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Force and laws of motion class 9 notes, CBSE Study Material

Important topics covered:


What is force?

Effect of force:

What are balanced and unbalanced forces?

Balanced force:

Unbalanced force:

Are there any other type of forces?

There are many forces which exist in nature:

What is motion?

Laws of motion:

Galileo’s observation on motion:

Newton’s law of motion:

Newton after observing Galelio’s studies and the research and expanding it, he proposed three laws of motion. These laws of motion resulted in the foundation of classical mechanics. These laws explains the relationship between body and the force acting upon it.

What are the three laws of motion?

Newton gave us three laws of motion. They are as follows:

Newton’s first law of motion (law of inertia):

Inertia and Mass:

Application of Newton’s First law of motion in everyday life:

Newton’s Second law of motion:

Application of Newton’s  second law of motion:

Mathematical formulation of Second law of motion:

Let the mass of any object be m and its initial velocity is u in a straight line. After time t with a constant acceleration a the object reaches to final velocity v.

Newton’s third law of Motion:

Applications of Newton’s Third law:

What is the law of conservation of momentum?

Conservation of momentum formula:

Conservation of momentum Example (class 9):

Q. How to find the Rate of Change of Momentum?

We can find the rate of change of momentum by simply applying Newton’s second law of motion. According to Newton’s second law of motion the rate of change of momentum is directly proportional to the applied force. 

Force and laws of motion examples:

Q. A Truck of mass 1200 kg is moving along a straight line with a uniform velocity of 90 km/h. Its velocity is reduced to 18 km/h in 4 s by an unbalanced external force. Calculate the acceleration and change in momentum.

Solution: Mass of the truck, m = 1200 kg 
Initial velocity (u) = 90 km/h = 25 m/s 
Final velocity (v) = 18 km/h = 5 m/s 
Time taken, t = 4 s 
According to the first equation of motion:
v = u + at
5 = 25 + a (4) 
a = – 5 m/s2 
Negative sign indicates that its a retarding motion i.e. velocity is decreasing. 
Change in momentum = mv – mu = m (v – u)
= 1200 (5 – 25) = – 24000 kg m s -1 
Force = Mass × Acceleration = 1200 × – 5 = – 6000 N 
Acceleration of the truck = – 5 m/s
Change in momentum of the truck = – 24000 kg m s -2

Q. why some of the leaves may get detached from a tree if we vigorously shake its branch?

Solution: When we shake its branches vigorously some leaves get detached because when the branches of a tree are shaken, it moves to and fro, but leaves have tendency to remain at rest. This is because the inertia of the leaves tend to resist the to and fro motion.

Q. How much momentum will a rod of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be 10 m s-2?

Solution:
Mass of the rod, m = 10 kg
Distance covered by the rod = 80 cm = 0.8 m
Acceleration (a) = 10 m/s2
Initial velocity of the rod, u = 0
Final velocity of the rod = v
According to the third equation of motion:
v2 = u2 + 2as
v2= 0 + 2 (10) 0.8
v= 4 m/s
Hence, the momentum with which the rod hits the floor is  = mv
= 40 kg m s-1

Q. A bike of mass 50 kg has a momentum of 1000 kg.m/s. Calculate the velocity of the bike.

Solution: Mass of the body, m = 50 kg
 Momentum p = 1000 kg.m/s
p = m x v
v= \(\frac{p}{m}\) = \(\frac{1000}{50}\) = 20 m/s
Hence, the velocity of bike = 20 m/s

Q. Why does a gunman get a jerk on firing a bullet?

Solution: When a bullet is fired from a gun, the force pushing the bullet forward is equal to the force sending the gun back but due to high mass of the gun, it moves only a little distance backwards giving a jerk to the gunman.

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