First Year Physics Notes: PTB Lahore
Chapter 3: Motion and Force
Topics:
3.1:
Displacement
d
= r2-r1
The change in the position of a body from its
initial to final position is known as displacement
3.2:
Velocity
The rate of change of displacement is called
velocity
v
= Dd/Dt
or
The rate of change of position of a body is called
velocity or
v
= xf -
xi / tf - ti
Displacement covered by a body in unit time is
called velocity
vav
= d/t
vins
= limDt®0
Dd/Dt
If te instantanous velocity does not change then the
the body moves with constant or uniform velocity
3.3:
Acceleration:
The rate of change of velocity is called
acceleration
a
= vf -
vi / tf - ti
a
= Dv/Dt
ains
= limDt®0
Dv/Dt
3.4:
Velocity-Time Graph
The graph between
velocity on y-axis and time on x-axis is called velocity-time graph. The slope
of this graph gives us average acceleration
3.5:
Review of Equations of Uniformly Accelerated Motion
s = (vf + vi)/2´t
1. vf = vi + at
2. s = vit
+ ½ at2
3. 2as = vf2
-
vi2
3.6:
Newton’s Laws of Motion
Frames
of Reference: There are two types of frames of
reference
Inertial frame of
Reference and Noninertial Frame of Reference
Inertial
Frame: A frame of reference that is at rest or moving
with constant velocity in straight line
v
= 0 or v = constant
or
A nonaccelerated frame
of reference is called inertial frame
a
= 0
Noninertial Frame of
Reference: A frame of reference that is accelerating or rotating or both
accelerating and rotating is called noninertial frame of reference
Newton’s
First Law
In the absence of any
unbalanced force, if a body is is at rest it will remain at rest and if it is
moving then it will continue its motion with constant velocity in straight line
Newton’s
2nd Law of Motion
When an unbalanced
force acts on a body the body is accelerated in the direction of applied force.
The magnitude of acceleration is directly proportional to the applied force and
inversely proportional to its mass
F
= ma
Newton’s
3rd Law of Motion
Action and reaction are equal in magnitude but in
opposite direction
F
= -
F
3.7:
Linear Momentum P
The Quantity of motion is called momentum or
The product of
mass and velocity is called momentum
P
= mv
Relation
Between Momentum and 2nd Law of Motion:
We know that acceleration is equal to the rate of
change of velocity
a
= vf -
vi / t
multiplying by mass m on both sides
ma = mvf -
mvi / t
But F = ma, mvf = Pf and
mvi
= Pi so
F
= Pf -
Pi / t
F
= DP / t
Newton’s 2nd law is defined as the rate
of change of linear momentum is equal to force
Impulse
Ft
= DP
The force applied for a very short time is called
impulse or sudden force
Law
of Conservation of Linear Momentum
In the absence of any external force, linear
momentum of two or more interactiong bodies of an isolated system remamains
constant.
Explanation:
consider two bodies of masses m1 and m2 are moving with
velocities v1 and v2 before collision and after collision
their velocities become v1¢ and v2¢
then by the law of conservation of momentum
Momentum before collision = momentum after collision
m1v1 + m2v2
= m1v1¢+ m2v2¢
3.8:
Elastic and Inelastic Collision
Elastic
Collision: The collision in which linear momentum and kinetic
energy before and after collisision
m1v1 + m2v2
= m1v1¢+ m2v2¢
and
½ m1v12 +1/2 m2v22
= ½ m1v1¢2+1/2 m2v2¢2
Elastic
Collision in one Dimension
We know that linear momentum and kinetic energy
befoere and after collision remains same
(1) m1v1
+ m2v2 = m1v1¢+
m2v2¢
m1v1 - m1v1¢
= m2v2¢- m2v2
(2) ½ m1v12
+1/2 m2v22 = ½ m1v1¢2+1/2
m2v2¢2
Simplifying equation (1) and (2)
(3) m1 (v1 - v1¢)
= m2 (v2¢- v2)
½ m1 (v12
-
v1¢2)
= ½ m2 (v2¢2
-v22)
(4) m1 (v1 -
v1¢)
(v1 + v1¢) = m2 (v2¢
-v2)
(v2¢
+v2)
Now dividing equation(4) by equation (3)
m1 (v1 -
v1¢)
(v1 + v1¢)/ m1 (v1 - v1¢)
= m2 (v2¢ -v2)
(v2¢
+v2)/
m2 (v2¢- v2)
(v1 + v1¢)
= (v2¢ +v2)
v1 - v2 =
v2¢-
v1¢ taking -1 common on right side only
(5) (v1
-
v2) = - (v1¢
-
v2¢)
This equation shows that relative velocity of
approach is equal to the relative velocity of recession
In order to find the values of v1¢
and v2¢, considering equation
v2¢ = v1
-
v2 + v1¢
and using in equation (1)
m1v1 + m2v2
= m1v1¢+ m2 (v1 -
v2 + v1¢)
m1v1 + m2v2
= m1v1¢+ m2v1 -
m2v2 + m2v1¢
m1v1¢+
m2v1 - m2v2 + m2v1¢
= m1v1 + m2v2
m1v1¢
+ m2v1¢ = m1v1 + m2v2
-
m2v1 + m2v2
(m1 + m2 )v1¢
= (m1 - m2 )v1 + 2m2v2
(6) v1¢ = [m1 - m2 /m1
+ m2 ]v1 + [2m2 /(m1 + m2 )]v2
Similarly, we get
(7) v2¢ = [2m1 /m1 + m2 ]v1 +
[m2 - m1 /(m1
+ m2 )]v2
Special
Cases:
Case
I: when m1= m2 = m, using in (6) and
(7), we get
v1¢ = v2
and
v2¢ = v1
It means both bodies exchange their velocities
Case
II: when m1= m2 = m and v2
-0, using in (6) and (7), we get
v1¢ = 0 and
v2¢ = v1
Body m1 comes to rest and body m2
gains the velocity of m1
Case
III: when a lighter body collides with a massive body at
rest i.e.
m1<< m2 =
m and v2= 0, using in (6) and (7), we get
v1¢ = -
v1 and
v2¢ = 0
Case
Iv: when
a massive body collides with a lighter body at rest i.e.
m2<< m1 =
m and v2= 0, using in (6) and (7), we get
v1¢ = v1
and
v2¢ = 2v1
3.9:
Force Due To Water Flow
F
= -
(m/t)v = - mass per second ´change
in velocity
The reaction force dueb to newton’s third law of
motion
F
= -
(-m/t)v = (m/t)v
3.10:
Momentum and Explosive Forces
Bullet
and gun system
MV¢
=
-mv
V¢
=
-
(m/M)v
3.11:
Rocket Propulsion
a
= mv/Mt
a
= mv/M when t =
1 second
3.12:
Projectile Motion
A body thrown at some
angle q
above or below the horizontal and moving freely under the action of gravity is
called projectile.
Horizontal distance
x = vx ´ t
Vertical Distance
y = ½ gt2
Horizontal Component of Velocity
Horizontal component of velocity remains constant
throughout the motion so
vix = vfx = vi cosq
Vertical component of velocity changes at each
instant so
vfy = vi sinq
-
gt
Magnitude
of Velocity
v = Ö (vfx)2 +(vfy)2
tanq = vfy/ vfx
Horizontal
Component of Acceleration
Since horizontal component of velocity remains
constant throught the motion so there is no change in velocity therefore no
acceleration
ax
= 0
Vertical
Component of Acceleration
In downwaqrd or upward direction, component of
velocity changes at each point so there exists acceleration that is
acceleration due to gravity g
ay
= g
Height
of The Projectile
In order to find height of projectile, we consider 3rd
equation of motion
2as = vf2 -
vi2
At the highest point vfy = 0, viy
= vi, a = - g, s = h, using in (1)
-2gh
= 0 -
vi2
2gh = vi2 but viy = visinq
h = vi 2sin2q/2g
Time
of Flight
Time taken by the projectile from the initial to
final position is called time of flight. From 2nd equation of motion
s = vi t + ½ gt2
but s = h = 0, using in (1)
0 = visinqt -½
gt2
½ gt2 =
visinqt
t = 2visinq/g
Range
of The Projectile
The maximum distance covered by the projectile along
the horizontal direction is called range of the projectile. It is denoted by R
R = vix ´ t
But vix
= vicosq
and t = 2visinq/g
R = vicosq ´2visinq/g
R = 2vi 2cosqsinq/g
R = vi 2 (2cosqsinq)/g
R = vi 2 (sin2q)/g (2cosqsinq)
=sin2q
Maximum
Range:
The range in which a projectile falls at maximum
distance is called maximum range
Rmax = vi 2/g When sin2q
= sin90
or q = 45
or q = 45
Common
Point of Fall of Two Projectiles:
If two projectiles are projected at different angles such that
the sum of their angles is 90
then they will fall at
the same point e.g.
then they will fall at
the same point e.g.
30 and 60
40 and 50, 10 and 80 etc.
and 6040 and 50, 10 and 80 etc.
Application
To Ballistic Missile
An unpowered and unguided missile is called
ballistic missile. Air friction does count for accuracy. For long distance,
ballistic missile do not reach at the target due to friction of air, spin
motion of earth
For long range and greater precision powered and
remote control guided missiles are used.
Special
Questions:
Q:
What do you mean by kinematics?
Ans:
The study of motion without mass or charge is called kinematics
Q:
What are the equations of kinematics?
Ans:
kinematical equations do not contain mass or charge
s = (vf + vi)/2´t
vf
= vi + at
s = vit
+ ½ at2
2as = vf2 -
vi2
Q:
What is dynamics?
Ans:
The study of motion with reference to mass or charge is called dynamics
Q:
Can we derive dynanamics from kinematics
?
Ans:
If we multiply kinematical equations with mass then we get dynamics
Considering first equation of motion which is
kinematical
vf = vi + at
multiplying by mass m on both sides
mvf = mvi + mat
Ft = mvf
-
mvi
Ft = Pf
-
Pi
F
= Pf
-
Pi /t
The above equation is dynamical equation
Q:
Displacement is defined as the change in posion. What is the cause of
displacement?
Ans:
The cause of displacement is time which is not mensioned in the literature.
Also the formula doesn’t show the involvement of time
If we write displacement in this way then time is
shown
d
= vt
d
= (x/t)t (v = x/t)
Q:
What is the mathematical formula of Newton’s first law?
Ans: Mathematical Formula For Newton’s First Law of Motion
It is very strange to note that, we have
mathematical formula for Newton’s second law of motion F = ma, but no formula
for Newton’s first law.
Q: how will you derive 2nd law of motion from first
law?
Ans: Simple Derivation of Newton’s Ist law From 2nd
Law
We know the relation between force and linear
momentum
F
= (Pf -
Pi)/tf -ti
F
= DP/Dt
(Rate of change of linear momentum)
As an analogy
P
= m(xf -
xi)/tf -ti
P
= mDx/Dt
P = mv {v = Dx/Dt}
Momentum of a body is equal to the rate of change of
change of position of a body of mass m which is nothing but Newton’s first law
If the position is not changing then its momentum is
zero means the body is at rest P = 0
Q: How to derive Newton’s First law?
Ans: Usual Derivation of of Newton’s First Law
In the absence of an unbalanced force, linear
momentum of a body is directly proportional to the its velocity and inversely
proportional to its mass
(1) vµ
P
(2) vµ
1/m
Combining (1) and (2), we have
vµ
P/m
v
= k P/m
when k =1 then
v
= P/m
P = mv
Linear momentum is equal to the rate of change of
position of a particle of mass m
Q:
What is the relation between linear momentum and kinetic energy?
Ans:
The relation between momentum and kinetic energy is given by
KE = ½ mv2
KE = ½ mv.v but P = mv
KE = ½ P.v
or
KE = P2/2m
Q:
SHORT
QUESTIONS AND ANSWERS
3.1:
What is difference between uniform and variable velocity?
Ans:
Uniform
Velocity: If a body covers equal distance in equal interval
of time then its velocity is called uniform velocity
Variable
Velocity: If a body doesn’t covers equal distance in equal
interval of time then its velocity is called variable
3.2:
An objectis thrown vertically upward. Discuss the sign of acceleration due to
gravity, relative to velocity, while bthe object is in air?
Ans:
During upward motion velocity decreases or negative and becomes zero at the
highest point and starts increasing while falling downward but acceleration due
to gravity remains constant
3.3:
Can the velocity of an object reverse the direction when acceleration is
constant? If so, give an example
Ans:
Velocity reverses its direction while an object moves in upward direction
and returns to downward direction
3.5:
A man standing on the top of a tower
throws a ball straight up with initial velocity vi and at the
same time throws a second ball straight downward with the same speed. Which
ball will have larger speed when it strikes the ground? Ignore air friction
Ans:
Ignoring the air friction, both bodies will have the same speed while striking
the ground
v = gt
3.6:
Explain the circumstances in which the velocity v and acceleration of a car are
(i). parallel (ii). Anti-parallel (iii). Perpendicular (iv). v = 0 but a 
(v). a = 0 but v
Ans:
i. While acceleration is positive then v and a are
parallel to each other
ii. When brakes are applied then v and a are
anti-parallel to each other
iii. When the car is moving on a circular path then
linear velocity v and centripetal acceleration are perpendicular to each other
iv. When brakes are applied at any instant, the body
stops so its velocity is zero but acceleration is not zero
v. when the body moves with constant velocity then
its acceleration becomes zero
3.7:
Find the change in momentum for an object subjected to a given force for a given
time and state law of motion in terms of momentum?
Ans:
We know that acceleration is equal to the rate of
change of velocity
a
= vf -
vi / t
multiplying by mass m on both sides
ma = mvf -
mvi / t
But F = ma, mvf = Pf and
mvi
= Pi so
F
= Pf -
Pi / t
F
= DP / t
Newton’s 2nd law is defined as the rate
of change of linear momentum is equal to force
3.9:
Define impulse and show that how it is related to linear momentum?
Ans:
The force applied for a very short time is called impulse or the impulse is
equal to the change in momentum
Ft
= DP
3.10:
State the law of conservation of linear momentum pointing out the importance of
isolated system?
Ans:
In the absence of any external force, the momentum of two or more interacting
bodies of an isolated system remains constant
Momentum before collision = momentum after collision
m1v1 + m2v2
= m1v1¢+ m2v2¢
3.11:
Explain the difference between elastic and inelastic collision?
Ans:
Elastic
Collision: The collision in which momentum and kinetic energy
before and after the collision remains same
Inelastic
Collision: The
collision in which momentum before and after collision remains same
kinetic energy before and after collision doesn’t remain same is called
inelastic collision
3.12:
Explain what is meant by projectile motion? Show that the range of projectile
is maximum when projectile is thrown at an angle of 45 with the horizontal?
with the horizontal?
Ans:
when a body is thrown at some angle then the body moves under the action of
gravity on a parabolic path. This type of motion is called projectile motion.
The formula for range of projectile motion is
R = vi 2 (sin2q)/g
For maximum range (sin2q)
must give maximum value that is equal to 45 so that it becomes sin90 = 1 that gives us
so that it becomes sin90 = 1 that gives us
Rmax = vi 2/g
3.13:
At what point or points in its path does
a projectile its minimum speed, its maximum speed?
Ans:
At the highest point of projectile motion the velocity is minimum and maximum
before striking the ground
NUMERICAL
PROBLEMS FROM EXERCISE
3.1:
Ahelicopter is ascending vertically at the rate of 19.6 ms-1.
when it is at the height of 156.8 m above the ground, a stone is dropped. How
long does the stone take to reach the ground ?
Data:
vi = 19.6 ms-1, h = 156.8 m, g = 9.8 ms-2, t = ? vf = 0
Solution:
2gS = vf 2 -
vi 2
-2
´9.8
´156.8
= -19.6´19.6
156.8 = 19.6t + 4.9t2
32 = 4t + t2
t2 +4t -32 = 0
t2 +8t-4t -32
= 0
t(t +8) -4(t
+8) = 0
(t +8)(t -4) = 0
t = -8 s , t =4 s
t
= 4s
3.3:
A proton moving with the speed of 1.0´
107 ms-1 passes
through a 0.020 cm thick sheet of paper and emerge with a speed of 2.0´ 106
ms-1.
Assuming uniform deceleration, find retardation and time taken to pass through
the paper?
Data:
Vi = 1.0´ 107
ms-1, s = 2.0´ 10-4m, vf = 2.0´
106 ms-1
a = ?,
t = ?
Solution:
2aS = vf 2 -
vi 2
a = vf
2 -
vi 2/2s
a = [(2.0´
106 )2 - (1.0´ 107 )2
/ 2 ´2.0´
10-4
]ms-2
a = [4.0´
10 12 - 1.0´ 1014
/ 4´
10-4
]ms-2
a = [10 12
(4 -
100) / 4´
10-4
]ms-2
a = 10 16
´
-
96 / 4 ]ms-2
a
= -
2.4´
10 17ms-2
t = vf
-
vi/a
t = 2.0´ 106 - 1.0´
107/- 2.4´ 10 17
t = 106 (2 - 10)/-
2.4´
10 17
t = - 8´106 /-
2.4´
10 17
t
= 3.33 ´10- 11 sec
3.4:
Two masses m1 and m2 are initially at rest with a spring
compressed between them. What is the ratio of the magnitudes of their
velocities after the spring has been released?
Ans:
Initially both bodies are at rest so
Initial momentum Pi = 0
After releasing the spring both bodies move in
opposite direction so
Final momentum Pf = -
m1v1 + m2v2
By the law of conservation of momentum
Pi = Pf
0 = - m1v1 + m2v2
m1v1 = m2v2
v1
/v2 = m2/ m1
3.8:
A truck weighing 2500 kg travelling at 21 ms-1
collides with a stationary car weighing 1000 kg. The truck and the car move
together after the impact. Calculate their common velocity?
Data:
m1 = 2500 kg, v1 = 21 ms-1, m2 = 1000 kg, v2 = 0
velocity of car and truck after collision = V = ?
Solution:
By the law of conservation of momentum
m1v1 + m2v2
= m1v1¢+ m2v2¢
m1v1 + m2v2
= m1V+ m2V
m1v1 + m2v2
= (m1 + m2 )V
V = m1v1 / (m1 + m2
)
V = 2500 ´ 21/ (2500 +1000
) ms-1
V = 2500 ´ 21/ 3500 )
ms-1
V
= 15 ms-1
3.9:
Two blocks of masses 2 kg and 0.5 kg are attached to the ends of a compressed
spring. The elastic potential energy stored in the spring is 10 J. Find the
velocities of the blocks if the spring delivers its energy to the blocks when
released?
Data:
m1 = 2 kg, v1 = ?, m2 = 0.5kg, v2 = ?, PE = 10 J
Solution:
Initial Momentum = Final Momentum
0 = m1v1 + m2v2
v2 = - m1 /
m2v1
v2 = - 2 /
0.5v1
v2 = - 4v1
By law of conservation of energy
½ m1v12 +1/2 m2v22
= 10
m1v12 + m2v22
= 20
2v12 + ½ v22
= 20
4v12 + 16 v12
= 40
20 v12 = 40
v12 = 2
v1
= Ö2
ms-1
v2 = - 4v1
= -
4Ö2
ms-1
v2
= -
5.6
ms-1
3.10:
A football is thrown upward at an angle of 30 with respect to the horizontal. To throw a 40
m pass what must be the initial speed of the ball?
with respect to the horizontal. To throw a 40
m pass what must be the initial speed of the ball?
Data:
q
= 30
, R = 40 m,
vi = ?
, R = 40 m,
vi = ?
Solution:
The formula of range is given by
R = vi 2 (sin2q)/g
40 = vi 2 (sin60)/9.8
)/9.8
vi 2 = 40 ´9.8/
sin60
vi
= 21.3
ms-1
3.11:
a ball is thrown horizontally from a height of 10 m with the velocity of 21ms-1.
How far off it hit the ground and with what velocity?
Data:
Height h = 10 m
vix = 21 ms-1
viy = 0 ms-1
Distance x = ?
Velocity v = ?
Solution:
h = viyt +1/2 gt2
10 = 0 t + ½ ´9.8´t2
4.9´t2 = 10
t2 = 10/4.9
t
= 1.414 s
x = vix t
x = 21´1.414 m
x
= 30 m
3.12:
A bomber dropped a bomb at a height of 490 m. when the velocity along the
horizontal was 300 107 kmh-1
(a) how long it was in air/
(b) At what distance from the point vertically below
the bomber at the instant that bomb was dropped, did it strike the ground?
Data:
Height h =y =
490 m
Horizontal velocity vix = 300kmh-1
= 83.3ms-1
Vertical velocity viy = 0
X = ?
t = ?
Solution:
Y = viyt + ½ gt2
490 = 0 t + ½ ´9.8t2
490 = 4.9t2
t2 = 100
t
= 10 s
x = vix ´t
x = 83.3 ´10 m
x
= 833 m
3.13:
Find the angle of projection of a projectile for which its maximum height and horizontal
range are equal?
Data:
Angle of projection q = ?
Maximum height = horizontal range
vi 2sin2q/2g
= vi 2sin2q/g
sin2q/2 = 2sinq cosq
sinq =
4 cosq
sinq/ cosq = 4
tanq
= 4
q
= tan-1(
4)
q = 76
3.14:
prove that the angle of projection, which exceed or fall short of 45 by an equal amounts, the ranges are equal?
by an equal amounts, the ranges are equal?
Data:
For Maximum range q = 45
R1 = R2 = ?
According to given condition
q1
= 45 + 15= 60 q2
= 45 - 15= 30
+ 15= 60 q2
= 45 - 15= 30
R1 =
vi 2sin2q1/g R2
= vi 2sin2q2/g
R1 =
vi 2sin120/g R2
= vi 2sin60/g
/g R2
= vi 2sin60/g
R1 = (0.866) vi 2/g R2
= (0.866)vi 2/g
This implies
R1
= R2
3.15:
A SLBM is fired from a distance of 3000 km. If the earth is considered as flat
and the angle of launch is 45 with horizontal. Find the velocity with which
the missile is fired and the time taken by the SLBM to hit the target
with horizontal. Find the velocity with which
the missile is fired and the time taken by the SLBM to hit the target
Data:
R = 3000km = 3´106 m
Angle of launch q = 45
Velocity of missile v = ?
Time taken t = ?
Solution:
R = vi 2sin2q/g
vi 2 = gR/sin2q
vi 2 = 9.8 ´3´106/sin90
vi = 5.42´103 ms-1
vi
= 5.42 kms-1
t = 2visinq/g
t = 2 ´5.42´103
´sin45/9.8
/9.8
t
= 782 s or 13 minutes
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