Piri Reis University 2011-2012/ Physics -I
Physics for Scientists &
Engineers, with Modern
Physics, 4th edition
Giancoli
Piri Reis University 2011-2012/ Physics -I
2011- 2012 PHYSICS 101 LECTURE AND LAB. PROGRAMME
Week Dates
Chapters
Subjects
Experiment/Problem Session
3-7 Oct.
Introduction
Basics: units, numbers, dimensional analysis, maths
reminders
Basic Measurement + Talk about Project
2
10-14 Oct.
1-D Kinematics
Speed, velocity, acceleration, Reference frames
Motion with Constant Acceleration
3
17-21 Oct.
2-D Kinematics
Vectors, 2-D kinematics, projectile motion
Projectile Motion
4
24- 28 Oct.
Newton’s Laws
Newton’s Laws and Gravity.
Problem Session
5
31 Oct.-4 Nov.
Linear Dynamics
Mass, Weight, Force, Friction. Application of Newton’s
Laws.
Simple Harmonic Motion/Simple Pendulum
6
7- 11 Nov.
Linear Dynamics
Work, energy, power. Conservation Laws
Friction Experiment
7
14-18 Nov.
Linear Dynamics
Momentum and motion of systems. Collisions.
Problem Session/1.Midterm
8
21- 25 Nov.
Rotational Dynamics
Rotation and Angular Momentum
Elastic and inelastic collisions
28 Nov.-2 Dec.
Equilibrium
Mechanics
Static equilibrium of rigid bodies
Archimedes' principle/Simple Pressure Experiment
10
5-9 Dec.
Fluids
Density, Buoyancy force, Archimedes' Principle
Problem Session
11
12-16 Dec.
Thermodynamics
Heat, temperature, the expansion of solids and gases,
Gases, Heat transfer.
Problem Session
12
19- 23 Dec.
Materials
The physical phase changes. Vapours, cooling.
Specific Heat experiment
26- 30 Dec.
SHM & Waves
Simple harmonic motion and waves
Problem Session+ Present the Projects
(5min for presentation+3 min for questions)
2- 6 Jan.
Travelling Waves
Sound, Light
Problem Session
+ Present the Projects
(5min for presentation+3 min for questions)
1
9
13
14
Piri Reis University 2011-2012/ Physics -I
Chapter 3
Kinematics in Two Dimensions; Vectors
Piri Reis University 2011-2012
Lecture III
I. Vectors and Scalars
II. Addition of Vectors- Graphical Methods
III. Substraction of Vectors and Multiplication of a Vector by a Scalar
IV. Adding Vectors by Components
V. Unit Vectors
VI. Vector Kinematics
VII. Projectile Motion
VIII. Solving Problems Involving Projectile Motion
IX. Relative Velocity
X. Homeworks
I. Vectors and Scalars
 A vector has magnitude as well as
direction.
Some vector quantities:
 Displacement(Δx)
 Velocity (V)
 Force (F)
 Momentum (P)
 A scalar has only a magnitude.
Some scalar quantities:
Mass (m)
Time(t)
Temperature(T)
II. Addition of Vectors- Graphical Methods
For vectors in one dimension, simple addition and subtraction are all that is
needed.
You do need to be careful about the signs, as the figure indicates.
A person walks 8 km East and
then 6 km East.
Displacement = 14 km East
A person walks 8 km East and
then 6 km West.
Displacement = 2 km
II. Addition of Vectors- Graphical Methods
If the motion is in two dimensions, the situation is somewhat more complicated.
Here, the actual travel paths are at right angles to one another; we can find the
displacement by using the Pythagorean Theorem.
II. Addition of Vectors- Graphical Methods
Graphical Method of Vector Addition Tail to Tip Method

V1

VR

V2

V3
II. Addition of Vectors- Graphical Methods
Graphical Method of Vector Addition Tail to Tip Method

V1

V2

V1

V3

VR

V2

V3
II. Addition of Vectors- Graphical Methods
Parallelogram Method
II. Addition of Vectors- Graphical Methods
Example
A person walks 10 km East and 5.0 km North



D R D1 D 2
2
DR
DR
2
D1
(10 . 0 km )
2
D2
( 5 . 0 km )
2
11 . 2 km
D2
sin
DR
sin
1
(
D2
DR
)
sin
1
(
5 . 0 km
)
26 . 5
0
11 . 2 km
Adding the vectors in the opposite order
gives the same result:
Order doesn’t matter
III. Substraction of Vectors
Negative of vector has same
magnitude but points in the
opposite direction.
For subtraction, we add the negative vector.
III. Multiplication of a Vector by a Scalar
A vector V can be multiplied by a scalar c; the result is a vector cV
that has the same direction but a magnitude cV. If c is negative,
the resultant vector points in the opposite direction.
IV. Adding Vectors by Components
Any vector can be expressed as the sum of two other vectors,
which are called its components. Usually the other vectors
are chosen so that they are perpendicular to each other.
Trigonometry Review
Hypotenus
e
Opposite
Adjacent
Opposite
sin
Hypotenuse
cos
Adjacent
Hypotenuse
tan
Opposite
sin
Adjacent
cos
IV. Adding Vectors by Components
If the components are
perpendicular, they can be
found using trigonometric
functions.
sin
cos
tan
Opposite
Vy
Hypotenuse
V
Adjacent
Vx
Hypotenuse
V
Opp
sin
Adj
cos
Vy
V sin
Vx
V cos
IV. Adding Vectors by Components
The components are effectively one-dimensional, so they can be
added arithmetically:
Signs of Components
y
Rx
Rx
Ry
Ry
x
Rx
Ry
Rx
Ry
IV. Adding Vectors by Components
Adding vectors:
1. Draw a diagram; add the vectors graphically.
2. Choose x and y axes.
3. Resolve each vector into x and y components.
4. Calculate each component using sines and cosines.
5. Add the components in each direction.
6. To find the length and direction of the vector, use:
sin
Vy
V
V. Unit Vectors
VI. Vector Kinematics
VI. Vector Kinematics
y (north)
10
If a particle starts from position r1 at time t = t1
and moves to position r2 at time t = t2
r = r2-r1
then its displacement vector is
r = r2-r1
which it moved in time
t = t2 – t1
5
r1
r2
5
10
X (east)
VI. Vector Kinematics
y (north)
10
Its average velocity is then defined as
v= r/ t
r = r2-r1
5
r1
Velocity is a vector quantity,
• its magnitude IvI is the speed of motion
• its direction is the direction of motion.
r2
5
10
X (east)
VI. Vector Kinematics
y (north)
10
As the change becomes small, part of a path
then
r = r2-r1
5
becomes the instantaneous velocity
v = dr/dt
r = rd-r1
r1
v= r/ t
which is the derivative of displacement with
respect to time.
rd
5
10
X (east)
VI. Vector Kinematics
A man pushing a mop across a floor causes it to undergo two
displacements. The first has a magnitude of 150 cm and makes a angle of 1200
with the positive x-axis. The resultant displacement has a magnitude of 140 cm
and is directed at an angle of 35.00 to the positive x axis. Find the magnitude
and direction of the second displacement.
Example

A
120

R
0
35

R

A


B

A
150 cm
B
R
A
R
140 cm
B
X
R
X
A
X
0
Bx
(140 cm ) cos 35
Bx
B
y
By
By

(150 cm ) cos 120
190 cm
R
y
A
y
(140 cm ) sin 35
49 . 6 cm

(150 cm ) sin 120


Alternative Solution. In the solution below, the angles for vector
A are measured from the negative x axis. In this case, we have to
assign the signs for the components. The answer is the same.
 

A
150 cm
R A B


B

B
A

R
60
B

R
X
A
R
X
R
A
140 cm
X
0
35
0
Bx
(140 cm ) cos 35
Bx
B
y
By
By

( 150 cm ) cos 60
190 cm
R
y
A
y
(140 cm ) sin 35
49 . 6 cm

(150 cm ) sin 60


Example (Continued)A man pushing a mop across a floor causes
it to undergo two displacements. The first has a magnitude of 150
cm and makes a angle of 1200 with the positive x-axis. The
resultant displacement has a magnitude of 140 cm and is directed
at an angle of 35.00 to the positive x axis. Find the magnitude and
direction of the second displacement.
2
B
B
(190 cm )
x
B
2
B
y
2
( 49 . 6 cm )
49 . 6 cm
sin
196 cm
sin
1
49 . 6
196
14 . 6

2
196 cm
VII. Projectile Motion
A projectile is an object moving in
two dimensions under the
influence of Earth's gravity; its
path is a parabola.
VII. Projectile Motion It can be understood by analyzing the horizontal and
vertical motions separately.
• Neglect air resistance
• Consider motion only after release
and before it hits
• Analyze the vertical and horizontal
components separately (Galileo)
• No acceleration in the horizontal, so
velocity is constant
• Acceleration in the vertical is – 9.8
m/s2 due to gravity and thus velocity
is not constant.
• Object projected horizontally will
reach the ground at the same time as
one dropped vertically
VII. Projectile Motion
vy0
If an object is launched at an initial angle of θ0 with
the horizontal, the analysis is similar except that the
initial velocity has a vertical component.
v 0 sin
vx0
v 0 cos
• If the ball returns to the y = 0 point, then the velocity at that point
will equal the initial velocity.
• At the highest point, v0 y = 0 and v = vx0
VIII. Solving Problems Involving Projectile Motion
Projectile motion is motion with constant acceleration in two dimensions, where
the acceleration is g and is down.
Example : A football is kicked at an angle of 50.00 above the
horizontal with a velocity of 18.0 m / s. Calculate the maximum
height. Assume that the ball was kicked at ground level and lands
at ground level.
v

x0
v
vy
at top:
t
v
y0
v y0
max
9 . 80 m
y max
y max
0
0
0
13 . 8 m
y0
y
gt
v t
yo
1
s
s
1 . 41 s
2
gt
2
2
(13 . 8 m )(1 . 41 s )
s
9 .7 m
s
13 . 8 m

g
y
11 . 6 m
(18 . 0 m )(cos 50 . 0 )
s
(18 . 0 m )(sin 50 . 0 )
s
1
2
(9 .8 m
s
2 )(1 . 41 s )
2
s
VII. Projectile Motion
Level Horizontal Range
• Range is determined by time it takes for ball to return to ground
level or perhaps some other vertical value.
• If ball hits something a fixed distance away, then time is
determined by x motion
• If the motion is on a level field, when it hits: y = 0
y
y0
vy0 t
Solving we find
1
gt
2
0
0
v yo t
2
1
gt
2
2
2 v y0
t
g
We can substitute this in the x equation to find the range R
2v
R
x
vx0 t
v xo (
g
y0
)
2 vx0 v
g
2
yo
2 v 0 sin
0
g
cos
0
VII. Projectile Motion
Level Horizontal Range
We can use a trig identity
2 sin
cos
sin 2
2
R
v 0 sin 2
g
• Greatest range:
= 450
• = 300 and 600 have same
range.
(
45
0
0
15 )
Caution– the range formula
has limited usefulness. It is
only valid when the
projectile returns to the same
vertical position.
VII. Projectile Motion
Example : A football is kicked at an angle of 50.00 above the
horizontal with a velocity of 18.0 m / s. Calculate the range.
Assume that the ball was kicked at ground level and lands at
ground level.
Assume time down = time up
For Range: t ( 2 )(1 . 41 s )
R
x x
33 m
0
v t
x0
2 . 82 s
0
(11 . 6 m )( 2 . 82 s )
s
Could also use range formula
2
R
v 0 sin 2
g
2
0
( 18 m / s ) sin ( 2 ) ( 50 )
9 .8 m / s
2
33 m
Example 4. (a) : A football is kicked at an angle of 50.00 above
the horizontal with a velocity of 18.0 m / s. The football hits a
window in a house that is 25.0 m from where it was kicked. How
high was the window above the ground.
v
(18 . 0 m )(cos 50 . 0 )
s
(18 . 0 m )(sin 50 . 0 )
s

x0
v
11 . 6 m
13 . 8 m

y0
x
Time to hit the window:
v
y
y
y
x0
y
0
s
v t
x0
25 . 0 m
11 . 6 m
s
x
t
s
v t
0
y0
1
2 . 16 s
gt
2
2
(13 . 8 m )( 2 . 16 s )
s
6 .9 m
1
2
(9 .8 m
s
2 )( 2 . 16 s )
2
Example 4. (b) : What is the final velocity and angle of the
football that hit the window in Example (a).
t
v
2 . 16 s
v
y
gt
y0
(13 . 8 m )
s
vy
v
11 . 6 m
x
(11 . 6 m )
s
v
tan
tan
1
v
y
v
x
7 . 37
11 . 6
(9 .8 m
s
2
) ( 2 . 16 s )
7 . 37 m
s
2
( 7 . 37 m )
s
32 . 4

2
13 . 7 m
below x axis
s
s
Example 5. A rescue plane wants to drop supplies to isolated mountain climbers
on a rocky ridge 235 m below. If the plane is traveling horizontally with a speed
of 250 km /h (69.4 m / s) how far in advance of the recipients (horizontal
distance) must the goods be dropped as shown in Figure? .
v y0
0
vx0
69 . 4 m / s
Coordinate system is 235 m
below plane
y
235 m
0
1
g t
2
0
2
t
( 2 ) ( 235 m )
9 .8 m / s
2
6 . 93 s
x
x0
v xo t
x
481 m
0
( 69 . 4 m / s ) ( 6 . 93 s )
Projectile Motion is Parabolic
In order to demonstrate that projectile
motion is parabolic, the book derives y as
a function of x. When we do, we find that
it has the form:
This is the
equation for a
parabola.
IX. Relative Velocity
•Will consider how observations made in different
reference frames are related to each other.
A person walks toward the front of a train at 5 km / h
(VPT). The train is moving 80 km / h with respect to the
ground (VTG). What is the person’s velocity with respect
to the ground (VPG)?

V PG
V PG

V PT
5 km / h

V TG
80 km / h
85 km / h
IX. Relative Velocity
• Boat is aimed upstream
so that it will move
directly across.
• Boat is aimed directly
across, so it will land at a
point downstream.
• Can expect similar
problems with airplanes.
Example 6 An airplane is capable of flying at 400 mi/h in still air.
At what angle should the pilot point the plane in order for it to
travel due east, if there is a wind of speed 50.0 mi/h directed due
south? What is the speed relative to the ground?

V PA

V AG
sin
V AG
V PA

V PG
sin
1
V AG
sin
50 . 0 mi
1
400 mi
V PA
V PG
( 400 mi
h
) cos ( 7 . 18
0
)
h
7 . 18
North of East
h
397 mi
0
h
HOMEWORK
Giancoli, Chapter 3
3, 5, 7,10, 11, 16, 18, 24, 27, 30
References
o “Physics For Scientists &Engineers with Modern Physics” Giancoli 4th edition, Pearson
International Edition