles, positions, and electron capture. Alpha particle: same as a helium nucleus. Beta particle: an electron.

Gamma ray: electromagnetic radiation (has no charge or mass.)

Electron capture: an electron is taken from the shell and the proton is converted into a neutron. www.acegamsat.com

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The following table shows the symbols of each type of particle and can be used as a guide to determine the change in mass and atomic number after an emission of radiation:

Table 3 – Particle symbols and their changes in mass and atomic number.

Nuclear Reactions and Radioactive Decay A nuclear reaction is characterised by changes in the composition of the nucleus. It is common in the GAMSAT that students are given a product and reactant of a radioactive decay reaction. Students are then required to identify the type of particle emitted as radiation.

Alpha decay

In alpha decay, the nucleus emits an alpha particle; an alpha particle is a helium nucleus (two protons and two neutrons).

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An example of an alpha decay involves uranium-238:

Beta decay A beta particle is often an electron. If an electron is involved, the number of neutrons in the nucleus decreases by one and the number of protons increases by one. For example:

Positron emission The emission of a positron occurs when a proton is transformed into a neutron. During this transformation, a positron is emitted. For example:

Electron capture Electron capture occurs when an electron is captured and merges with a proton to create a neutron. A proton is destroyed and a neutron is created: www.acegamsat.com

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Gamma ray emission Gamma ray emission usually occurs with other types of radioactive decay. It is the release of a high-frequency photon. Remember: The sum of the lower (or higher) numbers on the equation = the sum of the lower (or higher) numbers on the equation.

Half-life The predictable rate of decay can be determined in terms of half-life (t1/2). A half-life is defined as the length of time necessary for one-half of a substance to decay. Half-life related questions are extremely common in the GAMSAT. Students are commonly asked to determine how many half-lives a compound has gone through to reach a certain given quantity. This would be a matter of dividing the initial amount by two until arriving at the final amount.

Video: Nuclear Half-Life: Calculations

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Appendix Translational Motion Video: Introduction to Vectors and Scalars Video: Adding Vectors Video: Acceleration

Video: Deriving Displacement as a Function of Time, Acceleration, and Initial Velocity Video: Plotting Displacement, Acceleration, and Velocity Video: Projectile Height Vs Time

Video: Deriving max projectile displacement given time Video: Impact velocity from given Height One-Dimension Projectile Motion Video: One-Dimensional Motion Force Motion and Gravitation Video: Newton’s first law of motion

Video: Newton’s first law of motion concepts Video: Newton’s second law of motion Video: Newton’s third law of motion Free Fall Motion Video: Free Fall Motion Problem

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Projectile Motion Video: Visualizing vectors in 2 dimensions Video: Projectile at an angle

Video: Different ways to determine time in air

Video: Horizontally launched projectile problems Inclined Planes Video: Inclined plane force components

Video: Ice accelerating down an incline Circular and Centripetal Force Video: Constant speed around curve

Video: Centripetal force and acceleration intuition

Video: Visually understanding the centripetal acceleration formula Hooke’s Law Video: Introduction to springs and Hooke’s Law Tension Video: The force of Tension

Video: Introduction to Tension Pulley Systems Video: Pulley systems 1(hard way)

Video: Pulley systems 2 (easy way)

Video: Two masses hanging from a pulley www.acegamsat.com

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Torque and Static Equilibrium Video: Centre of mass

Video: Introduction to torque Video: Moments

Video: Moments (part 2) Momentum and Collisions Video: Introduction to momentum Energy and Work Video: Introduction to work and energy Video: Work and energy (part 2)

Video: Work as the transfer of energy Video: Work example problems Video: Conservation of energy Video: Power

Fluids and Solids Video: Specific gravity

Video: Pressure and Pascal’s principle 1 Video: Pressure and Pascal’s principle 2 Video: Pressure at a depth in a fluid

Video: Archimedes’ principle and buoyant force Video: Buoyant force example problems

Video: Volume flow rate and equation of continuity Video: Surface tension and adhesion

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Wave Characteristics and Periodic Motion Video: Introduction to waves

Video: Amplitude, period, frequency and wavelength of periodic waves Resonance Video: Wave resonance Standing Waves Video: Standing waves Harmonics Video: Harmonics Sound

Video: Production of sound Video: Sound properties Video: Decibel Scale Electrostatics Video: Triboelectric effect and charge Video: Coulomb’s Law

Video: Conductors and Insulators Video: Electric field Potential Energy Video: Electric potential energy www.acegamsat.com

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Electromagnetism Video: Magnetic field created by a current carrying a wire Electricity: Electric Circuits Video: Introduction to circuits and Ohm’s law Video: Resistors in series

Video: Resistors in parallel

Video: Complex resistor circuit

Video: Resistor circuit with two batteries Video: Resistivity and conductivity Video: Voltmeter and ammeters

Video: Capacitors and capacitance Video: Dielectrics in capacitors Kirchhoff’s Laws Video: Kirchhoff’s Laws Power Video: Power practice problems Light and Optics Video: Electromagnetic waves and the electromagnetic spectrum Video: Geometric optics series

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Images Video: Types of images Snell’s Law Video: Snell’s law and refraction Video: Snell’s law example 1 Video: Snell’s law example 2 Atomic and Nuclear Structure Video: Protons, neutrons, electrons, and isotopes Video: Fission and fusion Half-Life Video: Nuclear Half-Life: Calculations

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