Chapter Force and Laws of Motion for Class 9th

2. Introduction
In our daily life, we find that an effort is required to move a body at rest or to stop a moving body.
MUSCULAR EFFORT
FORCE

3. What is Force?
An external effort(Push, Pull and Hit) which may move a body at rest or stop a moving body or change the
speed of a body or change the direction of a moving body or change its shape and size.

4. Effects of Force
• Change states of motion (rest, speed up
or slow down)
• Change the direction of a moving body
• Change the shape and size of a body

5. Effects
of Force
Change in
Shape
Change in
direction of
moving object
Stops a moving
object or
makes resting
object in
motion
Change in
speed of the
moving object

7. Newton’s Laws of Motion
Third
Law of
Motion
Second
Law of
Motion
First
Law of
motion

8. Newton’s First Law of Motion
Newton's first law of motion is often stated as
“An object at rest stays at rest and an object in motion stays in motion
with the same speed and in the same direction unless acted upon by an
unbalanced force.”
According to Newton's first law, an object in motion continues in
motion with the same speed and in the same direction unless acted upon
by an unbalanced force. It is the natural tendency of objects to keep on
doing what they're doing. All objects resist changes in their state of
motion. In the absence of an unbalanced force, an object in motion will
maintain its state of motion. This is often called the law of inertia.
What would happen if the passengers were not wearing the seat belt?
The use of the seat belt assures that the forces necessary for accelerated and decelerated motion exist.
Yet, if the seat belt is not used, the passengers are more likely to maintain its state of motion. The
passengers would likely be propelled from the car and be hurled into the air.

9. Galileo and the Concept of Inertia
Galileo, a premier scientist in the
seventeenth century, developed the
concept of inertia. Galileo
reasoned that moving objects
eventually stop because of a force
called friction.
In experiments using a pair of inclined planes
facing each other, Galileo observed that a ball
would roll down one plane and up the opposite
plane to approximately the same height.
Galileo reasoned that any difference
between initial and final heights was due to
the presence of friction. Galileo postulated
that if friction could be entirely eliminated,
then the ball would reach exactly the same
height.
Galileo further observed that
regardless of the angle at which the
planes were oriented, the final
height was almost always equal to
the initial height. If the slope of the
opposite incline were reduced, then
the ball would roll a further
distance in order to reach that
original height.
if the opposite incline were elevated
at nearly a ZERO degree angle, then
the ball would roll almost forever in
an effort to reach the original height.

10. Newton’s Second Law of Motion
According to Newton, an object will only accelerate if there is a net or unbalanced force acting upon it. The
presence of an unbalanced force will accelerate an object - changing its speed, its direction, or both its speed
and direction.
Objects at equilibrium (the condition in which all forces balance) will not accelerate.
Newton's second law of motion can be formally
stated as follows:
The acceleration of an object as produced by a net
force is directly proportional to the magnitude of
the net force, in the same direction as the net force,
and inversely proportional to the mass of the
object.
a = Fnet / m
The net force is equated to the product of the mass times the acceleration.
Fnet = m x a
One Newton is defined as the amount of force required to give a 1-kg mass an acceleration of 1
m/s2.

11. Practice Time!!!!!!
5 m/s2
10 m/s2
5 m/s2
10 N
1 kg

12. Q: A truck hauls a car cross-country. The truck's mass is 4.00x103 kg and the car's mass is
1.60x103 kg. If the force of propulsion resulting from the truck's turning wheels is 2.50x104 N,
then determine the acceleration of the car (or the truck) and the force at which the truck pulls
upon the car. Assume negligible air resistance forces.
For the system: Fnet = 2.50x104 N and msystem = 5.60x103 kg. So
a = Fnet/m = (2.50x104 N) / (5.60x103 kg) = 4.4643 m/s2
For the individual object analysis on the car: m = 1.60x103 kg and a = 4.46 m/s2 (from above); so the Fnet is m
x a or 7143 N.

13. Newton’s Third Law of Motion
According to Newton, whenever objects A and B interact with each other, they exert forces upon
each other.
When you sit in your chair, your body exerts a downward force
on the chair and the chair exerts an upward force on your body.
There are two forces resulting from this interaction - a force on
the chair and a force on your body. These two forces are called
action and reaction forces and are the subject of Newton's third
law of motion.
Newton's third law is:
For every action, there is an equal and opposite reaction.
Faction = - Freaction

14. Examples of Newton’s Third Law
The baseball forces the bat to the left; the bat forces the ball to
the right. Together, these two forces exerted upon two different
objects form the action-reaction force pair.
The ball pushes the hand to the leftward and the hand pushes the
baseball rightward.
The air filled inside the balloon exert force on the balloon outward
and the Balloon wall pushes enclosed air particles inwards.
Fball-on-bat = - Fbat-on-ball
Fball-on-hand = - Fhand-on-ball
Fair-on-balloon = - Fballoon-on-air

15. Momentum
All objects have mass; so if an object is
moving, then it has momentum - it has
its mass in motion.
The standard metric unit of momentum
is the kg m/s
Momentum is a vector quantity

16. Q: Calculate the momentum of a bullet of 25 g when it is fired from a gum with a velocity of 100m/s.
Velocity of the bullet (v) = 100m/s
Mass of the bullet (m) = 25 g = 25/1000 kg = 0.025kg
Q: A vehicle is running with a velocity of 5m/s. If the momentum of the vehicle is 5000 kg m/s, what is its
mass?
Momentum (p) = 5000 kg m/s
Velocity (v) = 5m/s
p = m x v = 0.025 x 100 = 2.5 kg m/s
p= mx v or
m= p / v = 5000 / 5 = 1000 kg
Q: A bullet of 25 g is when fired from a piston gets a momentum of 50 kg m/s. Calculate the velocity of
bullet.
Momentum (m) = 50kg m/s
Mass (m) = 25 g = 25/1000 kg = 0.025 kg
Momentum (p) = Mass (m) x Velocity (v)
p / m = v
50 / 0.025 = v
2000 m/s = v

17. Law of Conservation of
Momentum
The states that -
“The total momentum of a collection of objects (a system) is conserved - that is, the total
amount of momentum is a constant or unchanging value.”
The animation below portrays the collision between a 1.0-kg cart and a 2-kg dropped brick. It will be assumed that
there are no net external forces acting upon the two objects involved in the collision. The only net force acting upon
the two objects (the cart and the dropped brick) are internal forces - the force of friction between the cart and the
dropped brick. The before- and after-collision velocities and momentum are shown in the data tables.
Before the collision, the momentum of the cart is 60 kg cm/s and the momentum of the dropped brick is 0 kg
cm/s; the total system momentum is 60 kg cm/s. After the collision, the momentum of the cart is 20.0 kg cm/s
and the momentum of the dropped brick is 40.0 kg cm/s; the total system momentum is 60 kg cm/s.

18. Mathematical Expression for Law of
Conservation of Momentum
Consider two objects A and B having masses M1
and M2 moving with velocities V1 and V2
respectively.
Force on B by A = FAB
Force on A by B = FBA
A B
B
B
A
A Momentum of the object A = M1U1
Momentum of the object B = M2U2
Before Collision
On Collision
Force of objects A and B on each
other,
FAB = - FBA
After Collision
Momentum of the object A = M1U1
Momentum of the object B = M2U2

19. Acceleration of object A before and after collision,
a1 = v1 – u1 / t
Acceleration of object B before and after collision,
a = v2 – u2 / t
As we know,
FAB = - FBA
M1 a1 = - M2 a2
M1 (v1 – u1 / t) = - M2 (v2 – u2 / t)
M1 v1 – M1u1 = - M2 v2 + M2u2
__________ ___________
t t
M1 v1 – M1u1 = - M2 v2 + M2u2
M1 v1 + M2 v2 = M1v1 + M2u2
Mathematical Expression for Law of
Conservation of Momentum

20. Examples of Conservation of
Momentum

21. Examples of Conservation of
Momentum

22. Practice This !!!!!
Q: Find the recoil velocity of a gun having mass equal to 5 kg, if a bullet of 25gm acquires the velocity of
500m/s after firing from the gun.
Mass of bullet (m1) = 25 gm = 0.025 kg
Velocity of bullet before firing (u1) = 0
Velocity of bullet after firing (v1) = 500 m/s
Mass of gun (m2) = 5 kg
Velocity of gun before firing, (u2) = 0
Velocity of gun after firing = ?
Q: A bullet of 5 gm is fired from a pistol of 1.5 kg. If the recoil velocity of pistol is 1.5 m/s, find the velocity
of bullet.
Mass of bullet, m1 = 5 gm = 5/1000 kg = 0.005 kg
Mass of pistol, m2 = 1.5 kg
Recoil velocity of pistol v2 = 1.5 m/s
Velocity of bullet v1 =?