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NCERT Most Important Questions For Class-9 Chapter-9 Force and Laws of Motion (Physics)

The start of class 9 marks the beginning of the foundation for class 11 and class 12. It is very important study the basics in order to understand each and every chapter properly. In this page, we have provided all the important question for cbse class 9 that could be asked in the examination. Students also need to study the ncert solutions for class 9 in order to gain more knowledge and understanding the lessons. Questions and Answers are way to learn the new things in a proper way. NCERT textbooks downloads for class 9 in pdf are also available for the students if they need more help. By downloading these books, they can study from it. Our experts also prepared revision notes for class 9 so that students should see the details of each and every chapters. Class 9 important questions are the best to revise all the chapters in the best way.

1. Give one point of difference between balanced and unbalanced forces.
Answer
When forces acting on a body from all sides are equal, they cancel effect of each other and are known as balanced forces. On the other hand, when forces acting on a body are not equal/do not cancel each other are called unbalanced forces.

2.Mass of a body is doubled. How does its acceleration change under a given force?
Answer
Acceleration becomes half.

3.Mention any two kinds of changes that can be brought about in a body by force.
Answer
Change in speed/change of direction/change of shape.

4.State the SI unit of pressure. Mention the unit which we use to measure pressure exerted by a gas. What do you understand by normal atmospheric pressure?

Answer

Pascal Atmosphere (atm)
Atmospheric pressure at sea level = 1 atm

5.Define SI unit of force. A force of 2N acting on a body changes its velocity uniformly from 2 m/s to 5m/s in 10s. Calculate the mass of the body.

Answer

6.Derive Newton’s first law of motion from the mathematical expression of the second law of motion.

Answer

Newton’s first law states that a body stays at rest if it is at rest and moves with a constant velocity unit if a net force is applied on it. Newton’s second law states that the net force applied on the body is equal to the rate of change in its momentum.
F= ma
F = m(v-u)/t
Ft = mv-mu

That is, when F = 0, v = u for whatever time, t is taken. This means that the object will continue moving with uniform velocity, u throughout the time, t. If u is zero than v will also be zero, i.e., object will remain at rest.
 
7.Why is it easier to stop a tennis ball in comparison to a cricket ball moving with the same speed?

Answer

Tennis ball is lighter (less mass) than a cricket ball. Tennis ball moving with same speed has less momentum (mass × velocity) than a cricket ball. It is easier to stop tennis ball having less momentum.
 
8.

What type of force is acting in the cases given above ?

Answer

(i) Accelerating unbalanced force.
(ii) No force.
(iii) Retarding unbalanced force.
 
9.What are the effects of the following on inertia of a body?
(i) If force is doubled
(ii) If density is halved
(iii) If volume is reduced to one third.

Answer
 
(i) No effect.
(ii) No effect.
(iii) No effect.
Because inertia depends on mass of body only
 
10. What is meant by ‘inertia’? What are different types of inertia? Give two examples in each case.

Answer

Inability of the body to change by itself its state of rest or state of uniform motion is called inertia.
Types: Inertia of rest: e.g. :
(i) When a card is flicked with a finger the coin placed over it falls in the tumbler.
(ii) Only the carom coin at the bottom of a pile is removed when a fast moving carom striker hits it.
Inertia of motion: e.g. :
(i) When a moving bus stops suddenly, the luggage might slide towards the front side of the bus and fall.
(ii) We tend to fall forward when a bus suddenly stops.
 
11.(i) Define momentum. Write its S.I. unit.
(ii) How much momentum will an object of mass 10 kg transfer to the floor, if it falls from a height of 5 m (g =10 m/s2).
(iii) Explain how a karate player can break a pile of tiles with a single blow of his hand.

Answer
 
(i) Momentum is the product of mass and velocity.
SI unit of momentum is – kg m/s.
(ii) v2=u2+2gh
v2= (0)2+2(10)(5)
v2=100
∴v=10 m/s

(ii) Momentum=m×v
=10×10=100 kg m/s

(iii) The karate player strikes the pile of tiles with his hand very fast. In doing so, the large momentum of fast moving hand is reduced to zero in a very short time. This exerts a very large force on the pile of tiles which is sufficient to break them.
 
12.
 (i) Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 m/s and 1 m/s respectively. They collide and after the collision, the first object moves at a velocity of 1.67 m/s. Determine the velocity of the second object.
(ii) If a man jumps out from a boat, the boat moves backward. Why?

Answer
 
Let
(i) m1=100=0.1 kg
m2=200 g=0.2 kg
u1=2 m/s
u2=1m/s
v1=1.67 m/s

According to law of conservation of momentum
m1u1+m2u2= m1v1+ m2v2
0.1×2 + 0.2×1 = 0.1 ×1.67 + 0.2×v2
0.2 + 0.2 = 0.167 + 0.2 v2
v2 =1.165 m/s

(ii) It is based on Newton’s third law of motion. As boat is floating and is not fixed, so it moves backward.
 
13.Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.

Answer

Some leaves of a tree get detached when we shake its branches vigorously because branches comes in motion while the leaves tend to remain at rest due to inertia of rest.
 
14.Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest?

Answer

In a moving bus, a passenger moves with the bus due to inertia of motion. As the driver applies brakes, the bus comes to rest. But, the passenger tries to maintain to inertia of motion. As a result, a forward force is exerted on him.
Similarly, the passenger tends to fall backwards when the bus accelerates from rest because when the bus accelerates, the inertia of rest of the passenger tends to oppose the forward motion of the bus. Hence, the passenger tends to fall backwards when the bus accelerates forward.
 
 
Physics in Animation: The Law of Inertia - Animator Island
When Bus Suddenly accelerate
Physics in Animation: The Law of Inertia - Animator Island
When Bus Suddenly Stop

15.From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 m s−1. Calculate the initial recoil velocity of the rifle.

Answer

Mass of the rifle, m1= 4 kg

Mass of the bullet, m2= 50g= 0.05 kg
Recoil velocity of the rifle= v1

Bullet is fired with an initial velocity, v2= 35m/s
Initially, the rifle is at rest.
Thus, its initial velocity, v= 0
Total initial momentum of the rifle and bullet system= (m1+m2)v= 0

Total momentum of the rifle and bullet system after firing:

= m1v1 + m2v2 = 0.05 × 35 = 4v1 + 1.75

According to the law of conservation of momentum:
Total momentum after the firing = Total momentum before the firing  4v1 + 1.75= 0
v1 = -1.75/4 = -0.4375 m/s
The negative sign indicates that the rifle recoils backwards with a velocity of 0.4375 m/s.

16.Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 ms−1 and 1 ms−1, respectively. They collide and after the collision, the first object moves at a velocity of 1.67 ms−1. Determine the velocity of the second object.

Answer

Mass of one of the objects, m1 = 100 g = 0.1 kg
Mass of the other object, m2 = 200 g = 0.2 kg
Velocity of m1 before collision, v1= 2 m/s

Velocity of m2 before collision, v2= 1 m/s

Velocity of m1 after collision, v3= 1.67 m/s

Velocity of m2 after collision= v4


According to the law of conservation of momentum:
Total momentum before collision = Total momentum after collision

Therefore, m1v1 + m2v2 = m1v3 + m2v4
                  2(0.1) + 1(0.2) = 1.67(0.1) + v4(0.2)
                  0.4 = 0.167 + 0.2v4
                  v4= 1.165 m/s

Hence, the velocity of the second object becomes 1.165 m/s after the collision.

17.When a carpet is beaten with a stick, dust comes out of it. Explain.

Answer

When the carpet is beaten, it is suddenly set into motion. The dust particles tend to remain at rest due to inertia of rest, therefore the dust comes out of it.

18.A motorcar of mass 1200 kg is moving along a straight line with a uniform velocity of 90 km/h. Its velocity is slowed down to 18 km/h in 4 s by an unbalanced external force. Calculate the acceleration and change in momentum. Also calculate the magnitude of the force required.

Answer

Mass of the motor car, m = 1200 kg
Initial velocity of the motor car, u = 90 km/h = 25 m/s
Final velocity of the motor car, 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 motor car = −5 m/s2
Change in momentum of the motor car = −24000 kg m s−1
Hence, the force required to decrease the velocity is 6000 N.
(Negative sign indicates retardation, decrease in momentum and retarding force)

19.An object of mass 100 kg is accelerated uniformly from a velocity of 5 m s−1 to 8 m
s−1 in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.

Answer

Initial velocity of the object, u = 5 m/s
Final velocity of the object, v = 8 m/s
Mass of the object, m = 100 kg
Time take by the object to accelerate, t = 6 s
Initial momentum = mu = 100 × 5 = 500 kg ms−1
Final momentum = mv = 100 × 8 = 800 kg ms−1
Force exerted on the object, F = (mv – mu)/ t
= m (v-u)/t
= 800 – 500
= 300/6
= 50 N

Initial momentum of the object is 500 kg ms−1.
Final momentum of the object is 800 kg ms−1.
Force exerted on the object is 50 N.

20. How much momentum will a dumbbell 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.

Answer

Mass of the dumbbell, m = 10 kg
Distance covered by the dumbbell, s = 80 cm = 0.8 m
Acceleration in the downward direction, a = 10 m/s2
Initial velocity of the dumbbell, u = 0
Final velocity of the dumbbell (when it was about to hit the floor) = 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 dumbbell hits the floor is
= mv = 10 × 4 = 40 kgms−1

NCERT Quick revision Notes of Chapter-9 Force & Laws of Motion

NCERT Solution of Chapter-9 Force & Laws of Motion

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