# Chapter 3 : Laws of Motion - Frank Solutions for Class 9 Physics ICSE

## Chapter 3 - Laws of Motion Exercise 113

Question 1

What do you mean by inertia of motion?

Solution 1

The property by which a body neither changes its present state of rest or of uniform motion in a straight line nor tends to change the present state,is known as inertia.

Question 2

Give one example each of inertia of rest and inertia of motion.

Solution 2

A book lying on a table will remain placed at table unless it is displaced by some external force. This is an example of inertia of rest.

A ball rolling on the ground will continue to roll unless the external force , the force of friction between the ball and the ground stops it.

A ball rolling on the ground will continue to roll unless the external force , the force of friction between the ball and the ground stops it.

Question 3

The greater is the -------- the greater is the inertia of an object.

Solution 3

The greater is the MASS , the greater is the inertia of the object.

Question 4

Name the different kinds of inertia an object can possess. Give an example of each.

Solution 4

An object possess two kind of inertia, inertia of rest and inertia of motion.A book lying on a table will remain placed at table unless it is displaced by some external force. This is an example of inertia of rest.

A ball rolling on the ground will continue to roll unless the external force , the force of friction between the ball and the ground stops it.

A ball rolling on the ground will continue to roll unless the external force , the force of friction between the ball and the ground stops it.

Question 5

Define one Newton. How much maximum acceleration can it produce in a mass of 1 kg?

Solution 5

1 Newton is the force which when applied to a body of unit mass produces a unit acceleration in it. 1 newton would produce acceleration of 1 ms

^{-2}in a mass of 1 kg.Question 6

The acceleration produced by a force in an object is directly proportional to the applied ------------ And inversely proportional to the ----------- Of the object.

Solution 6

The acceleration produced bya force in an object is directly proportional to the applied FORCE and inversely proportional to the MASS of the object.

Question 7

What is the SI unit of force?

Solution 7

SI unit of force is Newton (N).

Question 8

Name the physical quantity associated with N kg

^{-1}.Solution 8

Acceleration is the physical quantity associated with N kg

^{-1}.Question 9

Solution 9

1 N = 10

^{5}Dyne.Question 10

When a sports car and a loaded van are travelling at a speed of 50 km/h, which vehicle requires more force to stop?

Solution 10

As mass of loaded van is greater than sports car so it would require more force to stop.

Question 11

What is the acceleration produced by a force of 12 N exerted on an object of mass 4 kg?

Solution 11

We know force = mass X acceleration.

a= f/m = 12 N / 4 kg. = 3 ms

so acceleration of the body would be 3 ms

a= f/m = 12 N / 4 kg. = 3 ms

^{-2}so acceleration of the body would be 3 ms

^{-2}.Question 12

What is the ratio of SI to CGs units of force?

Solution 12

SI unit of force is Newton whereas CGS unit of force is dyne.

1 newton / 1dyne = 10

1 newton / 1dyne = 10

^{5}.Question 13

Write the SI unit of momentum.

Solution 13

SI unit of momentum is kgms

^{-1}.Question 14

Define momentum of a body.

Solution 14

Momentum is defined as the amount of motion contained in the body. It is given by the product of the mass of the body and its velocity.

Question 15

Name the physical quantity associated with the motion of a body.

Solution 15

Momentum is the physical quantity associated with the motion of the body.

Question 16

Momentum is possessed by bodies in ---------------

Solution 16

Momentum is possessed by bodies in MOTION.

Question 17

Which has more momentum: a fast pitched soft ball or a soft pitched soft ball?

Solution 17

A fast pitched soft ball has more momentum.

Question 18

What is the ratio of SI units to CGS units of momentum?

Solution 18

SI unit of momentum is kgms

And their ratio is = 1000 X 100 g ms

^{-1}and CGS unit of momentum is g cms^{-1}.And their ratio is = 1000 X 100 g ms

^{-1}=^{ }1:10.Question 19

What will be the momentum of a body at rest?

Solution 19

A body at rest has zero momentum as its velocity is zero.

Question 20

State Newton's third law of motion.

Solution 20

According to Newton's third law, for every action there is always an equal and opposite reaction.

Question 21

What do you mean by action?

Solution 21

When a force acts on a body then this is called an action.

Question 22

Do action and reaction act on the same body?

Solution 22

No, action and reaction never act on a same body they always act simultaneously on two different bodies.

Question 23

Name the law of motion which gives the definition of force.

Solution 23

2nd law of motion gives the definition of force.

Question 24

A boy jumps to a dock from a boat. The boat moves away from the dock. Which law explains the statement?

Solution 24

Newton's third law explains this statement.

Question 25

Is force a scalar or a vector quantity?

Solution 25

Force is a vector quantity.

Question 26

If a number of forces act on a stationary body at the same point, then what do you conclude from it?

Solution 26

This means these forces are balanced forces.

Question 27

Why do passengers tend to fall sideways when the bus takes a sharp turn?

Solution 27

Passengers tend to fall sideways when the bus takes a sharp turn due to the inertiaof direction.

Question 28

Why are passengers thrown in the forward direction when a running bus stops suddenly?

Solution 28

Passengers are thrown in the forward direction as the running bus stops suddenly because due to their inertia of motion, their upper body continues to be in the state of motion even though the lowerbody comes to rest when the bus stops.

Question 29

Why do passengers tend to fall backward when it starts suddenly?

Solution 29

Passengers tends to fall in backward direction when bus starts suddenly because due to their inertia of rest, as soon as the bus starts, their lower body comes in motion but the upper body continues to be in the state of rest.

Question 30

Can internal forces change the velocity of a body?

Solution 30

No, internal forces cannot change the velocity of a body.

Question 31

Why do the dust particles come out of a hanging carpet when it is beaten with a stick?

Solution 31

When a hanging carpet is beaten using a stick, the dust particles will start coming out of the carpet because the part of the carpet where the stick strikes, immediately comes in motion while the dust particle sticking to the carpet remains at rest . Hence a part of the carpet moves ahead alongwith the stick, and the dust particles fall down due to the earth's pull.

Question 32

Why is a tree shaken to get its fruit down?

Solution 32

When we shake the branches of a tree, the fruits and leaves remain in state of rest while branches comes in rest so fruits and leaves are detached from the tree.

Question 33

Which would require a greater force: accelerating a 10 g mass at 5 ms

^{-2}, or accelerating a 20 g mass of 2 ms^{-2}?Solution 33

We know force = mass X acceleration

F

F

So first body require more force.

F

_{1}= 10 X 5 = 50 dyne.F

_{2}= 20 X 2 = 40 dyne.So first body require more force.

Question 34

Solution 34

## Chapter 3 - Laws of Motion Exercise 114

Question 1

An object undergoes an acceleration of 8 ms-2 starting from rest. Find the distance travelled in 5 seconds.

Solution 1

initial velocity of the object = 0 ms

Acceleration of the object = 8 ms

Time = 5 s.

Distance covered would be S = ut + 1/2 at

S = 1/2 X 8 X 5 X5 = 100 m.

^{-1}Acceleration of the object = 8 ms

^{-2}.Time = 5 s.

Distance covered would be S = ut + 1/2 at

^{2}.S = 1/2 X 8 X 5 X5 = 100 m.

Question 2

A truck rolls down a hill with constant acceleration after starting from rest. It travels a distance of 100 m in 10 s. find its acceleration. Find the force acting on it, if its mass is 5 metric tons. (tonne)

Solution 2

Initial velocity of the truck = 0 ms

Distance covered by truck = 100 m

Time taken to cover this distance = 10 s.

We know Distance covered would be S = ut + 1/2 at

100 =1/2 Xa X100

a= 2 ms

Mass of truck = 5 metric tons = 5000 kg.

Force acted on truck = mass X acceleration

Force = 5000 X 2 = 10000 N.

^{-1}Distance covered by truck = 100 m

Time taken to cover this distance = 10 s.

We know Distance covered would be S = ut + 1/2 at

^{2}.100 =1/2 Xa X100

a= 2 ms

^{-2}.Mass of truck = 5 metric tons = 5000 kg.

Force acted on truck = mass X acceleration

Force = 5000 X 2 = 10000 N.

Question 3

Name the physical entity used for quantifying the motion of a body.

Solution 3

Momentum is used for quantifying the motion of body.

Question 4

Why do you feel a backward jerk on your shoulder when you fire a gun?

Solution 4

When we fire a gun, a force is exerted in the forward in the forward direction as the bullets comes out; in reaction to which an equal and opposite force is act in the backward direction and hence, we feel a backward jerk on the shoulder.

Question 5

How does a person move forward during swimming?

Solution 5

A person applies force on water in backward direction and water according to third law of motion water apply an equal and opposite force in forward direction which helps a person to swim.

Question 6

Name the principle involved in the working of a jet engine.

Solution 6

Newton's third law of motion is involved in the working of a jet plane.

Question 7

Is the statement correct?

A rocket can propel itself in vacuum.

A rocket can propel itself in vacuum.

Solution 7

Yes, a rocket can propel itself in a vacuum once it is given initial velocity.

Question 8

Since action and reaction forces are always equal in magnitude and opposite in direction, how can anything be ever accelerated?

Solution 8

Action is equal and opposite to reaction but they act on different bodies and object moves as movement requires an unbalanced force and these are provided once inertia is overcome.

## Chapter 3 - Laws of Motion Exercise 125

Question 1

Who stated the law of gravitation?

Solution 1

Sir Isaac Newton stated the law of gravitation.

Question 2

State Newton's law of gravitation.

Solution 2

Every object in the universe attracts every other object with a force directly proportional to the product of their masses and inversely proportional to the square of distance between them.

Question 3

Distinguish between gravity and gravitation.

Solution 3

Gravity is the force of attraction between the object and the earth whereas gravitation refers to the force of attraction that exists between any two bodies that possess mass.

Question 4

Define acceleration due to gravity.

Solution 4

Acceleration due to gravity is the acceleration experienced by a body during free fall.

Question 5

Write the relation between g and G.

Solution 5

g = GM/R

^{2}.Question 6

Define the constant of gravitation.

Solution 6

We know that law of gravitation.

F = G ( m

Here G is universal constant and is called constant of gravitation. It doesnot depend upon on the value of m

Its value is same between any two objects in the universe.

F = G ( m

_{1 }X_{ }m_{2})/R^{2}.Here G is universal constant and is called constant of gravitation. It doesnot depend upon on the value of m

_{1},m_{2}or R.Its value is same between any two objects in the universe.

Question 7

What are the SI units of constant of gravitation?

Solution 7

SI unit of constant of gravitation is Nm

^{2}kg^{-2}.Question 8

How will the force of gravitation between two objects change if the distance between them is

(a) Halved,

(b) Doubled,

(c) Made four times,

(d) Infinite

(e) Almost zero?

(a) Halved,

(b) Doubled,

(c) Made four times,

(d) Infinite

(e) Almost zero?

Solution 8

we know that law of gravitation.

F = G ( m

(a) If distance between them is halved then put R = R/2.

F = 4 X G( m

F

(b) If distance between them is doubled then put R = 2R.

F = G( m

F

(c) If distance between them is made four times then put R = 4R.

F = G( m1 X m2)/16 R

F

(d) If distance between them is infinite then put R = infinite.

F = G( m

F

(e) If distance between them is almost zero then put R = 0.

F = G( m

F

F = G ( m

_{1}X m_{2})/R^{2}.(a) If distance between them is halved then put R = R/2.

F = 4 X G( m

_{1}X m_{2})/ R^{2}.F

_{1}= 4 F.(b) If distance between them is doubled then put R = 2R.

F = G( m

_{1}X m_{2})/ 4R^{2}.F

_{1}= F/4.(c) If distance between them is made four times then put R = 4R.

F = G( m1 X m2)/16 R

^{2}.F

_{1}= F/16.(d) If distance between them is infinite then put R = infinite.

F = G( m

_{1}X m_{2})/ R^{2}.F

_{1}= 0.(e) If distance between them is almost zero then put R = 0.

F = G( m

_{1}X m_{2})/ 0.F

_{1}= infinite.Question 9

All objects in the universe attract each other along the line joining their -------------

Solution 9

All objects in the universe attract each other along the line joining their CENTRES.

Question 10

The force of attraction between any two material objects is called --------------.

Solution 10

The force of attraction between any two material objects is called FORCE OF GRAVITATION.

Question 11

The gravitational force of the earth is called earth's---------.

Solution 11

The gravitational force of the earth is called earth's GRAVITY.

Question 12

The gravity is a particular case of -----------.

Solution 12

The Gravity is a particular case of GRAVITATIONAL FORCE OF EARTH.

Question 13

The value of G is extremely --------- (Small/large/infinite)

Solution 13

The value of G is extremely SMALL.

Question 14

Is the law of gravitation applicable in case of the sun and the moon?

Solution 14

Yes the law of gravitation is also applicable in case of the sun and moon.

Question 15

Solution 15

we know that law of gravitation.

F = G ( m

Mass of earth = 6X10

Mass of the person = 100 kg.

G = 6.7 X10

Radius of earth = 6.4 X 10

F = (6.7 X10

Force of gravity due to earth acting on a 100 kg person is 981.4 N.

F = G ( m

_{1}Xm_{2})/R^{2}.Mass of earth = 6X10

^{24}kg.Mass of the person = 100 kg.

G = 6.7 X10

^{-11}Nm^{2}kg^{-2}.Radius of earth = 6.4 X 10

^{14}.F = (6.7 X10

^{-11}X 100 X 6 X10^{14})/ (6.4 X6.4 X10^{12}) = 981.4NForce of gravity due to earth acting on a 100 kg person is 981.4 N.

Question 16

Why do objects fall towards the earth?

Solution 16

Objects fall towards the earth due to force of gravitation.

Question 17

Why don't you feel the force of attraction between your friend sitting close to you and yourself?

Solution 17

Because the masses of persons are not large enough to overcome the value of small constant of gravitation so the force of gravitation is very small and negligible to feel.

Question 18

A ball is thrown up with a speed of 4.9 ms

(a) Calculate the maximum height it would gain before it begins to fall.

(b) Also, calculate the time it takes to reach this height.

(c) Prove that the time of ascent is equal to the time of descent.

^{-1}.(a) Calculate the maximum height it would gain before it begins to fall.

(b) Also, calculate the time it takes to reach this height.

(c) Prove that the time of ascent is equal to the time of descent.

Solution 18

Initial speed of ball is = 4.9 ms

Acceleration due to gravity = -9.8 ms

(a) We know v

At highest point final velocity is zero so

0 - 4.9 X 4.9 = 2 X (-9.8) S

S = 1.125 m

(b) We know v = u + at

0 = 4.9 - 9.8 t

T = 0.5 sec.

(c) for highest point initial velocity is zero

Acceleration due to gravity is = 9.8 ms

Final velocity at ground is v

V

V = 4.9 ms

Time taken to reach ground from highest point

V = u + at

4.9 = 0 + 9.8 t

T = 4.9/9.8 = 0.5 sec.

So time of ascent is equal to time descent.

^{-1}.Acceleration due to gravity = -9.8 ms

^{-2}.(a) We know v

^{2}- u^{2}=2asAt highest point final velocity is zero so

0 - 4.9 X 4.9 = 2 X (-9.8) S

S = 1.125 m

(b) We know v = u + at

0 = 4.9 - 9.8 t

T = 0.5 sec.

(c) for highest point initial velocity is zero

Acceleration due to gravity is = 9.8 ms

^{-2}.Final velocity at ground is v

V

_{2}- 0 = 2 X9.8 X 1.125V = 4.9 ms

^{-1}.Time taken to reach ground from highest point

V = u + at

4.9 = 0 + 9.8 t

T = 4.9/9.8 = 0.5 sec.

So time of ascent is equal to time descent.

Question 19

What is the relation between 'g' and 'G'?

Solution 19

g = GM/R

^{2}.Question 20

What is the value of g at the surface of the earth?

Solution 20

Value of the g at the surface of the earth is 9.8ms

^{-2}.Question 21

When some rock is brought to the earth from the surface of the moon, Will its mass and weight change?

Solution 21

Mass of the body is constant at all positions so mass will not change. But weight will change as gravity on the surface of earth is almost 6 times than on the surface of the moon, so its weight will increase almost 6 times on the surface of earth.

Question 22

Where will you weigh more: at the centre of the earth or at the surface of the earth?

Solution 22

We will weigh more on the surface of the earth.

Question 23

Name an instrument used to measure the mass of a body.

Solution 23

Beam balance is used to measure the mass of a body.

Question 24

Name the instrument by which the weight of a body can be measured.

Solution 24

Spring scale is used to measure the weight of a body.

Question 25

Where do you weigh more: at the poles or at equator?

Solution 25

The weight is greater at the poles than the equator.

Question 26

How are kg wt and N related to each other?

Solution 26

Newton 1N = 9.8 kgwt.

Question 27

Where will you weigh more: at the moon's surface or at the earth's surface?

Solution 27

We will weigh more on earth surface as value of g is greater on earth surface.

Question 28

Does the force of gravitation between two objects depend on the medium between them?

Solution 28

No, the force of gravitation between two objects does not depend on the medium between them.

Question 29

What will happen to the gravitational force of attraction between two objects if the mass of each is doubled and the distance between them is also doubled?

Solution 29

we know that law of gravitation.

F = G ( m

Now m

m

R = 2 R

F

F

So force between them remains same.

F = G ( m

_{1}Xm_{2})/R^{2}.Now m

_{1}= 2 m_{1}m

_{2}= 2 m_{2}R = 2 R

F

_{1}= G ( 2m_{1}X2 m_{2})/4R^{2}.F

_{1}= FSo force between them remains same.

## Chapter 3 - Laws of Motion Exercise 126

Question 1

In vacuum, all freely falling objects have the same force. Is it true?

Solution 1

Yes, in absence of gravity all freely falling body have same force acting on them.

Question 2

Solution 2

g= GM/R

it means acceleration due to gravity is directly proportional to the mass of body and inversely proportional to the square of distance between earth and object.

^{2}it means acceleration due to gravity is directly proportional to the mass of body and inversely proportional to the square of distance between earth and object.

Question 3

Neglecting air resistance, a body falling freely near the earth's surface has a constant acceleration. (True/False)

Solution 3

Yes a body falling freely near the earth surface has a constant acceleration.

Question 4

At which of the following locations, the value of g is the largest?

(i) On top of the Mount Everest

(ii) On top of Qutub Minar

(iii) At a place on the equator

(iv) A camp site in Antarctica

(i) On top of the Mount Everest

(ii) On top of Qutub Minar

(iii) At a place on the equator

(iv) A camp site in Antarctica

Solution 4

As we know

g= 1/R

so value of g is more at poles than equator so value of g is maximum near a camp site in Antarctica as this lie on the pole.

g= 1/R

^{2}so value of g is more at poles than equator so value of g is maximum near a camp site in Antarctica as this lie on the pole.

Question 5

At what height above the earth's surface would the value of acceleration due to gravity be half of what it is on the surface? Take the radius of earth to be R.

Solution 5

## Chapter 3 - Laws of Motion Exercise 128

Question 1

What is a force?

Solution 1

Force is that external agency which tends to change the state of rest or the state of motion of a body.

Question 2

Define one Newton.

Solution 2

1 Newton is the force which when applied to a body of unit mass produces a unit acceleration in it.

Question 3

What is the relation between Newton and dyne?

Solution 3

Newton is the SI unit of force whereas dyne is the CGS unit of force.

1 N = 10

1 N = 10

^{5}dyne.Question 4

Is force a scalar quantity?

Solution 4

No, force is a vector quantity.

Question 5

A force can produce ------. In an object at rest. It can ------ an object and change its ------- of motion.

Solution 5

A force can produce MOTION in an objectat rest. It can ACCELERATE an object and can change its DIRECTION of motion.

Question 6

What does a force do in the following cases?

(a) You pull the skin of your arm.

(b) You twist a piece of rubber.

(c) You apply brakes to a running car.

(d) You catch a kicked ball.

(a) You pull the skin of your arm.

(b) You twist a piece of rubber.

(c) You apply brakes to a running car.

(d) You catch a kicked ball.

Solution 6

(a) force changes the shape of skin.

(b) force produces stretching in the rubber.

(c) force provides retardation to the car and finally stops the car.

(d) force decreases the momentum of ball and finally stops the ball.

(b) force produces stretching in the rubber.

(c) force provides retardation to the car and finally stops the car.

(d) force decreases the momentum of ball and finally stops the ball.

Question 7

Can every force produced motion in every type of body?

Solution 7

No, every force does not produce motion in every type of body.

Question 8

The amount of inertia of a body depends on its ----------.

Solution 8

The amount of inertia of a body depends on its MASS.

Question 9

You can change the direction in which an object is moving by -----------.

Solution 9

You can change the direction in which an object is moving by APPLYING FORCE ON IT.

Question 10

A man riding on a car has -----------. Inertia.

Solution 10

A man riding on a car has INERTIA of motion.

Question 11

What do you mean by inertia of rest?

Solution 11

When a body is at rest , it will continue to remain at rest unless some external force is applied to change its state of rest. This property of body is called inertia of rest.

Question 12

Name and state the action and reaction in the following cases:

(i) A book lying on a table,

(ii) A person walking on the ground,

(iii) Hammering a nail,

(iv) Firing a bullet from a gun,

(v) Pushing a wall.

(i) A book lying on a table,

(ii) A person walking on the ground,

(iii) Hammering a nail,

(iv) Firing a bullet from a gun,

(v) Pushing a wall.

Solution 12

(i) Weight of the book is action and normal force applied by table on book is reaction.

(ii) Force applied by man on ground is action and force of friction is the reaction.

(iii) Force applied by hammer on nail is action and normal force applied by nail on hammer is reaction to this force.

(iv) Firing of bullet is the action and recoiling of gun is the reaction.

(v) Force applied by us on wall is action and opposite force applied by wall on us or we can say that resistance of wall to our force is reaction.

(ii) Force applied by man on ground is action and force of friction is the reaction.

(iii) Force applied by hammer on nail is action and normal force applied by nail on hammer is reaction to this force.

(iv) Firing of bullet is the action and recoiling of gun is the reaction.

(v) Force applied by us on wall is action and opposite force applied by wall on us or we can say that resistance of wall to our force is reaction.

Question 13

Give two examples of each:

(i) Inertia of rest, and

(ii) Inertia of motion

(i) Inertia of rest, and

(ii) Inertia of motion

Solution 13

A book lying on a table will remain placed on table unless it is displaced by some external force. This is an example of inertia of rest.

A ball rolling on the ground will continue to roll unless the external force , the force of friction between the ball and the ground stops it.This is an example of inertia of motion.

A ball rolling on the ground will continue to roll unless the external force , the force of friction between the ball and the ground stops it.This is an example of inertia of motion.

Question 14

What causes motion in a body?

Solution 14

Unbalance external force causes motion in the body.

Question 15

What do you mean by linear momentum of a body?

Solution 15

Linear Momentum is defined as the amount of motion contained in a body. It is given by the product of the mass of the body and its velocity.

Question 16

Write the SI unit of momentum.

Solution 16

SI unit of momentum is kgms

^{-1}.Question 17

Give qualitative definition of force on the basis of Newton's first law of motion.

Solution 17

According to Newton's first law force is that external agency which tends to change the state of rest or the state of motion of a body.

Question 18

State Newton's first law of motion.

Solution 18

According to Newton's first law, everybody continues in its state of rest or in uniform motion in a straight line unless compelled by some external force to act otherwise.

## Chapter 3 - Laws of Motion Exercise 129

Question 1

Which of the following has the largest inertia?

(i) A car

(ii) A truck

(iii) A cricket ball

(iv) A ball pen

(i) A car

(ii) A truck

(iii) A cricket ball

(iv) A ball pen

Solution 1

Out of all these, as mass of truck is greatest and mass is measure of inertia so a truck has maximum inertia.

Question 2

Why is it advantageous to turn before taking a long jump?

Solution 2

It is advantageous to run before taking a long jump because after running we get motion of inertia which helps in long jumping.

Question 3

Why does a ball moving on a table top eventually stops?

Solution 3

Ball moving on a table top stops eventually due to force of friction between the ball surface and table surface.

Question 4

Name the physical quantity which equals the rate of change of linear momentum.

Solution 4

Force is equal to the rate of change of linear momentum.

Question 5

State Newton's second law of motion. Is Newton's first law of motion contained in Newton's second law of motion?

Solution 5

According to newton second law of motion, when a force acts on a body, the rate of change in momentum of a body is equal to the product of mass of the body and acceleration produced in it.

Yes, Newton's first law is contained in the second law as if force is zero then acceleration would be zero which means body would remain in its state of rest or in state of constant motion.

Yes, Newton's first law is contained in the second law as if force is zero then acceleration would be zero which means body would remain in its state of rest or in state of constant motion.

Question 6

Define one newton.What is the ratio of SI to CGS units of force?

Solution 6

1 Newton is the force which when applied to a body of unit mass produces a unit acceleration in it.

1 newton / 1dyne = 10

1 newton / 1dyne = 10

^{5}.Question 7

Prove that 1 N = 10

^{5}dyne. Is 1 N same as 1 kgms^{-2}?Solution 7

1 newton = 1 kg X 1 ms

1 dyne 1 g X 1 cms

So 1 newton = 10

^{-1}= 1000 g X 100 cms^{-1}= 105 cms^{-1}.1 dyne 1 g X 1 cms

^{-1}= 1cms^{-1}.So 1 newton = 10

^{5}dyne.Question 8

Two equal and opposite forces acting at the same point on a stationary body. Will the body move? Give reason to explain your answer.

Solution 8

No, the body will not move as the two forces are equal and opposite and they constitute balanced forces.

Question 9

Two equal and opposite forces act on a moving body. How is its motion affected?

Solution 9

As these forces are balanced so they will not affect the motion and motion of the body will remain unaffected.

Question 10

State Newton's third law of motion. Explain, the motion of a rocket with the help of Newton's third law of motion.

Solution 10

According to Newton's third law, for every action there is always an equal and opposite reaction. Rocket works on the same principle. The exhaust gases produced as the result of the combustion of the fuel are forced out at one end of the rocket. As a reaction , the main rocket moves in the opposite direction.

Question 11

A boy pushes a wall with a force of 30 N towards east. What force is exerted by the wall on the boy?

Solution 11

According to Newton's third law, every action has equal and opposite reaction so force exerted by the wall on the boy is 30 N.

Question 12

Name the scientist who first stated the law of inertia.

Solution 12

Newton stated the law of inertia.

Question 13

State the law of gravitation. Why is it called universal?

Solution 13

Every object in the universe attracts every other object with a force directly proportional to the product of their masses and inversely proportional to the square of distance between them.

Law of gravitation is called universal because it applies to all bodies of universe.

Law of gravitation is called universal because it applies to all bodies of universe.

Question 14

What is the difference between gravity and gravitation?

Solution 14

Gravity is the force of attraction betwen the object and the earth whereas gravitation refers to the force of attraction that exists between any two bodies that possess mass.

Question 15

Where will a person weigh more: at Simla or at Delhi?

Solution 15

Person will weigh more at Delhi as we know that gravity decreases with increase in height. Now as Shimla is at a height from Delhi so weight is less in Shimla and more in Delhi.

Question 16

Name the instrument used to measure the weight of a body.

Solution 16

Spring scale is used to measure the weight of a body.

Question 17

Gravity is another kind of --------- . It exerts all through the ------------. The Sun's gravity keeps the --------- in their orbits. Gravity can only be felt with very large----------.

Solution 17

Gravity is another kind of FORCE. It exerts all through the UNIVERSE. The sun's gravity keeps the PLANETS in their orbits. Gravity can only be felt with very large MASS.

Question 18

Explain why:

(a) Objects fall to the earth.

(b) The atmosphere does not escape.

(c) A moon rocket needs to reach a certain velocity.

(a) Objects fall to the earth.

(b) The atmosphere does not escape.

(c) A moon rocket needs to reach a certain velocity.

Solution 18

(i) Objects fall on the earth due to gravitational force between the earth and object.

(ii) Atmosphere doesnot escape because molecules of atmosphere are attracted by earth due to gravitational force of earth.

(iii) A moon rocket needs to reach a certain velocity because during its motion earth attracts the rocket towards it by its gravitational force.

(ii) Atmosphere doesnot escape because molecules of atmosphere are attracted by earth due to gravitational force of earth.

(iii) A moon rocket needs to reach a certain velocity because during its motion earth attracts the rocket towards it by its gravitational force.

Question 19

Explain the difference between g and G.

Solution 19

'g' is acceleration due to earth's gravity and 'G' is universal gravitational constant.

Question 20

What do you understand by the free fall?

Solution 20

Free fall means motion of a body under the gravity of earth only.

Question 21

Is there a gravitational attraction between you and the book? Explain.

Solution 21

Yes, we have a gravitational force of attraction between us and a book. But our mass is very small so the force between us and book is very small almost negligible.

Question 22

Does the force of the earth's gravitation affect the motion of the moon? Explain your answer with reasons.

Solution 22

Yes, the force of gravitation of earth affects the motion of moon, because moon is revolving around earth and centripetal force for this revolution is provided by earth's gravitation.

Question 23

Differentiate between gravitational mass and inertial mass.

Solution 23

Inertial mass is measure of inertia of the object. According to second law of motion F = m X a

m= F/a and this mass is called as inertial mass.

Newton law of gravitation gives another definition of mass.

F = (G m

Thus m

m= F/a and this mass is called as inertial mass.

Newton law of gravitation gives another definition of mass.

F = (G m

_{1}m_{2})/R^{2}Thus m

_{2}is the mass of the body by which another body of mass m_{1}attracts it towards it by law of gravitation. This mass is called gravitational mass.Question 24

State Newton's law of gravitation. What is the difference between

(i) Gravity and gravitation

(ii) g and G?

(i) Gravity and gravitation

(ii) g and G?

Solution 24

Newton law of gravitation is that Every object in the universe attracts every other object with a force directly proportional to the product of their masses and inversely proportional to the square of distance between them.

(i) Gravity is the force of attraction between the object and the earth whereas gravitation refers to the force of attraction that exists between any two bodies that possess mass.

(ii) 'g' is acceleration due to earth's gravity and 'G' is universal gravitational constant.

(i) Gravity is the force of attraction between the object and the earth whereas gravitation refers to the force of attraction that exists between any two bodies that possess mass.

(ii) 'g' is acceleration due to earth's gravity and 'G' is universal gravitational constant.

Question 25

An apple falls towards the earth due to its gravitational force. The apple also attracts the earth with the same force. Why do we not see the earth rising towards the apple? Explain.

Solution 25

Yes, it is true that apple attracts the earth towards it with same force but the mass of earth is so huge that acceleration produced in it due to this force is very much small and negligible to notice.

Question 26

In fig. 1, a block of weight 20 N is hanging from a rigid support by a string. Find:

(a) The force exerted by block on the string.

(b) The force exerted by string on the block.

(a) The force exerted by block on the string.

(b) The force exerted by string on the block.

Solution 26

(i) The force exerted by the block on is the weight of box and that is equal to 20N.

(ii) The force exerted by string on block is equal to the tension in the string and this is also equal to the 20N.

(ii) The force exerted by string on block is equal to the tension in the string and this is also equal to the 20N.

Question 27

A force of 4 N gives a mass m

_{1}an acceleration of 8 ms^{-2}and a mass m_{2}an acceleration of 20 ms^{-2}. What would be the acceleration if the same force acts on both the masses tied together?Solution 27

we know F = m X a

m= F/a

so we can calculate mass of each body

Mass of body 1 m

Mass of body 2 m

Total mass when two masses are tied together M = 0.5 + 0.2 = 0.7 kg.

Now as force is acting on total mass so acceleration produced is

a= 4/0.7 = 5.71 ms

m= F/a

so we can calculate mass of each body

Mass of body 1 m

_{1}= 4/8 = 0.5 kg.Mass of body 2 m

_{2}= 4/20 = 0.2 kg.Total mass when two masses are tied together M = 0.5 + 0.2 = 0.7 kg.

Now as force is acting on total mass so acceleration produced is

a= 4/0.7 = 5.71 ms

^{-2}.## Chapter - Exercise

Solution 1

Solution 2

Solution 3

## Chapter 3 - Laws of Motion Exercise 130

Question 1

Solution 1

Question 2

A force of 0.9 N acting on a body increases its velocity from 5 ms

^{-1}to 8 ms^{-1}in 2 s. Calculate the mass of the body.Solution 2

Initial speed of body = 5 ms

Final speed of body = 8 ms

Time taken to acquire this speed = 2 s.

Acceleration of body = ( v- u)/t

a= (8- 5)/2 = 1.5 ms

Force applied on body = 0.9 N.

we know F = m X a.

m = f/a = 0.9/1.5 = 0.6 kg

mass of the body is 600 gm.

^{-1}Final speed of body = 8 ms

^{-1}Time taken to acquire this speed = 2 s.

Acceleration of body = ( v- u)/t

a= (8- 5)/2 = 1.5 ms

^{-2}.Force applied on body = 0.9 N.

we know F = m X a.

m = f/a = 0.9/1.5 = 0.6 kg

mass of the body is 600 gm.

Question 3

Solution 3

Question 4

What do you mean by an impulsive force?

Solution 4

The force that acts on a body for a very short time but produces a large change in its momentum, is known as impulsive force.

Question 5

For how long should a force of 100 N act on a body of mass 20 kg so that it acquires a velocity of 100 ms

^{-1}?Solution 5

initial velocity of body = 0 ms

Final velocity of body = 100 ms

Mass of body = 20 kg.

Force applied = 100N.

We know that

F X t = m (v - u)

100 t = 20 (100 -0)

T = 2000/100 = 20 s.

^{-1}.Final velocity of body = 100 ms

^{-1}.Mass of body = 20 kg.

Force applied = 100N.

We know that

F X t = m (v - u)

100 t = 20 (100 -0)

T = 2000/100 = 20 s.

Question 6

Write the SI unit of retardation.

Solution 6

SI unit of retardation is ms

^{-2}.Question 7

What is the relationship between force and acceleration?

Solution 7

Force applied is equal to the product of mass and acceleration produced in the body.

F = mass X acceleration.

F = mass X acceleration.

Question 8

State Newton's second law of motion.

A body of mass 400 g is resting on a frictionless table. Find the acceleration of the body when acted upon by a force of 0.02 N.

A body of mass 400 g is resting on a frictionless table. Find the acceleration of the body when acted upon by a force of 0.02 N.

Solution 8

According to Newton's second law of motion, when a force acts on a body, the rate of change in momentum of a body equals the product of mass of the body and acceleration produced in it due to that force, provided the mass remains constant.

Mass of body = 400 g = 0.4kg

Force applied on body = 0.02 N

Acceleration = force/mass = 0.02/0.4 = 0.05 ms

Mass of body = 400 g = 0.4kg

Force applied on body = 0.02 N

Acceleration = force/mass = 0.02/0.4 = 0.05 ms

^{-2}.Question 9

What do you mean by linear momentum of a body? A force causes an acceleration of 10 ms

^{-2}in a body of mass 1 kg. What acceleration will be caused by the same force in a body of mass 4 kg?Solution 9

Linear Momentum is defined as the physical quantity which is associated with bodies in linear motion. It is given by the product of the mass of the body and its velocity.

Mass of body = 1 kg

Acceleration produced = 10 ms

Force applied would be = 1 X 10 N = 10 N.

Mass of second body = 4 kg.

As same force has to be applied on second body so force = 10N.

Acceleration produced is = F/M =10/4 = 2.5 ms

Mass of body = 1 kg

Acceleration produced = 10 ms

^{-2}.Force applied would be = 1 X 10 N = 10 N.

Mass of second body = 4 kg.

As same force has to be applied on second body so force = 10N.

Acceleration produced is = F/M =10/4 = 2.5 ms

^{-2}.Question 10

Two bodies P and Q, of same masses m and 2m are moving with velocities 2 v and v respectively. Compare their (i) Inertia, (ii) Momentum, and (iii) The force required to stop them in the same time.

Solution 10

Mass of P is m

Velocity of P is v

Mass of Q is m

Velocity of Q is v

(i) inertia of P/inertia of Q = m

So ratio of inertia of two bodies is 1:2.

(ii) Momentum of P/momentum of Q = m

So ratio of momentum of two bodies is 1:1.

(iii) As force required to stop them is equal to change in their momentum from moving to rest.

So ratio would be same as the ratio of their momentum i.e 1: 1.

_{1}= m.Velocity of P is v

_{1}=2 vMass of Q is m

_{2}= 2mVelocity of Q is v

_{2}= v.(i) inertia of P/inertia of Q = m

_{1}/m_{2}= 1/2.So ratio of inertia of two bodies is 1:2.

(ii) Momentum of P/momentum of Q = m

_{1}v_{1}/m_{2}v_{2}= 1So ratio of momentum of two bodies is 1:1.

(iii) As force required to stop them is equal to change in their momentum from moving to rest.

So ratio would be same as the ratio of their momentum i.e 1: 1.

Question 11

Show that the rate of change of momentum = mass X acceleration. Under what condition does this relation hold?

Solution 11

According to newton second law

F = m X a

a= (v - u)/t.

F = m(v -u)/t

F = (mv - mu)/t

As F= m X a

ma = (mv - mu)/t

so rate of change of momentum = mass X acceleration.

This relation holds good when mass remains constant during motion.

F = m X a

a= (v - u)/t.

F = m(v -u)/t

F = (mv - mu)/t

As F= m X a

ma = (mv - mu)/t

so rate of change of momentum = mass X acceleration.

This relation holds good when mass remains constant during motion.

Question 12

What do you mean by the conservation of momentum? Briefly, explain the collision between two bodies and the conservation of momentum.

Solution 12

Question 13

State Newton's third law of motion. Give an experimental demonstration of Newton's third law.

Solution 13

According to newton third law, for every action there is always an equal and opposite reaction.

To demonstrate newton third law blow a balloon and hold its neck tightly facing downwards. When we release the balloon, the balloon will moves up instead of falling to the ground. As air is releasing from bottom of balloon and this air apply equal and opposite force to the balloon and this force helps balloon to move upwards.

To demonstrate newton third law blow a balloon and hold its neck tightly facing downwards. When we release the balloon, the balloon will moves up instead of falling to the ground. As air is releasing from bottom of balloon and this air apply equal and opposite force to the balloon and this force helps balloon to move upwards.

Question 14

A force acts for 0.1 s on a body of mass 2.0 kg initially at rest. The force is then withdrawn and the body moves with a velocity of 2 ms

^{-1}. Find the average force applied by the player.Solution 14

time for which force is applied = 0.1 s.

Mass of body = 2 kg

Initial velocity of body = 0 ms

Final velocity of body = 2 ms

We know FX t = m (v - u)

F X 0.1 = 2 (2 - 0)

F = 4 /0.1 = 40 N.

Mass of body = 2 kg

Initial velocity of body = 0 ms

^{-1}Final velocity of body = 2 ms

^{-1}.We know FX t = m (v - u)

F X 0.1 = 2 (2 - 0)

F = 4 /0.1 = 40 N.

Question 15

A cricket ball of mass 500 g moving at a speed of 30 ms 1 is brought to rest by a player in 0.03 s. Find the average force applied by the player.

Solution 15

mass of ball = 500g = 0.5 kg.

Initial speed of the ball = 30 ms

Final speed of ball = 0 ms

Time taken by player to stop the ball = 0.03 s.

We know FX t = m (v - u)

F X 0.03 = 0.5 (0 - 30)

F = - 1.5 / 0.03 = - 500 N

(- ) sign shows that player has to apply force in opposite direction of the motion of the ball.

Initial speed of the ball = 30 ms

^{-1}Final speed of ball = 0 ms

^{-1}Time taken by player to stop the ball = 0.03 s.

We know FX t = m (v - u)

F X 0.03 = 0.5 (0 - 30)

F = - 1.5 / 0.03 = - 500 N

(- ) sign shows that player has to apply force in opposite direction of the motion of the ball.

Question 16

A force acts for 0.1 s on a body of mass 3.2 kg initially at rest. The force then ceases to act and the body moves through 3 m in the next one second. Calculate the magnitude of force.

Solution 16

Time for which force is applied =0.1 s.

Mass of the body = 3.2 kg.

Initial speed of body = 0 ms

We know FX t = m (v - u)

F X 0.1 = 3.2 (3 -0)

F = 9.6/0.1 = 96 N.

So applied force is 96 N.

Mass of the body = 3.2 kg.

Initial speed of body = 0 ms

^{-1}After removal of forces body covers a distance of 3m in 1 second so final speed of body = 3/1 = 3ms^{-1}.We know FX t = m (v - u)

F X 0.1 = 3.2 (3 -0)

F = 9.6/0.1 = 96 N.

So applied force is 96 N.

Question 17

Solution 17

## Chapter 3 - Laws of Motion Exercise 131

Question 1

A force of 10 N acts on a body of mass 2 kg for 3 s, initially at rest. Calculate:

(i) The velocity acquired by the body, and

(ii) Change in momentum of the body.

(i) The velocity acquired by the body, and

(ii) Change in momentum of the body.

Solution 1

Time for which force is applied =3 s.

Mass of the body = 2 kg.

Initial speed of body = 0 ms

Force applied = 10 N.

(i) We know F X t = m (v - u)

10 X3 = 2 (v- 0)

v = 15 ms

Final velocity is 15 ms

(ii) As m(v - u) is change in momentum and this is equal to the F X t so change in momentum is equal to the 30 kgms

Mass of the body = 2 kg.

Initial speed of body = 0 ms

^{-1}Force applied = 10 N.

(i) We know F X t = m (v - u)

10 X3 = 2 (v- 0)

v = 15 ms

^{-1}.Final velocity is 15 ms

^{-1}.(ii) As m(v - u) is change in momentum and this is equal to the F X t so change in momentum is equal to the 30 kgms

^{-1}.Question 2

Use Newton's second law to explain the following:

(i) We always prefer to land on sand instead of hard floor while taking a high jump.

(ii) While catching a fast moving ball, we always pull our hands backwards.

(i) We always prefer to land on sand instead of hard floor while taking a high jump.

(ii) While catching a fast moving ball, we always pull our hands backwards.

Solution 2

(i) We always prefer to land on sand instead of hard floor while taking a high jump because sand increases the time of contact.As FX t = m ( v - u ) and our change in momentum is constant so if time increases then force experienced would decrease.

(ii) Again while catching a fast moving ball, we always pull our hands backwards to increase reaction time so force experienced would decrease.

(ii) Again while catching a fast moving ball, we always pull our hands backwards to increase reaction time so force experienced would decrease.

Question 3

A stone is dropped from a cliff 98 m high.

(a) How long will it take to fall to the foot of the cliff?

(b) What will be its speed when it strikes the ground?

(a) How long will it take to fall to the foot of the cliff?

(b) What will be its speed when it strikes the ground?

Solution 3

Height of cliff = 98 m.

Initial velocity of stone = 0 ms

Acceleration due to gravity = 9.8 ms

(i) We know H = ut + 1/2 gt

98 = 1/2 X 9.8X t

t

t= 4.47 sec.

(ii) Final velocity when it strikes the ground

V

V

V

V= 44.6 ms

Initial velocity of stone = 0 ms

^{-1}.Acceleration due to gravity = 9.8 ms

^{-2}.(i) We know H = ut + 1/2 gt

^{2}.98 = 1/2 X 9.8X t

_{2}.t

^{2}= 98X2/9.8 = 20t= 4.47 sec.

(ii) Final velocity when it strikes the ground

V

^{2}- u^{2}= 2 g HV

^{2}= 2X9.8X98V

^{2}= 1920V= 44.6 ms

^{-1}.Question 4

A stone is thrown vertically upward with a velocity of 9.8 m/s. When will it reach the ground?

Solution 4

Initial speed of ball is = 9.8 ms

Acceleration due to gravity = -9.8 ms

Final speed at maximum height = 0 ms

We know v = u + at

0 = 9.8 - 9.8 t

T = 1 sec.

We know v

At highest point final velocity is zero so

0 - 9.8 X 9.8 = 2 X (-9.8) S

S = 4.9 m.

for highest point initial velocity is zero

Acceleration due to gravity is = 9.8 ms

Final velocity at ground is v

V2 - 0 = 2 X9.8 X 4.9

V = 9.8 ms

Time taken to reach ground from highest point

V = u + at

9.8 = 0 + 9.8 t

T = 9.8/9.8 = 1 sec.

Total time = time of ascent + time of descent.

Total of flight = 1+ 1 = 2 seconds.

^{-1}.Acceleration due to gravity = -9.8 ms

^{-2}.Final speed at maximum height = 0 ms

^{-1}.We know v = u + at

0 = 9.8 - 9.8 t

T = 1 sec.

We know v

^{2}- u^{2}=2asAt highest point final velocity is zero so

0 - 9.8 X 9.8 = 2 X (-9.8) S

S = 4.9 m.

for highest point initial velocity is zero

Acceleration due to gravity is = 9.8 ms

^{-2}.Final velocity at ground is v

V2 - 0 = 2 X9.8 X 4.9

V = 9.8 ms

^{-1}.Time taken to reach ground from highest point

V = u + at

9.8 = 0 + 9.8 t

T = 9.8/9.8 = 1 sec.

Total time = time of ascent + time of descent.

Total of flight = 1+ 1 = 2 seconds.

Question 5

A ball is thrown vertically downward with an initial velocity of 10 m/s. What is its speed 1 s later and 2 s later?

Solution 5

Initial speed of ball = 10 ms

Acceleration due to gravity on ball = - 9.8 ms

We know that from first equation of motion

v= u + gt.

After 1 sec

v= 10 - 9.8 X1

v= 0.2 ms

so velocity after 1 sec would be 0.2 ms

Velocity after 2 seconds

v= 10 - 9.8X2 = 10 - 19.6 = -9.6 ms

Here negative sign shows that velocity is in downward direction and magnitude is 9.6 ms

^{-1}.Acceleration due to gravity on ball = - 9.8 ms

^{-2}We know that from first equation of motion

v= u + gt.

After 1 sec

v= 10 - 9.8 X1

v= 0.2 ms

^{-1}so velocity after 1 sec would be 0.2 ms

^{-1}.Velocity after 2 seconds

v= 10 - 9.8X2 = 10 - 19.6 = -9.6 ms

^{-1}.Here negative sign shows that velocity is in downward direction and magnitude is 9.6 ms

^{-1}.Question 6

A ball is thrown upward and reaches a maximum height of 19.6 m. Find its initial speed?

Solution 6

Maximum Height attained by ball = 19.6 m

Let initial speed of ball = u ms

Acceleration applied on ball due to gravity = -9.8 ms

Final speed of ball at maximum height = 0 ms

We know that from second equation of motion

V

0 -u

u

u= 19.6 ms

so initial speed of ball to attain maximum height of 19.6 m should be 19.6ms

Let initial speed of ball = u ms

^{-1}.Acceleration applied on ball due to gravity = -9.8 ms

^{-2}.Final speed of ball at maximum height = 0 ms

^{-1}.We know that from second equation of motion

V

^{2}- u^{2}= 2as0 -u

^{2}= 2 X(-9.8)X19.6u

^{2}= 19.6 X 19.6u= 19.6 ms

^{-1}so initial speed of ball to attain maximum height of 19.6 m should be 19.6ms

^{-1}.Question 7

A stone is dropped from a tower 98 m high. With what speed should a second stone be thrown 1 s later so that both hit the ground at the same time?

Solution 7

Height of tower = 98 m

Acceleration due to gravity on stone = 9.8 ms

Initial speed of ball= 0 ms

Let initial speed of second stone is v ms

We know from second equation of motion

S = ut + 1/2 a Xt

98 = 0 + 1/2 X9.8Xt

t

t= 4.47 sec.

As second stone is thrown 1 sec later so time taken by second body to cover distance of 98 m is = 4.47 - 1 = 3.47sec.

So again put t= 3.47 sec and S = 98 m in second equation of motion we get

98 = vX3.47 + 1/2 X9.8X3.47X3.47.

98 = 3.47Xv + 59

3.47X v = 98 - 59

v= 39/3.47 = 11.23 ms

Initial speed of second stone should be 11.23 ms

Acceleration due to gravity on stone = 9.8 ms

^{-2}.Initial speed of ball= 0 ms

^{-1}.Let initial speed of second stone is v ms

^{-1}.We know from second equation of motion

S = ut + 1/2 a Xt

^{2}.98 = 0 + 1/2 X9.8Xt

^{2}.t

^{2}= 20t= 4.47 sec.

As second stone is thrown 1 sec later so time taken by second body to cover distance of 98 m is = 4.47 - 1 = 3.47sec.

So again put t= 3.47 sec and S = 98 m in second equation of motion we get

98 = vX3.47 + 1/2 X9.8X3.47X3.47.

98 = 3.47Xv + 59

3.47X v = 98 - 59

v= 39/3.47 = 11.23 ms

^{-1}.Initial speed of second stone should be 11.23 ms

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