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
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.
Answer
(i) Accelerating unbalanced force.
(ii) No force.
(iii) Retarding unbalanced force.
(i) If force is doubled
(ii) If density is halved
(iii) If volume is reduced to one third.
Answer
(ii) No effect.
(iii) No effect.
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.
(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
SI unit of momentum is – kg m/s.
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.
(ii) If a man jumps out from a boat, the boat moves backward. Why?
Answer
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.
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.
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.
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
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
= 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 m1 after collision, v3= 1.67 m/s
According to the law of conservation of momentum:
Total momentum before collision = Total momentum after collision
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
NCERT MCQs of Chapter-9 Force & Laws of Motion
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