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NCERT Solution for Class 9 Science Chapter 9 : Force and Laws of Motion

Force and Laws of Motion
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You have studied the concept of Force in class 8, now in class 9, you will study it in depth. You will learn about balanced and unbalanced forces. You will understand all three laws of motion and will get to know physicist Galileo Galilei. This chapter will enhance your conceptual understanding of Force and Motion.

Solution Class 9 Science of Chapter 9 includes all Intext exercises 1, 2, 3, and an NCERT exercise and additional exercise for you. The NCERT exercise includes solutions all the 18 questions given in your NCERT Textbook. These questions will help you to testa and enhance your application skills, numerical ability, and memory.

Ncert solutions

NCERT Solution for Class 9 Science Chapter 9 : Force and Laws of Motion Ex 9.1

 

NCERT 9th Science Chapter 9 1

Exercise 9.1

Question 1:
Which of the following has more inertia: (a) a rubber ball and a stone of
the same size? (b) a bicycle and a train? (c) a five-rupee coin and a onerupee coin?

 

Solution:
Inertia of an object depend upon the mass of that object, the object having greater mass will have greater inertia whereas the object having smaller mass has smaller inertia. The objects having greater inertia due to their greater mass are:
(a) Stone
(b) Train
(c) Five-Rupee coin

 

Question 2:
In the following example, try to identify the number of times the velocity of
the ball changes: “A football player kicks a football to another player of his
team who kicks the football towards the goal. The goalkeeper of the
opposite team collects the football and kicks it towards a player of his own
team”. Also identify the agent supplying the force in each case.

 

Solution:
Let the two football teams be team A and team B respectively. Initially, the football is at rest. Later, the football is kicked by a player on team A to another player of team A. (The velocity of the ball has changed 1 time so far). This kick to the ball brings the change in velocity of the ball due the force applied by the player.
Now, the football is again kicked by the other player of team A towards the goal. This kick again changes the velocity of the ball thus the velocity of the ball is changed twice so far. The goalkeeper of team B stops the ball and brings the ball to rest. The velocity of the ball changes for the third time due to the force applied by the goalkeeper in the opposite direction. In the end, the goalkeeper of team B kicks the ball towards another player of team B. the velocity of the ball changes for the fourth and final time. Therefore, the velocity of the ball changes 4 times in this example.

 

Question 3:
Explain why some of the leaves may get detached from a tree if we
vigorously shake its branch.

 

Solution:
When we shake the branch of the tree, the branch moves to-and-fro due to the force applied. However, the inertia of the leaves that are attached to the branch resists this motion of the branch. Due to this resistance of inertia the leaves that are weakly attached to the branch fall off due to their own inertia whereas the leaves firmly attached to the branch remain attached.

 

Question 4:
Why do you fall in the forward direction when a moving bus brakes to a
stop and fall backwards when it accelerates from rest?

 

Solution:
When the bus starts accelerating in the forward direction from rest, the passengers experience a force in the backward direction because their inertia opposes the forward motion of the bus.
Once the bus starts moving, the passengers of the bus also get in a state of motion in the forward direction.
Now, when the driver applies brakes, the bus comes to the position of rest due to the force applied in backward direction. But, a force in the forward direction is exerted on the passengers of the bus because again the inertia of the moving passenger resists the change in the motion of the bus. This results the passengers is moving forwards when the brakes are applied to stop the bus.

 

Exercise 9.2

Question 1:
If action is always equal to the reaction, explain how a horse can pull a
cart.

Solution:
When the horse walks in the forward direction while the cart being attached to it, the horse xerts a force on the Earth in the backward direction. Earth applies an equal force in the opposite direction (forward direction) on the horse. This force applied by the earth moves the horse and the cart forward.
The velocity of the horse depends upon the mass of the horse along with the mass of the cart attached to it. The heavier the cart, the slower will be the motion of the horse. If the cart is very heavy, the force exerted by the horse will be insufficient to overcome the inertia of the cart at rest thus the horse will not be able to pull the cart.

 

Question 2:
Explain why is it difficult for a fireman to hold a hose, which ejects large
amounts of water at a high velocity.

 

Solution:
For ejecting water at high velocities by a hose, a force must be applied with the help of a pump or motor on the water. But, water applies an equal and opposite force back on the hose. To overcome the force applied by the water on the hose the fireman must apply a force to hold this hose. The greater the quantity and velocity of the water coming out of the hose, the greater will be the amount of force applied by the fireman to hold it steady.

 

Question 3:
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.

 

Solution: Refer pdf.

 

Question 4:
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.

 

Solution: Refer pdf.

 

Exercise Chapter 9

Question 1:
An object experiences a net zero external unbalanced force. Is it possible
for the object to be traveling with a non-zero velocity? If yes, state the
conditions that must be placed on the magnitude and direction of the
velocity. If no, provide a reason.

Solution:
Yes, it is possible. An object moving in some direction with constant velocity will continue to move with the same velocity in the same direction as long as an external unbalanced force acts upon it.

 

Question 2:
When a carpet is beaten with a stick, dust comes out of it. Explain.

 

Solution:
When we beat the carpet with a stick, the stick applies a force on the carpet which results in motion of the carpet. But the inertia of the dust particles on the carpet resists this motion of the carpet. Therefore, the forward motion of the carpet applies a force on the dust particles in the backward direction, they start moving in the opposite direction. This is the reason why dust comes out of the carpet when beaten.

 

Question 3:
Why is it advised to tie any luggage kept on the roof of a bus with a rope?

 

Solution:
When luggage is placed on the roof of a bus, the forward acceleration of the bus exerts a force in the opposite direction on the luggage. Similarly, when the bus initially in motion suddenly comes to rest as soon as brakes are applied, a forward force is exerted on the luggage. The magnitude of the force depends upon their inertia which is directly proportional to their mass. Tying up the luggage will help to overcome the force exerted by the bus and secure its position and thus prevent it from falling.

 

Question 4:
A batsman hits a cricket ball which then rolls on a level ground. After
covering a short distance, the ball comes to rest. The ball slows to a stop
because (a) the batsman did not hit the ball hard enough. (b) Velocity is
proportional to the force exerted on the ball. (c) There is a force on the ball
opposing the motion. (d) There is no unbalanced force on the ball, so the
ball would want to come to rest.

 

Solution:
When ball rolls on the ground, its motion is opposed by the force of friction between the ground and the ball. This force between ground and ball eventually stops the ball. So, the correct answer is (c).
If the surface of the ground is lubricated with oil, the friction between the ball and the ground will reduce largely, this lesser force will enable the ball to roll for a longer distance before stopping.

 

Question 5:
A truck starts from rest and rolls down a hill with a constant acceleration.
It travels a distance of 400 m in 20 s. Find its acceleration. Find the force
acting on it if its mass is 7 tonnes (Hint: 1 tonne = 1000 kg.)

 

Solution: Refer pdf.

 

Question 6:
A stone of 1 kg is thrown with a velocity of 20 ms-1
across the frozen surface of a lake and comes to rest after traveling a distance of 50 m. What is the force of friction between the stone and the ice?

 

Solution Refer pdf.

 

Question 7:
An 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a
horizontal track. If the engine exerts a force of 40000 N and the track offers
a friction force of 5000 N, then calculate: (a) the net accelerating force and
(b) the acceleration of the train

 

Solution:
(a)
Given:
Force exerted by the train (F) = 40,000 N
Force of friction = -5000 N (the negative sign indicates that the force is applied in the opposite direction)
Therefore, the net accelerating force = sum of all forces = 40,000 N + (-5000 N) = 35,000 N

(b)
Total mass of the train = mass of engine + mass of each wagon = 8000kg + 5 × 2000kg
The total mass of the train is 18000 kg.
According to second law of motion, F = ma (or: a = F/m)
Therefore, acceleration of the train = (net accelerating force) / (total mass of the train) = 35,000/18,000 = 1.94 ms-1
The acceleration of the train is 1.94 m.s-2

 

.
Question 8:
An automobile vehicle has a mass of 1500 kg. What must be the force
between the vehicle and road if the vehicle is to be stopped with a negative
acceleration of 1.7 ms-2 ?

 

Solution:
Given:
Mass of the vehicle (m) = 1500 kg
Acceleration (a) = -1.7 ms-2
As per the second law of motion, F = ma
F = 1500kg × (-1.7 ms-2) = -2550 N
Therefore, a force of 2550 N must act on the vehicle in a direction opposite to that of its motion.

 

Question 9:
What is the momentum of an object of mass m, moving with a velocity v?
(a)(mv)2
(b) mv2
(c) ½ mv2
(d) mv

 

Solution:
Momentum is defined as the product of mass and velocity of the object, the correct answer is (d), mv.

 

Question 10:
Using a horizontal force of 200 N, we intend to move a wooden cabinet
across a floor at a constant velocity. What is the friction force that will be
exerted on the cabinet?

 

Solution:
For velocity of the cabinet to be constant, the acceleration of the cabinet must be zero. For zero acceleration the effective force acting on it must also be zero. This means that the magnitude of opposing frictional force must be equal to the force exerted on the cabinet, which is 200 N. Therefore, the total frictional force exerted on the cabinet by floor is -200 N.

 

Question 11:
Two objects, each of mass 1.5 kg, are moving in the same straight line but
in opposite directions. The velocity of each object is 2.5 ms-1 before the
collision during which they stick together. What will be the velocity of the
combined object after the collision?

 

Solution:
Given:
Mass of the objects (m1 and m2) = 1.5kg
Initial velocity of the first object (u1) = 2.5 m/s
Initial velocity of the second object which is moving in the opposite direction (u2) = -2.5 m/s
When the two masses stick together, the resulting object has a mass of 3 kg (m1 + m2)
Velocity of the resulting object (v) =?
According to the law of conservation of momentum, the total momentum of object before the collision is equal to the total momentum of object after the collision.
Total momentum before the collision = m1u1 + m2u2
= (1.5kg) (2.5 m/s) + (1.5 kg) (-2.5 m/s) = 0
Therefore, total momentum after collision = (m1+m2) v = (3kg) v = 0
Therefore v = 0
This implies that the object formed after the collision has a velocity of 0 meters per second.

 

Question 12:
According to the third law of motion when we push on an object, the object
pushes back on us with an equal and opposite force. If the object is a
massive truck parked along the roadside, it will probably not move. A
student justifies this by answering that the two opposite and equal forces
cancel each other. Comment on this logic and explain why the truck does
not move.

 

Solution:
Due to the very high mass of the truck the static friction between the road and the truck is high. When we push the truck with a small force, this static frictional force between truck and road works in opposite direction to the applied force and the truck does not move. Therefore, the student’s logic is correct.

 

Question 13:
A hockey ball of mass 200 g traveling at 10 ms-1
is struck by a hockey stick
so as to return it along its original path with a velocity at 5 ms -1. Calculate the magnitude of change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.

 

Solution: Refer pdf.

Question 14:
A bullet of mass 10 g travelling horizontally with a velocity of 150 m s–1
strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the
distance of penetration of the bullet into the block. Also calculate the
magnitude of the force exerted by the wooden block on the bullet.

 

Solution: Refer pdf.

 

Question 15:
An object of mass 1 kg travelling in a straight line with a velocity of 10 ms-1
collides with, and sticks to, a stationary wooden block of mass 5 kg. Then
they both move off together in the same straight line. Calculate the total
momentum just before the impact and just after the impact. Also, calculate
the velocity of the combined object.

 

Solution:
Given:
Mass of the object (m1) = 1kg
Mass of the block (m2) = 5kg
Initial velocity of the object (u1) = 10 m/s
Initial velocity of the block (u2) = 0
Mass of the resulting object = m1 + m2 = 6kg
Velocity of the resulting object (v) =?
Total momentum before the collision = m1u1 + m2u2 = (1kg) × (10m/s) + 0 = 10 kg.m.s-1
According to the law of conservation of momentum, the total momentum of object before the collision must be equal to the total momentum of the object after the collision for conservation of momentum. Therefore, the total momentum post the collision is also 10 kg.m.s-1
Now, (m1 + m2) × v = 10kg.m.s-1
Therefore
10 kg m/s 1.66 m/s
6 kg
v

 
The resulting object moves with a velocity of 1.66 meters per second.

 

Question 16:
An object of mass 100 kg is accelerated uniformly from a velocity of 5 ms-1
to 8 ms-1 in 6 s. Calculate the initial and final momentum of the object. Also,
find the magnitude of the force exerted on the object.

 

Solution: Refer pdf.

 

.
Question 17:
Akhtar, Kiran, and Rahul were riding in a motorcar that was moving with
a high velocity on an expressway when an insect hit the windshield and got
stuck on the windscreen. Akhtar and Kiran started pondering over the
situation. Kiran suggested that the insect suffered a greater change in
momentum as compared to the change in momentum of the motorcar
(because the change in the velocity of the insect was much more than that
of the motorcar). Akhtar said that since the motorcar was moving with a
larger velocity, it exerted a larger force on the insect. And as a result the
insect died. Rahul while putting an entirely new explanation said that both
the motorcar and the insect experienced the same force and a change in
their momentum. Comment on these suggestions.

 

Solution:
Kiran’s suggestion is correct because the mass of the insect is very small when compared to the mass of the car. According to the law of conservation of momentum, the total momentum of insect and car before the collision between the insect and the car must be equal to the total momentum of both after the collision. Therefore, the change in the momentum for the insect is much greater than the change in momentum for the car to keep the total momentum
conserved. Akhtar’s suggestion is also correct because the mass of the car is very high as compared to insect, the force exerted on the insect during the collision with the car is also very high. Rahul’s suggestion is partially correct because according to the third law of motion, the force exerted by the car on the insects equal and opposite to the force exerted by the insect in the
car. But, his explanation that the change in the momentum of both insect and car is same contradicts the law of conservation of momentum.

 

Question 18:
How much momentum will a dumb-bell of mass 10 kg transfer to the floor
if it falls from a height of 80 cm? Take its downward acceleration to be 10
ms-2

 

.
Solution: Refer pdf.

Additional Exercises Chapter 9

Question 1:
The following is the distance-time table of an object in motion:
Time (seconds) Distance (meters)
0 0
1 1
2 8
3 27
4 64
5 125
6 216
7 343
(a) What conclusion can you draw about the acceleration? Is it constant,
increasing, decreasing, or zero? (b) What do you infer about the forces
acting on the object?

 

Solution:
(a) The distance covered by the object at a time interval is constantly increasing in the same amount of time. Therefore, it is clear that the acceleration of the object is increasing.
(b) According to the second law of motion, force = mass × acceleration. As the mass of the object remains constant for the entire time, the increase in acceleration implies that the effective force acting on the object is also increasing with time.

 

Question 2:
Two persons manage to push a motorcar of mass 1200 kg at a uniform
velocity along a level road. The same motorcar can be pushed by three
persons to produce an acceleration of 0.2 m s-2
. With what force does each
person push the motorcar? (Assume that all persons push the motorcar
with the same muscular effort).

 

Solution:
Given:
Mass of the car (m) = 1200kg
When the third person starts pushing the car, the acceleration of the car (a) is 0.2ms-2
.
Therefore, the force applied by the third person is given by:
F = ma
F = 1200kg × 0.2 ms-2 = 240N
The force applied by the third person on the car is 240 N. As all 3 people push with the same
muscular effort, the force applied by each person on the car is 240 N.

 

Question 3:
A hammer of mass 500 g, moving at 50 m s-1
, strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer?

 

Solution: Refer pdf.

 

Question 4:
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.

 

Solution:
Given:
Mass of the car (m) = 1200 kg
Initial velocity (u) = 90 km/hour = 25 meters/sec
Terminal velocity (v) = 18 km/hour = 5 meters/sec
Time period (t) = 4 seconds
Refer pdf.
.
Initial momentum of the car = m × u = (1200kg) × (25m/s) = 30,000 kg.m.s-1
Final momentum of the car = m × v = (1200kg) × (5m/s) = 6,000 kg.m.s.-1
Therefore, change in momentum (final momentum – initial momentum) = (6,000 – 30,000)
kg.m.s-1
= -24,000 kg.m.s-1
External force applied = mass of car × acceleration = (1200 kg) × (-5 ms-2
) = -6000N
Therefore, the magnitude of force required to slow down the vehicle to 18 km/hour is
6000 N.

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