nelson physics stage 6 hsc cvr - kookaburra.com.au · stuart ryan janet holmes c gn`j3` 9m`sp...

6PhysicsNelson

HSC

6PhysicsNelson

HSC

6Physics

Nelson

HSC

For learning solutions, visit cengage.com.au

Nelson Physics Stage 6 HSC is the second of two books in the Nelson Physics series. This book has been written by practising teachers to fully comply with the requirements of the Board of Studies Stage 6 Syllabus.

Key features of the student CD-ROM include:

* twenty auto-correcting multiple-choice questions for each module

* word checks to assist students in remembering key terms and defi nitions

* MP4 podcast revisions which students can download to a portable device and revise anytime, anywhere

* demonstration version of Logger Pro 3.6.

Key features of the Teacher CD-ROM include:

* worked solutions to all textbook questions

* solutions to Practical Physics for Senior Students Secondary Information and Problem Solving activities.

Key features of the text include:

* full-colour design* full coverage of the Stage 6 HSC Syllabus* full coverage of three Options: Medical Physics; Astrophysics; From

Quanta to Quarks* syllabus links at the beginning of each chapter* easy-to-read and easy-to-understand explanations of physics

concepts* margin glossaries throughout each chapter and an end-of-book

glossary* fully worked examples* extensive graded problem sets integrated throughout the text using

syllabus verbs* links to the accompanying student CD-ROM* margin links to Practical Physics for Senior Students HSC which

covers all mandated First Hand Investigations, Secondary Information and Problem Solving

* video links to take the students beyond the book to the wider world of physics

* ‘Did you know?’ boxes presenting interesting, topic-related material* end-of-chapter summaries and reviews.

Stuart RyanJanet Holmes

6PhysicsNelson

Preliminary

6Physics

Nelson

Preliminary

For learning solutions, visit cengage.com.au For learning solutions, visit cengage.com.au

Stuart RyanJanet Holmes

6PhysicsNelson

Preliminary

Nelson Physics Stage 6 Preliminary is the fi rst of two books in the Nelson Physics series. This book has been written by practising teachers to fully comply with the requirements of the Board of Studies Stage 6 Syllabus.



* word checks to assist students in remembering key terms and defi nitions







* full-colour design* full coverage of the Stage 6 Preliminary Syllabus* full coverage of three Options: Medical Physics; Astrophysics; From








6PhysicsNelson

HSC

6PhysicsNelson

HSC

6Physics

Nelson

HSC


Nelson Physics Stage 6 HSC is the second of two books in the Nelson Physics series. This book has been written by practising teachers to fully comply with the requirements of the Board of Studies Stage 6 Syllabus.



* word checks to assist students in remembering key terms and definitions







* full-colour design* full coverage of the Stage 6 HSC Syllabus* full coverage of three Options: Medical Physics; Astrophysics; From








Stuart Ryan Janet Holmes

Preliminary

Preliminary


Preliminary


Nelson Physics Stage 6 Preliminary is the first of two books in the Nelson Physics series. This book has been written by practising teachers to fully comply with the requirements of the Board of Studies Stage 6 Syllabus. Key features of the student CD-ROM include:* twenty auto-correcting multiple-choice questions for each module* word checks to assist students in remembering key terms and definitions* MP4 podcast revisions which students can download to a portable device and revise anytime, anywhere* demonstration version of Logger Pro 3.6.Key features of the Teacher CD-ROM include:* worked solutions to all textbook questions* solutions to Practical Physics for Senior Students Secondary Information and Problem Solving activities.

Key features of the text include:* full-colour design* full coverage of the Stage 6 HSC Syllabus* full coverage of three Options: Medical Physics; Astrophysics; From Quanta to Quarks* syllabus links at the beginning of each chapter* easy-to-read and easy-to-understand explanations of physics concepts* margin glossaries throughout each chapter and an end-of-book glossary* fully worked examples* extensive graded problem sets integrated throughout the text using syllabus verbs* links to the accompanying student CD-ROM* margin links to Practical Physics for Senior Students HSC which covers all mandated First Hand Investigations, Secondary Information and Problem Solving* video links to take the students beyond the book to the wider world of physics* ‘Did you know?’ boxes presenting interesting, topic-related material* end-of-chapter summaries and reviews. Stuart Ryan Janet Holmes

Practical Physics for Senior Students PreliminaryStudent Book ISBN 9780170135221

Practical Physics for Senior Students HSCStudent BookISBN 9780170135238

Nelson Physics Stage 6 PreliminaryStudent BookISBN 9780170177917

HSC


Nelson Physics Stage 6 HSCTeacher CD-RomISBN 9780170135221

Peter Roberson Stuart Ryan Richard Ward

HSC

Practical Physics

PRACTICAL MANUAL for senior studentsII

Practical Physics for senior students HSC provides students with all the core practical and problem-solving experiences required by the NSW Board of Studies Physics Stage 6 Syllabus.The workbook is divided into three chapters according to the syllabus: Chapter 1 SpaceChapter 2 Motors and generatorsChapter 3 From ideas to implementationEach chapter contains first-hand investigations, secondary information and problem-solving worksheets. These have been carefully designed to assist students in the development of skills in planning and conducting investigations, communicating information and understanding, scientific thinking and problem solving.The design and layout of the workbook make it easy for students to organise their studies whether they are working individually or in teams.Worked answers are provided at the back of the book for students to check and correct their understanding.

Practical Physics for senior studentsPreliminaryISBN 978 0 17 013522 1

Practical Physics HSC

Preliminary


Nelson Physics Stage 6 PreliminaryTeacher CD-RomISBN 9780170135221

ISBN 978-0-17-013523-8

Practical Physics forSenior StudentsPreliminaryStudent BookISBN 9780170135221

Practical Physics forSenior StudentsHSCStudent BookISBN 9780170135238

Nelson Physics Stage 6HSCStudent BookISBN 9780170177931

6PhysicsNelson

HSC

Teacher CD-ROM


Nelson PhysicsStage 6HSCTeacher CD-RomISBN 9780170135221


HS

C

Practical Physics

PRAC

TICAL M

ANU

AL for senior students

Activ

ities

I skil

ls I e

xam

prep

arat

ion

Practical Physics for senior students HSC provides students with all the core practical and problem-solving experiences required by the nsW Board of studies physics stage 6 syllabus.

The workbook is divided into three chapters according to the syllabus: Chapter 1 space

Chapter 2 Motors and generators

Chapter 3 From ideas to implementation

each chapter contains first-hand investigations, secondary information and problem-solving worksheets. these have been carefully designed to assist students in the development of skills in planning and conducting investigations, communicating information and understanding, scientific thinking and problem solving.

the design and layout of the workbook make it easy for students to organise their studies whether they are working individually or in teams.

Worked answers are provided at the back of the book for students to check and correct their understanding.

ISBN 978-0-17-013523-8

practical physics for senior studentspreliminaryisBn 978 0 17 013522 1

Practical P

hysics

for senior students H

SC

R

ob

erso

n • R

yan

• Wa

rd

6PhysicsNelson

Preliminary

Teacher CD-ROM


Nelson PhysicsStage 6PreliminaryTeacher CD-RomISBN 9780170135221

Practical Physics forSenior StudentsPreliminaryStudent BookISBN 9780170135221

Practical Physics forSenior StudentsHSCStudent BookISBN 9780170135238

Nelson Physics Stage 6PreliminaryStudent BookISBN 9780170177917


HS

C

Practical Physics

PRAC

TICAL M

ANU

AL for senior students

Ac

tiv

itie

s I s

kil

ls I e

xa

m p

re

par

atio

n

Practical Physics for senior students HSC provides students with all the core practical and problem-solving experiences required by the nsW Board of studies physics stage 6 syllabus.

The workbook is divided into three chapters according to the syllabus: Chapter 1 space

Chapter 2 Motors and generators

Chapter 3 From ideas to implementation

each chapter contains first-hand investigations, secondary information and problem-solving worksheets. these have been carefully designed to assist students in the development of skills in planning and conducting investigations, communicating information and understanding, scientific thinking and problem solving.

the design and layout of the workbook make it easy for students to organise their studies whether they are working individually or in teams.

Worked answers are provided at the back of the book for students to check and correct their understanding.

ISBN 978-0-17-013523-8

practical physics for senior studentspreliminaryisBn 978 0 17 013522 1

Practical P

hysics

for se

nio

r stud

en

ts H

SC

R

ob

erso

n • R

yan

• Wa

rd

Ryan Holmes

CD-ROM INSIDE!

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2Projectiles

Knowledge and understanding

v x 2 � u x

2 v � u � at v y

2 � u y 2 � 2ay �y

�x � ux t

�y � uy t � 1 __ 2

ay t2

• describe Galileo’s analysis of projectile motion.

When you have fi nished this chapter, you should be able to:

• describe the trajectory of an object undergoing projectile motion within the Earth’s gravitational fi eld in terms of horizontal and vertical components

• solve problems and analyse information to calculate the actual velocity of a projectile from its horizontal and vertical components using the equations:

FPO

02 PHYSICS STAGE 6 HSC 2PP.indd 1102 PHYSICS STAGE 6 HSC 2PP.indd 11 3/17/09 10:49:44 AM3/17/09 10:49:44 AM

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2.1 Projectile motionProjectile motion refers to the movement of objects through space or the atmosphere, moving freely under the infl uence of gravity. It includes a coin being dropped, a ball being kicked, a bullet being fi red and satellites in orbit around a planet. To study the motion of projectiles, it is necessary to break the motion up into its vector components and consider the horizontal motion separately to the vertical motion. Consider an object moving under the force of gravity. If it is a coin dropped from a height of, say, 1 metre, then it will start its motion with a velocity of zero and will gain in speed, as it accelerates in the Earth’s gravitational fi eld, until it hits the ground and comes to rest. If, on the other hand, it is fl ipped upwards, it will start its motion with the velocity imparted to it in the fl ip, and will gradually slow down under the infl uence of gravity until it reaches a velocity of zero, when it will start to accelerate

projectile motion

The movement of objects through space or the atmosphere, moving freely under the infl uence of gravity

Figure 2.1 Projectile motion is exhibited by fi reworks. (The effects of air resistance alter the paths of some projectiles.)

Figure 2.2 A skier undergoing projectile motion.


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ISBN 9780170177931

downwards, gaining speed until it reaches the ground and comes to rest. The only force acting on the motion of the coins in the two situations is the constant force of gravity. If the coin is now fl icked off a tabletop that is 1 metre above the ground, so that horizontal motion is imparted to it, will it reach the ground at the same time as the dropped coin? What if a coin and an apple are dropped from the same height – will they reach the ground at the same time? Galileo found that they did. Legend has him dropping cannonballs of different masses off the Tower of Pisa while employed by the University of Pisa (around 1590) to confi rm that they did indeed reach the ground at the same time. This was the start of our modern study of kinematics – the motion of objects along a trajectory. In Aristotle’s theory of projectile motion, projectiles were pushed along by an external force called ‘impetus’ that was transmitted through the air. This impetus caused the object to move in a straight line until the force was expended, at which point the object fell to the ground. With the spread of cannon warfare, the study of projectile motion took on a greater importance and it was realised through careful observation that projectiles do not behave in this Aristotelian manner. It was realised that projectiles followed the path of some sort of smooth curve, but it needed further study to determine the shape of this curve. Galileo had been studying balls rolling down smooth inclined planes to show that the force acting on them (gravity) was a constant and that no matter the mass of the balls, they rolled down the plane with the same acceleration. In other words, the time taken for balls of different masses to travel down the inclined plane was independent of their mass and only dependent on the force acting on them. Galileo now took his studies one step further. He allowed pre-inked balls to roll down an inclined plane on a table and off the edge of the table. The balls left a mark on a sheet of paper on the fl oor, indicating their landing position. This mark allowed the horizontal and vertical distances travelled by the ball to be measured. By varying the horizontal velocity and the vertical drop of the ball, Galileo was able to determine that the path of a projectile is parabolic. Galileo had reasoned that a projectile shot from a cannon is infl uenced not by one motion only, but by two at right angles to each other. Galileo was able to demonstrate using his inclined planes that a projectile is infl uenced by two independent motions – one horizontal and the other vertical and infl uenced by gravity.

2.2 Equations of motionThere are three equations of motion that we can apply to our study of projectiles. They describe the motion of objects under the infl uence of a gravitational fi eld, no matter how strong or weak that fi eld is. These equations also assume that there is no wind or air resistance, which is reasonable in most instances.

v � u � at

v 2 � u 2 � 2ar

r � ut � 1 __ 2

at 2

where v � fi nal velocity (m s�1) u � initial velocity (m s�1) t � time (s) a � acceleration (m s�2) r � displacement (m)

kinematics

The study of the motion of objects along a trajectory without regard to its causes

Figure 2.3 The two independent motions – vertical and vertical combined with the horizontal.

Conceptual physics — projectile motion


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The equations can be applied to any situation where the acceleration is constant or zero. They can be used to quantitatively describe the motion of a car accelerating at 0.5 m s�1, or a tennis ball as it follows its trajectory through the air under the influence of gravity. In both these situations, the acceleration that is applied may be considered to be constant. Gravity on Earth is taken to be a constant 9.8 m s�2 downwards as the differences in g at the Earth’s surface to do with latitude and altitude are minimal. Let’s go back to our coin and this time consider the situation in which it is dropped from a height of 5 m.

\\ WORKED EXAMPLE

Question 1

A coin is dropped from a height of 5.0 m.a At what speed is it travelling when it hits the ground?b How long does it take before it hits the ground?

Answer

Now it becomes very important to identify all of the data in the question.Data: a � g � �9.8 m s�2; r � �5.0 m; u � 0 m s�1; t � ?; v � ? (Note that a � or – simply defi nes the direction of the quantity.)Knowing a, r and u leads us to the equation v 2 � u � ar to answer part a.

a v 2 � u 2 � 2ar � 0 � 2 � �9.8 � �5 v � �

____ 98.0

� �9.9 m s�1

Velocity at the point of impact is 9.9 m s�1 downwards. (Our knowledge of gravity tells us that it must be downwards.)

b Data: a � g � �9.8 m s�2; r � �5.0 m; u � 0 m s�1; v � �9.9 m s�1; t � ?

v � u � at �9.9 � 0 � 9.8t

t � 9.9 ___ 9.8

� 1.0 s

The time taken for the coin to drop is 1.0 s.

Question 2

An object is thrown vertically in the air at a velocity of 50 m s�1. Calculate:a how high the object goesb the object’s velocity at 6 secondsc the object’s position at 6 seconds.

Answer

Start with the data.a Data: a � g � �9.8 m s�2; u � 50 m s�1; at its highest point it will have

a velocity of zero, so v � 0; r � ? v 2 � u 2 � 2ar

r � v 2 � u 2 _______

2a


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\\ DID YOU KNOW?

Equations of motionThe fi rst equation of motion is simply a rearrangement of the equation for acceleration studied in Preliminary Physics:

average acceleration � change in velocity

________________ time taken

a � v � u ______ t

at � v � u v � u � at (1)

A second equation begins with the expression for average velocity studied in Preliminary Physics:

average velocity � total displacement

_________________ time

vav � r _ t

Average velocity can also be found by summing the individual velocities:

vav � u � v _____

2

So now we have:

u � v _____ 2 � r _ t

Substituting in v � u � at, gives:

u � (u � at )

___________ 2 � r _ t

2u � at _______ 2 � r _ t

r � ut � 1 __ 2 at 2 (2)

The last equation is derived from the fi rst two equations:

v � u � at (1)

r � ut � 1 __ 2

at 2 (2)

r � 0 � 2500 ________ �19.6

� 127.6 m

Highest point reached is 127.6 m above its starting point.b Data: a � g � �9.8 m s�2; u � 50 m s�1; t � 6 s; v � ? v � u � at � 50 � 9.8 � 6 � �8.8 m s�1

At time 6 seconds the object is travelling with a velocity of 8.8 m downwards.

c Data: a � g � �9.8 m s�2; u � 50 m s�1; v � �8.8 m s�1; t � 6 s; r � ?

r � ut � 1 __ 2

at 2

� (50 � 6) � 1 __ 2 (�9.8 � 6 2)

� 300 � 176.4 � �123.6 m

The object will be at a position 123.6 m above its starting point.

Projectile motion (Part I)


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2.3 Motion in two dimensionsConsider a marble rolling off a table with an initial horizontal velocity (Figure 2.4). Its initial vertical velocity is 0, but it is subject to gravitational acceleration of g downwards (�g). (By convention, upwards is taken as positive.) The motion of this object can be described by taking components parallel to and at right angles to the direction of the constant force acting on it (in this case g). In fact, the trajectory taken by the marble is that of a parabola; however, we break it up and consider its horizontal and vertical components separately. Gravity only acts on an object in the vertical direction. The horizontal motion of the marble will behave exactly as if it were allowed to roll along a smooth and frictionless surface. Its initial velocity will be the same as its fi nal velocity. There is no acceleration acting on it in the horizontal direction. Its motion simply comes to a halt, because that is determined by what is happening in the vertical component of its motion. In the horizontal direction we fi nd that:

ux � u

x (initial horizontal velocity)

ax � 0 (as no force acts in this direction)

rx � x (horizontal displacement)

The three equations of motion become:

vx � u

x (1)

vx

2 � ux

2 (2)

x � uxt (3)

In the vertical direction we fi nd that:

uy � 0 (initially not moving in the vertical direction)

ay � �g (acceleration due to gravity)

ry � y (vertical displacement)

The three equations of motion become:

vy � �gt (1)

vy

2 � �2gr (2)

y � � 1 __ 2

gt 2 (3)

If the marble were projected horizontally with a speed of 3.0 m s�1, the horizontal and vertical components are as shown on the graphs in Figure 2.5.

Examining projectile motion, pages 12–16, Practical Physics for Senior Students, HSC

um

g

Figure 2.4 The motion of an object of mass m moving under gravity with an initial horizontal velocity u�.

\\\\\\\\\\\\\\\\\\\\\\\\\\\From equation (1):

t � v � u ______ a (1a)

Substituting equation (1a) into (2) gives:

r � u � v � u ______ a � � 1 __ 2 a � v � u ______ a �

2

r � uv � u 2 ________ a �

1 __ 2 a(v � u)(v � u)

______________ a 2

ar � uv � u 2 � 1 __ 2 v 2 � 1 __

2 u 2 � uv

ar � 1 __ 2 v 2 � 1 __

2 u 2

v 2 � u 2 � 2ar (3)

This is the third equation of motion.


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Question 3

An object is projected horizontally with a velocity of 5.0 m s�1.a When will its vertical displacement be 20 m below its starting point?b What will its horizontal displacement be at this time?c Find its velocity at this time.

Answer

a Vertically: Data: a � g � �9.8 m s�2; u � 0 m s�1; y � �20 m; t � ? (Remember

that downwards is negative.) This data enables the use of r � ut � 1 _

2 at 2 to fi nd t.

�20 � 0 � 1 __ 2 (�9.8)t 2

t 2 � �20 _____ �4.9

t � � ___

4.1 t � 2.0 s

The object will be 20 m below its starting point at 2.0 s.b Horizontally: Data: a � 0 m s�2; ux � vx � 5.0 m s

�1; t � 2.0 s; x � ? This data enables the use of r � ut � 1 _

2 at 2 to fi nd x, and in the horizontal

simplifi es to: x � ut � 5.0 � 2.0 � 10.0 m The object has travelled a distance of 10.0 m in the horizontal.

\\ WORKED EXAMPLE

Figure 2.5 (a) The horizontal and (b) the vertical components for the path of a ball projected horizontally with a speed of 3.0 m s�1.

Acc

eler

atio

n

1 2 3Time (s)

(a) Horizontal components

a = 0

Vel

oci

ty (

m s

–1)

4

2

1 2 3Time (s)

vh = 3.0 m s–1

9

6

3

Dis

pla

cem

ent

(m)

1 2 3Time (s)

x = 3.0

10

5

Acc

eler

atio

n (

m s

–2)

1 2 3Time (s)

(b) Vertical components

a

y

= 10 m s–2

Vel

oci

ty (

m s

–1)

30

20

10

1 2 3Time (s)

vv = 10t

45

30

15

1 2 3Time (s)

= 5t2

Dis

pla

cem

ent

(m)

Two-dimensional projectile motion (Part I)


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c The velocity is the vector sum of the horizontal and vertical components. Horizontally: vx � 5.0 m s

�1

Vertically: vy � uy � ayt vy � 0 � 9.8 � 2.0 � �19.6 m s�1

From the vector diagram (Figure 2.6):

v � � _______

v x 2 � v y

2

� � ___________

5.0 2 � 19.6 2

� � ______

409.16 � 20.2 m s�1

at an angle of:

tan � � 19.6 ____ 5

� � 76

The resultant velocity of the combined components is 20.2 m s�1 at an angle of 76° below the horizontal.

Question 4

A rock is thrown horizontally from an 80 m high cliff. It lands at a point on level ground 40 m from the base of the cliff.a How long does the rock take to reach the ground below the cliff?b At what speed was it thrown?

Answer

It is often helpful to draw a diagram of the problem.a Vertically: Data: a � g � �9.8 m s�2; u � 0 m s�1; y � �80 m; t � ?

This data enables us to use r � ut � 1 __ 2 at 2 to fi nd t.

y � uy � 1 __ 2 gt 2

�80 � 0 � 1 __ 2 (�9.8)t 2

t 2 � �80 _____ �4.9

t � � ____

16.3

� 4.0 s

The time taken to reach the ground is 4.0 s.b Horizontally: Data: a � 0 m s�1; t � 4.0 s; x � 40 m; ux � vy � ? This enables us to use r � ut � 1 __

2 at 2, which in the horizontal

simplifies to: x � ux t 40 � ux � 4.0 ux � 10.0 m s

�1

The rock was thrown horizontally with a speed of 10.0 m s�1.

19.6 m s−1v

5 m s−1

Figure 2.6 Vector addition triangle.

40 m

rock

80 m

cliff

Figure 2.7 A rock 40 m from base of an 80 m high cliff.


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Problem set 2A

Question 1

A rock is dropped from a height of 3 m.

a Calculate the speed that it attains.b How long is it in the air?

Question 2

Calculate the fi nal velocity of an object with an initial velocity of 35 m s�1 and an acceleration of 6 m s�2 over a period of 5 s.

Question 3

If v � 10 m s�1, a � 4 m s�2 and t � 12 s, fi nd u.

Question 4

A coin is fl ipped directly upwards with an initial speed of 0.2 m s�2.

a What is the maximum height that it reaches?b How long is it in the air, assuming that it is caught at exactly the

same height from which it was released?

Question 5

An object is launched horizontally with an initial velocity of 5.0 m s�1. What will be the equations for the:

a y - coordinate of its displacement after time t ?b x - coordinate of its displacement after time t ?

Question 6

An object is projected horizontally with a velocity of 2.0 m s�1.

a What will the horizontal displacement be after 2.0 s?b Find the vertical displacement at this time.c When will the horizontal displacement be 0.20 m?d What will the vertical displacement be when the horizontal

displacement is 0.20 m?

Question 7

An object is projected horizontally. After t seconds it has a horizontal displacement of 10.0 cm and a vertical displacement of 5.0 cm.

a What is the value of t ?b What is the magnitude of the initial velocity?

Question 8

An object is projected horizontally with an initial velocity of 3.0 m s�1.

a When will the vertical displacement be 5.0 m?b What will the horizontal displacement be at this time?

Question 9

A stone is thrown horizontally out to sea from the top of a cliff, with a velocity of 15 m s�1, and reaches the water below after 3.0 s.

a How high is the cliff?


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b How far from the base of the cliff is the stone when it reaches the water?

c What is the velocity of the stone just prior to striking the water?

Question 10

A multifl ash photograph is taken of an object that is projected horizontally. How can you tell from the photograph that the horizontal velocity component is constant and that the vertical velocity increases uniformly?

Question 11

A ball of mass 0.50 kg rolls off a horizontal ledge 4.0 m above the ground with a speed of 4.0 m s�1.

a How long does it take the ball to reach the ground below the ledge?

b How far, horizontally, from the ledge will the ball strike the ground?

c What is the horizontal component of the ball’s velocity when it strikes the ground?

d What is the vertical component of the ball’s velocity when it strikes the ground?

e What is the kinetic energy of the ball just before it strikes the ground?

f What is the speed of the ball as it strikes the ground?

Question 12

Two balls, A and B, each of mass 1.6 kg are rolling along a horizontal table 2.0 m above the fl oor. Both balls roll off the end of the table and reach the fl oor below. A has a horizontal velocity of 4.0 m s�1 as it leaves the table, while B has a horizontal velocity of 2.5 m s�1.

a What is the time taken for: i A to reach the fl oor? ii B to reach the fl oor?b What is the difference in the horizontal distances travelled by A

and B before they strike the fl oor?

2.4 Projectile motion at an angleObviously not all projectiles are either purely vertical, such as the dropped coin, or simple two-dimensional examples where there is only a horizontal velocity initially imparted to the object, such as in the case of the marble rolled off the flat table. Projectiles often start their motion at an angle, such as when a soccer ball is kicked. The soccer ball moves off at an angle to the horizontal and follows the trajectory of a parabola. Consider an object projected with speed u at an angle � as shown in Figure 2.8 (ignore air resistance).

Calculations of projectile motion, pages 9–12, Practical Physics for Senior Students, HSC


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Consider the vector triangle for the initial motion of this object. The horizontal ( u

x ) and vertical ( u

y ) components of the velocity are

independent and are found as follows. Horizontally:

cos � � u

x __ u

ux � u cos �

Vertically:

sin � � u

y __ u

uy � u sin �

So the horizontal component is given by u cos � and the vertical component by u sin �, where � is the angle between the initial velocity and the horizontal. The equations for the components of this motion are shown in Table 2.1. Consider an object projected at 50 m s�1 at such an angle that the horizontal component of the velocity is 30 m s�1 and the vertical component is 40 m s�1 in a region where the gravitational fi eld strength is 9.8 N kg�1. This means that the acceleration of a freely falling object is 9.8 m s�2 downwards. The position and velocity values at 1.0 s intervals are shown in Fig ure 2.9. These can be calculated from the equations for motion under constant acceleration. Note the following:

a The horizontal velocity (component) is constant at 30 m s�1.b The vertical velocity (component) changes by 9.8 m s�1 each second

(down).c The minimum velocity occurs at the top � it is not zero, it equals the

value of the horizontal component.d The acceleration is always 9.8 m s�2 down, even at the top position,

because the force of gravity acts at all times throughout the motion.

h

u

u cos θ x

u

θ

u sin θ

R

θ

0

Figure 2.8 The motion of a projectile with an initial velocity u at an angle of � to the horizontal.

u cos θ

u sin θu

θ

Figure 2.9 Vector triangle for the motion of an object with an initial velocity u at an angle of � to the horizontal.

Table 2.1

Horizontal Vertical vx � ux cos � vy � changing

ux � ux cos � uy � uy sin �

r � x r � y

a � 0 a � �g

t � t t � t


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Question 5

A soccer ball is kicked at an angle of 37.0° to the horizontal with a velocity of 25.0 m s�1 as shown in Figure 2.10. Calculate the:a maximum heightb time of travelc range of the soccer ball (horizontal distance travelled)d velocity at the maximum heighte acceleration at the maximum height.(Neglect air resistance and any spinning of the ball.)

37°

25.0

m s–

1

Figure 2.10

Answer

a Maximum height will be due to the initial vertical velocity and occurs when vertical velocity reaches a velocity of zero.

Vertically: uy � u sin � � 25.0 � sin 37 � 15.0 m s�1

vy � 0 a � g � �9.8 m s�2

y � ? Using v 2 � u 2 � 2ar gives:

y � v 2 � u 2 ______

2a

y � 0 � 15.02 ________

2(�9.8)

� �225 ______ �19.6

� 11.5 m

The maximum height reached is 11.5 m.b The time is also found using the vertical component and is the time taken

to go up to the maximum height and back down. Vertically: Data: a � g � �9.8 m s�2; uy � 15.0 m s

�1; vy � �15.0 m s�1; t � ?

Using v � u � at gives:

t � v � u _____ a

� �15.0 � 15.0 ____________ �9.8

� 3.1 s The total time of travel is 3.1 s. Note: This could also have been found by fi nding the time to reach its

maximum height and doubling that result.

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c To fi nd the range (horizontal distance travelled), the horizontal component of velocity is used in conjunction with the time taken for the object’s journey.

Horizontally: ux � u cos � � 25.0 � cos 37 � 20.0 m s�1

Data: a � 0 m s�1; ux � vx � 20.0 m s�1; t � 3.1 s; x � ?

Using r � ut � 1 __ 2

at 2 gives:

x � ux t � 20.0 � 3.1 � 62.0 m The range of the soccer ball is 62.0 m.d At the maximum height, the vertical component of the velocity is zero.

There is only the constant horizontal component. Hence, the velocity at the greatest height is 20.0 m s�1 in a horizontal direction.

e The acceleration at the maximum height is the same as at any other point in the parabolic path throughout the fl ight, namely 9.8 m s�2 downwards. Note that even though the vertical velocity component is zero at the maximum height, the ball is still in the Earth’s gravitational fi eld so that it still experiences a force of attraction.

\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

\\ DID YOU KNOW?

Air resistanceAir resistance is the name given to the friction force that opposes the motion of a projectile (or anything else) through the air. As an object moves through the air, it is constantly colliding with air molecules, and therefore transferring energy and momentum to them. The magnitude of this air resistance depends on the size and shape of the projectile, and its speed. The greater the object’s cross-sectional area at right angles to its direction of motion, the more molecules in the air it will collide with. Because of this, large objects experience more air resistance than do small objects. For example, a sheet of paper falls more quickly if it is crumpled into a ball. More kinetic energy is lost in a head-on collision than in a glancing collision. A streamlined object is shaped so that its leading surfaces strike the air at an angle, defl ecting the air rather than colliding head-on with it. A streamlined object therefore experiences less air resistance than one that is not streamlined. The faster an object is travelling, the more air molecules it will strike in a given time and, therefore, the greater the air resistance it will experience. In the discussion above, we have considered an object moving through still air, but the arguments apply equally to the force exerted on an object by a moving air stream. We have also considered the molecules in the air as being stationary, but in fact they are moving around in all directions at high speed. In the absence of wind, however, all these individual movements cancel out and we can ignore them. If you throw a cricket ball, a tennis ball and a table tennis ball, you will be aware of some of the effects of air resistance. A cricket ball has a


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Problem set 2B

Question 1

A cricket ball is thrown in the air at an angle of 60° to the horizontal at 12 m s�1.

a Calculate the initial: i horizontal velocity ii vertical velocity.b Calculate the maximum height that it attains.c Find how long it is in the air before it returns back to ground

level.d Calculate its distance travelled in the horizontal.

Question 2

A projectile is launched at 55 m s�1 at an angle of 30° to the horizontal.

a What is its initial horizontal velocity?b Find the vertical component of its velocity initially.c What is the horizontal component of its velocity after 3 s?d What is the vertical component of its velocity after 3 s?e What is its velocity after 3 s?

\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

higher mass to surface area ratio than a tennis ball and a table tennis ball, so it is easier to project it through the air. The path of any ball will not be parabolic in a strong wind. Objects dropped from moving aeroplanes would follow a parabolic path if they were not subjected to air resistance. The pilots of planes and helicopters dropping parcels and rescue equipment when in fl ight need to make allowance for the parabolic path of a projectile. The actual path will be affected by air resistance. For a ball or similar body in motion through the air, there will be two forces acting on it. The force of gravity will be acting vertically downwards and the force of air resistance will be acting against the motion. These forces are shown in Figure 2.11. The magnitude and the direction of the force of air resistance will change as the direction of motion and the speed change. However, the magnitude and direction of the gravitational force on the body will not change.

mg

mg

v

Fa

Fa

Fa

mg

Figure 2.11 The forces acting on an object moving through the air.


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Question 3

A shell is fi red from a cannon at 350 m s�1 at an angle of 45° to the horizontal.

a Calculate the maximum height that it reaches.b Find the time that it is in the air.c Calculate its range.d Find the fi nal: i horizontal velocity ii vertical velocity.e What is the velocity with which it hits the ground?

Question 4

An object is launched upwards at an angle of 55° to the horizontal with a velocity of 80 m s�1.

a How long before it hits the ground again?b How high does it go?

Question 5

A projectile is launched at 140 m s�1 at an angle of 50° to the ground.

a Where is it 5 s after launching?b Where does it strike the ground?c What is the greatest height that it reaches?

Question 6

A projectile is fi red at 20 m s�1 at 30° to the horizontal.

a Calculate its time of fl ight.b Calculate its range.c What is the maximum height reached?

Question 7

A projectile has a range of fl ight of 1120 m and a time of fl ight of 8.0 s. Calculate the:

a initial horizontal velocityb initial vertical velocityc initial velocity of the projectiled maximum height reached.

Question 8

A cannonball rises to a height of 90 m after being fi red at 60° to the horizontal. Find the:

a initial vertical velocityb initial horizontal velocityc range of the cannonball.

Question 9

A boy standing vertically on the tray of a lorry travelling at 20 km h�1 throws a ball vertically upwards with a speed of 7 m s�1 and catches it again at the same level. What distance horizontally does the ball move while it is in the air?


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Question 10

A cricket ball is hit with a velocity of 8.0 m s�1 at an angle of 60° with the horizontal. Calculate:

a its horizontal and vertical displacements after 0.50 s has elapsed

b the time taken to return to the level from which it was hit, and the horizontal distance travelled in this time.

Question 11

How far would a ball go if it was struck at an angle of 40° from the horizontal at 20.0 m s�1?

Question 12

A shell is fi red from a gun at 350 m s�1 at an angle of 15° to the horizontal. What would be the greatest height to which the shell would rise?

Question 13

A ball is thrown a distance of 75 m at an elevation of 23°. What was its original speed?

Question 14

On the planet Zag, an astronaut kicked a ball at an angle of 45°, giving the ball an initial speed of 50 m s�1. The ball fl ew 15 m before hitting the ground. What is the value of g on Zag?

Question 15

A projectile is fi red at 30° to the horizontal from the top of a cliff 150 m high. Its initial speed is 49 m s�1.

a Calculate its maximum height above the: i cliff ii ground.

b Calculate the total time for which the projectile is in the air.c Calculate the distance from the base of the cliff that the

projectile lands.


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Summary of projectiles• A projectile is any object moving under the infl uence of a gravitational fi eld.• A projectile follows the path of a parabola.• Projectiles on Earth move under the force of gravity.• Horizontal motion is independent of vertical motion in projectiles.• The three equations of motion are: v � u � at v 2 � u 2 � 2ar r � ut � 1 __ 2 at

2

• Horizontal motion may be considered to have no acceleration.• Vertical motion is affected by the pull of gravity (9.8 m s�2).

Review questions

Question 1In our calculations of projectile motion, what assumptions are we making?

Question 2How did Galileo infl uence the understanding of projectile motion?

Question 3a When a projectile is launched horizontally, which component of its velocity remains

constant?b When a projectile is launched horizontally, which component of its velocity changes

uniformly with time?

c If the initial horizontal velocity of the projectile is decreased, will the time taken for the projectile to fall to the ground be decreased, increased or remain the same? Explain.

Question 4Explain the strategy that you would use to calculate a projectile’s:

a maximum height

b time in the air

c velocity

d range.

Question 5A coin is fl ipped upwards with an initial velocity of 3.0 m s�1. Calculate the:

a maximum height that it attains

b time that it is in the air before it is caught again at the same height from which it was released.

Question 6A marble is rolled off a horizontal table of height 1.5 m with an initial speed of 10 m s�1. Determine the:

a time that elapses before the marble hits the ground

b distance from the table that it travels

c speed with which the marble hits the ground.


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Question 7A football is kicked from the ground with a velocity of 25 m s�1 at an angle of 60°. Calculate the:

a length of time that the ball is in the airb the horizontal distance that the football travels.

Question 8A stone is projected out to sea from the edge of a vertical cliff 125 m high. If the vertical projection is 10 m s�1 horizontally, fi nd the:

a time the stone takes to reach the waterb distance from the base of the cliff at which it strikes the waterc velocity at that instant.

Question 9A stone is thrown with a velocity of 20 m s�1 at an angle of 45° to the horizontal. Determine:

a its range horizontallyb the time for which it is above the level from which it is thrownc the greatest vertical height reached.

Question 10A stone is thrown horizontally from a cliff of height 50 m on the Moon’s surface. Given that the Moon has a gravitational acceleration of 1.67 m s�2, calculate the length of time to reach the bottom of the cliff.

Question 11A girl throws a ball vertically upwards so that it reaches a height of 20.0 m.

a What is the velocity with which the ball leaves the girl’s hand?b Find the time taken to reach the greatest height.c For how long does the ball remain in the air?d For what fraction of the total time is the ball above 15.0 m?

Question 12Sketch the velocity–time graph and the acceleration–time graph for:

a a freely falling object released from height hb an object projected vertically upwards from the ground.

Question 13A projectile is launched horizontally at 21 m s�1 at a height of 75 cm above the ground.

a How long does it take to reach the ground?b Where does it hit the ground?c What is its velocity just before it hits the ground?

Question 14A shell is fi red from a cannon at 350 m s�1 at an angle of 40° to the horizontal.

a What is the initial horizontal component of its velocity?b What is the initial vertical component of the shell’s velocity?c What is its horizontal displacement after 5 s?d How high off the ground is it after 5 s?e What is the displacement after 5 s?

Question 15A bullet is fi red from a gun with a horizontal component of velocity of 2000 m s�1. The bullet is to hit a target 6 m above the horizontal and 500 m away. Calculate the:

a time of fl ightb angle of elevation of the gun.


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Nelson Physics Stage 6 HSC CVRphysics stage 6 HSC e-sampler.pdf

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