Chapter : 02
Kinematics
Q 01:
Define rest and motion with examples ?
Ans:
Rest:
“When a body does not change its position with respect to its surroundings, it is said to
at rest”.
Example: A book lying on a table is at rest with respect to every object in the room.
Motion:
“When a body changes its position with respect to its surroundings, it is said to
in motion”.
Examples: A passenger in a moving car is in motion with respect to the people standing on
the pavement.
Q 02:
What do you know about position ?
Ans:
Position:
“The location of an object relative to some reference point or origin is called its
position”.
Explanation:
Consider a body is at some position ‘P’ in the x-y plane as shown in the
figure. Point O is the origin or reference point of the coordinate system.
y
P(x,y)
O
x
The position of point P is described by its distance from point O, i.e, the straight line OP.
Another way to find position coordinates of an object is to draw two perpendicular
lines from point P.
Q 03:
Explain types of motion with examples ?
Ans:
There are three types of motion explained below:
(1)
Translatory motion:
“A type of motion in which all particles of a body move parallel to
each other along any path straight or curved is called translatory motion”.
Examples:
(i)
Motion of falling bodies.
(ii)
Motion of a car.
(iii)
Motion of flying birds.
(2)
Rotatory motion:
“A type of motion in which each point of a body move around a fixed
point or axis is called rotatory motion”.
Examples:
(i)
Motion of a wheel of bicycle.
(ii)
Motion of hands of a clock.
(iii)
Motion of blades of a fan.
(3)
Vibratory motion:
“To and fro motion of a body over the same path about the mean position
is called vibratory motion”.
Examples:
(i)
Motion of a swing.
(ii)
Motion of a pendulum.
(iii)
Motion of mass attached to a spring
----------------------------------------------------------------------------------------
Q 04:
Define distance and displacement ?
Ans:
Distance:
“The length of the actual path followed by a body during its motion is called
distance”.
It is denoted by ‘S’. Its SI unit is meter (m). It is a scalar quantity.
Displacement:
“The shortest possible distance between any two points is called
displacement”.
It is also denoted by ‘S’. Its SI unit is meter (m). It is a vector quantity.
Q 05:
Define speed and its types ?
Ans:
Speed:
“The distance covered by a body in unit time is called speed”.
Mathematically,
speed 
or
v
distance
time
S
t
Speed is a scalar quantity. Its SI unit is meter per second ( m sec -1 ). Different kinds of speed
are given below:
(I)
Average speed:
“Total distance divided by the total time taken is called average speed”.
Mathematically,
Average speed 
total distance
total time
 v 
or
(II)
S
t
Instantaneous speed:
“The speed of a body at any particular instant is called instantaneous
speed”.
Mathematically,
vinst 
(III)
S
t
Uniform speed:
“If a body covers equal distances in equal intervals of time, then its speed is
uniform”.
Q 06:
Define velocity and its types ?
Ans:
Velocity:
“The rate of change of displacement is called velocity”.
Mathematically,
velocity 
displaceme nt
time

v  St

Velocity is a vector quantity. Its SI unit is meter per second ( m sec -1 ).
Types of velocity:
(I)
Average velocity:
“Total displacement covered divided by total time taken is called average
velocity”.
Mathematically,

S

v  
t

where  v is called average velocity.
(II)
Uniform velocity:
“If a body covers equal displacements in equal intervals of time, then the
velocity is called uniform velocity”.
OR
“If the speed and direction of a body does not change with respect to time, then the velocity
is called uniform velocity”.
(III)
Variable velocity:
“If the speed or direction or both changes with time, then velocity is
called variable velocity”.
Q 07:
Define acceleration and its types ?
Ans:
Acceleration:
“The rate of change of velocity is called acceleration”.
Mathematically,

a  vt

Where a is acceleration and Δ v is change in velocity. Acceleration is a vector quantity. Its
SI unit is meter per second square ( m sec-2 ).
Types of acceleration:
(I)
Positive acceleration:
“If the magnitude of velocity increases with time, then acceleration
is called positive acceleration”.
(II)
Negative acceleration:
“If magnitude of velocity decreases with time, then acceleration is
called negative acceleration”.
(III) Average acceleration:
It is defined as, “the change in velocity divided by the time
interval”.
Mathematically,
Average accelerati on 

or
Q 08:
 a 
change in velocit y
time

v
t
Define scalars and vectors with examples ?
Ans:
Scalars:
“Those physical quantities which are completely specified by their magnitude
only are called scalars”.
A magnitude is a number with a proper unit e.g, 5 Kg.
Examples:
Length, mass, time, speed, area, volume etc.
Vectors:
“Those physical quantities which are completely specified by their magnitude as
well as direction are called vectors”.
Examples:
Displacement, velocity, acceleration, force, momentum, torque etc.
Q 09:
How a vector can be represented ?
Ans:
A vector can be represented by following two methods:
(I)
Symbolic representation:
According to this method, a vector is represented by a letter
with an arrow head above or below it or the letter is written with a bold face.
For example, a vector A can be written symbolically as

A , A Or
A

The magnitude of a vector is represented by a modulus such that

A
(II)
or A
Graphical representation:
A vector is drawn graphically using the following steps.
a)
Draw NEWS.
b)
Select a suitable scale.
c)
Draw a line parallel to the given direction.
d)
Cut the line equal to the given magnitude according to the selected scale.
North
West
East
South
Q 10:
Derive the first equation of motion ?
Or Prove that
Ans:
v f = vi + at ?
Proof:
Consider the following figure.
Where
OA = vi (initial velocity)
BC = vf (final velocity)
AB = a (acceleration)
AD = OC = t (time)
It is clear from the figure that
BC = BD + DC
Or
BC = DC + BD
Or
BC = OA + BD
since ( DC = OA )
Putting values in above equation
vf = vi + BD ---------(1)
We know that
slope 
or
AB 
perpendicu lar
base
BD
AD
Multiplying AD on both sides,
AB  AD 
or
or
or
BD
 AD
AD
AB  AD  BD
BD  AB  AD
BD  a  t - - - - - - - -(2)
 AB  a & AD  t 
Put equation (2) in equation (1)
v f  vi  at
proved.
Q 11:
Derive second equation of motion ? OR Prove that
Ans:
Proof: Consider the following figure.
1
S  vi t  at 2 ?
2
Where
OA = CD = vi
BC = vf
AB = a
AD = OC = t
BD = at
Also consider, distance travelled = S
From the figure, it is clear that
S = Area of figure OABC
Or
S = Area of
OADC + Area of
ABD
Now we know that
Area of
OADC = OA x OC
Putting values in above equation
Area of
And we also know that
OADC = vi x t ---------------(2)
----------(1)
Area of
1
(AD x BD)
2
ABD =
Putting values in above equation
Area of
1
(t x at ) ------------(3)
2
ABD =
Put equations (2) and (3) in equation (1)
S  vi t 
or
Q 12:
1
t  at 
2
1
S  vi t  at 2
2
Derive third equation of motion?
OR Prove that 2aS  v f  vi
2
Ans:
proved.
2
?
Proof: Consider the following figure.
Where
OA = CD = vi
BC = vf
AD = OC = t
From the figure, it is clear that OABC is a trapezoid. We also know from the figure,
Distance travelled = Area of trapezoid OABC
Or
but
S = Area of trapezoid OABC -----------------(1)
Area of trapezoid 
1
sum of parallel sides  base
2
So equation (1) becomes
S
1
sum of p a r a l lseild es b a s e
2
From the figure, parallel sides are OA and BC, while the base is AD, so
S
1
OA  BC  AD
2
or
S
1
vi  vf  t
2
or
S
vi  v f   t
2
- - - - - - - -(2)
Now using first equation
vf  vi  at
or
v f  vi  at
or
at  vf  vi
Dividing both sides by ‘ a ‘
at vf - vi

a
a
t
or
vf - vi
- - - - - - - - - (3)
a
Put equation (3) in equation (2)
S
vi  v f v f - vi

2
a
Multiplying ‘2a ‘ on both sides,
2a  S 
or
v f  vi v f - vi   2a
2a
2aS  v f - v i
2
2
proved.
Q 13:
What is meant by gravitational acceleration ?
Ans:
Gravitational acceleration:
An Italian scientist Galileo stated that earth attract every body
with an acceleration equal to 9.8 m sec-2 . He arrived to this conclusion by dropping
several objects from the famous leaning tower at Pisa. He found that all objects reached
the earth surface at the same time.
Gravitational acceleration is denoted by ‘g’. Equations of
motion under gravity are written as
v f  vi  gt
1 2
gt
2
h  vi t 
2 gh  v f  vi
2
2
Exercise
Conceptual Questions
Q 01:
Ans:
Is it possible that displacement is zero but not the distance. Under what condition
displacement will be equal to distance ?
Yes, it is possible that displacement is zero but not the distance. Moreover, displacement is
equal to distance if and only if a body moves on a straight path.
Explanation:
(1)
Let a body move from point A to B, and then to C,D and finally reaches
again to A as shown in the figure. In this case, the displacement covered is zero but the
distance is non-zero.
A
B
D
(2)
C
If a body moves on a straight path from A to B as shown below, then its distance will be
equal to the displacement.
A
B
Q 02:
Does a speedometer measures a car’s speed or velocity ?
Ans:
A speedometer measures only speed of a car and not its velocity.
Explanation:
Since a speedometer gives no idea about the direction of a car. It will always
give speed of a car whether we are moving in any direction. Therefore, a speedometer only
measures the speed and not velocity.
Q 03:
Is it possible for an object to be accelerating and at rest at the same time ? Explain
with example ?
Ans:
Yes, it is possible that an object may be accelerating and at rest at the same time.
Explanation:
Consider a ball thrown vertically upward as shown in the figure. During its
entire motion, it has downward acceleration.
v=0
a = -g
v>0
a = -g
As the ball moves up, its velocity increases and it accelerates. When the ball reaches to the
top, it comes to rest ( v = 0 ) before moving downward. The speed or velocity becomes
zero at the top but the change in velocity is not zero. And we know

a  vt

Hence the ball will have acceleration at the top despite being at rest.
Q 04:
Can an object have zero acceleration and non-zero velocity at the same time ? Give
example ?
Ans:
Yes, an object may have zero acceleration and non-zero velocity simultaneously.
Explanation:
When an object is travelling with uniform or constant velocity, then there will be
no acceleration in the body. Because, acceleration is produced only when velocity of a
body changes.
For example, a car moving on a straight road with 70 km / h
constant velocity.
has no acceleration and
Q 05:
A person standing on the roof of a building throws a rubber ball down with a velocity
of 8.0 m / sec.
Ans:
What is the acceleration of the ball ?
The magnitude of acceleration in this case is 9.8 m / sec2
and its direction is always
downward ( toward the earth ).
Explanation:
Acceleration depends on force. Here the only force acting on the ball is the
gravity force. So the acceleration will only be due to gravity, which has a constant value of
9.8 m / sec2
Q 06:
near the surface of earth.
Describe a situation in which the speed of an object is constant while the velocity is
not ?
Ans:
An object performing uniform circular motion has constant speed but variable velocity.
Explanation:
Consider an object is moving in a circle with uniform speed as shown.
V
V
V
V
Its speed is constant throughout the motion but velocity is changing at every point due to
change in direction.
Q 07:
Can an object have a northward velocity and a southward acceleration. Explain ?
Ans:
Yes, an object having velocity in north direction may have its acceleration in the south
when it slows down.
Explanation:
A body slowing down will have acceleration in the opposite direction of
velocity. If we apply brakes to a car which is moving towards north, then its acceleration
will be in the south.
Q 08:
As a freely falling object speeds up, what is happening to its acceleration - does it
increase, decrease or stay the same ?
Ans:
If air resistance is negligible, the acceleration of a freely falling body stays the same as it
falls.
Explanation:
It must be noted that speed of an object increases in free fall, but since the
increase occurs at a constant rate, therefore, the acceleration will be constant. This
acceleration is called acceleration due to gravity and is given as
g = 9.8 m sec-2
Q 09:
A ball is thrown upward with an initial speed of 5 m / s. What will be its speed when it
returns to starting point ?
Ans:
Its speed will be the same ( that is 5 m / s ) when the ball returns to its starting point.
Explanation:
When a ball is thrown upward, it moves with a deceleration of 9.8 m sec-2
until it reaches to the top and stops there for a moment. When it falls, again it moves
under the same acceleration.
That is why when it comes to its starting point, it has the same
speed with which it was thrown upward.
Numerical Questions
Problem 01:
A squash ball makes contact with a squash racket and changes velocity from 15
m / s west to 25 m / s east in 0.10 second. Determine the vector acceleration of
the squash ball ?
Ans:
Given data
Initial velocity = vi = 15 m / s, W
Final velocity = vf = 25 m / s, E
Time = Δ t = 0.10 second

Required : accelerati on  a  ?
Solution:
As we know that

a  vt

or
 
 vf  vi
a
t
  
 v  v f - v i 
Putting values in above equation
 25 E  - 15 W 
a
0.10
Reversing the direction of
vi from west to east in above equation
 25 E   15 E 
a
0.10
or
or
 40 E 
a
0.10

a  400 m sec -2 , East
Ans.
Problem 02:
A golf ball initially traveling at 25 m / s hits a sand trap and slows down with an
acceleration of
Ans:
-20 m sec-2. Find its displacement after 2.0 seconds ?
Given data
Initial velocity = vi = 25 m / s
Final velocity = vf = 0 m / s ( because the ball finally stops )
Time = t = 2.0 second = 2 sec
Acceleration = a = - 20 m sec-2
Required: displacement = S = ?
Solution:
1
using S  vi t  at 2
2
Put values in above equation
S  25  2 
or S  50 
1
- 2022
2
1
- 204
2
or S  50 
1
- 80
2
or S  50 
1
80
2
or S  50  40
or S  10 m
Problem 03:
Ans.
A bullet accelerates the length of the barrel of a gun 0.750 m long with a
magnitude of 5.35 x 105 m sec-2. With what speed does the bullet exit the
barrel ?
Ans:
Given data
Length = l = S = 0.750 m
Acceleration = a = 5.35 x 105 m sec-2
Initial velocity = vi = 0 m / s
Required:
Final velocity = vf = ?
l=S
vi
vf
Solution:
using 2aS  v f  vi
2
2
Putting values in above equation
2  5.35 105  0.750  v f  02
2
or
8.025 105  v f
or
8.025 100000  v f
or
802500  v f
or
v f  802500
2
2
2
2
Taking square root on both sides
v f  802500
2
or
v f  895.8 m / sec
or
v f  896 m / sec
Ans.
Problem 04:
A driver is traveling at 18 m /s when she sees a red light ahead. Her car is
capable of decelerating at a rate of 3.65 m sec -2. If she applies brakes when
she is only 20 m from intersection when she sees the light, will she be able to
to stop in time ?
Ans:
Given data
Deceleration = a = -3.65 m sec-2
Distance between intersection and car = S = 20 m
Initial velocity = vi = 18 m sec-1
Final velocity = vf = 0
Required:
distance needed to stop = S = ?
Solution:
using 2aS   v f  vi
2
2
Dividing 2a on both sides
2aS  v f  vi

2a
2a
2
v f  vi
2
or
S 
2
2
2a
Putting values in above equation
02  18
S 
2 3.65
2
or
or
S 
 324
 7.3
S   44.38 m
Ans.
This means that the car will stop after covering a distance of 44.38 m, but the
intersection is at 20 m.
Hence, S > S , so the car will not be able to stop in time.
Problem 05:
An antelope moving with constant acceleration 2 m sec-2 crosses a point
where its velocity is 5 m / s. After 6 seconds how much distance it has covered
and what is its velocity ?
Ans:
Given data
Acceleration = a = 2 m sec-2
Time = t = 6 sec
Initial velocity = vi = 5 m / s
Required: A)
B)
distance = S = ?
final velocity = vf = ?
Solution:
A)
using second equation of motion
1
S  vi t  at 2
2
Putting values
1
2
S  5  6   2  6
2
S  30  36
or
S  66 m
or
B)
Ans.
using first equation of motion
v f  vi  at
Putting values
v f  5  2 6
or
v f  5  12
or
v f  17 m sec-1
Ans.
Problem 06:
With what speed must a ball be thrown vertically from ground level to rise to a
maximum height of 50 m ?
Ans:
Given data
Deceleration = a = - g = -9.8 m sec-2
Height = S = h = 50 m
Final speed = vf = 0
Required:
initial velocity = vi = ?
Solution:
using 2aS  v f  vi
2
2
Here a = - g ( negative sign is due to upward motion )
so 2 g h  v f  vi
2
or - 2 gh  v f  vi
2
2
2
Putting values
- 2  9.8  50  02  vi
or - 980  vi
2
or vi  980
2
Taking square root on both sides
vi  980
2
2
or vi  31.3 m sec -1
END
Ans.
Chapter : 03
Dynamics
Note: The word dynamics is derived from a Greek word “ dynami ” meaning power.
“It is the branch of physics which deals with the study of motion
of motion of objects with reference to the cause of motion (force).
Q 01:
Define force and its unit ?
Ans:
Force:
“It is a physical quantity which moves or tends to move an object”.
OR
“It is a physical quantity which stops or tends to stop a moving object”.
OR
“It is a physical quantity which tends to change the speed and direction of a moving object”.
The SI unit of force is Newton (N).
Newton:
“A force is said to be one newton if it produces an acceleration of 1m sec-2 in a body
of mass 1Kg”.
Q 02:
State and explain Newton’s first law of motion with example ?
Ans:
Statement:
“A body at rest will remain at rest and a body in motion will continue its
uniform motion unless an unbalanced force acts upon it”.
Mathematically,
If
F=0
Then
a=0
So
v=0
or
v = constant
Explanation:
This law has two parts which are discussed below:
(i)
The first part of this law is related to bodies at rest. It says that a static body will remain at
rest when there is no net force.
For example:
A ball lying on the floor will remain there forever unless someone move it
by applying a net force.
(ii)
The second part of the law is related to bodies moving with constant velocity. It says that
if there are no net forces ( that is, friction, gravitational force etc ) ,then a body moving
with uniform velocity will continue its motion forever.
Newton’s first law is also called “the law of inertia”.
Q 03:
Define and explain inertia ?
Ans:
Statement:
“The property of a body due to which it opposes any change in its state of
rest or uniform motion is called inertia”.
Explanation:
Inertia depends upon mass of a body. Hence greater the mass, greater will
be inertia and vice versa.
Examples:
(i)
When a stationary car suddenly starts, we are pushed backwards due to inertia.
(ii)
When a moving car suddenly stops, we are thrown forward due to inertia.
Q 04:
State and explain Newton’s second law ?
Ans:
Statement:
When a force is applied on a body, it produces an acceleration in the direction
of force which is directly proportional to the force and inversely proportional to the mass
of the body”.
Explanation:
Consider a force ‘ F ’ is applied on a body of mass ‘ m ‘ as shown in figure.
m
F
a
Mathematically,
a  F - - - - - - - - - - - - - (1)
and
a
1
m
- - - - - - - - - - - - - - - (2)
Combine equations (1) and (2)
a
or
F
m
ak
F
m
Where ‘ k ’ is a constant, whose value is unity (k = 1).
so
a 1
or
a
or
F
m
F
m
F  ma - - - - - - - - - - - -(3)
Equation (3) is the mathematical form of Newton’s second law of motion.
Q 05:
State and explain Newton’s third law of motion ?
Ans:
Statement:
“For every action, there is always an equal and opposite reaction”.
Explanation:
Consider two bodies ‘ A ‘ and ‘ B ‘ exerting forces on each other during
collision.
on
F AB
is the force exerted by A on B and
F BA
is the force exerted by B
A as shown in the figure.
FAB
FBA
A
B
According to third law,


FAB   FBA
(-ive sign shows opposite direction)
These forces are called action-reaction pair. Action and reaction can never cancel each
other because they act on two different bodies.
Examples:
(i)
A rocket engine expel hot gases downward, which is action. As a result, the rocket moves
upward, which is reaction.
(ii)
Firing a bullet is due to action force of the gun on bullet. Whereas the recoil of the gun is
reaction.
Q 06:
What is weight ?
Ans:
Weight:
“The force with which earth attracts a body towards its center is called weight”.
Mathematically,
W=mg
Where ‘ w ’ represents weight of the body.
Nature and unit of weight:
Since weight is the force acting on the body due to earth so its
nature and unit will be same as that of force. Hence weight is a vector quantity. Its SI unit
is newton (N).
Q 07:
Draw a comparison between mass and weight ?
Ans:
Mass
Weight
1) Mass is a property of matter.
1) Weight depends on the effect of gravity.
2) Mass can never be zero.
2) Weight becomes zero in absence of gravity.
3) Mass of a body is same everywhere.
3) Weight of a body may change due to
variation in gravity.
4) Mass is a scalar quantity.
4) Weight is a vector quantity.
5) Mass may be measured by an ordinary
5) Weight is measured with a spring balance.
balance.
6)
Its SI unit is kilogram.
6)
Its SI unit is newton.
Q 08:
What do you know about linear momentum ?
Ans:
Linear momentum:
“The product of mass and velocity of a body is called its linear
momentum”.
Mathematically,


p  mv
Where ‘ p ’ is the linear momentum of a body.
Nature and unit of momentum:
It is a vector quantity having same direction as that of
velocity. Its SI unit is kg m sec-1
or
N-sec.
Explanation:
Momentum measure the quantity of motion in a body. It depends on mass and
velocity of a body. If a car and a truck are moving with the same speed, the truck will have
greater momentum due to its greater mass.
Similarly, if two cars of same masses are moving with different speeds, then
the car with greater speed will have greater momentum.
Q 09:
What is the relation between force and change in momentum ?
OR
Prove that
F=Δp
?
Δt
Ans:
Relation between force and momentum:
According to Newton’s second law,


F  ma - - - - - - - - - - - -(1)

 v
but a 
t
 
 v f  vi
or
a
- - - - - - - - - (2)
t
  
 v  v f  vi 
Put equation (2) in equation (1)
put
 

v f  vi
F m
t


 mv f  mvi
or
F
t
 
mvi  pi
and
but
so


mv f  p f in above


 p f  pi
F
t



p f  pi  p
 p
F
t
- - - - - - - - - - - -(3)
Equation (3) shows that force is the rate of change of momentum.
Q 10:
Define isolated system ?
Ans:
Isolated system:
“A system which is completely independent of its environment is called
an isolated system”.
Example:
A perfectly insulated thermos flask can be treated as an isolated system.
Q 11:
What does the law of conservation of momentum mean ?
Ans:
Definition:
“In the absence of external force, the total momentum of a system remains
constant”.
OR
“In the absence of external force, the final momentum of a system must be equal to its
initial momentum”.
Explanation:
Since we know that
F
For an isolated system,
p
t
F=0
- - - - - - - - - -(1)
-----------(2)
Put equation (2) in equation (1)
0
p
t
Multiplying ‘ Δ t ‘ on both sides
0  t 
or
or
p
 t
t
0  p
p  0
 p  p f  pi 
or
p f  pi  0
or
p f  pi - - - - - - - - - -(3)
Equation (3) is the mathematical form of law of conservation of momentum.
Q 12:
Define collision ? Explain change in momentum in terms of collision ?
Ans:
Collision:
“A type of interaction in which two bodies physically touch each other is called as
collision”.
Change in momentum and collision:
Consider two bodies of masses m 1 and
moving with velocities u 1 and u 2 respectively as shown.
m2
u1
m1
u2
Before collision
m2
m1
During collision
m2
m1
v1
v2
m2
After collision
After collision both the bodies move in opposite direction with new velocities v 1 and v2.
Now we know that,
Momentum before collision  pi  m1u1  m2u2
Momentum after collision  p f  m1v1  m2 v2
According to the law of conservation
of momentum,
Initial momentum = final momentum
or
pi  p f
m1u1  m2u2  m1v1  m2v2
or
This shows that collisions obey the law of conservation of momentum.
Q 13:
Define explosion ? Explain change in momentum in terms of explosion ?
Ans:
Explosion:
“A type of interaction in which particles of a system move apart from each
other with intense release of energy”.
Change in momentum and explosion:
Consider a gun-bullet system in which,
m b = mass of bullet
And
m g = mass of gun
Before firing, the initial momentum of gun plus bullet is zero.
pi  0 - - - - - - - - - (1)
After firing, both the gun and bullet move in opposite directions with velocities v
and
vb
g
respectively. It can be shown that the final momentum of the gun-bullet system
is also zero.
p f  mb vb  mg v g - - - - - - - - - - - (2)
but
mb vb  mg vg - - - - - - - - - (3)
Put equation (3) in equation (2),
p f  mg vg  mg vg
or
p f  0 - - - - - - - - - -(4)
From equations (1) and (4),
pi  p f
This shows that the process of explosion also obey the law of conservation of momentum.
Q 14:
What is friction ? Explain microscopic basis of friction ?
Ans:
Friction:
“ Friction is a force that opposes the relative motion between two systems in
contact”.
It is denoted by ‘ f ‘
and its SI unit is newton (N). It is always directed in opposite
direction of motion. We perform our routine activities such as writing, walking etc, due to
friction.
Microscopic basis of friction:
It has been observed through the microscopic studies of
surfaces, that all surfaces are composed of microscopic ‘ hills ‘ and
‘ wells ‘. When one
surface is dragged over another, these small up-downs are interlocked, producing friction.
Friction has also its origins at the molecular level. The molecules of both surfaces in contact
exert attractive forces on each other. These inter-molecular forces also play a part in
producing friction.
Note:
Normal force:
“A contact force which is perpendicular to the surface and acts on the object
due to the surface is called normal force”.
In the following figure, F n
is the normal force.
Normal force ( F n )
Surface
Weight ( w )
Q 15:
Explain the types of friction ?
OR
Differentiate between static and kinetic friction. Also derive the co-efficient of static
and kinetic friction ?
Ans:
Types of friction:
There are two types of friction:
I) Static friction
I)
II) Kinetic friction
Static friction:
“The friction force acting on a body when it is at rest is called static
friction”.
Consider pushing a block over a horizontal surface by a force ‘ F ‘ as shown in the
figure. At first, the block does not move due to friction force ‘ f ‘.
Fn
F
f
w
As we increase the applied force ‘ F ‘, the friction force also increase by the same amount.
By gradually increasing ‘ F ‘, the friction force reaches a maximum limit ( f s ) max . After
crossing this limit, the block starts motion.
Co-efficient of static friction:
It has been observed that the maximum value of static friction
is directly proportional to the magnitude of normal force
( f s ) max
Fn.
Mathematically,
 f s max
 f s max   s Fn
or
or
Where
s 
 f s max
Fn
‘ μ s ‘ is a constant of proportionality known as co-efficient of static friction. The
value of
II)
 Fn
μs
depends on the nature of surfaces in contact.
Kinetic friction:
“ The friction force acting on a body during motion is called kinetic
friction”.
It is represented by ‘ f k ‘. If we continue to apply the same force to a body after it
has started motion, it will accelerate. This means that
f k  f s max
Co-efficient of kinetic friction:
It has been assumed that kinetic friction is proportional to the
normal force.
Mathematically,
f k  Fn
f k   k Fn
or
k 
or
fk
Fn
Where ‘ μ k ‘ is known as co-efficient of kinetic friction.
Q 16:
What are advantages and disadvantages of friction ? Also give methods to reduce
friction ?
Ans:
Following are some of the advantages and disadvantages of friction.
Advantages
Disadvantages
1) Friction helps us to walk and run.
1) Due to friction, there is wear and tear of
Objects.
2) Friction helps us in grabbing and holding.
2) It causes heat in machine parts.
3) Due to friction, the objects can be placed.
3) It slows down the motion of objects.
4) A match stick lit up due to friction.
4) Friction losses useful energy in machines.
Methods of reducing friction:
Various methods are used to reduce friction. Some methods are
given below:
1)
Polishing
2)
Application of lubricants ( oil, grease etc )
3)
Using ball bearings.
Q 17:
What is tension ?
Ans:
Tension:
“ The pulling force exerted by a rope, string etc on an object is called tension force”.
Consider an object of mass m is tied to a string as shown in figure. There are two forces
acting on mass m .
T
m
m
w
Its weight acting downward and tension along the string acting upward. If the body is at
rest, the magnitudes of weight and tension are equal. That is,
T=w
Q 18:
What is Atwood’s machine ? Derive the expression for tension and acceleration
in
Atwood’s machine ?
Ans:
Atwood’s machine: “ An arrangement in which two unequal masses are hung vertically
over a friction-less pulley is called Atwood’s machine”.
Calculation of acceleration:
The following figure shows an Atwood machine in which m 1 > m 2 .
T
T
w2
w1
Two forces are acting on m 1 :
I)
Its weight w 1
II)
Tension T in the string acting upward.
Since
acting downward.
m 1 is moving downward, so net force acting on
m1
is given by
Fnet  w1  T
Put
Fnet  m1a
w1  m1 g
and
in above
m1a  m1 g  T
T  m1g  m1a - - - - - - - - - -(1)
or
Similarly two forces, w 2
and
T
are acting on
m2
is moving upward, so net force on it will be
Fnet  T  w2
Put
Fnet  m2 a
and
w2  m2 g
m2 a  T  m2 g
in above
in opposite directions. But as m 2
m2 a  m2 g  T
or
T  m2a  m2 g - - - - - - - - - -(2)
or
Compare equations (1) and (2)
m2 a  m2 g  m1 g  m1a
or
m1a  m2a  m2 g  m1 g
or
m1a  m2a  m1 g  m2 g
or
m1  m2 a  m1  m2 g
Dividing both sides by ‘ m 1 + m 2 ‘
m1  m2
m  m2
a 1
g
m1  m2
m1  m2
a
or
m1  m2  g
m1  m2 
- - - - - - - - - (3)
Which is required.
Calculation of tension:
To calculate tension, put equation (3) in equation (2)
T  m2
Taking
m1  m2
g  m2 g
m1  m2
‘ m 2 g ‘ as common in above equation
 m  m2 
T  m2 g  1
 1
 m1  m2 
Now take
“ m 1 + m 2 “ as LCM inside the brackets
 m  m2  m1  m2 
T  m2 g  1

m1  m2


or
or
 m  m1 
T  m2 g  1

 m1  m2 
 2m1 
T  m2 g 

 m1  m2 
Rearranging the above equation
2m1m2 g
m1  m2
T
Which is required.
Q 19:
What is uniform circular motion ? Give a detailed description of centripetal
acceleration ?
Ans:
Uniform circular motion:
“ The motion of a body along a circular path with constant speed
is called uniform circular motion”.
Examples:
I)
A stone tied to a rope and being swung in circles.
II)
Motion of artificial satellites orbiting the earth at constant speed.
III)
A car turning through a curve in a race track.
Centripetal acceleration:
An object moving along a circular path will have an acceleration
directed towards the center of circular path as shown. Such acceleration is called centripetal
acceleration.
Mathematically,
ac 
Where
v2
- - - - - - - - - (1)
r
a c = centripetal acceleration
r = radius of circular path
r
v
ac
m
The negative sign in equation (1) shows the direction of
a c and
r which are always
opposite as shown in the figure. In uniform circular motion, the magnitude of velocity ( that is,
Speed ) is constant but its direction changes at every instant.
Factors affecting centripetal acceleration:
From equation (1), it is clear that centripetal
acceleration depends on
I)
Square of the velocity
II)
Radius of the circular path.
Q 20:
Ans:
What is centripetal force ? Explain some applications of centripetal force ?
Centripetal force:
“The force which acts on a body moving along a circular path and which
is directed towards the center is called centripetal force”.
Consider a body of mass m moving on a circular path of radius r as shown in the figure.
According to Newton’s second law,
Fc  mac - - - - - - - - - - - (1)
but
ac  
v2
- - - - - (2)
r
r
Put equation (2) in equation (1)
Fc  
mv 2
- - - - - - - (3)
r
Where Fc = centripetal force
r = radius of circular path
Fc
Negative sign in equation (3) shows that Fc and ‘ r ‘
are always opposite in direction.
v
m
Applications of centripetal force:
Some applications of centripetal force are given below:
1)
Banking of roads:
When a car move on a curved road, it bends to slip outward while friction
compels it to move inward. Sufficient centripetal force is required to keep the car on track. To
provide such force, the outer edge of the curved road is made a little higher as shown below.
F n ( normal force)
w ( weight )
ɵ
This elevation adds to the normal force of the vehicle, thus increasing the friction to keep the
car on the road .
2)
Centrifuge:
“A device which separates suspended particles in a liquid by spinning it quickly
around an axle”.
It consists of a wheel which rotates horizontally. Liquids containing the
suspended particles are kept in test tubes which are attached to the wheel. When the wheel
rotates at high speed, the particles aggregate together at the bottom of test tubes. These
particles are then collected and analyzed.
Exercise
Conceptual Questions
Q 01:
Ans:
Why does dust fly off, when a hanging carpet is beaten with a stick ?
The dust fly off due to inertia or Newtons first law of motion.
Explanation:
The dust stuck with the carpet is initially at rest. By beating, the carpet
suddenly starts motion. But the dust particles try to maintain their state of rest due to
inertia. That is why it flies off the carpet.
Q 02:
If your hands are wet and no towel is handy, you can remove some of the excess
water by shaking them. Why does this work ?
Ans:
The excess water is being removed due to inertia.
Explanation:
When we shake our wet hands, we see that some of the water is removed.
It is because of the fact that the water try to remain at rest. By shaking, those water drops
fly off due to inertia.
Q 03:
Why a balloon filled with air move forward , when its air is released ?
Ans:
It is because of reaction force that a balloon filled with air move forward by releasing its
air.
Explanation:
The principle at work, in this case is Newton’s third law. The air released
from the balloon is action. This released air provides the thrust to the balloon due to
which it moves forward.
Q 04:
Why does a hose pipe tends to move backward when the fireman directs a powerful
stream of water towards fire ?
Ans:
It is due to the reaction force that a hose pipe moves backwards when water is released
from it.
Explanation:
The principle applied here is Newton’s third law. When water from a hose
pipe is turned on with full speed. The fireman feels a backward force ( reaction ), which
results due to the ejection of water ( action ) .
Q 05:
Your car is stuck in wet mud. Some students see your predicament and help out by
sitting on the trunk of your car to increase its traction. Why does this help ?
Ans:
It increases the normal force on the car due to which friction is increased. This causes the
tires more sticky to wet mud which helps the car out of the mud.
Explanation:
When the car is stuck in the wet mud and students sit on the trunk. It
increases weight of the car which in turn increases normal force. As a result, friction is
increased due to the formula,
f s  Fn
This force of friction helps the car to get out of the mud.
Q 06:
How does friction help you walk ? Is it kinetic or static friction ?
Ans:
The friction sticks our feet on ground instead of going back by our force that we applied
to move. It is static friction.
Explanation:
When we push the ground with a force (action). As a reaction, the ground
pushes us with the same force. If there were no friction, then instead of walking, we
would go backward or even slip. It is static friction because our feet and the ground are
not moving relative to each other during contact.
Q 07:
The parking brake on a car causes the rear wheels to lock up. What would be the
consequence of applying the parking brake in a car that is in rapid motion ?
Ans:
The likely consequence in such case would be a car out of control resulting in a major
accident.
Explanation:
As the rear wheels of a car locks, they will rotate at their position instead of
moving forward. As a result, the car will be out of control and a serious accident may
happen.
This exercise is being performed by stuntmen. When the stuntmen applies the
hand brake, at high speed, the car rotates 360°.
Q 08:
Ans:
Why is the surface of a convey-er belt made rough ?
The surface of a convey-er belt is made rough to provide enough friction to baggage
carried upon it.
Explanation:
To avoid slipping of the luggage, the convey-er belt is made rough. By
doing so, friction tightly holds the bags on the surface.
Q 09:
Why does a boatman tie the boat to a pillar before allowing the passengers to step
on river bank ?
Ans:
The boatman does so as to stop the boat from motion. Because the water surface offers
less friction.
Explanation:
When the boat comes to the shore, the boatman ties the boat to a pillar
before allowing the passengers to step off the board. This is done to stop the boat from
slipping on water surface, and hence moving away from the shore.
Q 10:
In uniform circular motion, is the velocity constant ? Is the acceleration constant ?
Explain ?
Ans:
In uniform circular motion, the magnitude of velocity is constant but its direction
changes at every instant. This produces the centripetal acceleration which is also
constant.
Explanation:
We can not say that velocity is constant in uniform circular motion
because its direction changes continuously as shown below.
ac
v
However, the centripetal acceleration ( which is produced due to centripetal force ) remains
constant in uniform circular motion. The direction of velocity is tangent to the circle,
whereas, that of acceleration is towards the center of the circle.
Q 11:
You tie a brick to the end of a rope and whirl the brick around you in a horizontal
circle. Describe the path of the brick after you suddenly let go of the rope ?
Ans:
The brick will go out of the circular path because centripetal force seizes and it will fly in a
direction tangent to the circle.
Explanation:
Tension in the rope provides the necessary centripetal force to keep the brick
move in circular path. When we leave the rope, the tension and hence centripetal force
becomes zero. As a result, the brick will move in a direction tangent to the circle as shown in
the figure.
brick
Original
path
Fc
Q 12:
Why is the posted speed for a turn lower than speed limit on most highways ?
Ans:
The maximum speed limit for a turn is lower than normal otherwise the friction force will not
be sufficient to provide necessary centripetal force.
Explanation:
Increasing the speed from a certain value is prohibited in a turn. If we cross the
posted speed, the centripetal force will not be sufficient for the car to take a turn. And the car
will skid away from the road.
Numerical Questions
Problem 01:
A 1580 kg car is traveling with a speed of 15 m sec -1. What is the magnitude of
horizontal net force required to bring the car to a halt in a distance of 50 m ?
Ans:
Given data
Mass of car = m = 1580 kg
Initial velocity = v i = 15 m sec-1
Final velocity = v f = 0
Distance = S = 50 m
Required:
Force = F = ?
Solution:
As
F=ma
----------(1)
But first we have to find

2aS  v f  vi
2
‘a‘
2
Dividing ‘2S’ on both sides
2aS v f  vi

2S
2S
2
v f  vi
2
or
a
2
2
2S
Putting values
0 2  15
250 
2
a
 225
100
or
a
or
a  2.25 m. sec 2 - - - - - - - -(2)
Put equation (2) and value of mass in equation (1)
F  1580   2.25
or
Problem 02:
F  3555 N
Ans.
A bullet of mass 10 g is fired. The bullet takes 0.003 sec to move through the
barrel and leaves with velocity 300 m sec-1. Find the force exerted on the
bullet by the rifle ?
Ans:
Given data
Mass of bullet = m = 10 g
10
kg
1000
or
m
or
m  0.01 kg
Initial velocity of bullet = v i = 0
Final velocity of bullet = v f =
Time =
Required:
300 m sec-1
Δ t = 0.003 sec
Net force = F = ?
Solution:
As
F
p
- - - - - - - (1)
t
but
p  p f  pi
or
p  mv f  mvi
or
p  mv f  vi  - - - - - - - -(2)
Put equation (2) in equation (1)
F
mv f  vi 
t
Putting values
F
0.01300  0
0.003
F
F  1000 N
or
Problem 03:
3
0.003
Ans.
A 2200 kg vehicle traveling at 94 km h-1 ( 26 m sec-1 ) can be stopped in 21 sec
by gently applying brakes. It can be stopped in 3.8 sec if the driver slams on
brakes. What average force is exerted on vehicle in both of these stops ?
Ans:
Given data
Mass of vehicle = m = 2200 kg
Initial velocity = v i = 94 km h-1 = 26 m sec-1
Final velocity = v f = 0
Time to stop gently = Δ t1 = 21 sec
Time to stop abruptly = Δ t2 = 3.8 sec
Required:
Force to stop gently = F 1 = ?
Force to stop abruptly = F 2 = ?
Solution:
As
F
so
F1 
p mv f  vi 

t
t
mv f  vi 
t1
Putting values in above
F1 
F1 
22000  26
21
 57200
21
F1  2723.8 N
similarly
F2 
mv f  vi 
t 2
Again putting values
Ans.
F2 
F2 
22000  26
3.8
 57200
3.8
F2  15052.63 N
Problem 04:
Ans.
You want to move a 500 N crate across a floor. To start the crate moving, you
have to pull with 230 N horizontal force. Once the crate “ breaks loose “ and
starts to move, you can keep it moving at constant velocity with only 200 N.
What are the co-efficient of static and kinetic friction ?
Ans:
Given data
Weight of crate = w = 500 N
Static friction force = f s = 230 N
Kinetic friction force = f k = 200 N
Required:
Co-efficient of static friction = μ s = ?
Co-efficient of kinetic friction = μ k = ?
Solution:
As we know that
f s   s Fn
But F n = w
so
f s  s w
Dividing both sides by ‘ w ‘
f s s w

w
w
or
s 
fs
w
or
s 
230
500
or
s  0.46
Ans.
f k  k Fn
similarly
Problem 05:
 Fn  w
f k  k w
or
or
k 
fk
w
or
k 
200
500
or
k  0.4
Ans.
Two bodies of masses 3kg and 5kg are tied to a string which is passed over a
pulley. If the pulley has no friction, find the acceleration of bodies and
tension in the string ?
Ans:
Given data
m 1 = 3 kg
m 2 = 5 kg
g = 9.8 m sec-2
Required:
acceleration = a = ?
Tension = T = ?
Solution:
a
As
or
a
or
a
m1  m2 g
m1  m2
5  3 9.8
53
2  9.8
8
19.6
8
or
a
or
a  2.45 m. sec 2
Now to find tension in the string,
Ans.
T
T
or
Problem 06:
2m1m2 g
m1  m2
2  5  3  9.8
53
294
8
or
T
or
T  36.75 N
Ans.
Determine the magnitude of centripetal force exerted by the rim of a car
wheel on a 45kg tire. The tire has a 0.480m radius and is rotating at a speed of
30 m sec-1 ?
Ans:
Given data
Mass of tire = m = 45 kg
Radius of tire = r = 0.480 m
Speed of tire = v = 30 m sec-1
Required:
Centripetal force = F c = ?
Solution:
As
Fc 
mv 2
r
or
Fc 
45  30
0.480
or
Fc 
45  900
0.480
or
Fc 
40500
0.480
2
or
Problem 07:
Fc  84375 N
Ans.
A motorcyclist is moving along a circular wooden track of a circus of radius
5m at a speed of 10 m sec-1. If the total mass of motorcycle and rider is 150 kg.
Find the magnitude of centripetal force ?
Ans:
Given data
Radius = r = 5 m
Speed = v = 10 m sec-1
Mass = m = 150 kg
Required:
Centripetal force = F c = ?
Solution:
As
mv 2
Fc 
r
or
Fc 
150  10
5
2
or
or
Fc 
150 100
5
15000
5
Fc  3000 N
or
Problem 08:
Fc 
Ans.
A car of mass 1000 kg is running on a circular motorway interchange near
Sawabi with velocity 80 m sec-1, the radius of circular motorway is 800 m.
How much centripetal force is required ?
Ans:
Given data
Radius = r = 800 m
Velocity = v = 80 m sec-1
Mass = m = 1000 kg
Required:
Solution:
Centripetal force = F c = ?
mv 2
Fc 
r
As
1000  80 1000  6400

800
800
2
or
Fc 
Fc 
or
or
1000  64
8
Fc 
64000
8
or
Fc  8000
or
Fc  8 103 N
Ans.
END
Chapter : 04
Turning effect of forces
Q 01:
What are parallel forces ?
Ans:
Parallel forces:
“The forces whose direction is parallel, even in opposite sense, are said to
be parallel forces”.
Parallel forces are of two types:
1)
Like parallel forces:
“Parallel forces in the same direction are called like parallel”.
Forces F 1 and F 2 shown in figure are like parallel forces.
F2
F1
2)
Unlike parallel forces:
“Parallel forces in the opposite direction are called unlike parallel”.
Forces F and F shown in the figure are unlike parallel.
F
F
Q 02:
Explain addition of forces, in connection with head to tail rule ?
Ans:
Addition of forces:
“The process of obtaining a single force which produce the same effect
as produced by a number of forces acting together is called addition of forces”.
Addition of like parallel forces:
If the forces are like parallel, they are added by simply
adding their magnitudes as shown below.
2N
3N
5N
Addition of unlike parallel forces:
If the forces are unlike parallel, they are added by
subtracting their magnitudes as shown below.
3N
-2N
1N
And
2N
-3N
-1N
Addition of non parallel forces:
Non-parallel forces can be added by head to tail rule.
Suppose we have two non-parallel forces F A and F B making angles θ A and θ B shown
below.
y
y
FB
FA
θA
θB
O
x
O
x
These forces can be added by head to tail rule, according to the following steps :
1)
Draw the first vector F A according to a suitable scale.
2)
Put the tail of second force ( F B ) on the head of first ( F A ).
3)
Combine the tail of first force ( F A ) with the head of second force ( F B ) as shown in the
figure below.
y
FR
FB
FA
O
x
Where ‘F R ‘ is the resultant of ‘F A ‘ and ‘F B ‘ and is given by the following relation.



FR  FA  FB
Q 03:
What is resolution of forces ? Explain how forces can be resolved into rectangular
components ?
Ans:
Resolution of forces:
“The process of splitting a force into two or more forces is called
resolution of forces”.
It is the reverse process of addition of forces. The forces obtained after
resolution are called components of the original force. If components of a force are
perpendicular to each other, then they are called rectangular components.
Process of resolution of forces:
Consider a force ‘ F ’ represented by line OP in x-y axis. To
resolve this force into components, draw two perpendicular lines PQ and PS on x and y axis
respectively as shown in the figure.
y
S
Fx
P
Fy
Fy
θ
O
Fx
Q
x
Where OQ = PS = F x = horizontal component
And
OS = PQ = F y = vertical component
According to head to tail rule,
  
F  Fx  Fy - - - - - - - - - -(1)
Mathematical form of components F x and F y:
Consider triangle OPQ in above figure, in which
cos  
Put
Base
OQ

Hypotenuse OP
OQ  Fx
cos  
or
and
OP  F
Fx
F
Fx  F cos
- - - - - - - - - -(2)
sin  
Similarly
Put
QP  Fy
OP  F
and
sin  
or
Perpendicular QP

Hypotenuse
OP
Fy
F
Fy  F sin 
- - - - - - - - - -(3)
Equations (2) and (3) gives mathematical expressions of the components of force F.
Magnitude and direction of resultant force:
To find magnitude of resultant, use Pythagoras
Theorem which is
Hypotenuse 2  Base 2  Perpendicular 2
- - - - - - - -(4)
In triangl e OPQ, Hypotenuse  OP  F - - - - - (5)
and
Base  OQ  Fx - - - - - -(6)
and
Perpendicu lar  QP  Fy - - - - - -(7)
Put equations (5), (6) and (7) in equation (4)
F 2  Fx  Fy
2
2
Taking square root on both sides
F 2  Fx  Fy
2
or
F  Fx  Fy
2
2
2
- - - - - - - (8)
Equation (8) gives the magnitude of resultant force F.
Now to find the direction of resultant F, use the following
tan  
Perpendicu lar
Base
Put equations (6) and (7) in above equation
tan  
Fy
Fx
 Fy 
 - - - - - - - -(9)
 Fx 
  tan 1 
or
Equation (9) gives the direction of resultant force F.
Q 04:
Give a detailed account of center of mass and center of gravity ?
Ans:
Center of mass:
“A single point at which the whole mass of a body is imagined to be
concentrated
is called center of mass”.
Center of gravity:
“The point at which the whole weight of the body appears to act is called
center of gravity”.
Explanation:
In translatory motion, all points on a body move parallel to each other. But
when a body rotates or vibrates, there is only one point ( i.e, center of mass ) that moves in
the same way that a single particle would move under the influence of same external forces.
The motion of a body can be described as the motion of its center of mass. For example, in
the following figure, the center of mass of a rotating baseball bat follows the same path.
Location of CM or CG:
The center of mass and center of gravity may or may not lie inside the
body. For example, a banana has its CM or CG outside the mass distribution.
Similarly, the position of CM or CG will change if different parts of a
body change their relative position. Such as a stretched leg has its CM inside the body but when
the leg is bend, its CM shifts outside the body.
Q 05:
What is the difference between CM and CG ? How can we determine CM or CG of a
body ?
Ans:
Difference between CM and CG:
The center of mass is often confused with the center of
gravity. The two terms are so similar in many respects that one can use them
interchangeably.
In a uniform gravitational field (such as near the earth surface) the CM and
CG of a body lie at the same point. But if a body lie in a non-uniform gravitational field,
then CM and CG will be at two different locations.
For the purpose of this course, we will
take the CM and CG at the same position.
CM or CG for symmetric objects:
If a body is symmetrical and of uniform composition,
the CM or CG will be located at its geometrical center. For example:
I)
Center of mass of a sphere is at its center.
II)
Center of mass of a square is at the point of intersection of its diagonals.
III)
Center of mass of a circular cylinder lies at the middle point of central axis. ( see the figure
below.
CM
CM
Sphere
Square
CM
Circular cylinder
CM or CG for irregular shape objects:
The CM or CG for irregular objects or objects having
non-uniform density can be determined by hanging it from various positions.
Q 06:
Define moment of force ? Briefly explain ?
Ans:
Moment of force ( Torque ) :
“The turning effect produced in a body about a fixed point
due to applied force is called torque or moment of force”.
Mathematically,
  F  d - - - - - - - -(1)
where   torque
d  moment arm
and F  applied force
Moment arm is defined as, “the perpendicular distance between the axis of rotation and line
of action of force”.
Torque is a vector quantity. Its SI unit from equation (1) is newton - meter ( N m ).
Explanation:
Force can produce rotation in a body, for example, tightening a nut with
spanner as shown in the figure.
Where ‘ d ‘ is the moment arm. Torque cause changes in rotational motion just as force cause
changes in translational motion. It means that torque plays the same role in rotation as force
in translation.
Direction of torque:
Torque may have clockwise or anticlockwise direction depending on
the
direction of rotation of the body.
As a convention, clockwise torque is taken negative while
anticlockwise torque is taken positive.
Q 07:
Write the mathematical description of torque ? Explain the factors on which torque
depends ?
Ans:
Mathematical description:
Torque is given by
  F d
From above equation, following three cases arises.
Case 01:
If force and moment arm are parallel, then torque will be zero. Figure: 1 below
explains this case.
Figure: 1
Case 02:
If force and moment arm are perpendicular to each other, then torque will be
maximum. Figure : 2 below explain this case.
Figure: 2
Case 03:
If force is acting at the axis of rotation, then moment arm (d) and hence the
torque, both will be zero. Figure : 3 below shows this case.
Figure : 3
Factors on which torque depends:
Torque depends on the following three factors.
I)
II)
Applied force:
Torque increases by increasing applied force and vice versa.
Moment arm:
Increasing the moment arm increases torque and vice versa.
III)
Angle between force and moment arm:
Torque also depend on the angle between force F
and moment arm d.
when
F  d , then to rque is maximum.
when F II d , then to rque is zero.
Q 08:
What do you know about couple ? Briefly explain ?
Ans:
Couple:
“Two equal and opposite parallel forces acting along different lines on a body form
a couple”.
Explanation:
Consider two forces of equal magnitude ‘ F ‘ are applied on a rectangular bar
as shown in the figure.
F
A
B
F
The perpendicular distance between lines of action of forces is called the ‘arm of the
couple’. Hence, the distance between points A and B is known as arm of the couple. The
resultant force of a couple is zero but the net torque is non zero. Hence a couple produce
only rotation.
Examples:
(I)
Forces acting on steering wheel of a car form a couple.
(II)
Forces acting on a lug nut wrench to loose nuts of a car wheel also form a couple.
Q 09:
Define equilibrium ? Explain the types of equilibrium ?
Ans:
Equilibrium:
“The state of a body under the action of several forces when there is no
change in translational as well as rotational motion is called equilibrium”
Explanation:
It is clear from above definition that equilibrium prohibits accelerated motion.
This means that a body having acceleration, either in translatory motion or rotatory, will not
be in equilibrium.
Types of equilibrium: There are two types of equilibrium.
(I)
Static equilibrium:
“The type of equilibrium in which a body remains at rest under the
action of several forces is called static equilibrium”.
For example: A brick lying on a floor is
in static equilibrium. The two forces, weight ‘w’
Fn
and normal force ‘F n ‘ cancel out and there
is no acceleration. (see the figure below).
W
Dynamic equilibrium:
“The type of equilibrium in which a body is in uniform translational
or rotational motion is called dynamic equilibrium”.
It is further divided into two types:
(I)
Dynamic translational equilibrium:
“A body moving with uniform linear velocity is said to
be in dynamic translational equilibrium”.
For example: A paratrooper fallinf down with constant velocity.
(II)
Dynamic rotational equilibrium:
“A body rotating uniform angular velocity is in dynamic
rotational equilibrium”.
For example: A CD rotating in CD player with constant angular velocity.
Q 10:
Write the two conditions of equilibrium ?
Ans:
There are two conditions of equilibrium. For a body to be in complete equilibrium, it must
satisfy both of the conditions.
(I)
First condition of equilibrium:
According to first condition, the vector sum of all forces
acting on a body should be zero.
Mathematically,
  

F1  F2  F3        Fn  0
n

F 0
or
i
i 1
(II)
Second condition of equilibrium:
According to this condition, the vector sum of all torques
acting on a body must be zero.
Mathematically,




1   2   3        n  0
n


or
i
0
i 1
Q 11:
Explain the stability of objects with reference to the position of center of mass ?
Ans:
Stability:
“Stability is a measure of how hard it is to disturb equilibrium of a body or
system”.
Explanation:
The two conditions of equilibrium does not tell about the stability of an object.
It is the position of center of mass on which the stability of an object depends. This
knowledge is important in the design of many objects (toys, cars, etc.)
The stability of an object can be increased by
A)
lowering its center of mass
B)
increasing the area of base.
On the basis of stability, equilibrium is divided into three types
discussed below:
(I)
Stable equilibrium:
“If a body regains its previous position after removing the force that
displaced it, then it is in stable equilibrium”.
For example:
A cone lying on the table with its base as shown in the figure. Where C is the
center of mass of undisturbed cone and C
C
C
is that of disturbed cone. C
is above C.
(II)
Unstable equilibrium:
“If a body does not regain its previous position after removing the
force that displaced it, then it is in unstable equilibrium”.
For example: A cone balanced with its tip on a table. It is clear from the figure below
that C is below C.
C
C
(III)
Neutral equilibrium:
“When a body attains a new position similar to its previous
position after removing the force that displaced it, then it is in neutral equilibrium”.
For example:
A cone lying on a table in horizontal position. In this case, the center of
mass neither rises nor fall.
Q 12:
What is the principle of moments ? Explain ?
Ans:
Principle of moments:
“When an object is in equilibrium, the sum of clockwise torques
about a pivot must be equal to the sum of anticlockwise torques about the same pivot”.
Mathematically,


 clockwise   anticlockwise - - - - - - - - - - - (1)
The principle of moments is equivalent to second condition of equilibrium.
For example: In a balanced see saw, the principle of moments is satisfied.
Exercise
Conceptual Questions
Q 01:
Can the rectangular components of a vector be greater than the vector itself ?
Explain ?
Ans:
No, the rectangular components of a vector cannot be greater than the vector itself.
Explanation:
The rectangular components of a vector ‘A’ are given by
A
Ax  Ay
2
2
From the above equation, the rectangular components can only be equal to the vector if the
vector is along the direction of any of the components but can never be greater.
Q 02:
Explain why door handles are not put near hinges ?
Ans:
The door handles are not put near hinges because in such case, the moment arm and hence
the torque will become zero. And no rotation will be produced.
Explanation: Since torque is given by
  F d
From above equation, it is clear that by putting the handles near the hinges, moment arm
will be very small or zero. Hence, there will be no effective torque to rotate the door.
Q 03:
Can a small force ever exert a greater torque than a larger force ? Explain ?
Ans:
A small force may or may not produce a greater torque than a larger force. Because the
torque not only depends on force but also on moment arm and the angle between force and
moment arm.
Explanation:
If we apply a small force on the handle of a door, it will produce a greater
torque than a larger force applied on or near hinges. But if we apply both the forces at the
same point, then the larger force will produce greater torque, because
  F d
Q 04:
Why is it better to use long spanner rather than a short one to loosen a rusty nut ?
Ans:
Increasing length of the spanner will increase the amount of torque which will help to
loose a rusty nut.
Explanation: Since torque is given by
  F d
It is clear from above equation, the long spanner will have greater moment arm than the
short one. Hence, it will generate greater torque to loose a nut as compared to a short
spanner.
Q 05:
Gravitational force on satellite is always directed towards the earth’s center. Does
this force exert torque on the satellite ?
Ans:
No, the gravitational force does not exert any torque on the satellite because, it is
anti-parallel to radius (moment arm) of the orbit.
Explanation: The gravitational pull of earth on a satellite is always anti-parallel (at 180°)
to the radius as shown below.
F
R
v
Hence, there will be no torque due to gravitational force.
Q 06:
Can we have situations in which an object is not in equilibrium, even though the net
force on it is zero ? Give two examples ?
Ans:
Yes, we may have situations in which net force on an object is zero, but the body is not in
equilibrium.
Explanation:
For complete equilibrium, we need both conditions. If net torque acting on a
body is not zero, the body will not be in equilibrium even if there is no net force on it.
Examples:
(1)
Two forces of equal magnitude acting on a square block fixed at its center as shown
below.
-100 N
100 N
(2)
Two forces of equal magnitude ‘F’ acting on a steering wheel of a car as shown.
F
F
Q 07:
Why do tightrope walkers carry a long, narrow rod ?
Ans:
The narrow rod increases the rotational inertia of the walker, which helps in maintaining
stability while walking over the narrow rope.
Explanation:
Carrying a long narrow rod, while walking on a rope does two things:
(1)
It increases the rotational inertia of the walker, thus minimizing his body rotation around
the rope.
(2)
It increases the weight of the walker below the center of gravity or center of mass, which
also helps in increasing stability.
Q 08:
Why does wearing high-heeled shoes sometimes cause lower back pain ?
Ans:
High-heeled shoes push the center of mass a little forward from its normal position. To
bring the CM back to its normal position, the person must lean the shoulders back. This
effort causes fatigue in the muscles.
Explanation:
Normally the CM or CG of a human is about an inch below the navel in the
center of body. By wearing shoes of high heel, the CM or CG gets displaced from its
normal position. It disturbs equilibrium of the body and the person feel lower back pain.
Q 09:
Why is it more difficult to lean backwards ? Explain.
Ans:
It is because the body has to maintain its balance when we lean anyway.
Explanation:
The center of pressure (reaction of ground) must stay within the foot.
Otherwise, we may topple over. The normal position of center of pressure during standing is
about 5 to 6 cm in front of the ankle. It almost never goes behind the ankle, because the foot
is ‘L’ shaped rather than ‘T’ shaped. Greater portion of the foot in front of the ankle is in
contact with the ground, but only a small portion (i.e, the heels) are behind. This is why we
loose our balance when we try to lean backwards.
Q 10:
Can a single force applied to a body change both its translational and rotational
motion ? Explain ?
Ans:
Yes, a force applied away from the center of gravity will change translational and rotational
motion of a body.
Explanation:
A cricket ball bowled by a bowler has combined translational and rotational
motion. When this ball is hit by a batsman, at a point other than CM or CG, we can observe
a change in its translatory and rotatory motion.
Q 11:
Two forces produce same torque. Does it follow that they have the same magnitude ?
Explain ?
Ans:
It is not necessary for two forces of the same magnitude to produce equal torque.
Explanation:
This is because torque also depends on moment arm and the angle between
force and moment arm. i.e,
  F d
Hence two different forces with unequal moment arms could still give the same torque.
Numerical Questions
Problem 01:
To open a door, force of 15 N is applied at 30 ° to the horizontal, find the
horizontal and vertical components ?
Ans:
Given data
Force = F = 15 N
Angle = θ = 30 °
Required:
horizontal component = F x = ?
Vertical component = F y = ?
Solution:
F x = F cos θ
since
Putting values in above
F x = 15 cos 30 °
F x = 15 x 0.866
F x = 12.99 N
since ( cos 30°=0.866)
ANS.
similarly F y = F sin θ
Putting values in above
F y = 15 sin 30 °
F y = 15 x 0.5
F y = 7.5 N
Problem 02:
since ( sin 30°=0.5)
ANS.
A bolt on a car engine needs to be tightened with a torque of 40 N m. You use 25
cm long wrench and pull on one end perpendicularly. How much force do you
have to exert ?
Ans:
Given data
Torque = τ = 40 N m
Moment arm = d = 25 cm
d
or
Required:
25
 0.25m
100
force = F = ?
Solution:
  F d
As
Dividing both sides by ‘d’

d

F d
d
or
F
or
F
or
Problem 03:

d
40
0.25
F  160 N
ANS.
Sana, whose mass is 43 Kg, sits 1.8 m from the center of see saw. Faiz, whose
mass is 52 Kg, wants to balance Sana. How far from the center of see saw
should Faiz sit ?
Ans:
Given data
Mass of Sana = m s = 43 kg
Mass of Faiz = m f = 52 kg
Moment arm for Sana = d s = 1.8 m
Weight of Sana = w s = m s g
Weight of Faiz = w f = m f g
Required:
moment arm of Faiz = d f = ?
Solution:
using the principle of moments
Anticlockwise torque = Clockwise torque --------(1)
From the figure, it is clear that
Anti clockwise torque  w f  d f
and clockwise torque  ws  d s
So equation will become as
w f  d f  ws  d s
or
m f g  d f  ms g  d s
m f  d f  ms  d s
or
Dividing both sides by m f
mf d f
mf

ms d s
mf
df 
or
ms d s
mf
Putting values in above
Problem 04:
df 
43  1.8
52
or
df 
77.4
52
or
d f  1.48 m
or
d f  1.5 m
ANS.
Two kids weighing 300 N and 350 N are sitting at the ends of 6 m long see saw.
Where would a third kid sit so the see saw is in equilibrium in horizontal
position ? Weight of the third kid is 250 N ?
Ans:
Given data
Weight of first kid = w 1 = 300 N
Weight of second kid = w 2 = 350 N
Weight of third kid = w 3 = 250 N
Moment arm of first kid = d 1 = 3 m
Moment arm of second kid = d 2 = 3 m
Required:
Moment arm of third kid = d 3 = ?
Solution:
Let the third kid sit with the first kid. From the figure below, it is clear that w 1
and w 3 produce clockwise torque, and w 2 produce anticlockwise torque.
Applying the principle of moments
Clockwise torque = Anticlockwise torque
w1  d1  w3  d3  w2  d 2
or
or
w3  d3  w2  d 2  w1  d1
Dividing both sides by w 3
w3  d 3 w2  d 2  w1  d1

w3
w3
or
d3 
or
or
or
Problem 05:
d3 
w2  d 2  w1  d1
w3
350  3  300  3
250
d3 
1050  900 150

250
250
d3  0.6 m from the pivot.
ANS.
Two children push on opposite sides of a door. Both push horizontally and
perpendicularly. One child pushes with 20 N at 0.60 m from the hinge and the
second pushes at 0.50 m. What force must the second child exert to keep the
door from moving ? ( Ignore the friction).
Ans:
Given data
Force by first child = F 1 = 20 N
Moment arm for first child = d 1 = 0.60 m
Moment arm for second child = d 2 = ?
Required:
force by second child = F 2 = ?
Solution:
Let F 1 produce anticlockwise torque and F 2 produce clockwise torque as
shown below.
Applying principle of moments
Clockwise torque = Anticlockwise torque
or
F2  d2  F1  d1
Dividing both sides by d 2
F2 d 2 F1d1

d2
d2
F2 
or
or
or
F1d1
d2
F2 
F2  24 N
20  0.6 12

0.5
0.5
ANS.
Problem 06: A construction crane lifts building material of mass 1500 kg. Calculate moment
of force (torque) when the moment arm is 20 m. After lifting the crane arm,
which reduces the moment arm to 12 m. Calculate torque again ?
Ans:
Given data
Mass = m = 1500 kg
Moment arm = d 1 = 20 m
Moment arm = d 2 = 12 m
Required: Torque = τ1 = ?
Torque = τ2 = ?
Solution:
weight of the material = w = mg
Or
w = 1500 x 9.8 = 14700 N
Using
1  F  d1
or
1  w  d1
or
1  14700  20
or
1  294000 Nm
or
 1  2.94 105 Nm
here
F  w
ANS.
Now to find the torque ( τ2 ) when the arm of crane is lifted.
 2  F  d2
or
or
 2  w  d2
again
F  w
 2  14700 12
or
 2  176400 Nm
or
 2  1.76 105 Nm
END
ANS.
Chapter : 05
Gravitation
Q 01: State and explain law of universal gravitation ?
Ans:
Statement:
“Every body in the universe attract every other body with a force which is directly
proportional to the product of their masses and inversely proportional to the square of the
distance between their centers”.
Explanation:
Consider two bodies of masses m 1 and m 2 with their centers separated by
distance r as shown in the figure.
m2
m1
F
F
r
According to law of gravitation,
F  m1m2          (1)
F
1
r2
       (2)
Combine equations (1) and (2),
F
or
Where
F
m1m2
r2
Gm1m2
r2
- - - - - - - - - -(3)
G  6.67 1011 Nm2 kg 2
G is called universal gravitational constant. The force of gravitation is always attractive and it
does not depend on the medium between the bodies. Equation (3) shows that gravitational force
is an inverse square law.
Q 02:
Show that law of gravitation can be treated as an example of Newton’s third law of
motion ?
Ans:
Newton’s 3rd law and law of gravitation:
Consider the following figure. If F12 is the force exerted by
m 1 on m 2 and F21 is the force exerted by m 2 on m 1. Then by Newton’s third law,
F12
m2
m1
F21


F12   F21        (1)
Negative sign in equation (1) shows that both the forces are in opposite directions. Hence, the
magnitude of these two forces is equal, such that
F12  F21
They differ only in direction. They are also called action and reaction forces
because they act on two different bodies.
Q 03:
Using law of gravitation, determine the mass of earth ?
mo
Ans:
Mass of earth:
Consider the following figure in which
rE
m o = mass of object
m E = mass of earth
r E = distance between center of earth and object
mE
r E is also the radius of earth. Now applying the law of gravitation to the figure.
F
Gmo mE
2
rE
- - - - - - - - - (1)
Here the force of earth produces weight of the object such that
F=w
Thus
w=mog
--------(2)
Put equation (2) in equation (1)
mo g 
g
or
Gmo mE
2
rE
GmE
2
rE
2
Multiply rE on both sides of above equation
g  rE 
2
or
GmE
2
 rE
2
rE
g  rE  GmE
2
Dividing both sides by G in above equation
g  rE
GmE

G
G
2
or
g  rE
mE 
G
2
- - - - - - - -(3)
Put g  9.8m sec 2 , rE  6.4 106 m and
G  6.67 1011 Nm2 kg 2 in equation (3)

9.8  6.4 106
mE 
6.67 1011

9.8  40.96 1012
6.67 10 11
or
mE 
or
401.4 1012 1011
mE 
6.67
or
or
2
mE  60.11023
mE  6.011024 Kg.
Which is required mass of the earth.
Q 04:
What do you know about gravitational field and its strength ?
Ans:
Gravitational field:
“The region around a body within which its gravitational force can be
felt by other bodies is called gravitational field”.
It is a vector field because it has direction.
Gravitational field strength:
“The strength of earth’s gravitational field at a certain point is
the gravitational force exerted on a unit mass at that point”.
It is denoted by g and is given by the following relation
F
m
g
The direction of g is same as that of gravitational force F.
Q 05:
Show that weight of an object changes with location ?
Ans:
Variation in weight with location:
The weight of a body depends on the value of g, that is
w=mg
-----------(1)
The value of g also varies by changing altitudes from sea level. By law of gravitation
F
Gmo mE
2
rE
or
mo g 
or
g
Gmo mE
2
rE
 F  w  mo g 
GmE
- - - - - - - -(2)
2
rE
From equation (2), it is clear that value of g depends only on m E and r E. Since the earth is
not uniform. It has mountains and valleys, thus giving a different value of g at different
locations. This in turn will alter the weight of a body.
For example, the weight of a person at Murree will be little less than his weight at Karachi.
Q 06:
Derive and explain the expression for variation of ‘g’ with altitude ?
Ans:
Variation of ‘g’ with altitude:
The value of ‘g’ at the surface of earth is given by
g
GmE
- - - - - - - -(1)
2
rE
Now let a body of mass m o is raised to a height ‘h’ above the earth’s surface as shown in
the figure. Then equation (1) becomes
gh 
GmE
- - - - - - - -(2)
rE  h2
But from equation (1), we have
GmE  grE - - - - - - - - - -(3)
2
mo
h
rE+h
rE
mE
Put equation (3) in equation (2)
2
grE
gh 
- - - - - - - -(4)
rE  h2
Hence, the value of
Q 07:
‘g’ decreases with increasing height.
Derive an expression for the orbital speed of an artificial satellite ?
Ans:
Orbital speed of satellite:
Consider the following figure in which
m s = mass of satellite
h = height of satellite
v = orbital speed
r = distance between earth’s center and satellite
ms
h
v
r
rE
mE
r E = radius of earth
m E = mass of earth
Now the centripetal and gravitational forces on the satellite are given by following
equations (1) and (2) respectively
Fc 
Fg 
and
or
Gms mE
- - - - - - - - - -(2)
r2
Fc  Fg
But since
so
ms v 2
- - - - - - - - - (1)
r
ms v 2 Gms mE

r
r2
v2 
GmE
r
Taking square root of both sides, we get
GmE
r
v
But from the figure, r  rE  h
v
so
GmE
rE  h
- - - - - - - - - - - (3)
Equation (3) gives the orbital speed of a satellite. It shows that farther satellites will revolve
slowly than nearer satellites.
Exercise
Conceptual Questions
Q 01:
If there is an attractive force between all objects, why don’t we feel ourselves
gravitating toward nearby massive buildings ?
Ans:
This is because the mass of the buildings is very small to feel its gravitation.
Explanation:
F
Since
Gm1m2
- - - - - -(1)
r2
Now consider a building of mass ‘m 1’ having value of 100,000 kg, and a person of mass
‘m 2’ having value of 60 kg. Let the person is standing 1m away from the building. Putting
all of these values in equation (1), such that
F
6.67 10 11 105  60
12
By simplifying the above equation, we get
F  400.2 10 6
or
F  4.0 104 N
This force is too small to be felt. Hence, the gravitational force can only be felt in case of
astronomical masses.
Q 02:
Does the sun exert a larger force on earth than that exerted on the sun by earth ?
Explain.
Ans:
No, both the sun and earth exert the same amount of force on each other.
Explanation:
The gravitational forces exerted by the sun and earth on each other are
action and reaction. From Newton’s third law, action and reaction have equal magnitudes.
They differ only in direction. This is why the gravitational force between earth and sun
are equal in magnitude.
Q 03:
What is the importance of gravitational constant G ? Why is it difficult to calculate ?
Ans:
The gravitational constant is the constant of proportionality in the law of universal
gravitation. Its value is same everywhere in the universe.
Its value is very small, that is 6.67 x 10 -11 N m2 kg-2. So a very sensitive device is required
to calculate it. That is why it is difficult to calculate in normal conditions.
Q 04:
If earth somehow expanded to a larger radius, with no change in mass, how would
your weight be affected ? How would it be affected if earth instead shrunk ?
Ans:
Increasing the radius of earth will decrease our weight while decreasing its radius will
increase our weight.
Explanation:
As
and
w=mg
g
---------(1)
GmE
2
rE
So equation (1) becomes
w
GmmE
2
rE
Thus above equation shows that weight varies inversely with the square of radius. So if
earth expands, weight of a body will decrease. Conversely if the earth get shrunk, weight of
a body will increase.
Q 05:
What would happen to your weight on earth, if the mass of earth is doubled, but its
radius stays the same ?
Ans:
By doubling the mass of earth while its radius remains constant, the weight of a body will
also become double.
Explanation:
As weight of a body is given by
w
GmmE
2
rE
Put mE  2mE in above equation, the new weight becomes
w 
or
or
Gm2mE
2
rE
 GmmE
w  2
2
 rE




w  2w
It is clear from above equation that doubling the mass of earth will double the weight of a
body if radius of the earth is kept constant.
Q 06:
Ans:
Why light and heavier objects fall at the same rate towards the earth ?
This is because the value of ‘g’ is independent of the mass of falling body.
Explanation:
The value of ‘g’ at the surface of earth is given by
g
GmE
2
rE
From above equation, it is clear that ‘g’ only depends on the mass and radius of earth. It
does not depend on the mass of falling object. In the absence of air, a freely falling body is
only acted by gravity, which is the same ( that is 9.8 m sec -2 ) for every body.
This means that a ball and a feather will fall at the same rate if there is no air resistance.
Q 07:
The value of g changes with location on earth, however we take the same value of g as
9.8 m sec-2 for ordinary calculations. Why ?
Ans:
The value of ‘g’ is an approximation that is only valid at or near the earth’s surface. If we
go a few miles up, the change in the value of ‘g’ is not significant. We can use it in
ordinary calculations. It is only when we go very far from the earth’s surface, then we use
other values of ‘g’.
Q 08:
Moon is attracted by the earth, why it does not fall on earth ?
Ans:
The moon does not fall on earth due to a balance between the centripetal and centrifugal
forces on it.
Explanation: The moon revolves around the earth in circular orbit with a tangential
velocity ‘v’ as shown in figure.
Tangential velocity (v)
Earth
F
Moon
Due to this velocity, the moon always try to move away from the orbit. But the
gravitational force keep it in the orbit. The moon is always under the action of centripetal
and centrifugal forces. The centrifugal force keeps the moon from falling on earth.
Q 09:
Why for the same height larger and smaller satellites must have the same orbital
speed ?
Ans:
This is because the speed of satellite depends on the mass of earth and distance from the
center of earth to the satellite. And it does not depend on the mass of satellite.
Explanation:
The speed of a satellite is given by
v
GmE
GmE

r
rE  h
It is clear from above equation that speed of satellite does not depend on its mass. Hence,
two satellites of different masses at the same altitude (height) will have the same speed.
Numerical Questions
Problem 01:
Pluto’s moon Charon is unusually large considering Pluto’s size, giving them
the character of a double planet. Their masses are 1.25 x 1022 kg and 1.9 x 1021
kg, and their average distance from one another is 1.96 x 10 4 km. What is the
gravitational force between them ?
Ans:
Given data
Mass of Pluto = m p = 1.25 x 1022 kg
Mass of Charon = m c = 1.9 x 1021 kg
Distance between Pluto and Charon = r = 1.96 x 10 4 km
Or
r = 1.96 x 104 x 103 m
r = 1.96 x 107 m
Or
Required: F = ?
Solution:
F
As
Put m 1 = m p
and
F
or F 
Problem 02:
m 2 = m c in above equation
Gmp mc
r2
6.67 1011 1.25 1022 1.9 1021
or
or
Gm1m2
r2
1.96 10 
7 2
F
15.84 1032
3.84 1014
F  4.12 1018 N
ANS.
The mass of Mars is 6.4 x 1023 kg and having radius of 3.4 x 106 m. Calculate the
gravitational field strength ( g ) on Mars surface ?
Ans:
Given data
Mass of Mars = M = 6.4 x 1023 kg
Radius of Mars = R = 3.4 x 106 m
G = 6.67 x 10-11 N m2 kg-2
Required:
Gravitational field strength = g = ?
Solution:
As
g
GmE
2
rE
In case of Mars, the above equation becomes
g mars 
GM
R2
g mars 
or
g mars 
or
3.4 10 
6 2
42.68 1012
11.56 1012
g mars  3.69 m sec 2
or
Problem 03:
6.67 1011  6.4 1023
ANS.
Titan is the largest moon of Saturn and the only moon in the solar system
known to have a substantial atmosphere. Find the acceleration due to gravity
on Titan’s surface, given that its mass is 1.35 x 1018 kg and its radius is 2570
km ?
Ans:
Given data
Mass of Titan = m T = 1.35 x 1018 Kg
Radius of Titan = r T = 2570 km
Or
r T = 2570 x 1000 m
Or
r T = 2570000 m
Or
r T = 2.57 x 106 m
G = 6.67 x 10-11 N m2 kg-2
Required: Gravitational acceleration on Titan = g T = ?
Solution:
As
g
GmE
2
rE
In case of Titan, the above equation becomes
gT 
or
or
gT 
gT 
GmT
2
rT
6.67 1011 1.35 1018
2.57 10 
9.0045 107
6.6049 1012
6 2
or
or
Problem 04:
gT  1.36 10712
gT  1.36 105 m sec 2
ANS.
At which altitude above Earth’s surface would the gravitational acceleration
be 4.9 m sec-2 ?
Ans:
Given data
Gravitational acceleration = g h = ?
Mass of earth = m E = 6 x 1024 kg
Radius of earth = r E = 6.4 x 106 m
G = 6.67 x 10-11 N m2 kg-2
Required:
Altitude = h = ?
Solution:
2
As
gh 
grE
rE  h2
Multiply rE  h  on both sides
2
2
grE
2
g h  rE  h  
 rE  h 
2
rE  h
2
or gh  rE  h  grE
2
2
Divide ‘g h ‘ on both sides
g h  rE  h 
gr
 E
gh
gh
2
or
or
rE  h 2  grE
gh
Taking square root on both sides
2
2
rE  h 2
2
2
grE
gh
rE  h 
or
grE
gh

2
h
or
grE
 rE
gh
Putting values in above
9.8  6.4 106 
 6.4 106 
4.9
2
h
9.8  40.96 1012
h
 6.4 106
4.9

or


h  81.92 1012  6.4 106
or
Problem 05:
401.408 1012
 6.4 106
4.9
h
or
 81.92 







or
h
1012  6.4 106
or
2
h   9.05  106   6.4 106


or
h  9.05 106  6.4 106
 

 
or
h  9.05  6.4106
or
h  2.65 106 m


ANS.
Assume that a satellite orbits Earth 225 km above its surface. Given that the
mass of Earth is 6 x 1024 kg and its radius is 6.4 x 106 m. What is the satellite’s
orbital speed ?
Ans:
Given data
Height of satellite = h = 225 km
Or
h = 225 x 1000 m
Or
h = 225 000 m
Or
h = 0.225 x 106 m
Mass of earth = m E = 6 x 1024 kg
Radius of earth = r E = 6.4 x 106 m
G = 6.67 x 10-11 N m2 kg-2
Required:
Orbital speed = v = ?
Solution:
v
As
or
or
v
v
v
or
or
or
or
or
GmE
rE  h
6.67 1011  6 1024
6.4 106  0.225 106
40.02 1013
6.4  0.225106
40.02 1013
6.625 106
v  6.040 10136
v  6.040 107
v  60.40 106
v  60.40  106
 
2
or
v  7.77  103
or
v  7.77 103 m. sec 1
But 103 m  1km
so
Problem 06:
v  7.77 km. sec 1
ANS.
The distance from center of earth to center of moon is 3.8 x 10 8 m. Mass of
earth is 6 x 1024 kg. What is the orbital speed of moon ?
Ans:
Given data
Distance between centers of moon and earth = r = r E + h
r = r E + h = 3.8 x 108 m
Or
Mass of earth = m E = 6 x 1024 kg
G = 6.67 x 10-11 N m2 kg-2
Required:
Orbital speed of moon = v = ?
Solution:
v
As
or
v
v
or
GmE
GmE

rE  h
r
6.67 1011  6 1024
3.8 108
40.02 1013
3.8 108
or
v  10.53 10138
or
v  10.53 105
or
v  1.053 106
or
v  1.053  106
 
2
or
v  1.02  103
or
v  1.02 103 m. sec 1
But 103 m  1km
so
Problem 07:
v  1.02 km. sec 1
ANS.
The Hubble space telescope orbits the earth ( m E = 6 x 1024 kg ) with an orbital
speed of 7.6 x 103 m sec-1. Calculate its altitude above Earth’s surface ?
Ans:
Given data
Orbital speed of Hubble = v = 7.6 x 10 3 m sec-1
Mass of earth = m E = 6 x 1024 kg
Radius of earth = r E = 6.4 x 106 m
G = 6.67 x 10-11 N m2 kg -2
Required:
Altitude of Hubble = h = ?
Solution:
GmE
rE  h
v
As
Squaring both sides
 GmE 

v 2  

r

h
 E

v2 
2
GmE
rE  h
By cross multiplication, we get
rE  hv 2  GmE
Dividing ‘ v2 ‘ on both sides
rE  hv 2

GmE
v2
or
rE  h 
GmE
v2
or
h
v
or
or
2
h
GmE
 rE
v2
6.67 1011  6 1024
7.6 10 
3 2
40.02 1013
h
 6.4 106
6
57.76 10
or
h  0.692 10136  6.4 106
or
h  0.692 107  6.4 106
 6.4 106
h  6.92 106  6.4 106
or
or
h  6.92  6.4106
or
h  0.52 106
or
h  5.2 105 m
But 1m 
1
km
103
5.2 105
km
103
so
h
or
h  5.2 105 10-3 km
or
h  5.2 1053 km
or
h  5.2 10 2 km
or
h  5.2 100 km
or
h  520 km
END
ANS.