CBSE Class 9

These NCERT Solutions for Class 9 Science Chapter 9 Force and Laws of Motion Questions and Answers are prepared by our highly skilled subject experts to help students while preparing for their exams.

Force and Laws of Motion NCERT Solutions for Class 9 Science Chapter 9

Class 9 Science Chapter 9 Force and Laws of Motion InText Questions and Answers

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-rupees coin and a one-rupee coin?
Answer:
Inertia is the measure of the mass of the body. The greater is the mass of the body; the greater is its inertia and vice-versa.
(a) Mass of a stone is more than the mass of a rubber ball for the same size. Hence, inertia of the stone is greater than that of a rubber ball.
(b) Mass of a train is more than the mass of a bicycle. Hence, inertia of the train is greater than that of the bicycle.
(c) Mass of a five rupee coin is more than that of a one-rupee coin. Hence, inertia of the five rupee coin is greater than that of the one-rupee coin.

An online velocity calculator of physics specifically programmed to calculate velocity/speed of a moving object.

Question 2.
In the following example, by to identify the number of times the velocity of the ball changed:
“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.
Answer:
The velocity of the ball changes four times.
As a football player kicks the football, its speed changes from zero to a certain value. As a result, the velocity of the ball gets changed. In this case, the player applied a force to change the velocity of the ball. Another player kicks the ball towards the goal post. Asa result, the direction of the ball gets changed. Therefore, its velocity also changes. In this case, the player applied a force to change the velocity of the ball.

The goalkeeper collects the ball. In other words, the ball comes to rest. Thus, its speed reduces to zero from a certain value. The velocity of the ball has changed. In this case, the goalkeeper applied an opposite force to stop /change the velocity of the ball. The goal-keeper kicks the ball towards his team players. Hence, the speed of the ball increases from zero to a certain value. Hence, its velocity changes once again. In this case, the goalkeeper applied a force to change the velocity of the ball.

Question 3.
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. This is because when the branches of a tree are shaken, it moves to and fro, but its leaves tend to remain at rest. This is because the inertia of the leaves tend to resist the to and fro motion. Due to this reason, the leaves fall down from the tree when shaken vigorously.

Question 4.
Why do you fall in the forward direction when a moving bus breaks to a stop and fall backwards when it accelerates from rest?
Answer:
Due to the inertia of the passenger
Everybody tries to maintain its state of motion or state of rest. If a body is at rest, then it tries to remain at rest. If a body is moving, then it tries to remain in motion. In a moving bus, a passenger moves with the bus. As the driver applies brakes, the bus comes to rest. But, the passenger tries to maintain his state 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. This is because when the bus accelerates, the inertia of the passenger tends to oppose the forward motion of the bus. Hence, the passenger tends to fall backwards when the bus accelerates forward.

Question 5.
If action is always equal to the reaction, explain how a horse can pull a cart.
Answer:
A horse pushes the ground in the backward direction. According to Newton’s third law of motion, a reaction force is exerted by the Earth on the horse in the forward direction. As a result, the cart moves forward.

Question 6.
Explain, why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity.
Answer:
Due to the backward reaction of the water being ejected
When a fireman holds a hose, which is ejecting large amounts of water at a high velocity, then a reaction force is exerted on him by the ejecting water in the backward direction. This is because of Newton’s third law of motion. As a result of the backward force, tire stability of the fireman decreases. Hence, it is difficult for him to remain stable while holding the hose.

Question 7.
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 = 35 m/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 = 4(v1) + 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 = \(\frac{1.75}{4}\) = -0.4375 m/s
The negative sign indicates that the rifle recoils backwards with a velocity of 0.4375 m/s.

Question 8.
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
∴ m₁v₁ + m₂v₂ = m₁v₃ + m₂v₄
(0.1)2 + (0.2)1.67 + (0.2)v₄
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.

Class 9 Science Chapter 9 Force and Laws of Motion Textbook Questions and Answers

Question 1.
An object experiences a net zero external imbalanced force. Is it possible for the object to be travelling 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.
Answer:
Yes. Even when an object experiences a net zero external unbalanced force, it is possible that the object is travelling with a non-zero velocity. This is possible only when the object has been moving with a constant velocity in a particular direction. Then, there is no net unbalanced force applied on the body. The object will keep moving with a non-zero velocity. To change the state of motion, a net non-zero external imbalanced force must be applied on the object.

Question 2.
When a carpet is beaten with a stick, dust comes out of it. Explain.
Answer:
Inertia of an object tends to resist any change in its state of rest or state of motion. When a carpet is beaten with a stick, then the carpet comes to motion. But, the dust particles try to resist their state of rest. According to Newton’s first law of motion, the dust particles stay in a state of rest, while the carpet moves. Hence, the dust particles come out of the carpet.

Question 3.
Why is it advised to tie any luggage kept on the roof of a bus with a rope?
Answer:
When the bus accelerates and moves forward, it acquires a state of motion. However, the luggage kept on the roof, owing to its inertia, tends to remain in its state of rest. Hence, with the forward movement of the bus, the luggage tends to remain at its original position and ultimately falls from the roof of the bus. To avoid this, it is advised to tie any luggage kept on the roof of a bus with a rope.

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.
Answer:
(c) A batsman hits a cricket ball, which then rolls on a level ground. After covering a short distance, the ball comes to rest because there is frictional force on the ball opposing its motion.

Frictional force always acts in the direction opposite to the direction of motion. Hence, this force is responsible for stopping the cricket ball.

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 metric tonnes (Hint: 1 metric tonne = 1000 kg).
Answer:
Initial velocity, u = 0 (since the truck is initially at rest)
Distance travelled, s = 400 m
Time taken, t =20 s
According to the second equation of motion:
s = ut + \(\frac {1}{2}\)at²
Where,
Acceleration = a
400 = 0 + \(\frac {1}{2}\)a(20)²
400 = \(\frac {1}{2}\)a(400)
a = 2 m/s²
1 metric tonne = 1000 kg (Given)
∴ 7 metric tonnes = 7000 kg
Mass of truck, m = 7000 kg
From Newton’s second law of motion:
Force, F = Mass × Acceleration
F = ma = 7000 × 2 = 14000 N
Hence, the acceleration of the truck is 2 m/s² and the force acting on the truck is 14000 N.

Question 6.
A stone of 1 kg is thrown with a velocity of 20 m s^-1 across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?
Answer:
Initial velocity of the stone, u = 20 m/s Final velocity of the stone, v = 0 (finally die stone comes to rest)
Distance covered by the stone, s = 50 m
According to the third equation of motion: v² = u² + 2as
Where,
Acceleration, a
(0)² = (20)² + 2 × a × 50
a = -4 m/s²
The negative sign indicates that acceleration is acting against the motion of the stone.
Mass of the stone, m = 1 kg
From Newton’s second law of mouon:
Force, F = Mass × Acceleration
F = ma
F = 1 × (-4) = -4 N
Hence, the force of friction between the stone and the ice is -4 N.

Question 7.
A 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.
Answer:
(a) 35000 N
(b) 1.944 m/s2
(a) Force exerted by the engine, F = 40000 N
Frictional force offered by the track, F_f = 5000
Net accelerating force, F_a = F – F_f = 40000 – 5000 = 35000 N
Hence, the net accelerating force is 35000 N.

(b) Acceleration of the train = a
The engine exerts a force of 40000 N on all the five wagons.
Net accelerating force on the wagons, F_a = 35000 N
Mass of the wagons, m = Mass of a wagon × Number of wagons
Mass of a wagon = 2000 kg
Number of wagons = 5
∴ m = 2000 × 5 = 10000kg
Total mass of the train, M = m + mass of engine = 10000 + 8000 = 18000 kg

From Newton’s second law of motion:
F_a =Ma
a = \(\frac{\mathrm{F}_{a}}{m}\)
= \(\frac{35000}{18000}\)
= 1.944 ms^-2
Hence, file acceleration of the train is 1.944 m/s².

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 m s^-2?
Answer:
Mass of the automobile vehicle, m = 1500 kg
Final velocity, v = 0 (finally the automobile stops)
Acceleration of the automobile, a = -1.7 ms^-2
From Newton’s second law of motion:
Force = Mass × Acceleration = 1500 × (-1.7) = -2550 N
Hence, the force between tire automobile and the road is -2550 N, in the direction opposite to the motion of the automobile.

Question 9.
What is the momentum of an object of mass m, moving with a velocity v?
(a) (mv)²
(b) mv²
(c) \(\frac {1}{2}\)mv²
(d) mv
Answer:
(d) mv
Mass of the object = m
Velocity = v
Momentum = Mass × Velocity
Momentum = 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?
Answer:
A force of 200 N is applied in the forward direction. Thus, from Newton’s third law of motion, an equal amount of force will act in the opposite direction. This opposite force is the frictional force exerted on the cabinet. Hence, a frictional force of 200 N is exerted on the cabinet.

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 collision?
Answer:
Mass of one of the objects, m₁ = 1.5 kg
Mass of the other object, m₂ = 1.5 kg
Velocity of m1, before collision, v₁ = 2.5 m/s
Velocity of m₂, moving in opposite direction before collision, v₂ = -2.5 m/s
(Negative sign arises because mass m₂ is moving in an opposite direction)
After collision, the two objects stick together.
Total mass of the combined object = m₁ + m₂
Velocity of the combined object = v
According to the law of conservation of momentum: .
Total momentum before collision = Total momentum after collision m₁v₁ + m₂v₂ = (m₁ + m₂)v
1.5(2.5) + 1.5 (-2.5) = (1.5 + 1.5)v
3.75 – 3.75 = 3 v
v = 0
Hence, the velocity of the combined object after collision is 0 m/s.

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.
Answer:
The truck has a large mass. Therefore, the static friction between the truck and the road is also very high. To move the car, one has to apply a force more than the static friction. Therefore, when someone pushes the truck and the truck does not move, then it can be said that the applied force in one direction is cancelled out by the frictional force of equal amount acting in the opposite direction.

Therefore, the student is right in justifying that the two opposite and equal cancel each other.

Question 13.
A hockey ball of mass 200 g travelling 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 change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.
Answer:
Mass of the hockey ball, m = 200 g = 0.2 kg
Hockey ball travels with velocity, v₁ = 10 m/s
Initial momentum = mv₁
Hockey ball travels in the opposite direction with velocity, v₂ = -5 m/s
Final momentum = mv₂
Change in momentum = mv₁ – mv₂ = 0.2 [10 – (-5)] = 0.2 (15) = 3 kg ms^-1.
Hence, the change in momentum of the hockey ball is 3 kg m s^-1.

Question 14.
A bullet of mass 10 g travelling horizontally with a velocity of 150 ms^-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.
Answer:
Now, it is given that the bullet is travelling with a velocity of 150 m/s.
Thus, when the bullet enters the block, its velocity = Initial velocity, u = 150 m/s
Final velocity, v = 0 (since the bullet finally comes to rest)
Time taken to come to rest, t = 0.03 s
According to the first equation of motion, v = u + at
Acceleration of the bullet, a
0 = 150 + (a × 0.03 s)
a = \(\frac{-150}{0.03}\) = -5000 m/s²
(Negative sign indicates that the velocity of the bullet is decreasing.)
According to the third equation of motion: v² = u² + 2as
0 = (150)2 + 2 (-5000) s
s = \(\frac{-(150)^{2}}{-2(5000)}=\frac{22500}{10000}\) = 2.25 m
Hence, the distance of penetration of the bullet into the block is 2.25 m.
From Newton’s second law of motion:
Force, F = Mass × Acceleration
Mass of the bullet, m = 10 g = 0.01 kg
Acceleration of the bullet, a = 5000 m/s² .
F = ma = 0.01 × 5000 = 50 N
Hence, the magnitude of force exerted by the wooden block on the bullet is 50 N.

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.
Answer:
Mass of the object, m₁ = 1 kg
Velocity of the object before collision, v₁ = 10 m/s
Mass of the stationary wooden block, m₂ = 5
Velocity of the wooden block before collision, v₂ = 0 m/s
∴ Total momentum before collision = m₁v₁ + m₂v₂
= 1 (10) + 5 (0) = 10 kg ms^-1
It is given that after collision, the object and the wooden block stick together.
Total mass of the combined system = m₁ + m₂
Velocity of the combined object = v
According to the law of conservation of momentum:
Total momentum before collision = Total momentum after collision
m₁v₁ + m₂v₂ = (m₁ + m₂)v
1 (10) + 5 (0) = (1 + 5)u
v = \(\frac{10}{6}=\frac{5}{3}\) m/s
The total momentum after collision is also 10 kg m/s.
Total momentum just before the impact = 10 kg ms^-1
Total momentum just after the impact = (m₁ + m₂)v = 6 × \(\frac{5}{3}\) = 10 kg ms^-1
Hence, velocity of the combined object after collision = \(\frac{3}{5}\) m/s

Question 16.
An object of mass 100 kg is accelerated uniformly from a velocity of 5 m s^-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.
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 = 6s
Initial momentum = mu = 100 × 5 = 500 kg ms^-1
Final momentum = mv = 100 × 8 = 800 kg ms^-1
Force exerted on the object, F = \(\frac{m v-m u}{t}\)
\(=\frac{m(v-u)}{t}=\frac{800-500}{6}=\frac{300}{6}\) = 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.

Question 17.
Akhtar, Kiran and Rahul were riding in a motor car 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.
Answer:
According to the law of conservation of momentum:
Momentum of the car and insect system before collision = Momentum of the car and insect system after collision
Hence, the change in momentum of the car and insect system is zero.

The insect gets stuck on the windscreen. This means that the direction of the insect is reversed. As a result, the velocity of the insect changes to a great amount. On the other hand, the car continues moving with a constant velocity. Hence, Kiran’s suggestion that the insect suffers a greater change in momentum as compared to the car is correct. The momentum of the insect after collision becomes very high because the car is moving at a high speed. Therefore, the momentum gained by the insect is equal to the momentum lost by the car.

Akhtar made a correct conclusion because the mass of the car is very large as compared to the mass of the insect.

Rahul gave a correct explanation as both the car and the insect experienced equal forces caused by the Newton’s action-reaction law. But, he made an incorrect statement as the system suffers a change in momentum because the momentum before the collision is equal to the momentum after the collision.

Question 18.
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 ms^-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 dumb-bell hits the floor is
= mv = 10 × 4 = 40 kg ms^-1

Question 19.
The following is the distance-time table of an object in motion:

(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?
Answer:
(a) There is an unequal change of distance in an equal interval of time.
Thus, the given object is having a non-uniform motion. Since the velocity of the object increases with time, the acceleration is increasing.

(b) According to Newton’s second law of motion, the force acting on an object is directly proportional to the acceleration produced in the object. In the given case, the increasing acceleration of the given object indicates that the force acting on the object is also increasing.

Question 20.
Two persons manage to push a motorcar of mass 1200kg at a uniform velocity along a level road. The same motorcar can be pushed by three persons to produce an acceleration of 0. 2 ms^-2. With what force does each person push the motorcar? (Assume that all persons push the motorcar with the same muscular effort)
Answer:
Mass of the motor car = 1200 kg Only two persons manage to push the car. Hence, the acceleration acquired by the car is given by the third person alone.
Acceleration produced by the car, when it is pushed by the third person, a = 0.2 m/s²
Let the force applied by the third person be F. From Newton’s second law of motion:
Force = Mass × Acceleration
F = 1200 × 0.2 =240 N
Thus, the third person applies a force of magnitude 240 N.
Hence, each person applies a force of 240 N to push the motor car.

Question 21.
A hammer of mass 500 g, moving at 50 ms^-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?
Answer:
Mass of the hammer, m=500 g = 0.5 kg
Initial velocity of the hammer, u = 50 m/s
Time taken by the nail to the stop the hammer, t = 0.01 s
Velocity of the hammer, v = 0 (since the hammer finally comes to rest)
From Newton’s second law of motion:
Force, F= \(\frac{m(v-u)}{t}=\frac{0.5(0-50)}{0.01}\) = -2500N
The hammer strikes the nail with a force of -2500 N. Hence, from Newton’s third law of motion, the force of the nail on the hammer is equal and opposite, i.e., +2500 N.

Question 22.
A motorcar of mass 1200kg 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
Change in momentum = mv – mu = m (v – u)
= 1200 (5 – 25) = -24000 kg ms^-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 ms^-1
Hence, the force required to decrease the velocity is -6000 N.
(Negative sign indicates the retardation, decrease in momentum and retarding force respectively)

Class 9 Science Chapter 9 Force and Laws of Motion Additional Important Questions and Answers

Multiple Choice Questions
Choose the correct option:

Question 1.
Which of the following statement is not correct for an object moving along a straight path in an accelerated motion?
(a) Its speed keeps changing
(b) Its velocity always changes
(c) It always goes away from the earth
(d) A force is always acting on it
Answer:
(c) It always goes away from the earth

Question 2.
According to the third law of motion, action and reaction
(a) always act on the same body
(b) always act on different bodies in opposite directions
(c) have same magnitude and directions
(d) act on either body at normal to each other
Answer:
(b) always act on different bodies in opposite directions

Question 3.
A goalkeeper in a game of football pulls his hands backwards after holding the ball shot at the goal. This enables the goal keeper to
(a) exert larger force on the ball
(b) reduce the force exerted by the ball on hands
(c) increase the rate of change of momentum
(d) decrease the rate of change of momentum
Answer:
(b) reduce the force exerted by the ball on hands

Question 4.
The inertia of an object tends to cause the object
(a) to increase its speed
(b) to decrease its speed .
(c) to resist any change in its state of motion
(d) to decelerate due to friction
Answer:
(c) to resist any change in its state of motion

Question 5.
A passenger in a moving train tosses a coin which falls behind him. It means that motion of the train is
(a) accelerated
(b) uniform
(c) retarded
(d) along circular tracks
Answer:
(a) accelerated

Question 6.
An object of mass 2 kg is sliding with a constant velocity of 4 ms^-1 on a frictionless horizontal table. The force required to keep the object moving with the same velocity is
(a) 32 N
(b) 0 N
(c) 2 N
(d) 8N
Answer:
(b) 0 N

Question 7.
Rocket works on the principle of conservation of
(a) mass
(b) energy
(c) momentum
(d) velocity
Answer:
(c) momentum

Question 8.
A water tanker filled up to 2/3 of its height is moving with a uniform speed. On sudden application of the brake, the water in the tank would
(a) move backward
(b) move forward
(c) be unaffected
(d) rise upwards
Answer:
(b) move forward

Question 9.
Which of the following is the measure of inertia?
(a) Volume
(b) Velocity
(c) Momentum
(d) Mass
Answer:
(d) Mass

Question 10.
The S.I. unit of momentum is
(a) ms^-1
(b) ms^-2
(c) kg ms^-2
(d) kg ms^-1
Answer:
(d) kg ms^-1

Very Short Answer Type Questions

Question 1.
Define force.
Answer:
Force is defined as a pull or push that changes the position or shape or state of a body.

Question 2.
Define inertia.
Answer:
Inertia is the natural tendency of a body to stay at rest or to remain in motion with same velocity.

Question 3.
State the S.I. unit of inertia.
Answer:
Mass is the measure of inertia therefore S.I. unit of inertia is kilogram (kg).

Question 4.
What will be the acceleration of a body which is moving under the balanced force.
Answer:
If a body moves under the balanced force than its acceleration is zero.

Question 5.
Why a groove is povided in a saucer?
Answer:
A groove is provided in a saucer to prevent the toppling of cup due to a sudden jerk.

Question 6.
Why does a fan continue to move for sometime even after if has been switched off?
Answer:
When a fan is switched off, it tends to remain in motion due to inertia it gained during its motion.

Question 7.
Which of the two, a truck or trains has greater inertia? Why?
Answer:
A train has greater inertia because of its greater mass.

Question 8.
Define the SlI. unit of force.
Answer:
S.I. unit of force, Newton (N) is defined as force that produces an acceleration of 1 m/s² in a body of mass 1 kg.

Question 9.
How does the acceleration of a body changes on application of unbalanced force?
Answer:
The acceleration produced in a body is directly proportional to the applied unbalanced force.

Question 10.
According to third law of motion, action and reaction acts on same or different bodies?
Answer:
Action and reaction always act on two different bodies.

Question 11.
Give two examples in which force is applied from a distance.
Answer:
Non-contact forces include magnetic force and gravitational force.

Question 12.
What is the total momentum of the gun and bullet before firing of the bullet?
Answer:
With both gun and bullet in rest, the total momentum before firing of bullet is zero.

Question 13.
Which law of motion is also considered as the law of inertia?
Answer:
First law of motion is also considered as the law of inertia.

Question 14.
What is force of friction?
Answer:
Force of friction is defined as the force that acts against the direction of motion of body to oppose the motion.

Question 15.
What are the S.I unit of momentum and force?
Answer:
S.I. unit of momentum is kg ms~J and of force is kg ms^-2 (Newton).

Short Answer Type Questions

Question 1.
What are balanced and unbalanced forces?
Answer:
Balanced forces : When two or more forces act on a body and their resultant force is zero which do not change the state of rest or motion of a body, the forces are called balanced forces.

Example : When a wooden block is pulled from both sides with equal but opposite forces, the wooden block does not undergo any displacement, it remains at its original position.

Unbalanced forces: When the resultant of two or more forces acting on a body is not zero hence, bring the change in the position, state or shape of body, the forces are called imbalanced forces.
Example: Kicking of a football by player or with drawing of water from a well.

Question 2.
State the different possible effects of the ‘ application of force on an object.
Answer:
The different possible effects of the application of force on an object include:
(i) Displacement of object from its initial po-sition.
(ii) Change in the direction of motion of a body.
(iii) Change in the shape of a body.
(iv) Change in the state of a body.

Question 3.
Give reasons for the following:
(i) A foot ball when kicked flies away but not a solid metallic ball of same size.
(ii) A rubber ball covers a longer distance than a leather ball thrown with same force.
(iii) You travelling in car happened to be pushed to one side, when the car takes a sharp turn.
(iv) A passenger moving from a running bus ends up fall with his face downward.
(v) Driver and passengers of a car are advised to wear safety belts.
Answer:
(i) The mass of metallic ball is more than the foot ball hence, metallic ball has greater inertia than fooball. It has lesser tendency to bring change in its position.

(ii) A rubber ball has less mass and we know that acceleration is inversely proportional to mass, therefore rubber ball gains more acceleration and covers large distance than heavier leather ball on application of equal force.

(iii) When travelling in a car, the direction of motion of body is in the direction of the motion of car but when the car suddenly takes a sharp turn the direction of car changes but not of the passenger. Therefore, he is pushed to one side of the car.

(iv) When a passenger jumps out from a moving bus, his feet comes in state of rest on touching the ground but his upper body tends to remain in the state of motion due to inertia. Therefore, the passenger tends to fall with his face downward.

(v) The body of the driver and passenger in a moving car tends to be in the state of motion.

Therefore, to stop the body going forward on the application of sudden brakes, they are adviced to wear seat belt. The seat belts also called safety belts. Exerts a force on passengers body to make forward motion, due to inertia, much slower.

Question 4.
A student placed a water filled tumbler on a tray and turned around as fast as he could? What do you expect to happen to the tumbler? Why?
Answer:
The tumbler would topple down spilling the water.

Initially both tumbler and tray were in state of rest but as student after turning around came to stand still the tray came in rest but not the tumbler, it remained in motion because of inertia and therefore fell down, spilling the water.

Question 5.
Why does a sprinter keep running for sometime even after crossing the finish line?
Answer:
A sprinter keeps running for sometime to some distance even after crossing the finish line because of inertia that opposes the change in state of motion of his body.

Question 6.
What do you mean by momentum ? What is its S.I. unit?
Answer:
Momentum refers to the product of mass with die velocity of a body in motion. It’s a vector quantity which has both the magnitude and direction. The direction is same as the direction of the velocity.
Momentum = mass × velocity
or P = m × v
It’s S.I. unit is kg ms^-1 (kilogram meter per second).

Question 7.
State the three laws of motion.
Answer:
First law of motion : It states that an object at rest or in uniform motion will remain in rest or uniform motion unless unbalanced force acts on it.

Second law of motion: It states that the rate of change of momentum of an object is proportional to the applied force and always takes place in the direction of the force.
If P₁ and p₂ are the change in momentum in time ‘t’ an object of mass ‘m’, then

Third law of motion: It states that when an object exert a force on another object, then the second object exerts the force on the first object. These two forces are always equal in magnitude but in the opposite direction i.e. action and reaction forces are always equal in magnitude but opposite in the direction.

Question 8.
Deduce the first law of motion from second law of motion.
Answer:
When F is the applied force on a body of mass ‘m’ then if body gains acceleration ‘a’ then according to 2nd law of motion.
F = ma
F = m \(\frac{(v-u)}{t}\left(a=\frac{v-u}{t}\right)\)
or F_t = m(v – u) or mv – mu
When F = 0 with body at rest
mv – mu = 0
mv = mu
or v = u
This shows that the object would continue to move with same uniform velocity ‘u’ through the timer.

Question 9.
How does the acceleration of an object changes with the
(a) applied unbalanced force?
(b) mass of the object?
Answer:
(a) Acceleration of an object is directly proportional to the unbalanced force applied on the object. It increases with the increasing force.

(b) Acceleration of an object is inversely proportional to the mass of the object. Acceleration of an object decreases with increasing mass keeping the applied imbalanced force constant.

Question 10.
Explain why a Karate player breaks a slab of ice or brick or a pile of tiles by striking with his hand in a single blow?
Answer:
A Karate player strikes the slab of ice or a pile of tiles as fast as possible with his hand. In doing so, the entire momentum of the hand is reduced to zero in a short time. This causes the application of large force on pile of tiles or slab of ice to break.

Question 11.
Explain why a fielder in the ground moves his hand backward while catching a moving cricket ball?
Answer:
The fielder in the ground gradually pulls his hand backward when catching the ball so that ball is stopped in longer time. This allows the acceleration of ball to reduce gradually causing less impact on hand/palm to protect them from getting hurt.

Question 12.
Explain why in a high jumping ground, the athletes are made to fall either on a cushioned bed or on a sand bed?
Answer:
When an athlete after making a jump, falls on a cushioned bed or on a sand bed, the time of athletes’ fall after making a jump increases hence, reduces the rate of change of momentum and force to cause less impact on athlete.

Question 13.
Action and reaction force are equal in the magnitude, do they produce equal acceleration in the two bodies? Why?
Answer:
Action and reaction forces are equal in their magnitude but they may or may not produce equal acceleration in the two bodies because acceleration produced is inversely proportional to mass. The body with less mass will experience greater acceleration than body with greater mass.

Question 14.
Explain how you manage to walk easily on road but end up slipping if happened to step on the banana peel?
Answer:
When we walk on the ground, we push the road or ground below backwardly with a force. This force is the action. Then, in turn we get a reaction of the applied force which is equal in magnitude but opposite in direction to enable us to move forward.

When we happen to step on banana peel, the force of friction is very-very less and hence, we end up slipping.

Question 15.
Explain the principle of the working of jet engine.
Answer:
In a jet engine a large volume of the gases are produced by burning of fuel. These gases are allowed to escape throuh a jet or nozzle at the rear end of jet-engine at a very high speed in the backward direction forcing the body to go in forward direction with high velocity.

Question 16.
What do you mean by interaction? State the different types of interactions.
Answer:
When an object influences another object by the application of force, the first object is said to be interacted with second object.

The interaction may happen when two bodies come in direct contact of each other such as the muscular force applied by a person on an object or from a distance such as magnetic or electrical force. A magnet or an electrical charge exerts its force from a distance Le. without coming in contact of another object.

Question 17.
Where do you find easy to walk, on a sandy or tar road?
Answer:
It is easier to walk on a tar road than on a sandy road. When we step over a patch of sand, the sand particles get displaced by the foot take sometime to reach the sandy ground. Therefore the action from the sandy soil is reduced and so also the reaction in the feet. This reduced action and reaction makes it difficult to walk on sandy road than on tar road.

Question 18.
A student hits a floor and mattress lying on the same floor. In which case, he will hurt himself more? Why?
Answer:
When the student hits the cemented floor and a mattress lying on the floor, then he will feel more hurt on hitting the floor than mattress. The mattress being soft increases the impact time but decreases the rate of change of momentum.

Question 19.
A constant force acts on an object of 5 kg mass for a period of 2 s. It increases its velocity from 3 ms^-1 to 7 ms^-1. Find the magnitude of the applied force. If the force is applied for 4 seconds, what could be the final velocity?
Answer:
Mass of object = 5 Kg
Initial velocity (u) =3 ms^-1
Time taken for applied force = 2s
We know that F = \(\frac{m(v-u)}{t}=\frac{5(7-3)}{2}=\frac{5 \times 4}{2}\) = 10 N

If the force is appled for, 4 seconds, then Initial velocity, u = 3ms^-1, F = 10 N, v = 7
F = \(m \frac{(v-u)}{t}\)
10 = \(5 \frac{(v-3)}{4}\)
\(\frac {40}{5}\) = v – 3
v = 8 + 3
v = 11 ms^-1

Question 20.
Which would require a greater force accelerating a 2 kg mass at 5 ms^-2 or a 4 kg mass at 2 ms^-2?
Answer:
(i) Mass of the object = 2 kg
Acceleration, a = 5 ms^-2
Force = Mass × Acceleration = 2 × 5 = 10N

(ii) Mass of the object = 4 kg
Acceleration, a = 2 ms^-2
Force = Mass × Acceleration = 4 × 2 = 8N
Therefore, accelerating 2 kg mass would require more force.

Question 21.
A motor car is moving with a velocity of 108 Km/h and it takes 4 s to stop after the brakes are applied. Calculate the force exerted by the brakes on the motor car if its mass along with the passengers in motor car is 1000 kg.
Answer:
Initial velocity of car (u) = 108 km/h.
= \(\frac{108 \times 1000}{3600}\) = 30 ms^-1
Final velocity = 0 ms^-1
Time taken = 4 s
Mass of motor car = 1000 kg
∴ Magnitude of force applied, F = \(m \frac{(v-u)}{t}\)
= \(1000 \frac{(0-30)}{4}\)
= – 7500 N
The negative sign indicates that the force is acting in the direction opposite to the direction of motion of motor car.

Question 22.
A force of 5N gives an acceleration of 120 ms^-2 in mass m₂. What acceleration would it give if both the masses m, and m₂ are tied together?
Answer:
(i) Force F = 5N
Acceleration, a = 10 ms^-2
mass, m = ?
with F = m × a
5 = m × 10
or, m = \(\frac {5}{10}\) = 0.5 kg

(ii) Force, F = 5N
Acceleration, a =20 ms^-2
With F = m × a
5 = m × 20
m = \(\frac {5}{20}\)
= \(\frac {1}{4}\)
= 0.25 kg
When masses are tied together = 0.5 + 0.25 kg
= 0.75 kg.
Withforce F =5N
and mass m =0.75 kg
a = \(\frac{F}{m}=\frac{5}{0.75}\)
= 6.67 ms^-2

Question 23.
The velocity-time graph of a ball of mass 20 g moving along a straight line on a long table 15 shown in the figure 9.1 given below.

How much force does the table exert on the ball to bring it to rest?
Answer:
Initial velocity of ball (u) = 20cm^-1 = 0.2ms^-1
Final velocity of ball, (v) = 0
Time taken (t) = 10s
Mass of ball = 20g = 0.02kg

= -0.0004 N
The negative sign show that force is acting against the direction of motion.

Question 24.
What would be the force required to produce acceleration of 2ms^-2 on a body of mass 12 kg? What would be its acceleration, if the force applied is doubled to its initial value?
Answer:
Acceleration, a = 2ms^-2
mass, m = 12kg
∴ Force required = Mass × Acceleration
= 12 × 2 = 24 N
When force is doubled, than F = 24 × 2 = 48 N
Mass of body = 12 kg
∴ Acceleration = \(\frac{\text { Force }}{\text { mass }}=\frac{48}{12}\) = 4 ms^-2

Question 25.
A man pushes a box of 50 kg with a force of SON. What will be the acceleration of the box due to this force? If the mass of the box was halved what would be the acceleration?
Answer:
Mass of the box, m = 50 kg
Force applied, F = 80 N
Then acceleration, a = \(\frac{\mathrm{F}}{\mathrm{M}}\)
= \(\frac {80}{50}\) = 1.6 ms^-2
When the mass is healed = \(\frac {50}{2}\)
= 25kg.
Force applied = 80 N
Then acceleration, a = \(\frac{\mathrm{F}}{\mathrm{M}}\)
= \(\frac {80}{25}\) = 3.2ms^-2

Question 26.
A force exerted for 1.2s raises the velocity of an object from 1.8 ms^-1 to 4.2 ms^-1. If mass of object is 5.5 kg, calculate the force applied on the object?
Answer:
Initial velocity, u = 1.8ms^-1
Time taken, t = 1.2s
Final velocity, v = 4.2 ms^-1
Mas of the object = 5.5kg
Force applied, F = \(m \frac{(v-u)}{t}\)
= \(5.5 \frac{(4.2-1.8)}{1.2}=\frac{5.5 \times 2.4}{1.2}\) = 11 N

Question 27.
A bullet of mass 20 kg is horizontally fired from a pistol of mass 2kg with a horizontal velocity of 150 ms^-1. What is the recoil velocity of the pistol?
Answer:
Mass of bullet =20g = 0.02kg
Mass of pistol = 2kg
Before firing, momentum = 0
Therefore u₁ =1
and u₂ =0
Final velocity of bullet = 150ms^-1
Final velocity of pistol = v
Total momentum of pistol and bullet after firing = m₁v₁ + m₂v₂
Initial and final momentum are equal therefore
3.0 + 2v = 0
v = \(\frac {-3}{2}\)1.5ms^-1
The negative sign shows that the direction of the motion of bullet and pistol are in opposite direction.

Question 28.
A girl of mass 40 kg jumps with a horizontal velocity of 5ms-1 into a stationary cart with frictionless wheels. The mass of cart is 3kg. What is her velocity when she moves off the cart?
Answer:
Mass of- girl, m₁ =40kg
Mass of cart, m₂ = 3kg
Initial velocity, u₁ = 5 ms^-1
Initial velocity, u₂ = 0 ms^-1
The momentum of the girl and cart before interaction
= 40 × 5 + 3 × 0
= 200 kg ms^-1
After interaction, velocity of cart + girl = v.
Total momentum after interaction = (m₁, + m₂) v.
= (40 + 3) v = 43v
Total momentum before interaction = Total momentum after interaction.
430v = 200
v = 4.65 ms^-1

Question 29.
Two hockey players of opposite teams, while trying to hit a hockey ball in the ground colloid and immediately become entangled. One has mass of 80 kg and moving with a velocity of 5 0 ms^-1. While the other has mass of 70 kg and is moving with a velocity of 6 ms^-1 towards the first player. In which direction and with what velocity they will move after they combine together? Assume that the friction force acting between the ground and feet of two players is negligible.
Answer:
Mass of Is hockey player = 80kg
It’s initial velocity u = 5 ms^-1
Mass of 2nd hockey player = 70 kg
Initial velocity of 2nd player u₂ = – 6 ms^-1
Therefore total momentum before collision = m₁ + u₂ + m₂u₂
= 80 × 5 + 70 × (-6)
= 400 – 420 = -20 kg ms^-1
Let velocity of two entangled players after collision = v
Total momentum = (m₁ + m₂)v = (80 + 70) v = 150M.
According to law of conservation of momentum. Momentum before collosion Momentum after collision
1500 = -20
v = \(\frac {-20}{150}\) = 0.13 ms^-1
The entangled players will move indirection of the second player.

Question 30.
A motor car of mass 800 kg was moving with a uniform velocity of 60 km/h along a straight line when the driver applied the brakes after seeing a child crossing road to reduce its speed to half of its original speed. Calculate the change in momentum and force applied by the driver in brakes in 2 s.
Answer:
Mass of car =800 kg
Initial velocity, u = 60 km/h = 16.6 m/s
Final velocity of car v = 30.km/h = 8.3 m/s
Time taken =2s
so Acceleration = \(\frac{v-u}{t}=\frac{16.6-8.3}{2}\)
= \(\frac {6.3}{2}\) = 3.15 ms^-2
Change in momentum = mv – mu = m(v – u)
= 800 (16.6 – 8.3) = 6640 kg ms^-1
Force required, F = m × a = 800 × 3.15
= 2520N.

Long Answer Type Questions

Question 1.
Describe an experiment on the basis of which Galileo suggested that objects move with constant speed when no force acts on them.
Answer:
Galileo observed the motion of objects on inclined plane. He observed that when a marble rolls down an inclined plane, its velocity increases due to the unbalanced force of gravity. It attains a definite velocity tire time it reaches the bottom. It’s velocity decreases when it climbs upon an inclined plane as shown in the fig. 9.2. He further noted that on a frictionless plane inclined equally on both side, when a marble is released from one side it would roll down the slope and climb up to the opposite side to the same height from which it was released.

If the angle of inclination on one side is gradually decresed then marble would have to cover a large distance to reach the same height as shown in the fig. (c). If the one side of the plane ultimately is made horizontal (or slope is reduced to zero), the marble would continue to travel forever to reach the same height it was released. These imbalanced forces acting on the marble in this case is zero. I suggests that no imbalanced force is required to keep the marble moving with constant velocity. However, in practical situation, it is difficult to achieve a zero unbalanced force. This is because of the presence of friction for acting in opposite to the direction of motion. Thus, in practice marble stops after tavelling some distance. The effect of friction must be reduced by using smoother marble and smoother plane and providing a lubricant.

Question 2.
State and explain the first law of motion.
Answer:
First law of motion: According to the first law of motion body or an object at rest or in a uniform motion will remain at rest or in uniform motion unless an unbalanced force acts upon it.

Explanation: This law consists of two parts.
(i) A body at rest remains at rest unless a force is applied on it: A book lying on a table or a box lying on a floor remain at rest until someone tries to disturb it which means it will remain in its respective position of rest unless imbalanced force acts upon them.

(ii) The object in uniform motion will continue moving in straight line with same uniform motion unless unbalanced external force acts on it. This part seems to be contrary in our daily life experiences. Usually, we observe that the moving object slow down and come to rest after sometime. It occurs due to the force of friction that opposes the motion of the object.

The ball is released from a certain height on an inclined plane placed on a horizontal plane. Note the distance moved by the ball before it stops. Roll the same ball from the same height on inclined plane placed on a glass sheet and note the distance travelled by the ball before it stops. We will find that the distance covered by ball glass sheet will be longer than the rough surface. This shows that if the resistance of the air and friction force offered by the surface is reduced the ball on will cover more and more distance. If the surface becomes completely frictionless and their is no resistance of air the ball will continue moving with uniform motion in a straight line.

Question 3.
What is the intertia of rest? Demonstrate it with the help of an activity.
Or
How will you demonstrate that an object will remain at rest unless acted upon by an unbalanced force?
Answer:
Intertia of rest: A body always tends to remain in the position of rest or tendency of body to resist any change in its state of rest is known as inertia of rest. It can be demonstrated by the following activities.

Activity 1 : Take an empty glass tumbler and place a thick or stiff playing card on the top of the empty glass tumbler. Place a five rupees coin on the card as shown in the fig 9.3. Give the card a horizontal flick with a finger. The card shoots away allowing the coin to fall vertically into the glass tumbler due to its inertia of rest. The inertia keeps the coin to maintain its state of rest even when the card flow off.

Activity 2 : Make a pile of similar carom coin on a table as shown in the fig. 9.4. Now give a horizontal hit at the bottom of the pile using a striker. If the hit is strong enough then the bottom coins moves quickly and are removed from their place but the rest of the pile will not move horizontally but they fall vertical on the floor due to the inertia of the other coins.

Question 4.
Derive the relation “F = ma” by using Newtons second law of motion. How will you define the S.I. unit of force from this law?
Answer:
According to the second law of motion, the rate of change of momentum of a body is directly proportional to the imbalanced force applied on the body and always takes place in the direction of the force.

Let a body of mass ‘m’ is moving with an initial velocity along a straight line. It is uniformly accelerated to velocity V in the time T by the application of constant force F through the time t.

The initial momentum of the body = mu
Final momentum of the body = mv
Change of momentum ∝ mv – mu
∝ m (v – u)
Change in momentum per unit time = \(\frac{m(v-u)}{t}\)
But change in momentum per unit time = Applied force.
Therefore F ∝ \(\frac{m(v-u)}{t}\)
But \(\frac{(v-u)}{t}\) = a = acceleration produced produced in the body
F ∝ m.a.
F = k.m.a. (where k is constant of proportionality).
One unit of force is defined as the amount that produces an acceleation of 1m/s² in an object of mass 1 kg i.e.
1 (unit of force) = k. (1 kg) × 1 (ms^-2)
k = I
Putting the value of k in equation (2) we have
F =ma

S.I. Unit of force: We can derive the S.I. unit of force by taking the mass 1 kg and acceleration, a = 1ms^-2.
1 unit force = 1 kg × 1 ms^-2 = 1 kg ms^-2
This unit of force was given the name Newton and is denoted by ‘N’.

Question 5.
What is the third law of motion? Give some examples to illustrate it.
Answer:
Newton’s third law of motion: According to the third law of motion, when one object exerts a force on another object then the second object exert a force on the first. These two forces are always equal in magnitude but opposite in direction. This says that the forces always occurs in pairs as result of interaction between them, that is the force act on different objects and not on the same object.

Let two bodies A and B interact with each other. The force exerted by body A on B is called action and the force exerted by the body B on A which is equal to magnitude but opposite in direction is called reaction.

Example: 1. Take two spring balance connected together as shown in the fig. 9.5. The fixed end of the spring balance B attached to a rigid support like ar wall. When a force is applied through the fixed end of the spring balance it shows the same reading on their scale. It means the source exerted by spring balance B on A are equal.

2. A swimmer pushes the water in backward direction with a certain force with his hand (action) and in turn the swimmer is pushed forward due to the reaction of water with the same force.

These NCERT Solutions for Class 9 Science Chapter 8 Motion Questions and Answers are prepared by our highly skilled subject experts to help students while preparing for their exams.

Motion NCERT Solutions for Class 9 Science Chapter 8

Class 9 Science Chapter 8 Motion InText Questions and Answers

Question 1.
An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
Answer:
Yes. An object that has moved through a distance can have zero displacement. Displacement is the shortest measurable distance between the initial and the final position of an object. An object which has covered a distance can have zero displacement, if it comes back to its starting point, i.e., the initial position.

Consider the following situation. A man is walking in a square park of length 20 m (as shown in the following figure). He starts walking from point A and after moving along all the corners of the park (point B, C, D), he again comes back to the same point, i.e., A.

In this case, the total distance covered by the man is 20 m + 20 m + 20 + 20 m = 80 m.
However, his displacement is zero because the shortest distance between his initial and final position is zero.

Question 2.
A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds?
Answer:
The farmer takes 40 s to cover 4 × 10 = 40 m.
In 2 min and 20 s (140 s), he will cover a distance = \(\frac{40}{40}\) × 140 = 140 m
Therefore, the farmer completes \(\frac{140}{40}\) rounds (3 complete rounds and a half round) of the field in 2 min and 20 s.

That means, after 2 min 20 s, the farmer will be at the opposite end of the starting point.
Now, there can be two extreme cases.

Case I: Starting point is a corner point of the field.
In this case, the farmer will be at the diagonally opposite comer of the field after 2 min 20 s.
Therefore, the displacement will be equal to the diagonal of the field.
Hence, the displacement will be \(\sqrt{10^{2}+10^{2}}\) = 14.1 m

Case II: Starting point is the middle point of any side of the field.
In this case the farmer will be at the middle point of the opposite side of the field after 2 min 20 s.
Therefore, the displacement will be equal to the side of the field, i.e., 10 m.
For any other starting point, the displacement will be between 14.1 m and 10 m.

Question 3.
Which of the following is true for displacement?
(a) It cannot be zero.
(b) Its magnitude is greater than the distance travelled by the object.
Answer:
(a) Not true
Displacement can become zero when the initial and final position of the object is the same.

(b) Not true
Displacement is the shortest measurable distance between the initial and final positions of an object. It cannot be greater than the magnitude of the distance travelled by an object. However, sometimes, it may be equal to the distance travelled by the object.

Question 4.
Distinguish between speed and velocity.
Answer:
Speed:

1. It refers to the distance travelled by a body in a unit time.
2. It is a scalar quantity.
3. It is always positive.

Velocity:

1. It refers to the net displacement a body undergoes in a unit time.
2. It is a vector quantity.
3. It can be positive or negative.

Make use of free Average Velocity Calculator to find the average velocity. It will generate the accurate average velocity by taking initial, final velocities.

Question 5.
Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?
Answer:
If the total distance covered by an object is the same as its displacement; then its average speed would be equal to its average velocity.

Question 6.
What does the odometer of an automobile measure?
Answer:
The odometer of an automobile measures the distance covered by an automobile.

Question 7.
What does the path of an object look like when it is in uniform motion?
Answer:
An object having uniform motion has a straight line path.

Question 8.
During an experiment, a signal from a spaceship reached the ground station in five minutes. What was the distance of the spaceship from the ground station? The signal travels at the speed of light, that is, 3 × 10⁸ m s^-1.
Answer:
Time taken by the signal to reach the ground station from the spaceship
= 5 min = 5 × 60 = 300 s
Speed of the signal = 3 × 10⁸ m/s
Speed = Speed = \(\frac{\text { Distance travelled }}{\text { Time taken }}\)
∴ Distance travelled = Speed × Time taken = 3 × 10⁸ × 300 = 9 × 10¹⁰ m
Hence, the distance of the spaceship from the ground station is 9 × 10¹⁰ m.

Question 9.
When will you say a body is in (i) uniform acceleration? (ii) non-uniform acceleration?
Answer:
(i) A body is said to have uniform acceleration if it travels in a straight path in such a way that its velocity changes at a uniform rate, i.e., the velocity of a body increases or decreases by equal amounts in an equal interval of time.

(ii) A body is said to have non-uniform acceleration if it travels in a straight path in such a way that its velocity changes at a non-uniform rate, i.e., the velocity of a body increases or decreases in unequal amounts in an equal interval of time.

Question 10.
A bus decreases its speed from 80 km h^-1 to 60 km h^-1 in 5 s. Find the acceleration of the bus.
Answer:
Initial speed of the bus,
u = 80 km/h = 80\(\frac{5}{18}\) = 22.22 m/s
Final speed of the bus,
v = 60km/h = 60 × \(\frac{5}{18}\) = 16.66 m/s
Time take to decrease the speed, t = 5 s
Acceleration a = \(\frac{v-u}{t}=\frac{16.66-22.22}{5}\)
= -1.112 m/s²
Here, the negative sign- of acceleration indicates that the velocity of the car is decreasing.

Question 11.
A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h^-1 in 10 minutes. Find its acceleration.
Answer:
Initial velocity of the train, u = 0 (since the train is initially at rest)
Final velocity of the train,
v = 40 km/h = 40 × \(\frac{5}{18}\) = 11.11 m/s
Time taken, t = 10 min = 10 × 60 = 600 s
Acceleration,
a = \(\frac{v-u}{t}=\frac{11.11-0}{600}\) = 0.0185 m/s²
Hence, the acceleration of the train is 0.0185 m/s².

Question 12.
What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?
Answer:
The distance-time graph for uniform motion of an object is a straight line (as shown in the following figure).

The distance-time graph for non-uniform motion of an object is a curved line (as shown in the given figure).

Question 13.
What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?
Answer:
When an object is at rest, its distance-time graph is a straight line parallel to the time axis.

A straight line parallel to the x-axis in a distance-time graph indicates that with a change in time, there is no change in the position of the object. Thus, the object is at rest.

Question 14.
What can you say about the motion of an object if its speed-time graph is a straight line parallel to the time axis?
Answer:
Object is moving uniformly.

A straight line parallel to the time axis in a speed-time graph indicates that with a change in time, there is no change in the speed of the object. This indicates the uniform motion of the object.

Question 15.
What is the quantity which is measured by the area occupied below the velocity-time graph?
Answer:
Distance

The graph shows the velocity time graph of a uniformly moving body.
Let the velocity of the body at time (t) be v.
Area of the shaded region = length × breath
Where,
Length = t
Breath = v
Area = vt = velocity × time …….. (i)
We know,
Velocity = \(\frac{\text { Displacement }}{\text { Time }}\)
∴ Displacement = Velocity × Time, ……… (ii)
From equations (i) and (ii),
Area = Displacement
Hence, the area occupied below the velocity-time graph measures the displacement covered by the body.

Class 9 Science Chapter 8 Motion Textbook Questions and Answers

Question 1.
A bus starting from rest moves with a uniform acceleration of 0.1 m s^-2 for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.
Answer:
(a) 12 m/s (b) 720 m
(a) Initial speed of the bus, u = 0 (since the bus is initially at rest)
Acceleration, a = 0.1 m/s²
Time taken, t = 2 minutes = 120 s
Let v be the final speed acquired by the bus.
∴ a = \(\frac{v-u}{t}\)
0.1 = \(\frac{v-0}{120}\)

(b) According to the third equation of motion: v² – u² = 2as
Where, s is the distance covered by the bus (12)² – (0)² = 2(0.1)s
s = 720 m
Speed acquired by the bus is 12 m/s.
Distance travelled by the bus is 720 m.

Question 2.
A train is travelling at a speed of 90 km h^-1. Brakes are applied so as to produce a uniform acceleration of -0.5 m s^-2. Find how far the train will go before it is brought to rest.
Answer:
Initial speed of the train, K = 90 km/h = 25 m/s
Final speed of the train, v = 0 (finally the train comes to rest)
Acceleration = -0.5 m s^-2
According to third equation of motion:
v² = u² + 2 as
(0)² = (25)² + 2 (-0.5) s
Where, s is the distance covered by the train
s = \(\frac{(25)^{2}}{2(0.5)}\) = 625 m
The train will cover a distance of 625 m before it comes to rest.

Question 3.
A trolley, while going down an inclined plane, has an acceleration of 2 cm s^-2. What will be its velocity 3 s after the start?
Answer:
Initial velocity of the trolley, u = 0 (since the trolley was initially at rest)
Acceleration, a = 2 cm s^-2 = 0.02 m/s²
Time, t = 3 s
According to the first equation of motion: ν = u + at
Where, ν is the velocity of the trolley after 3 s from start
ν = 0 + 0.02 × 3 = 0.06 m/s
Hence, the velocity of the trolley after 3 s from start is 0.06 m/s.

Question 4.
A racing car has a uniform acceleration of 4 m s^-2. What distance will it cover in 10 s after start?
Answer:
Initial velocity of the racing car, u = 0 (since the racing car is initially at rest) Acceleration, a = 4 m/s²
Time taken, t = 10 s
According to the second equation of motion: S= ut + \(\frac {1}{2}\)at²
Where, s is the distance covered by the racing car
Hence, the distance covered by the racing car after 10 s from start is 200 m.

Question 5.
A stone is thrown in a vertically upward direction with a velocity of 5 m s^-1. If the acceleration of the stone during its motion is 10 m s^-2 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
Answer:
Initially, velocity of the stone, u = 5 m/s
Final velocity, ν = 0 (since the stone comes to rest when it reaches its maximum height)
Acceleration of the stone, a = acceleration due to gravity, g = 10 m/s²
(in downward direction)
There will be a change in the sign of acceleration because the stone is being thrown upwards.
Acceleration, a = -10 m/s²
Let s be the maximum height attained by the stone in time t.
According to the first equation of motion:
ν = u + at
0 = 5 + (-10) t
∴ t = \(\frac {-5}{10}\) = 0.5 S
According to the third equation of motion: ν² = u² + 2 as
(0)² = (5)² + 2 (-10)s
= S = \(\frac {52}{20}\) = 1.25
Hence, the stone attains a height of 1.25 m in 0.5 s.

Question 6.
An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?
Answer:
Diameter of a circular track, d = 200 m
Radius of the track, r = r = \(\frac{d}{2}\) = 100 m
Circumference = 2πr = 2π (100) = 200π m
In 40 s, the given athlete covers a distance of 200p m.
In 1 s, the given athlete covers a distance = \(\frac{200 \pi}{40} m\)
The athlete runs for 2 minutes 20 s = 140 s
∴ Total distance covered in
s = \(\frac{200 \times 22}{40 \times 7}\) × 140 = 2200m

The athlete covers one round of the circular track in 40 s. This means that after every 40 s, the athlete comes back to his original position. Hence, in 140 s he had completed 3 rounds of the circular track and is taking the fourth round.

He takes 3 rounds in 40 × 3 = 120 s. Thus, after 120 s his displacement is zero.
Then, the net displacement of the athlete is in 20 s only. In this interval of time, he moves at the opposite end of the initial position. Since displacement is equal to the shortest distance between the initial and final position of the athlete, displacement of the athlete will be equal to the diameter of the circular track.

∴ Displacement of the athlete = 200 m
Distance covered by the athlete in 2 min 20 s is 2200 m and his displacement is 200 m.

Question 7.
Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?
Answer:
(a) 2m/s, 2 m/s (b) 1.90 m/s, 0.95 m/s
(a) From end A to end B

Distance covered by Joseph while jogging from A to B = 300 m
Time taken to cover that distance = 2 min 30 seconds = 150 s
Average speed = \(\frac{\text { Total distance covered }}{\text { Total time taken }}\)
Total distance covered = 300 m
Total time taken = 150 s
Average velocity = \(\frac {300}{150}\) = 2m/s
Average velocity = \(\frac{\text { Displacement }}{\text { Time interval }}\)
Displacement = shortest distance between A and B = 300 m
Time interval = 150 s
Average velocity = \(\frac {300}{150}\) = 2m/s
The average speed and average velocity of Joseph horn A to B are the same and equal to 2 m / s.

(b) From end A to end C

Average speed = \(\frac{\text { Total distance covered }}{\text { total time taken }}\)
Total distance covered = Distance from A to B + Distance from B to C
= 300 + 100 = 400 m
Total time taken = Time taken to travel from A to B + Time taken to travel from B to C = 150 + 60 = 210 s
Average speed = \(\frac {400}{210}\) = 1.90 m/s
Average velocity = \(\frac{\text { Displacement }}{\text { Time interval }}\)
Displacement from A to C = AC = AB – BC = 300 – 100 = 200 m
Time interval = time taken to travel from A to B + time taken to travel from B to C
= 150 + 60 = 210 s
Average velocity = \(\frac {200}{210}\) = 0.95 tn / s
The average speed of Joseph from A to C is 1.90 m/s and his average velocity is 0.95 m/s.

Question 8.
Abdul, while driving to school, computes the average speed for his trip to be 20 km h^-1. On his return trip along the same route, there is less traffic and the average speed is 30 km h^-1. What is the average speed for Abdul’s trip?
Answer:
Case I: While driving to school
Average speed of Abdul’s trip = 20 km/h
Average speed = \(\frac{\text { Total distance }}{\text { Total time taken }}\)
Total distance = Distance travelled to reach school = d
Let total time taken = t₁
∴ 20\(\frac{d}{t_{1}}\)
t₁ = \(\frac{d}{20}\) ………(i)

Case II: While returning from school
Total distance = Distance travelled while returning front school = d
Now, total time taken = t₂
30 = \(\frac{d}{t_{2}}\) ……….(ii)
Average speed for Abdul’s trip = \(\frac{\text { Total distance covered in the trip }}{\text { Total timetaken }}\)

Where,
Total distance covered in the trip = d + d = 2d
Total time taken, t = Time taken to go to school + Time taken to return to school
= t₁ + t₂
∴ Average speed = \(\frac{2 d}{t_{1}+t_{2}}\)
From equations (i) and (ii),
Average Speed = \(\frac{2 d}{\frac{d}{20}+\frac{d}{30}}=\frac{2}{\frac{3+2}{60}}\)
Average Speed = \(\frac {120}{5}\) =24 m/s
Hence, the average speed for Abdul’s trip is 24 m/s.

Question 9.
A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s^-2 for 8.0 s. How far does the boat travel during this time?
Answer:
Initial velocity, u = 0 (since the motor boat is initially at rest)
Acceleration of the motorboat, a = 3 m/s²
Time taken, t = 8 s
According to the second equation of motion:
s = ut + \(\frac {1}{2}\)at²
Distance covered by the motorboat, s
s = 0 + \(\frac {1}{2}\)3(8)² = 96 m
Hence, the boat travels a distance of 96 m.

Question 10.
A driver of a car travelling at 52 km h^-1 applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km h^-1 in another car applies his brakes slowly and stops in 10 s. On the same graph paper, plot the speed versus time graphs for the two cars. Which of the two cars travelled farther after the brakes were applied?
Answer:
Case A:
Initial speed of the car, u₁ = 52 km/h = 14.4 m/s
Time taken to stop the car, t₁ = 5 s
Final speed of the car becomes zero after 5 s of application of brakes.

Case B:
Initial speed of the car, u₂ = 3 km/h = 0.833 m/ss ≅ 0.83 m/s
Time taken to stop the car, t₂ = 10 s
Final speed of the car becomes zero after 10 s of application of brakes.
Plot of the two cars on a speed-time graph is shown in the following figure:

Distance covered by each car is equal to the area under the speed-time graph.
Distance covered in case A,
s1 = \(\frac {1}{2}\) × OP × OR = \(\frac {1}{2}\)14.4 × 5 = 36 m
Distance covered in case B,
s2 = \(\frac {1}{2}\) × OS × OQ = \(\frac {1}{2}\) × 0.83 × 10 = 4.15m
Area of ΔOPR > Area of ΔOSQ
Thus, the distance covered in case A is greater than the distance covered in case B.
Hence, the car travelling with a speed of 52 km/h travels farther after brakes were applied.

Question 11.
Fig 8.11 shows the distance-time graph of three objects A, B and C. Study the graph and answer the following questions:

(a) Which of the three is travelling the fastest?
(b) Are all three ever at the same point on the road?
(c) How far has C travelled when B passes A?
(d) How far has B travelled by the time it passes C?
Answer:
(a) Object B
(b) No
(c) 5.714 km
(d) 5.143 km

(a) Speed = \(\frac{\text { Distance }}{\text { Time }}\)
slope of the graph = \(\frac{y-\text { axis }}{x- \text { axis }}=\frac{\text { Distance }}{\text { Time }}\)
∴ Speed = slope of the graph
Since slope of object B is greater than objects A and C, it is travelling the fastest.

(b) AD three objects A, B and C never meet at a single point. Thus, they were never at the same point on road.

(c)

On the distance axis:
7 small boxes = 4 km
∴ 1 small box = \(\frac {4}{7}\) km
Initially, object C is 4 blocks away from the origin.
∴ Initial distance of object C from origin = \(\frac {16}{7}\) km
Distance of object C from origin when B passes A = 8 km
Distance covered by C
= \(8-\frac{16}{7}=\frac{56-16}{7}=\frac{40}{7}\) = 5.714 km
Hence, C has travelled a distance of 5.714 km when B passes A.

(d)

Distance covered by B at the time it passes C = 9 boxes
= \(\frac {4}{7}\) × 9 = \(\frac {36}{7}\) = 5.143 km
Hence, B has travelled a distance of 5.143 km when B passes A.

Question 12.
A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 m s^-2, with what velocity will it strike the ground? After what time will it strike the ground?
Answer:
Distance covered by the ball, s = 20 m
Acceleration, a= 10 m/s
Initially, velocity, u = 0 (since the ball was initially at rest)
Final velocity of the ball with which it strikes the ground, v
According to the third equation of motion:
_ν2 = u₂ + 2 as
_ν2 = 0 + 2 (10) (20)
v = 20 m/s
According to the first equation of motion:
ν = u + at
Where,
Time, f taken by the baU to strike the ground is,
20 = 0 + 10(t)
t = 2s
Hence, the ball strikes the ground after 2 s with a velocity of 20 m/ s.

Question 13.
The speed-time graph for a car is shown is Fig. 8.12.

(a) Find out how far the car travels in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.
(b) Which part of the graph represents uniform motion of the car?
Answer:
(a)

The shaded area which is equal to represents the distance travelled by the car in the first 4 s.

(b)

The part of the graph in red colour between time 6 s to 10 s represents uniform motion of the car.

Question 14.
State which of the following situations are possible and give an example for each of these:
(a) an object with a constant acceleration but with zero velocity.
(b) an object moving in a certain direction with an acceleration in the perpendicular direction.
Answer:
(a) Possible:
When a ball is thrown up at maximum height, it has zero velocity, although it will have constant acceleration due to gravity, which is equal to 9.8 m/s².

(b) Possible:
When a car is moving in a circular track, its acceleration is perpendicular to its direction.

Question 15.
An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the earth.
Answer:
Speed = \(\frac{\text { Distance }}{\text { Time }}\)
r Time
Distance = 2πr = 2 × 3.14 × 42 × 42250 = 265330 km
Time = 24 h
Speed = \(\frac{265330}{24}\) = 11055.4 km/h

Class 9 Science Chapter 8 Motion Additional Important Questions and Answers

Multiple choice Questions

Question 1.
The change in velocity of an object per unit time is called
(a) Average velocity
(b) Acceleration
(c) Force
(d) Momentum
Answer:
(b) Acceleration

Question 2.
If the distance-time graph of a body ¡s parabola, then its motion is
(a) Uniform
(b) Non-uniform
(d) Body’s in rest
(d) Body is in vibratory motion
Answer:
(a) Uniform

Question 3.
Retardation is
(a) Positive acceleration
(b) Zero acceleration
(c) Non-uniform acceleration
(d) Zero acceleration
Answer:
(b) Zero acceleration

Question 4.
If the displacement of an object is proportional to square of time, then the object moves with
(a) uniform velocity
(b) uniform acceleration
(c) increasing acceleration
(d) decreasing acceleration,
Answer:
(b) uniform acceleration

Question 5.
Suppose a boy is enjoying a ride on a merry-go-round which is moving with a constant speed of 10 ms^-1. It implies that the boy is
(a) at rest
(b) moving with no acceleration
(c) in accelerated motion
(d) moving with uniform velocity
Answer:
(c) in accelerated motion

Question 6.
In which of the following cases of motions, the distance moved and the magnitude of displacement are equal?
(a) If the car is moving on straight road
(b) If the car is moving in circular path
(c) The pendulum is moving to and fro
(d) The earth is revolving around the Sun
Answer:
(a) If the car is moving on straight road

Question 7.
When a body completes one round in a circular path of radius ‘r’, its displacement is equal to
(a) r
(b) 2r
(c) 2nr
(d) Zero
Answer:
(d) Zero

Question 8.
The S.I. unit used to express acceleration is
(a) ms
(b) ms^-1
(c) Ms^-2
(d) m²m^-1
Answer:
(c) Ms^-2

Question 9.
A body falls freely from a point and reaches the ground in 1 second, then the height of the point from the earth is equal to
(a) 9.8 m
(b) 4.9 m
(c) 10m
(d) 19.6 m
Answer:
(b) 4.9 m

Question 10.
A body starts moving from rest with a uniform acceleration of 2ms^-2.
(a) 6 m
(b) 9m
(c) 12 m
(d) 18 m
Answer:
(b) 9m

Question 11.
Which of the following figures (Fig.) represents uniform motion of a moving object correctly?

Answer:

Question 12.
Which of the following is a vector quantity?
(a) Speed
(b) Distance
(c) Velocity
(d) energy
Answer:
(a) Speed

Very Short Answer Type Questions

Question 1.
How will you describe the position of an object?
Answer:
The position of an object can only be described with respect to a fixed point called reference point or origin.

Question 2.
When is the motion of a body said to be rectilinear motion?
Answer:
When a body moves in a straight line, its motion is said to be rectilinear motion.

Question 3.
Define velocity. Give its S.I. unit.
Answer:
Velocity is defined as the displacement produced per unit time. It’s S.I. unit is ms^-1.

Question 4.
Define acceleration. Give its S.I. unit.
Answer:
The change in velocity of an object per unit time is called its acceleration.
Acceleration = \(\frac{\text { Final velocity – Intitial velocity }}{\text { Time }}\)
It’s S.I. unit is ms^-2.

Question 5.
What does the slope of the following graphs indicate?
(a) Distance-time graph passing through the origin and making an angle with the time axis.
(b) Speed-time graph making an angle with lime axis and a straight line passing through the origin.
Answer:
(a) Speed of the object.
(b) Acceleration of the object.

Question 6.
Draw a velocity-time graph showing negative acceleration of the body.
Answer:

Question 7.
The graph of two friends A and B are shown in the figure 8.13 below:

(a) Which of the two friends is moving faster?
(b) Where and when will they meet?
Answer:
(a) A is moving faster than B because of its greater slope.
(b) The two will meet after 4 minutes at distance of 80 meters.

Question 8.
What does the following graph indicate?
(a) Distance-time graph is a straight line making an angle with time-axis
(b) Speed-time graph is a straight line making an angle with the time-axis and passing through the origin.
Answer:
(a) Uniform motion of body.
(b) Uniform positive acceleration.

Question 9.
Draw a graph describing the osillatory motion of a body.
Answer:

Question 10.
What is the nature of uniform circular motion?
Answer:
The uniform circular motions are always accelerated motion.

Short Answer Type Questions

Question 1.
When is a body said to be in motion? Explain with an example that the motion of a body is relative to the position of other objects.
Answer:
A body is said to be in motion if its position continues to change with the surroundings. Motion is relative i.e. an object/ body may appeard to be moving to some but not others. Such as two students travelling in a bus while sitting on same seat. For people outside they are in motion but with respect to each other, the two friends are in state of rest.

Question 2.
What do you mean by distance and displacement? How are they different?
Answer:
The length of the actual path travelled or covered between initial and final position of an object is called distance.
The change in position of an object is a given direction from its initial position is called displacement.

The differences between distance and displacements are:

Question 3.
A bus starts from point A and goes to point B after travelling 60 km on a straight road and then returns to point A.
(i) What is the total distance covered by the bus?
(ii) What is its displacement?
Answer:
(i) Total distance travelled by bus from A to B and back to point A, then
AB + BA = 60m + 60m = 120m
(ii) The net displacement is zero because the final position coincides with initial position.

Question 4.
What do you mean by uniform and non-uniform acceleration?
Answer:
Uniform motion: When a body covers equal distance in equal interval of time, how small time interval may be, the motion is called uniform motion.

If a body covers 60 km in an hour moving with a uniform motion, then it would cover 30 km in half hour and 15 km in 15 minutes.

Non-uniform motion: A body is said to be in non-uniform motion which it covers unequal distances in equal interval of time or regular changes its direction. Most of the motion we see in daily life like that of vehicles on city roads are non-uniform motion.

Question 5.
What is the speed and average speed of a body in motion?
Answer:
The distance covered by a body in motion in unit time is called its speed.
Speed = \(\frac{\text { Distance travelled }}{\text { Time }}\)
If a car covers 40 km in 1st hours, 60 km in 2nd hour and 50 km in 3rd hour, then its, average speed is
\(\frac{40+60+50}{3}\) = 50 km/h

Question 6.
On a stormy night, the ligtning is seen first and thunder is heard later, though they are produced simultaneously at same time. Why?
Answer:
The lightning is seen first and thunder is heard later because of the difference in speed of light and sound. The light travelling with speed of 3 × 10⁸ ms^-1 reaches the earth earlier than sound travelling with speed of 340 ms^-1.

Question 7.
On a rainy day, the sound of thunder is heard after 5 seconds of the flash of lightning. :
Calculate the distance of the nearest point of ligtning.
Answer:
In atmosphere, lightning and thunder occur at same time.
Time taken by sound to reach to earth = 5s
Speed of sound = 346 ms^-1
∴ Distance of nearest point of lightning = 346 × 5 = 1730 m

Question 8.
What do you mean by positive and negative acceleration? Give examples.
Answer:
When the acceleration of a body increases with time, it is called positive aceleration such as the free falling body experiences an increase in its velocity by 9.8 m/s, every second.

When the acceleration of a body decreases with time, it is called negative acceleration such as the stopping of car after application of breakes in time t, covering a distance of s metres.

Question 9.
Give examples of the following motion:
(a) Uniform motion in a straight line in which the velocity is increasing
(b) A body moving with constant speed in which velocity is changing at uniform rate.
(c) Non-uniform motion with constant acceleration in the direction of motion.
(d) Non-uniform motion in which acceleration is not constant.
Answer:
(a) The motion of a train between two stations on a straight track between certain interval of time when its velocity becomes constant is an example of uniform motion in which velocity remains constant.

(b) When a body is moving in a circular path with constant speed, its direction changes at every point hence, due to change in its direction, its velocity also changes simultaneously.

(c) The motion of a free falling body is an example of non-uniform motion with the constant acceleration in tire direction of motion as it fall downward with constant acceleration of 9.8 ms^-2 due to gravity of the earth.

(d) When a train starts moving from a railway station, it increases its speed by unequal amount in equal interval of time till it attains the maximum speed.

Question 10.
What is the importance of graphs in daily life.
Answer:
A graph provides a convenient method to represent pictorially the basic information about an event. A graph can be bar graph, line graph or a pie chart. These graphs are used in:

• Comparing and solving the linear motion and equations with two variable.
• Finding the motion of a body to be uniform or non-uniformotion.
• Calculating the distance travelled by a body from speed-time graph.

Question 11.
Draw the distance-time graph of a uniform motion. How will you calculate the speed of the object from this graph.

Answer:
Tire given graph is the distance time graph. To determine the speed of object between A and B, draw,a line from A, parallel to X-axis and an another line parallel to speed-axis from point B. These two lines meet at point C and form a triangle ABC. In ΔDBC, AC denotes the time interval and BC denotes the distance travelled. Hence,
Speed (V) = \(\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{S_{2}-S_{1}}{t_{2}-t_{1}}\) Sloped line A

Question 12.
The time of arrival and departure of a train at three station A, B and C and their distances from station A are given the table below. Plot the distance for graph for the motion of the tain assuming that its motion between the onation is uniform.

Answer:
The distance timetable for the motion of the train

Question 13.
Megha and his sister Anu goes to the school on bicycle. Both of them start at the same time but take different time to reach the school although they follow the same route. Table below shows the distance moved by both at different instant of time, plot the distance-time graph for their motion on the same graph and interpret the graph.

Answer:

The distance-time gaph shows both are moving with non-uniform motion. The average speed of Megha is greater than that of Anu.

Question 14.
The average walking speed of a boy while going to school is 4 km/h. It take half an hour to reach the school. Calculate the school’s distance from home.
Answer:
Average speed of the boy = 4 km/h
Time taken to reach school = \(\frac {1}{2}\) hour
Distance of school from house = Speed × Time
= 4 km/h × \(\frac {1}{2}\) hour = 2 km

Question 15.
The odometer of a car reads 2000 km from the start of a trip by John. The trip takes 8 hours and shows reading of 2400 km. What is the average speed of John’s car in km/h and m/ s?
Answer:
Distance covered by car = Final reading – Initial read
= 2400 – 2000 = 400 km
Time taken by John to cover 400 km = 8 hr
∴ Average speed = \(\frac{400 \mathrm{~km}}{8 \mathrm{hr}}\) = 50km/hr
Average speed (in ms^-11) \(\frac{50 \times 1000}{60 \times 60}\) = 13.9 ms^-1

Question 16.
Usha swims in a 90 m long swimming pool. She covers 180 m in one minute swimming from one end of pool and back to the pool along a straight path. Find the average speed and average velocity?
Answer:
Total distance covered by Usha in one minium = 180 m.

Question 17.
Narang is moving his car with a velocity of 90 km/h. How much distance will he cover in (a) one minute (b) one second.
Answer:
Velocity of car = 90 km/h
(a) Distance covered in one minute
= \(\frac{90}{60}=\frac{3}{2}=\frac{3 \times 1000}{2}\) = 1500 m
(b) Distance covered in one second
\(\frac{90}{60 \times 60} k m=\frac{90 \times 1000}{60 \times 60}\) = 25 m

Question 18.
An electric train is moving with an aver age velocity of 120 km/h. How much distance will it cover in 30s ?
Answer:
Velocity of train (V) = 120 km/h.
= \(\frac{120 \times 1000}{60 \times 60}=\frac{100}{3} \mathrm{~m} / \mathrm{s}\)
Time taken = 30s
∴ Distance covered in 30s
= Average × Time velocity = \(\frac {100}{3}\) × 30 = 1000 m

Question 19.
In a long distance race, the athletes were expected to take four rounds of the track such that the line of finish was same as the line of start. The length of track is 200 m.
(a) Calculate the total distance covered by the athletes.
(b) Displacement of athletes when they touch the finishing line.
(c) Is the motion of athletes uniform or non-uniform?
Answer:
(a) In four rounds, total distance covered
= 4 × length of track = 4 × 200 = 800 m
(b) Displacement upon touching the finishing line would be zero as track is a circular track.
(c) Motion of athletes would be non-uniform as they are not completing the race at the same time i.e. distance covered by them in equal interval of time is not equal.

Question 20.
A circular track has a circumference of 314 m with AB as one of its diameter. A cyclist traveels from A to B along the circular path in 30s. Find
(a) Speed of cyclist.
(b) Displacement of cyclist.

Answer:
(a) Circumference of circular track = 314
∴ Distance from A to B = \(\frac {314}{2}\) = 157 m.
Time taken by cyclist = 30 s.
Time = \(\frac{\text { Distance }}{\text { Time }}=\frac{157}{30}\) = 5.23 m/s

(b) Diplsacement of cyclist from A to B which represents diameter of circular track
2πr = 314
2 × 3.14 × r =314
r = \(\frac{314}{3.14 \times 2}\) = 50m
When r = 50 m, d = 2r = 50 × 2 = 100 m

Question 21.
A car travels a certain distance at an average speed of 30 km/h and returns back with an average speed of 50 km/h. Calculate the average speed of car.
Answer:
Let distance covered by car travelling at 30 km/h = x
∴ Time taken = \(\frac{x}{30}\)h
Similarly when coming back time taken = \(\frac{x}{50}\)h
Total time taken = \(\frac{x}{30}+\frac{x}{50}=\frac{5 x+3 x}{150}=\frac{8 x}{150} h\)
Total distance covered = x + x = 2x
Average speed of car = \(\frac{2 x}{8 x / 50}=\frac{2 x \times 150}{8 x}=\frac{75}{2}\) = 37.5 km/h

Question 22.
Rahul starting from rest paddles his bicycle to attain a velocity of 6 m s^-1 in 30s. Then he applies brakes so that the velocity of bicycle comes to 4ms^-1 in next 5 seconds. Calculate acceleration of bicycle in both cases.
Answer:
In first case, initial velocity (u) = 0
Final velocity (ν) =6 ms^-1
Time taken = 30s
Acceleration (a) = ?
We know that,
a = \(\frac{v-u}{t}\)
d = \(\frac{6-0}{30}=\frac{1}{5}\) = 0.2 ms^-1
In second case, u = 6ms^-1, ν = 0.2 ms^-1,
Time = 68
∴ a = \(\frac{4-6}{5}=\frac{-2}{5}\) = -0.4 ms^-2

Question 23.
A train starting from rest attains a velocity of 72 km/h in 5 minutes. Assuming that the acceleration is uniform, find (i) the acceleration and distance travelled by the train while attaining this velocity.
Answer:
Initial velocity, u = 0 final velocity (ν) = 72 km/h
Time (t) = 5 minutes = 5 × 60 = 300 s
Acceleration (a) = ?
a = \(\frac{v-u}{t}=\frac{20-0}{300}=\frac{1}{15} m s^{-2}\)
Let distance travelled by train = S
2as = ν² – u²
2 × \(\frac {1}{5}\) × 5 = (20)² (72 km/h = 20 mis)
S = 2 × \(\frac {1}{5}\) × 5 = 3000m = 3 km.

Question 24.
A can accelerates uniformly from 18 km/h to 36 km/h in 5 seconds. Calculate (i) The acceleration, (ii) Distance covered by car is that time.
Answer:
Initial velocity of car (h) = 18 km/h
= \(\frac{18 \times 100}{60 \times 60} 5 \mathrm{~m} / \mathrm{s}\)
Final velocity of car (ν) =36 km/h
= \(\frac{36 \times 100}{60 \times 60}=10 \mathrm{~m} / \mathrm{s}\)
Time (t) = 5s.
(i) ∴ Acceleration, a = \(\frac{v-u}{t}=\frac{10-5}{5} 1 \mathrm{~ms}^{-2}\)
(ii) Let distance covered by car = S
2as = ν² – u²
2 × 1 × s = (10)² – (5)²
2 × 1 × s = 100 – 25
s = \(\frac {75}{2}\) = 37.5m

Question 25.
The brakes applied to a car produce a negative acceleration of – 6 m/s². The car takes 2 seconds to stop after applying the brakes. Calculate the distance it would cover before coming to a stop.
Answer:
Acceleration of car, a = – 6 ms^-2
Time taken by car to stop = 2s
Final velocity ν = 0
We know that
ν = u + at
= 4 – 6 × 2
or u = 12m/s

Let distance covered = s
u = ut + \(\frac {1}{2}\)at^–
s = 12 × 2 + \(\frac {1}{2}\) × (6) × (2)^–
= 24 + 12 = 36 m.

Question 26.
Find the initial velocity of car which is stopped in 10 seconds after appplication of brakes. Also calculate the distance travelled by the car if its negative acceleation is – 2.5 ms^-2.
Answer:
Final velocity (ν) = 0
Let initial velocity =u
Acceleration (a) = -2.5 ms^-2
Time taken (t) = 10 seconds
We know that,
ν = u + at
0 = u – 2.5 × 10
u = 25 ms^-1

Let distance covered = s
s = ut + \(\frac {1}{2}\)at²
= 25 x 10 –\(\frac {1}{2}\) × 2.5(10)²
= 250 + 125 = 375 m

Question 27.
A ball is thrown upward with initial velocity of 90 km/h in upward direction. Calculate
(i) Maximum height reached.
(ii) Time taken to reach the maximum height.
Answer:
Initial velocity of ball, u = 90 km/h
= \(\frac{90 \times 1000}{3600}=25 \mathrm{~m} / \mathrm{s}\)
Final velocity of ball, ν = 0 (at max height)
Acceleration, a = -10ms^-2
Let the height gained by ball = s
2as = ν² – u²
2 × (-10) × s = 0 – (25)²
– 20S = – 625
s = \(\frac {625}{20}\) = 31.25 m
Let time taken to achieve the height = t
ν = u + at
0 = 25 – 10 × t
t = \(\frac {25}{10}\) = 2.5

Question 28.
Can the average speed of a moving object be zero? Can the average velocity of a moving object be zero? Give example.
Answer:
The average speed of a moving object cannot be zero while the average velocity of a moving object can be zero.
A car start from point A and travels 40 km in an hour and then returns to point A again in 40 minutes.

Then Average speed = \(\frac{(40+40) \mathrm{km}}{1.40 \mathrm{hr}}=\frac{80}{1.40} \mathrm{hr}=57.14 \mathrm{~km} / \mathrm{hr}\)
The displacement produced after the comple-tion of return to point A = 0.
Average velocity = \(\frac{\text { Total displacement }}{\text { Total time }}\) = 0.

Question 29.
A radar spots an enemy plane. A radio pulse emitted by radar and reflected by the plane reaches back in 0.5 x 10^-3 s. Find the distance between plane and radar.
(Speed of radar pulse = 3 × 10⁸ m/s)
Answer:
Total time taken for pulse to reach and get reflected = 0.5 × 10^-3 s
= \(\frac{0.5 \times 10^{-3}}{2}\)
Distance = Speed × Time
= 3 × 10⁸ × \(\frac{0.5 \times 10^{-3}}{2}\) = 0.75 × 10⁵m = 7.5 km

Question 30.
Draw a velocity time graph for a uniformly accelerated motion. How will you determine the distance covered by a body in a given interval of time using the graph.
Answer:

The velocity time graph for uniformly accelerated motion is a straight line curve passing through the origin and makes an angle with time axis. Using the graph, the distance travelled by a body between any given time interval can eas easily be determined.

Let the time interval be t₁ and t₂ The distance covered would be equal to area of ABCDE
Area of ABCDE = Area of ΔADE + Area of rectangle ABCD
= \(\frac {1}{2}\) (AD × DE) + (AB × BC)

Long Answer Type Questions

Question 1.
Draw a velocity time graph of a car moving with uniform velocity of 40 km/h. Using the graph determine the distance travelled by car in a given time interval of t₁ and t₂.
Answer:
The velocity-time graph of a car moving with the uniform velocity shown in the figure the graph is parallel to time-axis.

To determine the distance travelled between the time interval t1 and t2, perpendicular are drawn from point t₁ and t₂ on the graph.
Distance travelled = Velocity × time
= AC × CD = 40 × (t₂ – t₁) km
= Area of rectangle ABCD

Question 2.
Derive the relation V = u + at from the velocity-time graph for uniform motion.
Answer:
The velocity-time graph for a uniformly accelerated motion is given in figure. A represents

initial velocity and B represent final velocity in time t shown by OC.
Draw AD || OC, so that
BD + DC = BC or
BA + CA = BC
If BC = u, OA = u
Them ν = BD + u
ν – u = BD ….. (i)
But accleration (a) = \(\frac{\text { Change in velocity }}{\text { Time }}\)
= \(\frac{\mathrm{BD}}{\mathrm{AD}}=\frac{\mathrm{BD}}{\mathrm{BC}}\)
or a = \(\frac{\mathrm{BD}}{\mathrm{t}}\)
BD = at
or ν = u + at

Question 3.
Derive the relation s = ut + 1at² from velocity-time graph of a uniform accelerated motion.

Answer:
The velocity-time graph for a uniform accelerated motion is shown in figure. Initial velocity = u, at point A,
Final velocity = ν, at point B
Time = t
From point, perpendicular BL is drawn on time-axis and BE on velocity-axis respectively. OA represents initial velocity OE or BC represents final velocity in time interval, t or OC.

The distance travelled by object is given by the area enclosed with in OBC under the velocity time graph AB.
Area of OBC = Area of rectangle OADC + Area of ΔADB
= (OA × OC) + \(\frac {1}{2}\)(AD × BD)
Acceleration (a)

Question 4.
Derive the relation v² – u² = 2a from velocity-time graph of a uniformly accelerated motion.
Answer:
The velocity time graph of an acclerated motion is shown in the figure.
The distance covered by object moving under uniform accleration ‘a’ is given by area of trapezium OABC
∴ Distance s = \(\frac {1}{2}\)(OA + BC) × OC
OA = u, BC = ν and OC = t
s = \(\left(\frac{u+v}{a}\right) \times t\)
But we know ν = u + at

These NCERT Solutions for Class 9 Maths Chapter 15 Probability Ex 15.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 15 Probability Exercise 15.1

Question 1.
In a cricket match, a batswoman hits a boundary 6 times out of 30 balls she plays. Find the probability that she did not hit a boundary.
Solution:
Total number of balls she plays = 30
Total number of balls that she hits a boundary = 6
The total number of balls that she does not hit a boundary = 30 – 6 = 24 balls.
Let us denote the event that she does not hit a ball is E: So,
P(E) = \(\frac{24}{30}=\frac{4}{5}\)

Question 2.
1500 families with two children were selected randomly, and the following data were recorded.

Compute the probability of a family, chosen at random having.
(i) 2 girls
(ii) 1 girl
(iii) No girls
Also, check whether the sum of these probabilities is 1.
Solution:
Total number of families = 1500
(i) Total number of families which have 2 girls are 475.
∴ P(2 girls in a family) = \(\frac{475}{1500}=\frac{19}{60}\)

(ii) Total number of families which have 1 girl is 814.
∴ P(1 girl in a family) = \(\frac{814}{1500}=\frac{407}{750}\)

(iii) Total number of families which ahve no girl is 211.
∴ P(no girl in a family) = \(\frac{211}{1500}\)

Check:
Sum of probabilities = \(\frac{9}{60}+\frac{407}{750}+\frac{211}{1500}\)
= \(\frac{475+814+211}{1500}\)
= \(\frac{1500}{1500}\)
= 1

Question 3.
Refer to example 5, section 14.4, chapter 14 final the probability that a student of the class was born in August.
Solution:
Total number of students who were born in August = 6
Total number of students = 40
Therefore,
P(Number of students bom in August) = \(\frac{6}{40}\) = \(\frac{3}{20}\)

Question 4.
Three coins are tossed simultaneously 200 times with the following frequencies of different outcomes:

If the three coins are simultaneously tossed again, compute the probability of 2 heads coming up.
Solution:
Since the coin is tossed 200 times, so the total number of trials is 200.
Let us call the events of getting 2 heads is E.
Then the number of times E happens, i.e. two heads coming up is 72.
So, P(E) = \(\frac{72}{200}=\frac{9}{25}\)

Question 5.
An organization selected 2400 families at random and surveyed them to determine a relationship between income level and the number of vehicles in a home. The information gathered is listed in the table below:

Suppose a family is chosen. Find the probability that the family chosen is:
(i) earning Rs. 10,000 – 13,000 per month and owning exactly 2 vehicles.
(ii) earning Rs. 16000 or more per month and owning exactly 1 vehicle.
(iii) earning less than Rs. 7000 per month and does not own any vehicle.
(iv) earning Rs. 13,000 – 16,000 per month and owning more than 2 vehicles.
(v) owning not more than 1 vehicle.
Solution:
The total number of families = 2400.
(i) The number of families earning Rs. 10,000 – 13,000 per month and owning exactly two vehicles is 29.
So, P(family earning Rs. 10,000 – 13,000 per month and owning exactly two vehicles) = \(\frac{29}{2400}\)

(ii) The number of families earning Rs. 16,000 or more per month and owning exactly one vehicle is 579.
So, P(earning Rs. 16,000 or more per month and owning exactly one vehicle) = \(\frac{579}{2400}\)

(iii) The number of families earning less than Rs. 7000 per month and does not own any vehicle is 10.
So, P(earning less than Rs. 7000 per month and does not own any vehicle) = \(\frac{10}{2400}=\frac{1}{240}\)

(iv) The number of families earning Rs. 13,000 – 16,000 per month and owning more than two vehicles is 25.
So, P(earning Rs. 13,000 – 16,000 per month and owning more than two vehicles) = \(\frac{25}{2400}=\frac{1}{96}\)

(v) The number of families owning not more than 1 vehicle is 2062 (family owning exactly one vehicle + family owning no vehicle)
So, P(owning not more than one vehicle) = \(\frac{2062}{2400}=\frac{1031}{1200}\)

Question 6.
Refer to table 14.7, chapter 14
(i) Find the probability that a student obtained less than 20% in the mathematics test.
(ii) Find the probability that a student obtained marks 60 or above.
Solution:
(i) Total number of students = 90
Number of students who gets less than 20% marks in mathematics test = 7.
∴ P(student obtained less than 20% marks) = \(\frac{7}{90}\)

(ii) Number of student getting 60 or above marks is 23.
∴ P(student obtained 60 or above marks) = \(\frac{23}{90}\)

Question 7.
To know the opinion of the students about the subject statistics, a survey of 200 students was conducted. The data is recorded in the following table.

Find the probability that a student is chosen at random
(i) like statistics
(ii) does not like it.
Solution:
The total number of students = 200.
(i) Total number of students like statistics = 135
So, P(like statistics) = \(\frac{135}{200}=\frac{27}{40}\)

(ii) Total number of students does not like statistics = 65.
So, P(does not like statistics) = \(\frac{65}{200}=\frac{13}{40}\)

Question 8.
Refer to Q.2, Exercise-14.2. What is the empirical probability that an engineer lives:
(i) less than 7 km from her place of work?
(ii) more than or equal to 7 km from her place of work?
(iii) within \(\frac {1}{2}\) km. from her place of work?
Solution:
(i) Total number of female engineers whose residence to their place of work given is 40.
The number of engineers whose residence is less than 7 km from their place of work = 9.
∴ P(less than 7 km from her place of work) = \(\frac{9}{40}\)

(ii) Number of engineer who’s residence is more than or equal to 7 km from her place of work = 31.
∴ P(more than or equal to 7 km from place of work) = \(\frac{31}{40}\)

(iii) Number of engineer who’s residence is within \(\frac {1}{2}\) km. from her place of work = 0.
∴ P(within \(\frac {1}{2}\) km. from place of work) = \(\frac {0}{40}\) = 0

Question 9.
Activity: Note the frequency of two-wheelers, three-wheelers, and four-wheelers going past during the time interval, in front of your school gate. Find the probability that any one vehicle out of the total vehicles you have observed is a two-wheeler.
Answer:
Let the person saw the vehicles for half an hour.
The number of vehicles seen by the person.
Two wheelers = 5
Three wheelers = 10
Four wheelers = 15
So,total number of space = 5 + 10 + 15 = 30
Number of events when two wheeler is seen = 5
Probability = \(\frac{\text { no. of event }}{\text { no. of space }}\)
= \(\frac{5}{30}\)
= \(\frac{1}{5}\)

Question 10.
Activity: Ask all the students in your class to write a three-digit number. Choose any student from the room at random, what is the probability that the number written by her/him is divisible by 3? Remember that a number is divisible by 3 if the sum of its digits is divisible by 3.
Solution:
The total number of students in my class = 40.
Out of 40 students, 10 students write a three-digit number which is divisible by 3.
∴ P(student write a number which is divisible by 3) = \(\frac{10}{40}=\frac{1}{4}\)

Question 11.
Eleven bags of wheat flour, each marked 5 kg. actually contained the following weights of flour (in kg): 4.97, 5.05, 5.08, 5.03, 5.00, 5.06, 5.08, 4.98, 5.04, 5.07, 5.00
Find the probability that any of these bags chosen at random contains more than 5 kg. of flour.
Solution:
The total number of wheat flour bags is 11.
The total number of wheat flour bags which contain more than 5 kg. of flour is 7.
∴ P(bags contain more than 5 kg of flour) = \(\frac{7}{11}\)

Question 12.
In Q.5 Exercise 14.2, you were asked to prepare a frequency distribution table, regarding the concentration of sulphur dioxide in the air in parts per million of a certain city for 30 days. Using this table, find the probability of the concentration of sulphur dioxide in the interval 0.12 – 0.16 on any of these days.
Solution:
The total number of observations = 30.
Number of observation lie in the interval 0.11 – 0.16 is 2.
∴ P(The concentration of sulphur dioxide in the interval 0.12 – 0.16 = \(\frac{2}{30}=\frac{1}{15}\)

Question 13.
In Q. 1, Exercise 14.2, you were asked to prepare a frequency distribution table regarding the blood groups of 30 students of a class. Use this table to determine the probability that a student of this class, selected at random, has blood group AB.
Solution:
Total number of students whose blood group are recorded = 30.
Number of students which have blood group AB is 3
∴ P(student has blood group AB) = \(\frac{3}{30}=\frac{1}{10}\)

These NCERT Solutions for Class 9 Maths Chapter 14 Statistics Ex 14.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 14 Statistics Exercise 14.4

Question 1.
The following number of goals were scored by a team in a series of 10 matches:
2, 3, 4, 5, 0, 3, 3, 4, 3
Find the mean median and mode of these scores.
Solution:
We know that

or, Mean = 2.8
To obtain the median we arrange the goals in ascending order to get 0, 1, 2, 3, 3, 4, 4, 5
Since the number of observations is 10.
So, there are two middle terms, i.e. the \(\left(\frac{10^{\mathrm{th}}}{2}\right)\) and \(\left(\frac{10}{2}+1\right)^{\text {th }}\) i.e. the 5th and 6th term.
i.e. The median = \(\frac{3+3}{2}\) = 3
So, the median goat was scored is 3.
In the given observation, 3 occurs most frequently i.e. four times.
So, the mode is 3.

Question 2.
In a mathematics test given to 15 students, the following marks (out of 100) are recorded.
41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60
Find the mean, median and mode of the above marks.
Solution:
We know that

To obtain the median we arrange the marks in ascending order to get.
39, 40, 40, 41, 42, 46, 48, 52, 52, 54, 54, 60, 62, 96, 98
Since the number of observation is 15.
So their are one middle term i.e. \(\left(\frac{15+1}{2}\right)^{\mathrm{th}}\) i.e. 8th term.
i.e. the median = 52.
So, the median marks get by students is 52.
In the given observation, 52 occur most frequently i.e. 3 times. So the mode is 52.

Question 3.
The following observations have been arranged in ascending order. If the median of the data is 63, find the value of x.
29, 32, 48, 50, x, x + 2, 72, 78, 84, 95
Solution:
We have given that the data arranged in ascending order i.e. 29, 32, 48, 50, x, x + 2, 72, 78, 84, 95.
Since the number of observations is 10.
So there are two middle terms, i.e. the \(\left(\frac{10^{\mathrm{th}}}{2}\right)\) and \(\left(\frac{10}{2}+1\right)^{\text {th }}\) term.
i.e. the 5th and 6th term.
i.e. the median = \(\frac{x+(x+2)}{2}\)
⇒ 63= \(\frac{2 x+2}{2}\) (∵ median =63)
⇒ 63 = x + 1
⇒ x = 62
Therefore, the value of x is 62.

Question 4.
Find the mode of
14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18
Solution:
we arrange the data in the following form:
14, 14, 14, 14, 17, 18, 18, 18, 22, 23, 25, 28.
Here 14 occur most frequently i.e. four times.
So, the mode is 14.

Question 5.
Find the mean salary of 60 workers of a factory from the following table:

Solution:
First, we make the frequency distribution table of given data.

Question 6.
Give one example of a situation in which
(i) the mean an appropriate measure of central tendency.
(ii) the mean is not an appropriate measure of central tendency but the median is an appropriate measure of central tendency.
Solution:
Consider a situation when a student Vikas received their test copies.
The test had five questions, each carrying ten marks. His scores were as follows:

Vikas’s average score = \(\frac{42}{5}\) = 8.4
To arranged his scores in ascending order and found out the middle score as:

In this situation means is an appropriate measure of central tendency.
(ii) Consider another situation when a student Tanvi received their test copies. The test had again five questions, each carrying the marks. Her scores were as follows:

Tanvi’s average score = \(\frac{41}{5}\) = 8.2
To arranged her scores in ascending order and found out the middle score as:
Tanvi’s Score 4 7 10 10 10
In this situation meaning is not an appropriate measure of central tendency but the median is an appropriate measure of central tendency.