# Chapter 2 – Force and Motion | Chapter Solution Class 9

 Publisher : Santra publication pvt. ltd. Book Name : Madhyamik Physical Science And Environment Class : 9 (Madhyamik) Subject : Physical Science Chapter Name : Force and Motion

## Multiple Choice Question (MCQ)

[Each of Mark-1]

Question 1

A body whose momentum is constant must have constant

1. force
2. velocity
3. acceleration
4. all of these

velocity

Explanation

A body whose momentum is constant must have a constant velocity. This is known as the law of conservation of momentum. In the absence of external forces, the total momentum of a system remains constant. If the momentum of a body is constant, it means that there is no net external force acting on the body, and therefore the body will continue to move at a constant velocity.

Question 2

The motion of the rocket is based on the principle of conservation of

1. mass
2. kinetic energy
3. linear momentum
4. angular momentum

linear momentum

Explanation

The motion of a rocket is based on the principle of conservation of linear momentum. According to Newton’s third law of motion, every action has an equal and opposite reaction. In a rocket, the force of the gases expelled from the rocket engine creates an equal and opposite force in the opposite direction, which propels the rocket forward.

Question 3

A force of 250 N acts on a body and the momentum acquired is 125 kg m/s. What is the time for which the force acts on the body?

1. 0.2 s
2. 0.5 s
3. 0.4 s
4.  0.25 s

0.5 s

Explanation

The formula for calculating the momentum acquired by an object due to an applied force is given by:

p = F × t

where p is the momentum acquired, F is the force applied, and t is the time for which the force acts.

We are given that a force of 250 N acts on a body and the momentum acquired is 125 kg m/s. Substituting these values into the above formula, we get:

125 = 250 × t

Solving for t, we get:

t = 125 / 250

t = 0.5 s

Question 4

A scooter of mass 120 kg is moving with a uniform velocity of 108 km/h. The force required to stop the vehicle in 10 s is

1. 720 N
2. 180 N
3. 360 N
4. 216 N

360 N

Explanation

We can start by calculating the initial momentum of the scooter, which is given by:

p = m × v

where p is momentum, m is mass, and v is velocity.

Substituting the given values, we get:

p = 120× 108 ×{5 \over 18 }

To bring the scooter to a stop in 10 s, a net force must be applied in the opposite direction of motion. The magnitude of this force can be found using the equation:

F = Δp / t

where F is force, Δp is the change in momentum, and t is the time interval over which the change occurs.

Since the scooter comes to a stop, the final momentum is zero. Therefore, the change in momentum is equal to the initial momentum. Substituting the values, we get:

F = 36000 / 10 s  = 3600 N

Question 5

A body is moving at a constant speed in a straight line path. A force is not required to

1. increase its speed
2. change the direction
3. decrease the momentum
4. keep it moving with uniform velocity

keep it moving with uniform velocity

Explanation

A force is not required to keep a body moving at a constant speed in a straight line path. According to Newton’s first law of motion, a body will continue to move with uniform velocity in a straight line path unless acted upon by an external force. In the absence of external forces, the momentum of the body will remain constant, and therefore, the body will continue to move with a constant speed in a straight line path. Therefore, the correct answer is “keep it moving with uniform velocity”.

Question 6

A cricket player catches a ball of mass 0.1 kg moving with a speed of 10 m/s in 0.1 s. The force exerted by him is

1. 10 N
2. 5 N
3. 2 N
4. 1 N

10 N

Explanation

F = Δp/t

F = m(v-u)\over t

F = 0.1 \times (0-10)\over 0.1

F = -10 N

The negative sign indicates that the force is in the opposite direction to the motion of the ball.

Question 7

What force will change the velocity of a body of mass 1 kg from 20 m/s to 30 m/s in 2 s?

1. 25 N
2. 5 N
3. 10 N
4. 2 N

5 N

Explanation

F = Δp/t

F = m(v-u)\over t

F = 1 \times (30-20)\over 2

F = 1 \times (10)\over 2

F = 5 N

Question 8

A body of mass 2 kg is moving with a velocity of 8 m/s on a smooth surface. If it is to be brought to rest in 4 s, then the force to be applied is

1. 8 N
2. 4 N
3. 2 N
4. 1 N

4 N

Explanation

F = Δp/t

F = m(v-u)\over t

F = 2 \times (0-8)\over 2

F = 1 \times (-8)\over 2

F = – 4 N

The negative sign indicates that the force is in the opposite direction to the motion of the ball.

Question 9

The rocket engines lift a rocket from the Earth because hot gases with high velocity

1. push it against the Earth
2. push is against the air
3. react against the rocket and push it up
4. heat up the air which lifts the rocket

react against the rocket and push it up

Explanation

The rocket engines lift a rocket from the Earth because hot gases with high velocity react against the rocket and push it up. According to Newton’s third law of motion, every action has an equal and opposite reaction. The rocket engines expel hot gases with high velocity in the opposite direction of the desired motion of the rocket. As a result, the rocket experiences a reaction force in the opposite direction, which propels it upwards.

Question 10

A man is at rest in the middle of a pond on perfectly smooth ice. He can get himself to the shore by making use of Newton’s

1. first law
2. second law
3. third law
4. all the laws

third law

Explanation

The man can get himself to the shore by making use of Newton’s third law of motion, which states that every action has an equal and opposite reaction. By applying a force on the ice, he can cause the ice to react with an equal and opposite force, propelling him towards the shore.

## Fill in the blanks

[Each of Mark-1]

Question

1. To every ____ there is an ____ and ____ reaction.
2. The SI unit of momentum is ____.
3. In nature forces always occur in ____.
4. Newton’s first law holds good in the ____ frame.
5. ____ provides the measure of the inertia of a body.
6. A rocket works on the principle of conservation of ____
7. SI unit of force is ____
8. The ratio of the SI to the CGS unit of force is ____.
9. In any collision, the principle of conservation of  ____ along holds
10. Action and reaction always act on two ____ bodies.
11. In collision and explosions, the total ____ remains constant, provided that no external ____ acts.

1. action, equal, opposite
2. kgm/s
3. pairs
4. inertial
5. mass
6. momentum
7. Newton (N)
8. 105
9. momentum
10. different
11. momentum
12. force

## Answer in one word or in one sentence

[Each of Mark-1]

Question

1. Which physical quantity corresponds to the rate of change of momentum ?
2. What is the SI unit of force?
3. State the relation between the momentum and force acting on a body.
4. Name the principle on which a rocket works.
5. What is the relationship between force and acceleration?
6. What is the relation between dyne and newton?
7. Name the physical quantity which makes it easier to accelerate a small car than a large car.
8. What is the force which produces an acceleration of 1 m/s2 in a body of mass 1 kg ?
9. What is the SI unit of inertia ?
10. Which law of motion is called the ‘law of inertia’?
11. Are action and reaction act on the same body or different bodies?

1. Force
2. Newton
3. Force (F) ∝ Rate of change of momentum (Δp\over t)
4. Conservation of momentum
5. F = ma, where F is the force and a is the acceleration
6. 1 N = 105 dynes
7. Mass
8. 1 newton
9. Kilogram
10. Newton’s first law
11. Different bodies

Question 1

State Newton’s first law of motion. Define force from this law.

Newton’s first law of motion states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force. Force can be defined as a push or pull on an object that can cause a change in its motion, according to this law.

Question 2

State Newton’s second law of motion and use it to define the unit of force in CGS system.

Newton’s second law of motion states that the acceleration of a body is directly proportional to the force acting on it and inversely proportional to its mass.

Mathematically, F = ma, where F is force, m is mass, and a is acceleration.

In the CGS system, the unit of force is defined as the dyne, which is the force required to impart an acceleration of 1 cm/s2 to a body of mass 1 gram.

Question 3

‘Mass is a measure of inertia’-Explain.

Mass is a measure of inertia because it quantifies how much resistance an object has to changes in its state of motion. Inertia is the tendency of an object to maintain its state of motion. The more massive an object is, the more difficult it is to change its state of motion.

Question 4

Establish the connection between Newton’s second law of motion with the first law of motion.

Newton’s second law of motion is directly connected to the first law of motion because it provides a quantitative relationship between force, mass, and acceleration. It states that a force is required to change the state of motion of a body, which is consistent with the first law of motion. In fact, the first law can be seen as a special case of the second law, where the force acting on a body is zero.

Question 5

Define newton. Establish its relation with dyne.

The newton is the SI unit of force, defined as the force required to impart an acceleration of 1 m/s2 to a body of mass 1 kg.

1 Newton (N) = 1 kg m / s2

= 1000 × 100 g cm / s2

= 100000 dyne

= 105 dyne

Question 6

Explain why passengers are thrown forward from their seats when a speeding bus stops suddenly.

When a speeding bus stops suddenly, passengers are thrown forward from their seats because of inertia. Inertia is the tendency of objects to keep moving in a straight line unless acted upon by an external force. So when the bus suddenly stops, the passengers continue moving forward due to their inertia.

Question 7

State Newton’s third law of motion. Do action and reaction act on the same body?

Newton’s third law of motion states that for every action, there is an equal and opposite reaction. Action and reaction always act on two different bodies, not on the same body.

Question 8

What is the linear momentum of a body? Is it a scalar or a vector?

The linear momentum of a body is defined as the product of its mass and velocity. It is a vector quantity, as it has both magnitude and direction.

Question 9

If action and reaction are always equal and opposite, why do not they cancel each other?

Action and reaction do not cancel each other out because they act on different bodies. When a force is exerted on a body, it reacts by exerting an equal and opposite force on a different body, which can result in motion or deformation of that body.

Question 10

State the basic conservation principle used in rocket motion.

The basic conservation principle used in rocket motion is the principle of conservation of momentum. The total momentum of the rocket and the ejected gases remains constant, and therefore, the rocket can be propelled forward by expelling gases in the opposite direction to the desired motion.

Question 11

In a ‘Tug of war’ if both the parties exert a force of 7 dynes, what will be the tension of the string?

In a tug of war, the tension in the string will be equal to the force exerted by either party, which is 7 dynes each. The total tension in the string will be equal to the sum of these two forces, which is 14 dynes.

Question 12

Explain why cleaning a garment from dust particles, it is suddenly set into motion.

When a garment is cleaned from dust particles, it is suddenly set into motion due to the sudden removal of the static friction force between the dust particles and the garment. The garment tends to maintain its state of motion as per Newton’s first law of motion, which causes it to move suddenly.

Question 13

Explain why a gun receives a backward kick when a bullet is fired from it.

When a gun is fired, the bullet is accelerated in one direction, which results in a reactive force acting in the opposite direction on the gun. This causes the gun to experience a backward kick, as per Newton’s third law of motion.

Question 14

Write down two differences between distance and displacement.

Distance is a scalar quantity that measures the total path covered by an object, whereas displacement is a vector quantity that measures the change in position of an object from its initial position to its final position.

Question 15

Write down three differences between speed and velocity.

Speed is a scalar quantity that measures the rate of change of distance with respect to time, whereas velocity is a vector quantity that measures the rate of change of displacement with respect to time. Speed is always positive, while velocity can be positive or negative, depending on the direction of motion. Speed does not indicate the direction of motion, while velocity does.

Question 16

Explain why in the unit of acceleration the unit of time comes twice.

Acceleration is defined as the rate of change of velocity with respect to time. Since velocity is a vector quantity that has both magnitude and direction, it requires two units of time in its calculation to determine the change in velocity over a certain time interval. Therefore, the unit of acceleration is m/s2, where the unit of time is squared.

[Each of Mark-3]

Question 1

State Newton’s laws of motion. Establish the relation F = ma.

Newton’s laws of motion are:

1. An object will remain at rest or in uniform motion in a straight line unless acted upon by an external force.
2. The rate of change of momentum of an object is directly proportional to the applied force and takes place in the direction in which the force acts.
3. For every action, there is an equal and opposite reaction.

Rate of change of momentum = Δp/t

m(v-u)\over t [∵ Δp = m(v-u)]

= ma [∵ a = (v-u)\over t]

Now, according to Newton's second law of motion, the rate of change of momentum of an object is directly proportional to the applied force and takes place in the direction in which the force acts.

∴ F ∝ ma

or F = kma

where k is a constant. Here k = 1

Thus, F = ma

or, Force = mass × acceleration

Question 2

State Newton's second law of motion. Define dyne and newton. Establish a relationship between them.

Newton's second law of motion states that the force acting on an object is directly proportional to its mass and acceleration, and is given by the equation F = ma.

Dyne is a unit of force used in the CGS system of units. It is defined as the force required to impart an acceleration of 1 centimetre per second squared to a mass of 1 gram, and is denoted by the symbol "dyn".

Newton is the SI unit of force, defined as the force required to impart an acceleration of 1 meter per second squared to a mass of 1 kilogram, and is denoted by the symbol "N".

1 Newton (N) = 1 kg m / s2

= 1000 × 100 g cm / s2

= 100000 dyne

= 105 dyne

∴ 1 Newton (N) = 105 dyne

Question 3

State and explain the principle of conservation of linear momentum and establish it from Newton's third law of motion.

The principle of conservation of linear momentum states that the total momentum of a system of objects remains constant if no external forces act on the system. This principle can be established from Newton's third law of motion, which states that every action has an equal and opposite reaction. When two objects interact, they exert equal and opposite forces on each other, which means that the total momentum of the system remains constant. The forces acting on each object may be different in magnitude, but their effects on the momentum cancel out each other. This principle of conservation of linear momentum has wide applications in many areas of physics and has helped scientists to understand and explain many phenomena in the natural world.

Question 4

'All rest and motion are relative'- Explain. Define displacement, speed, velocity, acceleration and retardation.

"All rest and motion are relative" means that the state of motion of an object depends on the observer's frame of reference. In other words, an object may be considered to be at rest or in motion depending on the reference point of the observer. For example, a person sitting in a moving train may appear to be at rest to the other passengers but may be moving relative to an observer standing outside the train.

• Displacement is the change in position of an object from its initial position to its final position, measured in a specific direction. It is a vector quantity and is denoted by the symbol s.
• Speed is the rate of change of distance with respect to time. It is a scalar quantity and is denoted by the symbol v.
• Velocity is the rate of change of displacement with respect to time. It is a vector quantity and is denoted by the symbol v.
• Acceleration is the rate of change of velocity with respect to time. It is a vector quantity and is denoted by the symbol a.
• Retardation is the negative acceleration that opposes the motion of an object. It is also known as deceleration and is denoted by the symbol -a.

Question 5

Establish graphically :

1. v = u + at,
2. s = ut + 1⁄2 at2,
3. v2 = u2 + 2as.

(i) Derivation of v = u + at

Change in velocity in time interval t ⇒ BE = BD - ED

If AE be drawn parallel to OD, then from the graph,

BD = BE + ED = BD + OA

or, v = BE + u

or, BE = v - u

Now, acceleration, a = Change\ in\ velocity\over time

= BE\over AE

= BE\over OD

Putting OD = t  ⇒ a = BE\over t

⇒  BE = at = v - u

therefore,v = u + at

(ii) Derivation of s = ut + 1⁄2 at2

In the figure, the distance travelled by the body is given by the area of the space between the velocity-time graph AB and the time axis OC, which is equal to the area of the figure OABD.

Thus, Distance travelled = Area of the trapezium OABD

But, Area of the figure OABD

= Area of rectangle OAED + Area of triangle ABE

= Area of rectangle OAED + area of triangle ABE

Now, find out the area of rectangle OAED and the area of triangle ABE.

(1) Area of rectangle OAED

= (OA)×(OC)

= (u)×(t)

(2) Area of triangle ABE

= 1\over 2 AE×BE

= 1\over 2 t×at

= 1\over 2 at2

Distance travelled (s) is,

So, s = Area of rectangle OAED + Area of triangle ABE

= ut + 1\over 2 at2

(iii) Derivation of v2 = u2 + 2as.

In the given figure, the distance travelled (s)  by a body in time  (t) is given by the area of the figure OABC which is a trapezium.

Distance travelld = Area of the trapezium OABC

So, Area of trapezium OABC

= 1\over 2 ×sum of parallel sides×height

= 1\over 2 (OA + CB) × OC

Now, (OA + CB) = u + v and (OC) = t.

Putting these values in the above relation, we get:

s = (u+v)\over 2 t ---- (1)

Eliminate t from the above equation. This can be done by obtaining the value of t from the first equation of motion.
v = u + at

So, t = (v-u)\over a

Now, put this value of in equation (1), we get:

s = (v+u)(v-u)\over 2a

On further simplification,

2as = v2 – u2

Finally the third equation of motion.

v2 = u2+2as

## Numerical Problems

[Each of Mark-3]

Question 1

The speed-time graph of a particle moving along a fixed direction is shown in fig. Obtain the distance traversed by the particle between (a) t = 0 s to 10 s, (b) t = 2 s to 6 s. What is the average speed of the particle over the intervals in (a) and (b)?

[Ans. : (a) 60 m, 6 m/s, (b) 24 m, 6 m/s.]

Solution

(a) t = 0 s to 10 s

Distance = Area under the graph

= 1\over 2 × 10 ×12

= 5 × 12 = 60 m

Average speed = total\ distance\over time

= 60\over 10

= 6 m/s

(b) beyond the scope (Should not be in class 9)

Question 2

The velocity-time graph of a body moving in a straight line is shown in the Fig. Find the displacement and distance travelled by the body in 6 seconds. [Ans. : 8 m, 16 m.]

Solution

(a) Displacement = Area under the graph

= 4 × 2 - 2 × 2 + 2 × 2

=8 - 4 + 4 = 8 m

(b) Distance = Area under the graph

= 4 × 2 + 2 × 2 + 2 × 2

= 8 + 8 + 8 = 24 m

Question 3

The velocity-time graph of the motion of a car is given. Find the distance travelled by the car in the first six seconds. What is the retardation of the car during the last two seconds? [Ans. 90 m, 15 m/s2]

Solution

(a) Distance travelled in 6 s = Area under the graph from 0 to 6 s

= Area under the graph from 0 to 2 s + Area under the graph from 2 to 6 s

= 1\over 2 × 2 × 30 + (6 - 2) × 30

= 1\over 2 × 2 × 30 + 4 × 30

= 30 + 120 = 150 m

(b) Retardation (a) = (u - v)\over t

= (30 - 0)\over 2

= 30\over 2

= 15 m/s2

Question 4

Calculate the acceleration of a body of mass 5 kg when a force of 15 Newton acts on it.

[Ans: 3 m/s2]

Solution

Force (F) = 15 N

Mass (m) = 5 kg

Force = Mass × acceleration

or, acceleration = Force\over Mass

= 15\over 3 = 3 m/s2

Question 5

The velocity of a body of mass 5 kg changes from 20 m/s to 10 m/s. Calculate the change of momentum of the body.

[Ans : 50 kg-m/s]

Solution

Mass (m) = 5 kg

Initial velocity (u) = 20 m/s

Final velocity (v) = 10 m/s

Change of momentum = m (v - u)

= 5 (10 - 20)

= 5 × (-10) = - 50 kg-m/s

The negative sign indicates that the momentum is in the opposite direction to the motion of the body.

Question 6

A body of mass 100 g is at rest. A force acts on it for 3 seconds and the body attains a velocity of 15 cm/ s. Calculate the magnitude of force.

[Ans: 500 dyne]

Solution

Mass (m) = 100 g

Initial velocity (u) = 0 cm/s

Final velocity (v) = 15 cm/s

Time (t) = 3 second

Force = m×(v - u)\over t

= 100 × (15 - 0)\over 3

= 100 × 5

= 500 dyne

Question 7

A body of mass 10 kg is moving with a velocity of 100 m/s. Find the force needed to stop it in 5 seconds.

[Ans: 200 N]

Solution

Mass (m) = 10 kg

Initial velocity (u) = 100 m/s

Final velocity (v) = 0 m/s

Time (t) = 5 seconds

Force = m×(v - u)\over t

= 10 × (0 - 100)\over 5

= 10 × (- 100)\over 5

= - 200 N

The negative sign indicates that the Force is in the opposite direction to the motion of the body.

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