## Exercise 3(A) — Multiple Choice Type

#### Question 1

A force can change :

- the size or shape of a body
- the state of rest or of motion of a body
- the dimensions of a body
- All of the above.

**Answer**

All of the above

**Reason** — A force when applied on a body can produce the following two main effects:

- It can change the state of rest or of motion of the body.
- It can change the size or shape of the body i.e., it can change the dimensions of the body.

#### Question 2

Which of the following is a contact force —

- electrostatic force
- gravitational force
- frictional force
- magnetic force

**Answer**

frictional force

**Reason** — The forces which are applied on bodies by making a physical contact with them, are called contact forces. Hence, out of the given options, frictional force is a contact force.

#### Question 3

The non-contact force is —

- force of reaction
- force due to gravity
- tension in string
- force of the friction

**Answer**

force due to gravity

**Reason** — The forces experienced by bodies even without being physically in touch, are called non-contact forces. Hence, out of the given options, force due to gravity is a non-contact force.

#### Question 4

A ball placed on a table starts rolling down when the table is tilted. This is an example of :

- contact force
- non-contact force
- both (a) and (b)
- none of these

**Answer**

non-contact force

**Reason** — The force on a body due to earth's attraction is called the force of gravity or the weight of the body and it is a non-contact force.

#### Question 5

When a comb is rubbed on dry hair, it gets charged. This is an example of :

- Gravitational force
- normal reaction force
- magnetic force
- electrostatic force

**Answer**

electrostatic force

**Reason** — When a comb is rubbed on dry hair, it gets charged. If this comb is brought near the small bits of paper, opposite charges are induced on the bits of paper and they begin to move towards the comb. The motion of paper bits is due to the electrostatic force of attraction exerted between the unlike charges on the comb and the paper bits.

#### Question 6

The nature of gravitational force is always:

- repulsive
- attractive
- depends on the bodies involved
- None of the above

**Answer**

attractive

**Reason** — In the universe each particle attracts the other particle due to its mass. This force of attraction between them is called gravitational force.

#### Question 7

The magnitude of a non-contact force between two bodies ............... with an increase in their separation and ............... with a decrease in their separation.

- increases, decreases
- decreases, increases
- increases, increases
- decreases, decreases

**Answer**

decreases, increases

**Reason** — The magnitude of non-contact force between two bodies depends on the distance of separation between them. It decreases with the increase in separation and increases as the separation decreases.

#### Question 8

On double the separation between two bodies the magnitude of the non-contact force between them becomes :

- one-eighth
- one-half
- one-fourth
- remains the same

**Answer**

one-fourth

**Reason** — Magnitude of non-contact force varies inversely as the square of distance of separation i.e., on doubling the separation, the force becomes one-fourth.

#### Question 9

A spring balance works on the principle of :

- gravitational force
- magnetic force
- restoring elastic force
- frictional force

**Answer**

restoring elastic force

**Reason** — In a spring balance, a spring is fixed at one end with a hook attached to an object at the other end. When the spring is stretched, the spring exerts a force F opposite to the direction of displacement of its free end, the magnitude of this force is directly proportional to the magnitude of displacement. This force is restoring force.

#### Question 10

A horizontal spring with two object A and B attached to its ends are shown in the figure below. If the spring is compressed, it ............... each object and if it is stretched, it ............... each object.

- pushes, pulls
- pulls, pushes
- pushes, pushes
- pulls, pulls

**Answer**

pushes, pulls

**Reason** — As the spring has the tendency to regain its original form or shape due to restoring force, therefore, if the spring is compressed, it ** pushes away** each object and if it is stretched, it

**each object.**

*pulls in*## Exercise 3(A) — Very Short Answer Type

#### Question 1

Define a contact force.

**Answer**

The force which are applied on bodies by making a physical contact with them, are called contact forces.

#### Question 2

Define a non-contact force.

**Answer**

The force experienced by bodies even without being physically in touch, are called non-contact force or forces at a distance.

#### Question 3

Classify the following amongst contact and non-contact forces —

(a) frictional force

(b) normal reaction force

(c) force of tension in a string

(d) gravitational force

(e) electrostatic force

(f) magnetic force

**Answer**

(a) frictional force — Contact force

(b) normal reaction force — Contact force

(c) force of tension in a string — Contact force

(d) gravitational force — Non-contact force

(e) electrostatic force — Non-contact force

(f) magnetic force — Non-contact force

#### Question 4

Give one example in each case where —

(a) the force is of contact, and

(b) force is at a distance.

**Answer**

(a) The **force of contact** — When a body slides over a rough surface, a force starts acting on the body in a direction opposite to the motion of the body, along the surface in contact. This is frictional force and is an example of contact force.

(b) The **force at a distance** or non contact force — In the universe, each particle attracts the other particle due to it's mass. This force of attraction between them is called **gravitational force** and it is an example of a non contact force.

#### Question 5

State one factor on which the magnitude of a non-contact force depends. How does it depend on the factor stated by you?

**Answer**

The magnitude of non-contact forces between two bodies depends on the **distance of separation between them.**

It **decreases with the increase in separation and increases as the separation decreases.** It's magnitude varies inversely as the square of distance of separation i.e., on doubling the separation, the force becomes one fourth.

#### Question 6

The separation between two masses is reduced to half. How is the magnitude of gravitational force between them affected?

**Answer**

As we know that, force of attraction acting between two bodies is **inversely proportional to the square of the distance between them.**

Hence, magnitude of gravitational **force will become four times.**

#### Question 7

Give one example in each of the following cases where a force —

(a) stops a moving body.

(b) moves a stationary body.

(c) changes the size of a body.

(d) changes the shape of a body.

**Answer**

The examples are as follows —

(a) **Stops a moving body** — A fielder on the ground stops a moving ball by applying force with his hands.

(b) **Moves a stationary body** — A ball lying on the ground moves when it is kicked.

(c) **Changes the size of a body** — By loading a spring hanging from a rigid support, the length of the spring increases.

(d) **Changes the shape of a body** — On pressing a piece of rubber, it's shape changes.

## Exercise 3(A) — Short Answer Type

#### Question 1

Explain giving two examples each of —

(a) Contact forces, and

(b) Non-contact forces.

**Answer**

**(a) Contact Forces** — The forces which are applied on bodies by making a physical contact with them, are called **contact forces.**

Examples of contact force:

**Frictional force**— When a body slides (or rolls) over a rough surface, a force starts acting on the body in a direction opposite to the motion of the body, along the surface in contact. This is called**frictional force or force of friction.****Normal reaction force**— When a body is placed on a surface, the body exerts a force downwards, equal to it's weight, on the surface, but the body does not move (or fall) because the surface exerts an equal and opposite force on the body normal to the surface which is called normal reaction force.

**(b) Non-Contact forces** — The forces experienced by bodies even without being physically in touch, are called **non-contact forces or forces at a distance.**

Examples of non-contact force:

**Gravitational force**— In the universe, each particle attracts the other particle due to it's mass. This force of attraction between them is called**gravitational force.****Electrostatic Force**— Two like charges repel, while two unlike charges attract each other. The force between the charges is called**electrostatic force.**

#### Question 2

State the effects of a force applied on (i) a non-rigid, and (ii) a rigid body. How does the effect of the force differ in the two cases?

**Answer**

(i) **Non-rigid force** — A force when applied on a non-rigid object, **changes the inter - spacing** between it's constituent particles and therefore **causes a change in it's dimensions and can also produce motion in it.**

(ii) **Rigid-force** — A force when applied on a rigid object **does not change the inter-spacing** between it's constituent particles and therefore **it does not change the dimensions of the object, but causes only motion in it.**

#### Question 3

What is a frictional force? Draw a neat labelled diagram showing a frictional force acting on a body.

**Answer**

When a body slides (or rolls) over a rough surface, a force starts acting on the body in direction opposite to the motion of the body, along the surface in contact. This is called frictional force or the force of friction.

#### Question 4

What is a normal reaction force? Draw a neat labelled diagram showing a normal reaction force acting on a body.

**Answer**

When a body is placed on a surface, the body exerts a force downwards, equal to its weight, on the surface, but the body does not move (or fall) because the surface exerts an equal and opposite force on the body normal to the surface which is called normal reaction force.

## Exercise 3(A) — Long Answer Type

#### Question 1(a)

A ball is hanging by a string from the ceiling of the roof. Draw a neat labelled diagram showing the forces acting on the ball and the string.

**Answer**

The diagram below shows the forces acting on the ball and the string.

#### Question 1(b)

A spring is compressed against a rigid wall. Draw a neat labelled diagram showing the forces acting on the spring.

**Answer**

The diagram below shows the forces acting on the spring when it is compressed against a rigid wall.

#### Question 1(c)

A wooden block is placed on a table top. Name the forces acting on the block and draw a neat and labelled diagram to show the point of application and direction of these forces.

**Answer**

The diagram below shows the point of application of force and the direction of these forces when a wooden block is placed on a table top.

The forces acting are —

(i) The block exerts a force ( = weight ) on the table top, downwards.

(ii) The table top exerts an equal reaction force upwards normal to the top of the table.

## Exercise 3(B) — Multiple Choice Type

#### Question 1

Newton's first law includes:

- the definition of inertia
- the definition of force
- both (a) and (b)
- none of the above

**Answer**

both (a) and (b)

**Reason** — Newton's first law can be understood in two parts:

- Definition of inertia — In the first part, Newton's first law gives the definition of inertia, according to which an object cannot change its state by itself. If the object is in state of rest, it will remain in the state of rest and if it is moving it will continue to move with same speed and in same direction, unless an external force is applied.
- Definition of force — The second part of Newton's first law defines force, according to which force is that external cause which can move a stationary object or which can change the state of motion of a moving object.

#### Question 2

Force is a ............... quantity.

- scalar
- vector
- directionless
- none of these

**Answer**

vector

**Reason** — Force is a vector quantity as it has magnitude as well as direction.

#### Question 3

The property of an object by virtue of which it tends to retain its state of rest or of motion is called —

- friction
- inertia
- force
- mass

**Answer**

inertia

**Reason** — If the object is in the state of rest, it will remain in the state of rest and if it is moving in some direction, it will continue to move with the same speed in the same direction unless an external force is applied on it. This property is called **inertia**.

#### Question 4

The property of inertia is more in —

- a car
- a truck
- a horse cart
- a toy car

**Answer**

a truck

**Reason** — As we know, **more the mass, more is the inertia** of the body. As **truck is the heaviest, hence, inertia of truck will be maximum.**

#### Question 5

A force is needed to —

- change the state of motion or state of rest of the body
- keep the body in motion
- keep the body stationary
- keep the velocity of body constant

**Answer**

change the state of motion or state of rest of the body

**Reason** — According to Newton's first law of motion, if a body is in a state of rest, it will remain in the state of rest and if it is in the state of motion, it will remain moving in the same direction with the same speed unless an external force is applied on it.

Hence, to change the state of motion or state of rest, external force is required.

#### Question 6

A tennis ball and a cricket ball, both are stationary. To start motion in them —

- A less force is required for the cricket ball than for the tennis ball
- A less force is required for the tennis ball than for the cricket ball
- Same force is required for both the balls
- Nothing can be said.

**Answer**

A less force is required for the tennis ball than for the cricket ball

**Reason** — As we know, **less the mass, less is the inertia** of the body and because the mass of the tennis ball is less than the cricket ball, hence less force is required for the tennis ball than for the cricket ball to start motion.

#### Question 7

A student has four balls - table tennis ball, tennis ball, cricket ball and a football. The correct ascending order of inertia is :

- table tennis ball, tennis ball, football, cricket ball
- cricket ball, football, tennis ball, table tennis ball
- table tennis ball, football, cricket ball, tennis ball
- football, cricket ball, table tennis ball, tennis ball

**Answer**

table tennis ball, tennis ball, football, cricket ball

**Reason** — As the property of inertia is because of the mass of the body. Greater the mass, greater the inertia. Ascending order of mass is table tennis ball < tennis ball < football < cricket ball, hence ascending order of inertia is table tennis ball < tennis ball < football < cricket ball.

#### Question 8

Which one of the following is not an example of inertia of rest?

- When a train suddenly starts moving forward, the passenger standing in the compartment tends to fall backwards.
- When a hanging carpet is beaten with a stick, dust particles start falling out of it.
- On shaking the branches of a tree, the fruits fall down.
- When a passenger jumps out of a moving train, he falls down.

**Answer**

When a passenger jumps out of a moving train, he falls down.

**Reason** — When a passenger jumps out of a moving train, he falls down because inside the train, his whole body was in a state of motion. On jumping out of the moving train, as soon as his feet touches the ground, the lower part of the body comes to rest, while the upper part still remains in motion due to inertia of motion and not inertia of rest. Hence, the person falls down.

#### Question 9

An athlete often runs before taking a long jump. This is an example of :

- Newton's first law
- inertia of rest
- inertia of motion
- both (a) and (c)

**Answer**

both (a) and (c)

**Reason** — An athlete often runs before taking a long jump. The reason is that by running he brings his entire body in the state of motion. When the body is in motion, it becomes easier to take a long jump. Hence, it is an example of Newton's first law and inertia of motion.

#### Question 10

In inertia of motion, the body continues to be in a state of motion with ............... speed in ............... direction unless an external force is applied.

- variable, zigzag
- same, same
- same, variable
- variable, same

**Answer**

same, same

**Reason** — According to inertia of motion, a body in a state of motion, continues to be in the state of motion with the same speed in the same direction in a straight line unless an external force is applied on it to change its state.

## Exercise 3(B) — Very Short Answer Type

#### Question 1

Name the physical quantity which causes motion in a body.

**Answer**

**Force** is the physical quantity which causes motion in a body.

#### Question 2

Is force needed to keep a moving body in motion?

**Answer**

**No**, force is not needed to keep a moving body in motion.**Reason** — If a body is set in motion, it will remain in motion even when the force applied to set the body in motion is withdrawn, provided that there is no other force such as friction etc., to oppose the motion.

#### Question 3

A ball is moving on a perfectly smooth horizontal surface. If no force is applied on it, will it's speed decrease, increase or remain unchanged?

**Answer**

It's speed will remain unchanged.**Reason** — According to Newton's first law of motion, if a body is in a state of rest, it will remain in the state of rest and if it is in the state of motion, it will remain moving in the same direction with the same speed unless an external force is applied on it.

#### Question 4

What is Galileo’s law of inertia?

**Answer**

According to Galileo’s law of inertia — "An object, if once set in motion, moves with uniform velocity if no force acts on it."

Thus, a body continues to be in a state of rest or in a state of uniform motion unless an external force is applied on it.

#### Question 5

State Newton's first law of motion.

**Answer**

According to Newton's first law of motion, if a body is in a state of rest, it will remain in the state of rest and if it is in the state of motion, it will remain moving in the same direction with the same speed unless an external force is applied on it.

#### Question 6

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

**Answer**

The qualitative definition of force on the basis of Newton's first law of motion is —

**Force is that external cause which tends to change the state of rest or the state of motion of an object.**

**Example** — A **book lying on a table gets displaced from it's place when it is pushed.**

#### Question 7

Name the factor on which inertia of a body depends and state how it depends on the factor stated by you.

**Answer**

The factor on which inertia of a body depends is **mass**.

**More the mass, more is the inertia** of the body. Thus, a lighter body has less inertia than a heavier body. In other words, more the mass of a body, more difficult it is to move the body from rest (or to stop the body if it initially in motion).

Hence, **mass is a measure of inertia.**

#### Question 8

Name the two kinds of inertia.

**Answer**

The two kinds of inertia are —

- Inertia of rest
- Inertia of motion

#### Question 9

Give one example of each of the following —

(a) Inertia of rest, and (b) inertia of motion.

**Answer**

(a) **Example of inertia of rest** — When a train suddenly starts moving forward, the passenger standing in the compartment tends to fall backwards. The reason is that the lower part of the passenger's body is in close contact with the train. As the train starts moving, the person's lower part shares the motion at once. However, the upper part, due to the inertia of rest cannot share the motion simultaneously and so it tends to remain at the same place.

Hence, the lower part of the body moves ahead and the upper part is left behind, so the passenger tends to fall backwards.

(b) **Example of inertia of motion** — A cyclist riding along a level road does not come to rest immediately after he stops pedalling. The reason is that the bicycle continues to move due to inertia of motion even after the cyclist stops applying the force on the pedal.

#### Question 10

An aeroplane is moving uniformly at a constant height under the action of two forces (i) upward force (lift) and (ii) downward force (weight). What is the net force on the aeroplane.

**Answer**

When the aeroplane is acted upon by two opposing forces then the net force acting on the aeroplane is **zero** as the two forces are in opposite direction and they cancel each other.

## Exercise 3(B) — Short Answer Type

#### Question 1

A ball moving on a table top eventually stops. Explain the reason.

**Answer**

As the ball moves on the table top, **force of friction** comes into play and it opposes the motion of the ball. Hence, the ball stops.

A ball moving on a table top stops, as the **force of friction between the moving ball and the table top opposes the motion.**

#### Question 2

What is meant by the term inertia?

**Answer**

**The property of an object by virtue of which it tends to retain its state of rest or of motion is called inertia.**

If the object is in the state of rest, it will remain in the state of rest and if it is moving in some direction, it will continue to move with the same speed in the same direction unless an external force is applied on it.

Example — A book lying on a table top will remain placed at it's place unless it is displaced. Similarly, a ball rolling on a horizontal plane keeps on rolling unless the force of friction between the ball and the plane stops it.

#### Question 3

Give two examples to show that greater the mass, greater is the inertia of the body.

**Answer**

Below examples illustrate that mass is a measure of inertia i.e., greater the mass, greater is the inertia of the body:

- A cricket ball is more massive then a tennis ball. The cricket ball acquires much smaller velocity than a tennis ball when the two balls are pushed with equal force for the same duration.

In case when they are**moving with the same velocity, it is more difficult to stop the cricket ball (which has more mass) in comparison to the tennis ball (which has less mass).** - It is difficult (i.e., larger force is required) to set a loaded trolley (which has more mass) in motion than an unloaded trolley (which has less mass). Similarly, it is difficult to stop a loaded trolley than an unloaded one, if both are moving initially with the same speed.

#### Question 4

'More the mass, more difficult it is to move the body from rest'. Explain this statement by giving an example.

**Answer**

As we know, **more the mass, more is the inertia** of the body. So **mass is a measure of inertia.**

If we take the example of a loaded trolley, then we observe that, it is difficult (i.e., larger force is required) to set a loaded trolley (which has more mass) in motion than an unloaded trolley (which has less mass).

#### Question 5

Two equal and opposite forces act on a stationary body. Will the body move? Give reason to your answer.

**Answer**

**No**, the body will not move when two equal and opposite forces act on a stationary body. As the **net force on the body is zero, so the body will remain stationary due to inertia of rest.**

#### Question 6

Two equal and opposite forces act on a moving object. How is it's motion affected? Give reason.

**Answer**

When two equal and opposite forces are acting on a moving object, the **motion remains unaffected** because the **net force on the object is zero.**

#### Question 7

Why does a person fall when he jumps out from a moving train?

**Answer**

A person falls when he jumps out from a moving train, because inside the train, his whole body was in a state of motion with the train. On jumping out of the moving train, as soon as his feet touch the ground, the lower part of the body comes to rest, while the upper part still remains in motion.

As a result, he falls in the direction of motion of the train and gets hurt.

To avoid falling, as soon as the passenger's feet touch the ground he should start running on the ground in the direction of motion of the train for some distance.

#### Question 8

Why does a coin placed on a card, drop into the tumbler when the card is rapidly flicked with the finger?

**Answer**

A coin placed on a card, drop into the tumbler when the card is rapidly flicked with the finger, because when the card is flicked the momentary forces acts on the card, so it moves away. But the coin kept on it does not share the motion at once and it remain at it's place due to inertia of rest. The coin then falls down into the tumbler due to the pull of gravity.

#### Question 9

Why does a ball thrown vertically upwards in a moving train, come back to the thrower’s hand?

**Answer**

A ball thrown vertically upwards in a moving train, comes back to the thrower’s hand because when ball was thrown, it was in motion along with the person and the train. It remains in the same state of forward motion even during the time the ball remains in air.

The person, the inside air and the ball all move ahead by the same distance due to inertia and so the ball falls back into his palm on it's return.

#### Question 10

People often shake branches of a tree for getting down its fruits. Why?

**Answer**

The reason is that when the stem (or branches) of the tree are shaken, they come in motion, while the fruits due to inertia, remain in the state of rest. Thus, the massive and weakly attached fruits get detached from the branches and the fall down due to the pull of gravity.

## Exercise 3(B) — Long Answer Type

State and explain the law of inertia (or Newton’s first law of motion).

**Answer**

The law of inertia or Newton’s first law of motion states that —

**If a body is in a state of rest, it will remain in the state of rest and if it is in the state of motion, it will remain moving in the same direction with the same speed unless an external force is applied on it.**

It implies that,

(i) if a body is at rest, it remains at rest unless a force is applied on it.

(ii) If a body is moving, it will continue to move with the same speed in the same direction unless a force is applied on it.

#### Question 2(a)

Explain the following:

When a train suddenly moves forward, the passenger standing in the compartment tends to fall backwards.

**Answer**

The reason is that the lower part of the passenger's body is in close contact with the train. As the train starts moving, his lower part shares the motion at once, but the upper part due to inertia of rest cannot share the motion simultaneously and so it tends to remain at the same place.

Consequently, the lower part of the body moves ahead and the upper part is left behind, so the passenger tends to fall backwards.

#### Question 2(b)

Explain the following:

When a corridor train suddenly starts, the sliding doors of some compartments may open.

**Answer**

The reason is that the frame of sliding door being in contact with the floor of the train also comes in motion on start of train, but the sliding door remains in it's position due to inertia.

Thus, the frame moves ahead with the train while door slides opposite to the direction of motion of the train. Thus, the door may open.

#### Question 2(c)

Explain the following:

After alighting from a moving bus, one has to run for some distance in the direction of bus in order to avoid falling.

**Answer**

One needs to run for some distance in the direction of the bus after alighting from it because inside the bus, his whole body was in a state of motion with the bus. On jumping out of the moving bus, as soon as his feet touch the ground, the lower part still remains in motion due to inertia of motion. As a result he falls in the direction of motion of the bus.

Hence in order to avoid falling, as soon as the passenger's feet touch the ground he should start running on the ground in the direction of motion of the bus for some distance.

#### Question 21(d)

Explain the following:

Dust particles are removed from a carpet by beating it.

**Answer**

The reason is that the part of the carpet where the stick strikes, comes in motion at once, while the dust particles settled on it's fur, remain in position due to inertia of rest. Thus, the part of the carpet moves ahead with the stick, leaving behind the dust particles which fall down due to the earth's pull.

#### Question 21(e)

Explain the following:

It is advantageous to run before taking a long jump.

**Answer**

It is advantageous to run before taking a long jump because by running a person brings his entire body in the state of motion. When the body is in motion, it becomes easier to take a long jump.

## Exercise 3(C) — Multiple Choice Type

#### Question 1

The force needed to stop a moving body in a definite time depends on:

- mass of the body
- velocity of the body
- both (a) and (b)
- none of these

**Answer**

both (a) and (b)

**Reason** — The force needed to stop a moving body in a definite time depends on the product of mass and velocity, which is called linear momentum of the moving body. Thus, **p = mv**.

#### Question 2

The linear momentum of a body of mass m moving with velocity v is —

- v/m
- m/v
- mv
- 1/mv

**Answer**

mv

**Reason** — Linear momentum of a body is defined as the **product of its mass and velocity.**

#### Question 3

The unit of linear momentum is —

- N s
- kg m s
^{-2} - N s
^{-1} - kg
^{2}m s^{-1}

**Answer**

N s

**Reason** — As we know,

Value of N = kg m s^{-2} and

Unit of linear momentum = kg m s^{-1} = kg m s^{-1} x ($\dfrac{s}{s}$)

= (kg m s^{-2}) x s

= **N s**

#### Question 4

Linear momentum is a :

- scalar quantity
- vector quantity
- directionless quantity
- none of these

**Answer**

vector quantity

**Reason** — Linear momentum is a **vector quantity** in the direction of the motion of the body (i.e., along the velocity of the body).

#### Question 5

Newton's first law of motion defines force :

- quantitatively
- qualitatively
- both (a) and (b)
- none of these

**Answer**

qualitatively

**Reason** — Newton's first law of motion defines force only qualitatively. A force is that physical cause which changes the state of motion of a body when it is applied on it. It means that force produces acceleration in the body.

#### Question 6

According to Newton's second law, the acceleration produced in a body is:

- inversely proportional to the force applied on it
- directly proportional to the force applied on it
- not dependent on force
- none of these

**Answer**

directly proportional to the force applied on it

**Reason** — Experimentally, Newton found that the acceleration produced in a body is directly proportional to the force applied on it.

#### Question 7

The correct form of Newton's second law is —

F = $\dfrac{Δp}{Δt}$

F = m $\dfrac{Δv}{Δt}$

F = v $\dfrac{Δm}{Δt}$

F = mv

**Answer**

F = $\dfrac{Δp}{Δt}$

**Reason** — As we know,

Force = Rate of change of momentum

F = $\dfrac{Δp}{Δt}$

= $\dfrac{Δ(mv)}{Δt}$

= $\dfrac{m(Δv)}{Δt}$

= ma

Thus, Newton’s second law will be **F = $\dfrac{Δp}{Δt}$**

#### Question 8

The mathematical expression of Newton's second law of motion is:

- F = m x v
- F = m x a
- F = pv
- F = m/a

**Answer**

F = m x a

**Reason** — On the basis of experiments Newton concluded that :

(i) The acceleration produced in a body of given mass is directly proportional to the force applied on it. i.e., a ∝ F (if mass remains constant)

(ii) The force needed to stop a given acceleration in a body is directly proportional to the mass of the body. i.e., F ∝ m (if acceleration remains the same).

Hence, F ∝ ma or F = Kma

Here, K is a constant and we get,

**F = m x a**

#### Question 9

For a given force F, the graph plotted for acceleration against mass is:

**Answer**

**Reason** — If a given force is applied on bodies of different masses, the acceleration produced in them is inversely proportional to their masses i.e., a ∝ $\dfrac{1}{\text{m}}$ (for a given force F). A graph plotted for acceleration a against mass m is a curve (hyperbola).

#### Question 10

The acceleration produced in a body by a force of given magnitude depends on —

- size of the body
- shape of the body
- mass of the body
- None of these

**Answer**

mass of the body

**Reason** — As we know,

**Force (f) = mass (m) x acceleration (a)**

Hence, acceleration produced in a body by a force of given magnitude depends on mass of the body.

#### Question 11

The S.I. unit of force is:

- dyne
- g cm s
^{-2} - Newton
- Joule

**Answer**

Newton

**Reason** — As F = m x a, on this basis, the S.I. unit of force is **newton**.

One newton is the force which when acts on a body of mass 1 kg, produces an acceleration of 1 ms^{-2}. i.e., 1 newton = 1 kg x 1 ms^{-2}.

#### Question 12

1 Newton is equal to :

- 10
^{3}dyne - 10
^{8}dyne - 10
^{10}dyne - 10
^{5}dyne

**Answer**

10^{5} dyne

**Reason** — 1 newton = 1 kg x 1 ms^{-2}

= 1000 g x 100 cms^{-2}

= 10^{5} g x cms^{-2}

= 10^{5} dyne.

Thus, 1 N = 10^{5} dyne.

#### Question 13

The force required to produce an acceleration of 5 ms^{-2} in a body of mass 2 kg is:

- 2.5 N
- 20 N
- 10 N
- 25 N

**Answer**

10 N

**Reason** — Given, a = 5 ms^{-2} and m = 2 kg

Using, F = ma

we get, F = 5 x 2 = 10 N

Hence, **force required = 10 N**

#### Question 14

According to Newton's second law of motion, the change in momentum takes place in a direction :

- opposite to the direction of force applied
- perpendicular to the direction of the force applied
- similar to the direction of the force applied
- none of these

**Answer**

similar to the direction of the force applied

**Reason** — According to Newton's second law of motion, the rate of change of momentum of a body is directly proportional to the force applied on it and the change in momentum takes place **in the direction in which the force is applied.**

#### Question 15

For a relation F = $\dfrac{m(Δv)}{Δt}$, to hold true, the condition(s) necessary is/are:

- velocities are equal to the velocity of light
- velocities are much smaller than the velocity of light
- mass remains constant
- both (b) and (c)

**Answer**

both (b) and (c)

**Reason** — It is observed that the mass of a particle increases with increase in velocity but it becomes perceptible only when the velocity v of the particle is comparable with the speed of light c. At velocities v << c, the change in mass is not perceptible. At such velocities (v << c ), mass m can be considered to be constant. Then Newton's second law takes the form F = $\dfrac{m(Δv)}{Δt}$ = ma.

Thus, for F = $\dfrac{m(Δv)}{Δt}$ = ma to hold true, two conditions are required :

- velocities are much smaller than the velocity of light
- mass remains constant.

#### Question 16

Two balls of masses 4m and 8m are in motion with velocities 2v and v respectively. The ratio of their momentum would be :

- 2 : 1
- 1 : 2
- 1 : 1
- 1 : 4

**Answer**

1 : 1

**Reason** — Given,

m_{1} = 4m m_{2} = 8m

v_{1} = 2v

v_{2} = v

As, p = mv

Hence, $\dfrac{\text{p}_1}{\text{p}_2}$ = $\dfrac{\text{4\text{m}2\text{v}}}{\text{8\text{m}\text{v}}}$ = $\dfrac{\text{8\text{mv}}}{\text{8\text{m}\text{v}}}$ = 1 : 1

Hence, ratio of their momentum = **1 : 1**

## Exercise 3(C) — Very Short Answer Type

#### Question 1

Define linear momentum and state it's S.I. unit.

**Answer**

Linear momentum of a body can be defined as the **product of its mass and velocity.**

For a body of mass m, moving with velocity v, linear momentum p is expressed as

**p = mv**

The S.I. unit of linear momentum is **kg ms ^{-1}**

#### Question 2

A body of mass m moving with a velocity v is acted upon by a force. Write expression for change in momentum in each of the following cases (i) when v << c (ii) when v → c, and (iii) when v << c but m does not remain constant. Here c is the speed of light.

**Answer**

(i) When v << c

i.e., if the velocity of the moving particle is much smaller than the velocity of light (c) , then —

**Change in momentum (Δp) = m Δv**

It happens when the velocity of particle is of the order of 10^{6} m s^{-1} or less than this, then the variation in mass with velocity is small enough and mass can be considered to be constant.

(ii) when v → c

**Change in momentum (Δp) = Δ(mv)**

(iii) when v << c but m does not remain constant.

In case of atomic particles moving with velocity comparable to the velocity of light c, it was observed that mass of the particle does not remain constant, but it increases with increase in velocity, according to the relation

m = $\dfrac{m_0}{\sqrt {1-\big(\dfrac{v}{c}\big)^2}}$

where, m_{0} is the mass of the particle when it is at rest (i.e., v = 0)

Then,

**Change in momentum (Δp) = Δ(mv)**

#### Question 3

Two bodies A and B of same mass are moving with velocities v and 2v respectively.

Compare (i) their inertia (ii) their momentum.

**Answer**

(i) The factor on which inertia of a body depends is **mass**.

**More the mass, more is the inertia** of the body. Thus, if mass of two bodies is same their inertia will also be same.

Hence, inertia of A and B will be same as mass of both A and B are same.

Hence, ratio of their inertia is **1 : 1**

(ii) As we know, momentum of a body (p) = mass (m) x velocity (v)

Given, mass of A and B are equal.

**For A**,

P_{A} = mv_{A}

**For B**,

P_{B} = mv_{B}

Ratio between the two is —

$\dfrac{{P}_A}{{P}_B} = \dfrac{m{v}_A}{m{v}_B} \\[0.5em]$

As, mass of A = mass of B , hence,

$\dfrac{{P}_A}{{P}_B} = \dfrac{{v}_A}{{v}_B} \\[0.5em]$

Substituting the values, we get,

$\dfrac{{P}_A}{{P}_B} = \dfrac{v}{2v} \\[0.5em] \Rightarrow \dfrac{{P}_A}{{P}_B} = \dfrac{1}{2} \\[0.5em]$

Hence, ratio between the momentum of A and B is **1 : 2**

#### Question 4

Two balls A and B of masses m and 2m are in motion with velocities 2v and v respectively.

Compare

(i) their inertia, (ii) their momentum, and (iii) the force needed to stop them in same time.

**Answer**

(i) Given,

Mass of A = m

Mass of B = 2m

The factor on which inertia of a body depends is **mass**.

**More the mass, more is the inertia** of the body.

Therefore,

$\dfrac{\text {Inertia of A}}{\text {Inertia of B}} = \dfrac{\text {mass of A}}{\text {mass of B}} \\[0.5em]$

Substituting the values, we get,

$\dfrac{\text {Inertia of A}}{\text {Inertia of B}} = \dfrac{\text {m}}{\text {2m}} \\[0.5em] \dfrac{\text {Inertia of A}}{\text {Inertia of B}} = \dfrac{\text {1}}{\text {2}} \\[0.5em]$

Hence, ratio of their inertia = **1 : 2**

(ii) As we know, momentum of a body (p) = mass (m) x velocity (v)

Given,

v_{A} = 2v

v_{B} = v

Ratio between the two is —

$\dfrac{{P}_A}{{P}_B} = \dfrac{(mv)_A}{(mv)_B} \\[0.5em]$

Substituting the values, we get,

$\dfrac{{P}_A}{{P}_B} = \dfrac{m \times 2v}{2m \times v} \\[0.5em] \dfrac{{P}_A}{{P}_B} = \dfrac{2mv}{2mv} \\[0.5em] \Rightarrow \dfrac{{P}_A}{{P}_B} = \dfrac{1}{1} \\[0.5em]$

Hence, ratio between the momentum of A and B is **1 : 1**

(iii) According to Newton's second law of motion, the rate of change of momentum of a body is directly proportional to the force applied on it and as the ratio of momentum between A and B is 1 : 1, hence, ratio of force needed to stop A and B is also **1 : 1**

#### Question 5

Name the S.I. unit of force and define it.

**Answer**

The S.I. unit of force is **newton.**

**One newton is the force which when acts on a body of mass 1 kg, produces an acceleration of 1 m s ^{-2}.**

Hence, 1 newton = 1 kg x 1 m s^{-2}

The standard symbol of newton is **N**.

#### Question 6

What is the C.G.S unit of force ? How is it defined ?

**Answer**

In C.G.S system, unit of force is **dyne**.

One dyne is the force which when acts on a body of mass 1 g, produces an acceleration of 1 cm s^{-2}

i.e.,

Hence, **1 dyne = 1 g x 1 cm s ^{-2}**

#### Question 7

Name the S.I. and C.G.S units of force. How are they related ?

**Answer**

The S.I. unit of force is **newton (N)** and the C.G.S. unit of force is **dyne**.

Relationship between newton and dyne —

1 newton = 1 kg x 1 m s^{-2}

= 1000 g x 100 cm s^{-2}

= 10^{5} g cm s^{-2}

= 10^{5} dyne.

Thus, **1 newton = 10 ^{5} dyne**.

## Exercise 3(C) — Short Answer Type

#### Question 1

Name the two factors on which the force needed to stop a moving body in a given time, depends.

**Answer**

The two factors on which the force needed to stop a moving body in a given time depends are:

- Mass
- It's Velocity

#### Question 2

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

**Answer**

When a force is applied on a moving body, it's velocity changes. Due to change in velocity of the motion, it's momentum also changes.

Let a force (F) be applied on a body having mass (m) for time (t) due to which it's velocity changes from u to v.

Then,

Initial momentum of the body = mu

Final momentum of the body = mv

Change in momentum of the body in (t) seconds

= mv - mu = m (v - u)

Then, rate of change of momentum

= $\dfrac{\text {Change in velocity}}{\text {Time}}$

= $\dfrac{\text {m(v - u)}}{\text {t}}$

And we know,

a = $\dfrac{\text {(v - u)}}{\text {t}}$

Therefore,

**Rate of change of momentum = ma = mass x acceleration.**

**This relation holds true when mass of the body remains constant.**

#### Question 3

State Newton’s second law of motion. What information do you get from it?

**Answer**

Newton’s second law of motion states that —

**The rate of change of momentum of a body is directly proportional to the force applied on it and the change in momentum takes place in the direction in which the force is applied.**

Newton’s second law of motion **provides the quantitative value of force, i.e., it relates force to the measurable quantities like acceleration and mass.**

#### Question 4

How does Newton’s second law of motion differ from first law of motion?

**Answer**

**Newton's first law of motion defines force only qualitatively.** A force is that physical cause which changes the state of motion of a body when it is applied on it. It means that the force produces acceleration in the body i.e., **the force is the cause of acceleration.**

Whereas, Newton's second law of motion gives the quantitative value of force, i.e., it relates force to the measurable quantities like acceleration and mass.

#### Question 5

Write the mathematical form of Newton’s second law of motion. State condition if any.

**Answer**

The mathematical form of Newton’s second law of motion is —

**F = ma**

where,

F = force

m = mass and

a = acceleration

The conditions for the relation to hold true are —

(i) velocities must be much smaller than the velocity of light and

(ii) mass of the body should remain constant.

#### Question 6

State Newton’s second law of motion. Under what condition does it take the form F = ma ?

**Answer**

Newton’s second law of motion states that —

**The rate of change of momentum of a body is directly proportional to the force applied on it and the change in momentum takes place in the direction in which the force is applied.**

The mathematical form of Newton’s second law of motion is —

**F = ma**

where,

F = force

m = mass and

a = acceleration

The conditions for the relation to hold true are —

- velocities must be much smaller than the velocity of light.
- Mass of the body should remain constant.

#### Question 7

How can Newton’s first law of motion be obtained from the second law of motion?

**Answer**

To obtain Newton’s first law of motion from second law of motion —

From Newton’s second law, F=ma

If F = 0, then a = 0

This means that if no force is applied on the body, it's acceleration will be zero. If the body is at rest, it will remain at rest and if it is moving, it will remain moving in the same direction with the same speed.

Thus, a **body not acted upon by any external force, does not change it's state of rest or of motion.**

This statement is **Newton's first law of motion.**

#### Question 8

Why does a glass vessel break when it falls on a hard floor, but it does not break when it falls on a carpet ?

**Answer**

When a glass vessel falls from a height on a hard floor, it comes to rest almost instantaneously (i.e., in a very short time) so the floor exerts a large force on the vessel and it breaks.

But, if the glass vessel falls on a carpet, the time duration in which the vessel comes to rest, increases and so the carpet exerts a less force on the vessel and it does not break.

#### Question 9

Use Newton’s second law of motion to explain the following —

(a) A cricketer pulls his hands back while catching a fast moving cricket ball.

(b) An athlete prefers to land on sand instead of hard floor while taking a high jump.

**Answer**

(a) Let u be the velocity of the ball of mass m, when it reaches the hands of the player catching it.

The initial momentum of the ball = mu

If the cricketer does not pull his hands and stops the ball as soon as it touches his hands, he uses very little time (t_{1}) to stop the ball.

Then, the force exerted by the ball on the hands of the cricketer is —

F = $\dfrac{\text {Change in momentum}}{\text {Time interval}}$ = $\dfrac{mu}{t_1}$

But if the cricketer pulls back his hands along with the ball, he takes a much longer time t_{2} to stop the ball.

The force now exerted by the ball on his hands is F_{2} = $\dfrac{mu}{t_2}$

Since, t_{2} > t_{1}, therefore, F_{2} < F _{1} or force exerted on the hands of cricketer by the fast moving ball is less when he withdraws his hands. Thus cricketer avoids the chances of injury to his palms by withdrawing his hands along with the ball while catching it.

(b) When an athlete lands from a height on a hard floor, he may hurt his feet because his feet comes to rest almost instantaneously (i.e., in a very short time) so a very large force is exerted by the floor of his feet.

On the other hand, when he lands on sand, his feet push the sand for some distance, therefore the time duration in which his feet comes to rest, increases.

As a result, the force exerted on his feet decreases and he is saved from getting hurt.

## Exercise 3(C) — Long Answer Type

#### Question 1

Draw graphs to show the dependence of (i) acceleration on force for a constant mass, and (ii) force on mass for a constant acceleration.

**Answer**

(i) The acceleration produced in a body of given mass is directly proportional to the force applied on it. i.e.,

**a ∝ F** (if mass remains constant)

The graph plotted for acceleration on force for a constant mass is as shown below —

(ii) The force needed to produce a given acceleration in a body is proportional to the mass of the body. i.e.,

**F ∝ m** (if acceleration remains the same)

The graph plotted for force on mass for a constant acceleration is shown below —

#### Question 2

How does the acceleration produced by a given force depend on mass of the body? Draw a graph to show it.

**Answer**

When a force is applied on bodies of different masses, the acceleration produced in them is inversely proportional to their masses

i.e.,

**a ∝ $\dfrac{1}{m}$** (for a given force F).

The graph plotted for acceleration against mass m is a hyperbola and is shown below,

#### Question 3

How does acceleration vary with the force applied for a given body? Show it with the help of a graph.

**Answer**

Acceleration produced in a body of given mass is directly proportional to the force applied on it. i.e., **a ∝ F** (if mass remains constant)

The graph plotted for acceleration a against force F is a straight line as shown below:

## Exercise 3(C) — Numericals

#### Question 1

A body of mass 5 kg is moving with velocity 2 m s^{-1}. Calculate it's linear momentum.

**Answer**

As we know,

**linear momentum (p) = mass (m) x velocity (v)**

Given,

**v = 2 m s ^{-1}**

**m = 5 kg**

Substituting the values in the formula above we get,

$p = 5 \times 2 \\[0.5em] p = 10 \\[0.5em]$

Hence, **linear momentum = 10 kg m s ^{-1}**

#### Question 2

The linear momentum of a ball of mass 50 g is 0.5 kg m s^{-1}. Find it's velocity.

**Answer**

As we know,

**linear momentum (p) = mass (m) x velocity (v)**

Given,

**p = 0.5 kg m s ^{-1}**

m = 50 g

Converting g to kg,

$1000 g = 1 kg \\[0.5em] 50 g = (\dfrac{1}{1000}) \times 50 \\[0.5em] \Rightarrow 50 g = 0.05 \text {kg} \\[0.5em]$

Hence, **m = 0.05 kg**

Substituting the values in the formula above we get,

$0.5 = 0.05 \times v \\[0.5em] \Rightarrow v = \dfrac{0.5}{0.05} \\[0.5em] \Rightarrow v = \dfrac{0.1}{0.01} \\[0.5em] \Rightarrow v = 10 \text {m s}^{-1}\\[0.5em]$

Hence, **velocity = 10 m s ^{-1}**

#### Question 3

A force of 15 N acts on a body of mass 2 kg. Calculate the acceleration produced.

**Answer**

As we know,

**Force (f) = mass (m) x acceleration (a)**

Given,

**F = 15 N**

**m = 2 kg**

Substituting the values in the formula above we get,

$15 = 2 \times a \\[0.5em] \Rightarrow a = \dfrac{15}{2} \\[0.5em] \Rightarrow a = 7.5 \text {m s}^{-2} \\[0.5em]$

Hence,

**Acceleration produced by the body = 7.5 m s ^{-2}**

#### Question 4

A force of 10 N acts on a body of mass 5 kg. Find the acceleration produced.

**Answer**

As we know,

**Force (f) = mass (m) x acceleration (a)**

Given,

**F = 10 N**

**m = 5 kg**

Substituting the values in the formula above we get,

$10 = 5 \times a \\[0.5em] \Rightarrow a = \dfrac{10}{5} \\[0.5em] \Rightarrow a = 2 \text {m s} ^{-2} \\[0.5em]$

Hence,

**Acceleration produced by the body = 2 m s ^{-2}**

#### Question 5

Calculate the magnitude of force which when applied on a body of mass 0.5 kg produces an acceleration of 5 m s ^{-2}.

**Answer**

As we know,

**Force (f) = mass (m) x acceleration (a)**

Given,

**a = 5 m s ^{-2}**

**m = 0.5 kg**

Substituting the values in the formula above we get,

$F = 0.5 \times 5 \\[0.5em] \Rightarrow F = 2.5 N \\[0.5em]$

Hence,

**Magnitude of force = 2.5 N**

#### Question 6

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.

**Answer**

As we know,

**Force (f) = mass (m) x acceleration (a)**

Given,

**f = 10 N**

**m = 2 kg**

**t = 3 s**

**u = 0**

Substituting the values in the formula above we get,

$10 = 2 \times a \\[0.5em] \Rightarrow a = \dfrac{10}{2} \\[0.5em] \Rightarrow a = 5 \text {m s}^{-2} \\[0.5em]$

Hence,

**Acceleration produced by the body = 5 m s ^{-2}**

The 1^{st} equation of motion states that;

**v = u + at**

Substituting the values in the formula above we get,

$v = 0 + (5 \times 3) \\[0.5em] \Rightarrow v = 15 \text {m s}^{-1}$

Hence, **velocity of the body is 15 m s ^{-1}**

(ii) As we know,

**Change in momentum = Final momentum - Initial momentum**

Initial momentum = mu = 2 x 0 = 0

Final momentum = mv = 2 x 15 = 30 kg m s^{-1}

Substituting the values in the formula above we get,

Change in momentum = 30 - 0 = **30 kg m s ^{-1}**

#### Question 7

A force acts for 10 s on a stationary body of mass 100 kg after which the force ceases to act. The body moves through a distance of 100 m in the next 5 s. Calculate (i) the velocity acquired by the body, (ii) the acceleration produced by the force, and (iii) the magnitude of the force.

**Answer**

(i) Velocity (v) = $\dfrac{\text {distance (S)}}{\text {time (t)}}$

Given,

**s = 100 m in 5 s**

Substituting the values in the formula above we get,

$v = \dfrac{100}{5} \\[0.5em] v = 20 \\[0.5em]$

Hence, **velocity acquired by the body = 20 m s ^{-1}**

(ii) As we know,

**v ^{2} - u^{2} = 2as**

Given,

s = 100 m

u = 0

v = 20 m s^{-1}

Substituting the values in the formula above we get,

$20^2 - 0^2 = 2 \times a \times 100 \\[0.5em] 400 = 200 \times a \\[0.5em] \Rightarrow a = 2 \text {m s}^{-2} \\[0.5em]$

Hence, a = 2 m s^{-2}

(iii) As we know,

**Force (f) = mass (m) x acceleration (a)**

Given,

**m = 100 kg**

**a = 2 m s ^{-2}**

Substituting the values in the formula above we get,

$F = 100 \times 2 \\[0.5em] \Rightarrow F = 200 N \\[0.5em]$

Hence,

**Magnitude of force = 200 N**

#### Question 8

Figure shows the velocity-time graph of a particle of mass 100 g moving in a straight line. Calculate the force acting on the particle.

(**Hint** — acceleration = slope of v-t graph)

**Answer**

As we know,

**Force (f) = mass (m) x acceleration (a)**

Given,

m = 100 g

Converting g to kg,

1000 g = 1 kg

100 g = ($\dfrac{1}{1000}$) x 100 kg

100 g = 0.1 Kg

Hence, **m = 0.1 Kg**

As,

acceleration = slope of v-t graph = $\dfrac{v}{t}$

= $\dfrac{20}{5}$

Hence, **a = 4 m s ^{-2}**

Substituting the values in the formula above we get,

$F = 0.1 \times 4 \\[0.5em] F = 0.4 \\[0.5em]$

Hence, **force acting on the particle = 0.4 N**

#### Question 9

A force causes an acceleration of 10 m s^{-2} in a body of mass 500 g. What acceleration will be caused by the same force in a body of mass 5 kg ?

**Answer**

As we know,

**Force (f) = mass (m) x acceleration (a)**

Given,

m = 500 g

Converting g to kg,

1000 g = 1 kg

500 g = ($\dfrac{1}{1000}$) x 500 kg

100 g = 0.5 Kg

Hence, **m = 0.5 Kg**

**a = 10 m s ^{-2}**

Substituting the values in the formula above we get,

$F = 0.5 \times 10 \\[0.5em] \Rightarrow F = 5 \text {N} \\[0.5em]$

Hence,

**Magnitude of force = 5 N**

If, **m = 5 Kg , f = 5 N**

Substituting the values in the formula above we get,

$5 = 5 \times a \\[0.5em] \Rightarrow a = \dfrac{5}{5} \\[0.5em] \Rightarrow a = 1 \text {m s}^{-2} \\[0.5em]$

Hence, **a = 1 m s ^{-2}**

#### Question 10

A cricket ball of mass 150 g moving at a speed of 25 m s^{-1} is brought to rest by a player in 0.03 s. Find the average force applied by the player.

**Answer**

The 1^{st} equation of motion states that;

**v = u + at**

Given,

m = 150 g

Converting g to kg,

1000 g = 1 kg

150 g = ($\dfrac{1}{1000}$) x 150

150 g = 0.15 kg

Hence, **m = 0.15 kg**

**u = 25 m s ^{-1}**

**v = 0**

**t = 0.03 s**

Substituting the values in the formula above we get,

$0 = 25 + (a \times 0.03) \\[0.5em] \Rightarrow a = -\dfrac{25}{0.03} \\[0.5em] \Rightarrow a = - 833.33 \\[0.5em]$

Hence, **acceleration of the ball = - 833.33 m s ^{-2}**`

Now,

**Force (f) = mass (m) x acceleration (a)**

Substituting the values in the formula above we get,

$F = 0.15 \times (- 833.33) \\[0.5em] \Rightarrow F = -125 N \\[0.5em]$

Hence,

**Magnitude of force = -125 N**

#### Question 11

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 m s^{-1}. Find the magnitude of force.

**Answer**

The 1^{st} equation of motion states that;

**v = u + at**

Given,

**t = 0.1 s**

**m = 2.0 Kg**

**u = 0**

**v = 2 m s ^{-1}**

Substituting the values in the formula above we get,

$2 = 0 + (a \times 0.1) \\[0.5em] \Rightarrow a = \dfrac{2}{0.1} \\[0.5em] \Rightarrow a = 20 \text {m s}^{-2} \\[0.5em]$

Hence, **acceleration of the body = 20 m s ^{-2}**`

Now,

**Force (f) = mass (m) x acceleration (a)**

Substituting the values in the formula above we get,

$F = 2 \times 20 \\[0.5em] \Rightarrow F = 40 N \\[0.5em]$

Hence,

**Magnitude of force = 40 N**

#### Question 12

A body of mass 500 g, initially at rest, is acted upon by a force which causes it to move a distance of 4 m in 2 s. Calculate the force applied.

**Answer**

As we know,

**S = ut + ($\dfrac{1}{2}$) at ^{2}**

Given,

m = 500 g

Converting g to kg,

1000 g = 1 kg

500 g = ($\dfrac{1}{1000}$) x 500 kg

100 g = 0.5 Kg

Hence,

**m = 0.5 Kg**

**u = 0**

**s = 4 m**

**t = 2 s**

Substituting the values in the formula above we get,

$4 = (0 \times 2) + (\dfrac{1}{2} \times a \times 2^2) \\[0.5em] 4 = 0 + (\dfrac{1}{2} \times a \times 4) \\[0.5em] 4 = 2a \\[0.5em] \Rightarrow a = \dfrac{4}{2} \\[0.5em] \Rightarrow a = 2 \text {m s}^{-2} \\[0.5em]$

Hence,

**a = 2 m s ^{-2}**

Now,

**Force (f) = mass (m) x acceleration (a)**

Substituting the values in the formula above we get,

$F = 0.5 \times 2 \\[0.5em] \Rightarrow F = 1 N \\[0.5em]$

Hence,

**Magnitude of force = 1 N**

#### Question 13

A car of mass 480 kg moving at a speed of 54 km h^{-1} is stopped by applying brakes in 10 s. Calculate the force applied by the brakes.

**Answer**

We know that,

acceleration = $\dfrac{v - u}{t}$

Given,

m = 480 kg

u = 54 km h^{-1}

converting km h^{-1} to m s^{-1}

$54 \text {km h}^{-1} = 54 \times (\dfrac{1000 \text {m}}{60 \times 60 \text {s}}) \\[0.5em] 54 \text {km h}^{-1} = 54 \times (\dfrac{10 \text {m}}{36 \text {s}}) \\[0.5em] 54 \text {km h}^{-1} = \dfrac{540 \text {m}}{36 \text {s}} \\[0.5em] 54 \text {km h}^{-1} = 15 {\text {m s}^{-1}} \\[0.5em]$

Hence, u = 15 m s^{-1}

**v = 0**

**t = 10 s**

Substituting the values in the formula above we get,

$a = \dfrac{0 - 15}{10} \\[0.5em] \Rightarrow a = -1.5 \text{m s}^{-2} \\[0.5em]$

Hence, **a = - 1.5 m s ^{-2}**

The negative sign shows that it is retardation.

Now,

**Force (f) = mass (m) x acceleration (a)**

Substituting the values in the formula above we get,

$F = 480 \times (1.5) \\[0.5em] \Rightarrow F = 720 N \\[0.5em]$

Hence,

**Magnitude of force applied by the brakes = 720 N**

#### Question 14

A car is moving with a uniform velocity of 30 m s^{-1}. It is stopped in 2 s by applying a force of 1500 N through it's brakes. Calculate (a) the change in momentum of car, (b) the retardation produced in car, and (c) the mass of the car.

**Answer**

(i) We know that,

acceleration = $\dfrac{v - u}{t}$

Given,

**v = 0**

**u = 30 m s ^{-1}**

**t = 2 s**

**f = 1500 N**

Substituting the values in the formula above we get,

$a = \dfrac{0 - 30}{2} \\[0.5em] \Rightarrow a = - 15 \text{m s}^{-2} \\[0.5em]$

The negative sign shows that it is retardation. So **retardation = 15 m s ^{-2}**

Now,

**Force (f) = mass (m) x acceleration (a)**

Substituting the values in the formula above we get,

$1500 = m \times (15) \\[0.5em] \Rightarrow m = 100 \text{kg} \\[0.5em]$

Hence,

**Mass of the car = 100 kg**

As we know,

**Change in momentum = Final momentum - Initial momentum**

Initial momentum = mu = 100 x 30 = 3000 kg m s^{-1}

Final momentum = mv = 100 x 0 = 0

Substituting the values in the formula above we get,

Change in momentum = 0 - 3000 = **- 3000 kg m s ^{-1}**

Hence,

(i) change in momentum = 3000 kg m s^{-1}

(ii) The retardation produced in the car = **15 m s ^{-2}**

(iii) The mass of the car = **100 Kg**

#### Question 15

A bullet of mass 50 g moving with an initial velocity of 100 m s^{-1}, strikes a wooden block and comes to rest after penetrating a distance 2 cm in it. Calculate (i) initial momentum of the bullet (ii) final momentum of the bullet, (iii) retardation caused by the wooden block, and (iv) resistive force exerted by the wooden block

**Answer**

Given,

m = 50 g

Converting g to Kg

1000 g = 1 Kg

50 g = ($\dfrac{1}{1000}$) x 50 kg

50 g = 0.05 Kg

Hence, **m = 0.05 Kg**

**u = 100 m s ^{-1}**

**S = 2 cm**

Converting cm to m

100 cm = 1 m

2 cm = ($\dfrac{1}{100}$) x 2 = 0.02 m

(i) **Initial momentum of the bullet = mass (m) x initial velocity (u)**

Substituting the values in the formula above we get,

Initial momentum = 0.05 x 100 = 5 kg m s^{-1}

Hence, **Initial momentum = 5 kg m s ^{-1}**

(ii) **Final momentum = mass (m) x final velocity (u)**

Substituting the values in the formula above we get,

Final momentum = 0.05 x 0 = 0 kg m s^{-1}

Hence, **Final momentum = Zero**

(iii) As we know,

**v ^{2} - u^{2} = 2as**

Substituting the values in the formula above we get,

$0^2 - 100^2 = 2 \times a \times 0.02 \\[0.5em] - 10000 = 0.04 \times a \\[0.5em] \Rightarrow a = - \dfrac {10000}{0.04} \\[0.5em] \Rightarrow a = - \dfrac {1000000}{4} \\[0.5em] \Rightarrow a = - 250000 \text{m s}^{-2}\\[0.5em]$

Hence,

**Retardation caused by wooden block = 2.5 x 10 ^{5} m s^{-2}**

(iv) Now,

**Force (f) = mass (m) x acceleration (a)**

Substituting the values in the formula above we get,

$F = 0.05 \times (2.5 \times 10^5) \\[0.5em] \Rightarrow F = 0.125 \times 10^5 \text{N} \\[0.5em]$

Hence,

**Resistive force exerted by wooden block = 12500 N**

## Exercise 3(D) — Multiple Choice Type

#### Question 1

Newton's third law —

- defines the force qualitatively
- defines the force quantitatively
- explains the way a force acts on a body
- gives the direction of force.

**Answer**

explains the way a force acts on a body

**Reason** — Newton's third law **explains the way a force acts on a body** and states that, **to every action there is always an equal and opposite reaction.**

#### Question 2

Which law of motion explains how a force acts on a body or an object?

- Newton's third law
- Newton's second law
- Newton's first law
- All of the above.

**Answer**

Newton's third law

**Reason** — Newton's third law of motion states that : In an interaction of two bodies A and B, the magnitude of reaction (i.e., the force F_{AB} applied by the body A) is equal in magnitude to the action (i.e., the force F_{BA} applied by the body A on the body B), but they are in directions opposite to each other.

Hence, the law explains how a force acts on a body or an object.

#### Question 3

To every action there is always :

- an equal reaction
- an equal and opposite reaction
- an equal reaction in the same direction
- none of these

**Answer**

an equal and opposite reaction

**Reason** — Newton's third law of motion states that to every action there is always an equal and opposite reaction.

#### Question 4

Action and reaction act on the —

- same body in opposite directions
- different bodies in opposite directions
- different bodies, but in same direction
- same body in same direction.

**Answer**

different bodies in opposite directions

**Reason** — As per the Newton’s third law of motion —

In an interaction of two bodies A and B, the magnitude of reaction (i.e., the force F_{AB} applied by the body A) is equal in magnitude to the action (i.e., the force F_{BA} applied by the body A on the body B), but they are in directions opposite to each other.

#### Question 5

The motion of a boat moving away from the shore is an example of :

- Newton's first law
- Newton's second law
- Newton's third law
- not related with Newton's law

**Answer**

Newton's third law

**Reason** — To move a boat ahead in water, the boatman pushes **(action)** the water backwards with his oar and at the same time, the water exerts an equal and opposite force **(reaction)** in the forward direction on the boat due to which the boat moves ahead. Hence, it is an example of **Newton's third law**

#### Question 6

In an interaction of two bodies A and B, which of the following statements is correct ?

- $\overrightarrow{\text{F}}_{\text{AB}} = \overrightarrow{\text{F}}_{\text{BA}}$
- $\overrightarrow{\text{F}}_{\text{BA}} \ge \overrightarrow{\text{F}}_{\text{AB}}$
- $\overrightarrow{\text{F}}_{\text{AB}} = -\overrightarrow{\text{F}}_{\text{BA}}$
- $\overrightarrow{\text{F}}_{\text{AB}} \gt \overrightarrow{\text{F}}_{\text{BA}}$

**Answer**

$\overrightarrow{\text{F}}_{\text{AB}} = -\overrightarrow{\text{F}}_{\text{BA}}$

**Reason** — According to Newton’s third law, **to every action there is always an equal and opposite reaction.** In magnitude, F_{AB} = F_{BA} but they are in opposite directions. Therefore, $\overrightarrow{\text{F}}_{\text{AB}} = -\overrightarrow{\text{F}}_{\text{BA}}$.

#### Question 7

A footballer hits a football with a force of 10 N. What force does the footballer experience ?

- 5 N
- 10 N
- 20 N
- 15 N

**Answer**

10 N

**Reason** — According to Newton’s third law **to every action there is always an equal and opposite reaction.** Hence, the footballer will also experience a reaction force of 10 N.

#### Question 8

Out of the following statements, which one is correct ?

- the sum of action and reaction on a body is zero.
- action and reaction act on different bodies in opposite direction.
- action and reaction are always in opposite direction on the same body.
- action and reaction act on different bodies in the same direction.

**Answer**

action and reaction act on different bodies in opposite direction.

**Reason** — The action and reaction never act on the same body, but they always act simultaneously on two different bodies i.e., the forces on interaction are always present in a pair.

#### Question 9

Which of the following are examples of Newton's third law of motion?

(i) While catching a ball, the cricketer withdraws his hands along with the ball.

(ii) Athletes often land on sand after taking a high jump.

(iii) Motion of a man on ground

- (iii)
- (ii)
- (i)
- all of the above.

**Answer**

(iii)

**Reason** — When a man applies a force F **(action)** backward by his foot on the ground against the force of friction, the ground exerts an equal and opposite force R **(reaction)** forward on his foot. The horizontal component of the force of reaction enables the man to move forward.

Obviously, it will be difficult to move on a slippery road where friction is less.

## Exercise 3(D) — Very Short Answer Type

#### Question 1

State Newton's third law of motion.

**Answer**

Newton's third law of motion states that to every action there is always an equal and opposite reaction.

#### Question 2

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

(a) firing a bullet from a gun,

(b) hammering a nail,

(c) a book lying on a table,

(d) a moving rocket,

(e) a person walking on the floor,

(f) a moving train colliding with a stationary train.

**Answer**

(a) **Firing a bullet from a gun** — When a man fires a bullet from a gun, a force F is exerted on the bullet **(action)** and the gun experiences an equal recoil R **(reaction)**.

(b) **Hammering a nail** — When we hammer a nail, the hammer exerts a force F **(action)** on the nail and the nail in turn also exerts an equal and opposite force R **(reaction)** on the hammer.

(c) **A book lying on a table** — When a book is placed on a table top, the book exerts a force equal to it's weight **(action)** on the table in downward direction and the table balances it by an equal force called the **(reaction)** acting upwards on the book.

(d) **A moving rocket** — The rocket exerts a force F **(action)** on gases (produced by burning of fuel) to expel them through a nozzle backwards. The outgoing gases exert an equal and opposite force R **(reaction)** on the rocket due to which it moves in the forward direction.

(e) **A person walking on the floor** — When a man applies a force F **(action)** backward by his foot on the ground against the force of friction, the ground exerts an equal and opposite force R **(reaction)** forward on his foot.

(f) **A moving train colliding with a stationary train** — The moving train exerts a force F (action) on the stationary train and the stationary train in turn exerts a force R (reaction) on the moving train.

#### Question 3

'The action and reaction both act simultaneously.' Is this statement true?

**Answer**

**Yes**, action and reaction both act simultaneously.

**Example** — To move a boat ahead in water, the boatman pushes **(action)** the water backwards with his oar and at the same time, the water exerts an equal and opposite force **(reaction)** in the forward direction on the boat due to which the boat moves ahead.

#### Question 4

'The action and reaction are equal in magnitude'. Is this statement true?

**Answer**

**Yes**, action and reaction are equal in magnitude.

**Example** — The spring of balance A pulls the spring of balance B due to which we get some reading in balance B. The same reading is seen in balance A because the spring of balance B also pulls the spring of balance A by the same force.

The pull on the spring B by the spring A is the action F_{BA} and the pull on the spring A by the spring B is the reaction F_{AB}.

This demonstrates that **"to every action, there is an equal and opposite reaction" (i.e., in magnitude F _{AB} = F_{BA} but they are in opposite directions. )**

#### Question 5

Comment on the statement 'the sum of action and reaction on a body is zero'.

[Hint: The statement is wrong]

**Answer**

The statement 'the sum of action and reaction on a body is zero' is **wrong**.

As per the **Newton’s third law of motion** —

In an interaction of two bodies A and B, the magnitude of reaction (i.e., the force F_{AB} applied by the body A) is equal in magnitude to the action (i.e., the force F_{BA} applied by the body A on the body B), but **they are in directions opposite to each other**.

Hence, **the two forces can't add up to zero.**

## Exercise 3(D) — Short Answer Type

#### Question 1

State the usefulness of Newton’s third law of motion.

**Answer**

Newton's first law and second law does not explain **how the force acts on the object.** This question is answered by Newton's third law, which states —

**"To every action there is always an equal and opposite reaction".**

**Example** — While moving on the ground, we exert a force by our feet to push the ground backwards, the ground exerts a force of the same magnitude on our feet forward which makes us to move forward.

Hence, the **force exerted by our feet on the ground is the force of action and the force exerted by the ground on our feet is the force of reaction.**

#### Question 2

State and explain the law of action and reaction, by giving two examples.

**Answer**

The Newton’s third law or the law of action and reaction states —

**“To every action there is always an equal and opposite reaction”.**

**Examples**

**Motion of boat in water**— To move a boat ahead in water, the boatman pushes**(action)**the water backwards with his oar and the water exerts an equal and opposite force (reaction) in the forward direction on the boat due to which the boat moves ahead.**Catching a ball**— While catching a ball, the ball exerts a force (action) on the palm of cricketer and the cricketer exerts an equal force (reaction) on the ball to stop it.

#### Question 3

Explain the motion of a rocket with the help of Newton’s third law

**Answer**

Newton’s third law states that **for every action there is always an equal and opposite reaction.**

In a rocket, fuel burning inside the rocket and the burnt gases at high pressure and high temperature are expelled out of the rocket through a nozzle. Thus, rocket exerts a force **(action)** on gases to expel them through a nozzle backwards.

The outgoing gases exert an equal and opposite force R **(reaction)** on the rocket due to which it moves in the forward direction.

Hence, Newton's third law is obeyed.

#### Question 4

When a shot is fired from a gun, the gun gets recoiled. Explain.

**Answer**

The Newton’s third law states —

**“To every action there is always an equal and opposite reaction”.**

So, when a bullet is fired from a gun, a force F is exerted on the bullet **(action)** and the gun experiences an equal recoil R **(reaction)** as shown.

Hence, Newton's third law is obeyed.

#### Question 5

When you step ashore from a stationary boat, it tends to leave the shore. Explain.

**Answer**

In order to get out of the boat we exert a force **(action)** on the board of the boat. This creates a force **(reaction)** which enables us to step out of the boat.

At the same instant, the boat tends to leave the shore due to the force exerted by us (i.e., **action**).

For the safety of the passengers the boatman ties the boat to the pole on the shore so that it does not move away.

#### Question 6

When two spring balances joined at their free ends, are pulled apart, both show the same reading. Explain.

**Answer**

The spring of balance A pulls the spring of balance B due to which we get some reading in balance B. The same reading is seen in balance A because the spring of balance B also pulls the spring of balance A by the same force.

The pull on the spring B by the spring A is the action F_{BA} and the pull on the spring A by the spring B is the reaction F_{AB}.

This demonstrates that **"to every action, there is an equal and opposite reaction" (i.e., in magnitude F _{AB} = F_{BA} but they are in opposite directions. )**

#### Question 7

To move a boat ahead in water, the boatman has to push the water backwards by his oar. Explain.

**Answer**

To move a boat ahead in water, the boatman pushes **(action)** the water backwards with his oar and the water exerts an equal and opposite force **(reaction)** in the forward direction on the boat due to which the boat moves ahead.

#### Question 8

A person pushing a wall hard is liable to fall back. Give reason.

**Answer**

When a person exerts a force **(action)** on a wall by pushing the palm of his hand against it, he will experience a force **(reaction)** exerted by the wall on his palm and hence, he may fall back.

#### Question 9

A light ball falling on ground, after striking the ground rises upwards. Explain the reason.

**Answer**

When a light ball strikes the ground it exerts a force on the ground **(action)** and the ground in turn exerts an equal amount of force R **(reaction)** on the ball, due to which the ball rises up.

## Exercise 3(D) — Numericals

#### Question 1

A boy pushes a wall with a force of 10 N towards east. What force is exerted by the wall on the boy?

**Answer**

As we know, Newton’s third law states that **for every action there is always an equal and opposite reaction.**

Hence, when the boy pushes the wall with a force of 10 N towards east then the wall will also push the boy with a force of **10 N towards west.**

#### Question 2

In figure, a block of weight 15 N is hanging from a rigid support by a string. What force is exerted by —

(a) block on the string,

(b) string on the block.

Name them and show them in the diagram.

**Answer**

(a) The force exerted by the block on the string is **15 N acting downwards due to the weight of the block.**

(b) The force exerted by the string on the block is **15 N acting upwards because of the tension generated.**

## Exercise 3(E) — Multiple Choice Type

#### Question 1

The gravitational force between two bodies is —

- always repulsive
- always attractive
- attractive only at large distances
- repulsive only at large distances.

**Answer**

always attractive

**Reason** — The force of attraction between between two particles because of their masses, is called the gravitational force of attraction. Hence, the gravitational force between two bodies is **always attractive**.

#### Question 2

The gravitational force of attraction between two particles is because of:

- their masses
- their shapes
- their sizes
- none of these

**Answer**

their masses

**Reason** — Each mass particle of the universe attracts other mass particles. The force of attraction between two particles because of their masses, is called **gravitational force of attraction**

#### Question 3

The law of gravitation was given by :

- Thomas Alva Edison
- Alexander Graham Bell
- Sir Isaac Newton
- Albert Einstein

**Answer**

Sir Isaac Newton

**Reason** — For the magnitude of gravitational force of attraction, **Sir Isaac Newton** gave a law, known as the law of gravitation.

#### Question 4

The forces of attraction acting between two bodies is :

- directly proportional to the product of their masses
- directly proportional to the square of the distance between them
- inversely proportional to the square of the distance between them
- both (a) and (c)

**Answer**

both (a) and (c)

**Reason** — According to Newton, the force of attraction acting between two bodies is :

- directly proportional to the product of their masses.
- inversely proportional to the square of the distance between them.

Hence, we get, **F = $\dfrac{Gm_1m_2}{r^2}$**

#### Question 5

The value of gravitational constant G depends on :

- the nature of the particle
- the temperature of the particle
- medium
- none of these

**Answer**

none of these

**Reason** — The value of gravitational constant G remains same at all places, and is independent of the nature of particle, temperature, medium, etc.

#### Question 6

The value of G is —

- 9.8 N m
^{2}kg^{-2} - 6.7 x 10
^{-11}N m^{2}kg^{-2} - 6.7 x 10
^{-11}m s^{-2} - 6.7 N kg
^{-1}

**Answer**

6.7 x 10^{-11} N m^{2} kg^{-2}

**Reason** — The numerical value of gravitational constant G is —

**6.67 x 10 ^{-11} N m ^{2} kg^{-2}**

#### Question 7

The value of gravitational constant G changes with the change in :

- height
- place
- distance
- it remains unchanged

**Answer**

it remains unchanged

**Reason** — The value of gravitational constant G remains same at all places, and is independent of height, place, distance, etc.

#### Question 8

The gravitational force between two masses is :

- directly proportional to the product of the masses
- inversely proportional to the square of the separation between them
- significant between heavenly bodies, but is insignificant between atomic bodies.
- all of the above

**Answer**

all of the above

**Reason** — The gravitational force between two masses is :

- is always attractive
- directly proportional to the product of the masses
- inversely proportional to the square of the separation between them
- significant between heavenly bodies, but is insignificant between atomic bodies because of small magnitude of G.

#### Question 9

The force of attraction between two masses each of 1 kg kept at a separation of 1 m is —

- 9.8 N
- 6.7 N
- 980 N
- 6.7 x 10
^{-11}N

**Answer**

6.7 x 10^{-11} N

**Reason** — As we know,

F = G $\dfrac{m_1 m_2}{r^2}$ and

Given,

r = 1 m

weight of masses = 1 kg each

G = 6.7 x 10 ^{-11} N m ^{2} kg^{-2}

Substituting the values in the formula above, we get,

$F = \dfrac {(6.7 \times 10^{-11}) \times 1 \times 1}{1^2} \\[0.5em] \Rightarrow F = 6.7 \times 10^{-11} \\[0.5em]$

Hence,

the **force of attraction = 6.7 x 10 ^{-11} N**

#### Question 10

The gravitational force of attraction between two bodies of masses 60 kg and 40 kg separated by a distance 10 m is : (where G = 6.7 x 10^{-11} Nm^{2}kg^{-2} )

- 1.60 x 10
^{-9}N - 2.8 x 10
^{-11}N - 1.0 x 10
^{-11}N - 1.6 x 10
^{-11}N

**Answer**

1.60 x 10^{-9} N

**Reason** — As we know,

$\text{F} = \text{G}\dfrac{\text{m}_1\text{m}_2}{\text{R}^2}$

Given,

**masses 60 kg and 40 kg**, when **separation = 10 m**,

Hence,

$\text{F} = \dfrac{{6.7 \times 10^{-11} \times 60 \times 40}}{\text{10}^2} \\[0.5em] \text{F} = 1.60 \times 10^{-9}$

Hence, F = 1.60 x 10^{-9} N

#### Question 11

The S.I. unit of g is :

- m s
^{-1} - m s
^{-2} - cm s
^{-1} - cm s
^{-2}

**Answer**

m s^{-2}

**Reason** — The acceleration due to gravity is denoted by the letter g. Its S.I. unit is m s^{-2}. It is a vector quantity.

#### Question 12

The relationship between g and G is :

**g = $\dfrac{GMm}{R^2}$****G = $\dfrac{gm}{R^2}$****g = $\dfrac{GM}{R^2}$****G = $\dfrac{MR^2}{g}$**

**Answer**

**g = $\dfrac{GM}{R^2}$**

**Reason** —

**g = $\dfrac{GM}{R^2}$**

where,

g = acceleration due to gravity

G = gravitational constant

m = mass of earth

R = radius of earth.

Hence, we can say that **acceleration due to gravity (g) is directly proportional to universal gravitational constant (G)**.

#### Question 13

The value of g on a planet depends on :

- the mass of the planet
- the radius of the planet
- both (a) and (b)
- composition of the planet

**Answer**

both (a) and (b)

**Reason** — As, **g = $\dfrac{GM}{R^2}$** hence, the value of g on a planet depends on the mass and radius of that planet (or satellite).

#### Question 14

A body is projected vertically upward with an initial velocity u. If acceleration due to gravity is g, the time for which it remains in air, is —

- u/g
- ug
- 2u/g
- u/2g

**Answer**

2u/g

**Reason** — At the maximum height, v = 0

maximum height = $\dfrac{u^2}{2g}$ (from equation v^{2} = u^{2} - 2gh)

time taken by the body to reach the maximum height (t) = $\dfrac{u}{g}$ (from equation v = u - gt).

Hence, the same will be the time it takes to get back to the initial point from the highest point. So, total time of journey = 2t = $\dfrac{2u}{g}$.

∴ The time for which the body remains in air = **$\dfrac{2u}{g}$**

#### Question 15

An object falling freely from rest reaches ground in 2 s. If acceleration due to gravity is 9.8 m s^{-2} , the velocity of the object on reaching the ground will be —

- 9.8 m s
^{-1} - 4.9 m s
^{-1} - 19.6 m s
^{-1} - Zero

**Answer**

19.6 m s^{-1}

**Reason** — As we know,

Velocity of the object = g x t

Given,

u = 0,

g = 9.8 m s^{-2}

t = 2 s

v = u + gt

Substituting the values in the formula above, we get,

Velocity of the object = 0 + 9.8 x 2 = 19.6 m s^{-1}

#### Question 16

The body of mass 25 kg is taken from the earth to the moon. If the value of g on moon is 1.6 ms^{-2} then the weight of the body on moon is :

- 50 N
- 100 N
- 40 N
- 60 N

**Answer**

40 N

**Reason** — As we know,

**Weight (W) = mass (m) x acceleration due to gravity (g)**

Given,

**mass = 25 kg**

**g on moon = 1.6 ms ^{-2}**

W_{m} = ?

Substituting the values, we get,

W_{m} = 25 x 1.6 = 40 N

Hence, **weight of body on moon = 40 N**.

#### Question 17

The relationship between weight and mass is :

- W = m
- W = mG
- W = mg
- m = Wg

**Answer**

W = mg

**Reason** — The weight of the body is related to the mass as follows :

Weight = mass x acceleration due to gravity or W = mg

#### Question 18

The S.I. unit of weight is :

- kg
- Joule
- Newton
- dyne

**Answer**

Newton

**Reason** — The weight of a body is the force with which the earth attracts it. In other words, weight of a body is the force of gravity on it.

Weight is a **vector** quantity. It's direction is downwards towards the centre of the earth.

S.I. unit of weight is **Newton (N)**.

#### Question 19

The mass of a body :

- changes with a change of place
- changes with direction
- changes when the velocity of the body is close to the velocity of light
- none of these

**Answer**

changes when the velocity of the body is close to the velocity of light

**Reason** — The mass of a body increases with its velocity but this change is perceptible only when the velocity of the body v becomes more than 10^{6} ms^{-1} i.e., reaches close to the speed of light c = (3 x 10^{8} ms^{-1}), so for a body moving with velocity less than 10^{6} ms^{-1}, its mass is taken to be constant.

## Exercise 3(E) — Very Short Answer Type

#### Question 1

State Newton’s law of gravitation.

**Answer**

Newton’s law of gravitation states that, the force of attraction acting between two bodies is —

(i) **directly proportional to the product of their masses** and

(ii) **inversely proportional to the square of the distance between them**. This force acts along the line joining the two particles.

#### Question 2

State whether the gravitational force between two masses is attractive or repulsive ?

**Answer**

The gravitational force between two masses is always **attractive.**

#### Question 3

Write an expression for the gravitational force of attraction between two bodies of masses m_{1} and m_{2} separated by a distance r.

**Answer**

F = $\text{G}\dfrac{m_1 m_2}{r^2}$

where G is the constant of proportionality known as the gravitational constant.

#### Question 4

How is the gravitational force between two masses affected if the separation between them is doubled ?

**Answer**

As we know,

F = G $\dfrac{m_1 m_2}{r^2}$

If separation between two bodies is doubled then,

F = G $\dfrac{m_1 m_2}{(2r)^2}$

Hence,

F = G $\dfrac{m_1 m_2}{4r^2}$

Therefore, when **the separation between the bodies is doubled the gravitational force reduces to one-fourth**.

#### Question 5

Define gravitational constant G.

**Answer**

The gravitational constant G is numerically equal to the magnitude of attraction between two masses each of 1 kg placed at a separation of 1 m.

#### Question 6

Write the numerical value of gravitational constant G with its S.I. unit.

**Answer**

G = 6.67 x 10^{-11} N m^{2} kg^{-2}

#### Question 7

Define the term acceleration due to gravity ? Write it's S.I. unit.

**Answer**

The rate at which the velocity of a freely falling body increases, is called the acceleration due to gravity. In other words, it is the acceleration produced in a freely falling body due to the gravitational force of attraction of the earth.

It's S.I. unit is **m s ^{-2}**.

#### Question 8

Write down the average value of g on the earth's surface.

**Answer**

The average value of g on the surface of the earth is **9.8 m s ^{-2}**.

#### Question 9

How are g and G related ?

**Answer**

As we know,

**g = $\dfrac{GM}{R^2}$**

where,

g = acceleration due to gravity

G = gravitational constant

m = mass of earth

R = radius of earth.

Hence, we can say that **acceleration due to gravity (g) is directly proportional to universal gravitational constant (G)**.

#### Question 10

A body falls freely under gravity from rest and reaches the ground in time t. Write an expression for the height fallen by the body.

**Answer**

As we know,

**h = ut + $\dfrac{1}{2}$ gt ^{2}**, where,

g = acceleration due to gravity,

t = time

h = height fallen by the body.

Given,

u = 0

Hence, on substituting the values we get,

h = (0 x t) + $\dfrac{1}{2}$ gt^{2}

= 0 + $\dfrac{1}{2}$ gt^{2}

= $\dfrac{1}{2}$ gt^{2}

Therefore, **maximum height attained by the body = $\dfrac{1}{2}$ gt ^{2}**

#### Question 11

A body is thrown vertically upwards with an initial velocity u. Write an expression for the maximum height attained by the body.

**Answer**

As we know,

**v ^{2} = u^{2} - 2gh**

Initial velocity = u

Final velocity = 0

Substituting the values in the formula above we get,

$0^2 = u^2 - 2gh_{max} \\[0.5em] 2gh_{max} = u^2 \\[0.5em] \Rightarrow h_{max} = \dfrac{u^2}{2g} \\[0.5em]$

Hence, **h _{max} = $\dfrac{u^2}{2g}$**

#### Question 12

State the S.I. units of (a) mass and (b) weight.

**Answer**

The S.I. unit of mass is **kilogram (kg)**

The S.I. unit of weight is **newton (N)**

#### Question 13

The value of g at the center of the earth is zero. What will be the weight of a body of mass m kg at the center of the earth ?

**Answer**

Zero

**Reason** — As we know,

weight of a body of mass m kg = mg

At the center of the earth, acceleration due to gravity g = 0.

Substituting the value in the formula above we get,

W = m x 0 = 0

#### Question 14

Which of the following quantity does not change by change of place of a body — mass or weight ?

**Answer**

**Mass** of a body is constant and does not change with the change in place whereas weight varies from place to place.

## Exercise 3(E) — Short Answer Type

#### Question 1

How does the gravitational force of attraction between two masses depend on the distance between them ?

**Answer**

The gravitational force of attraction between two masses is **inversely proportional to the square of the distance between the masses**. It acts along the line joining the two particles.

F ∝ $\dfrac{1}{r^2}$

#### Question 2

What is the importance of the law of gravitation ?

**Answer**

The importance of the law of gravitation is that **Newton used this law to explain, the motion of planets around the sun, the motion of the moon (satellite) around the earth and the motion of a freely falling body**.

#### Question 3

What do you understand by the term force due to gravity ?

**Answer**

According to the law of gravitation, the earth attracts each object around it, towards its center. The force with which the earth attracts a body is called the **force due to gravity** on the body, which can be taken to act vertically downwards at the centre of gravity of the body.

The force due to gravity on a body of mass m kept on the surface of earth of mass M and radius R, is equal to the force of attraction between the earth and that body.

It is given as —

$\text{F} = \text{G}\dfrac{\text{M}\text{m}}{\text{r}^2}$

and the **numerical value of force due to gravity = 9.8 N**.

#### Question 4

Write an expression for the force due to gravity on a body of mass m and explain the meaning of the symbols used in it.

**Answer**

The force due to gravity ‘F’ on a body of mass 'm' kept on the surface of earth of mass 'M' and radius 'R', is equal to the force of attraction between the earth and that body.

Hence, the expression is

$\text{F} = \text{G}\dfrac{\text{M}\text{m}}{\text{R}^2}$

The value of G remains same at all places and it is independent of the nature of the particles , temperature, medium, etc.

Therefore, it is a universal constant and is known as **Universal Gravitational Constant.**

#### Question 5

How is the acceleration due to gravity on the surface of the earth related to it's mass and radius ?

**Answer**

The value of 'g' on earth depends on the value of mass and radius of earth.

Let, g be the acceleration due to gravity of earth of mass M and radius R.

By newton's law of motion, the force due to gravity on a body of mass m on it's surface will be

F = mass x acceleration due to gravity

or F = mg **[Equation 1]**

By Newton's gravitational law, this attractive force is given by

F = $\text{G}\dfrac{\text{M}\text{m}}{\text{R}^2}$ **[Equation 2]**

From eqn. 1 and 2 we get,

$\text{G}\dfrac{\text{M}\text{m}}{\text{R}^2} = \text{m}\text{g}$

⇒ acceleration due to gravity g = $\dfrac{\text{G}\text{M}}{\text{R}^2}$.

#### Question 6

Define the terms mass and weight.

**Answer**

**Mass** — The **mass of a body is the quantity of matter it contains**.

It is a **scalar** quantity and it's S.I. unit is **kg**.

**Weight** — The **weight of a body is the force with which the earth attracts it. In other words, weight of a body is the force of gravity on it**.

Weight is a **vector** quantity. It's direction is downwards towards the centre of the earth.

Unit of weight — S.I. unit of weight is **newton (N)**.

#### Question 7

Distinguish between mass and weight.

**Answer**

Mass | Weight |
---|---|

It is a measure of the quantity of matter contained in the body, at rest. | It is the force with which the earth attracts a body. |

It is a scalar quantity. | It is a vector quantity. |

It's S.I. unit is kg. | It's S.I. unit is newton (N). |

It is measured by a physical balance. | It is measured by a spring balance which is calibrated to read in newton. |

It is a constant for a body and does not change with the change in place. | It is not constant for a body, but varies from place to place due to the change in the value of g. |

#### Question 8

Explain the meaning of the following statement '1 kgf = 9.8 N'.

**Answer**

When we say that 1 kgf = 9.8 N, we mean that we have to exert a force of 9.8 N to hold a mass of 1 kg.

## Exercise 3(E) — Numericals

#### Question 1

The force of attraction between two bodies at a certain separation is 10 N. What will be the force of attraction between them if the separation is reduced to half ?

**Answer**

As we know,

$\text{F} = \text{G}\dfrac{\text{M}\text{m}}{\text{R}^2}$

Given,

**F = 10 N**, when **separation = R**,

Hence,

F_{1} = $\dfrac{G M m}{R^2}$ = 10 N **[Equation 1]**

If **separation = $\dfrac{R}{2}$** then,

Substituting the values in the formula above, we get,

$F_2 = \dfrac{G M m}{\big(\dfrac{R}{2}\big)^2} \\[0.5em] F_2 = \dfrac{4 G M m}{R^2} \\[0.5em]$

$F_2 = 4 \times \big(\dfrac{G M m}{R^2}\big)$ **[Equation 2]**

Substituting the value of equation 1 in equation 2 we get,

F_{2} = 4 x 10 N = **40 N**

#### Question 2

Write the approximate weight of a body of mass 5 kg. What assumption have you made ?

**Answer**

As we know,

**Weight = mg**

Given,

**m = 5 kg**,

**Assumption : g = 10 m s ^{-2}**

Substituting the values in the formula above, we get,

$W = 5 \times 10 \\[0.5em] W = 50 N \\[0.5em]$

Hence, weight of the body = **50 N**.

#### Question 3

Calculate the weight of a body of mass 10 kg in (a) kgf and (b) newton. Take g = 9.8 m s^{-2}

**Answer**

(i) As we know,

**Weight = mg**

Given,

**m = 10 kg**

g = 9.8 m s^{-2}

As m s^{-2} can be written as N kg^{-1},

So, **g = 9.8 N kg ^{-1}**

Substituting the values in the formula above, we get,

$W = 10 \times 9.8 \\[0.5em] W = 98 N\\[0.5em]$

As 9.8 N = 1 kgf,

Hence, **98 N = 10 kgf**

Therefore, **weight in kgf = 10 kgf**

(ii) **Weight in newton = 9.8 N**

#### Question 4

State the magnitude and direction of the force of gravity acting on a body of mass 5 kg. Take g = 9.8 m s^{-2}

**Answer**

As we know,

**Force of gravity (F) = mass (m) x acceleration due to gravity (g)**

Given,

**m = 5 kg**

**g = 9.8 m s ^{-2}**

Substituting the values in the formula above, we get,

$F = 5 \times 9.8 \\[0.5em] \Rightarrow F = 49 N \\[0.5em]$

Hence, the **force of gravity = 49 N**, acting **vertically downwards**.

#### Question 5

The weight of a body is 2.0 N. What is the mass of the body ? (g = 10 m s^{-2})

**Answer**

As we know,

**Weight (W) = mass (m) x acceleration due to gravity (g)**

Given,

**w = 2 N**

**g = 10 m s ^{-2}**

Substituting the values in the formula above, we get,

$2 = m \times 10 \\[0.5em] \Rightarrow m = \dfrac{2}{10} \\[0.5em] \Rightarrow m = 0.2 kg \\[0.5em]$

Hence, the **mass of the body = 0.2 kg**.

#### Question 6

The weight of a body on earth is 98 N where the acceleration due to gravity is 9.8 m s^{-2}. What will be it's (a) mass and (b) weight on moon where the acceleration due to gravity is 1.6 m s^{-2} ?

**Answer**

(a) As we know,

**Weight (W) = mass (m) x acceleration due to gravity (g)**

Given,

**w = 98 N**

**g = 9.8 m s ^{-2}**

Substituting the values in the formula above, we get,

$98 = m \times 9.8 \\[0.5em] \Rightarrow m = \dfrac{98}{9.8} \\[0.5em] \Rightarrow m = 10 kg \\[0.5em]$

Hence, the **mass of the body = 10 kg**.

(b) As mass is constant and it's value remains same on moon (m) as on earth (e), hence,

**m _{e} = m_{m} = 10 kg**

**g _{moon} = 1.6 m s^{-2}**

Substituting the values in the formula above, we get,

$W = 10 \times 1.6 \\[0.5em] \Rightarrow W = 16 N \\[0.5em]$

Hence, the **weight on moon = 16 N**.

#### Question 7

A man weighs 600 N on earth. What would be his approximate weight on moon ? Give reason for your answer ?

**Answer**

As we know,

**Weight (W) = mass (m) x acceleration due to gravity (g)**

Given,

**W _{e} = 600 N**

W_{m} = ?

and

**g _{m} = $\dfrac{1}{6}$g_{e}**

Hence, **W _{m} = $\dfrac{1}{6}$W_{e}**

Substituting the values, we get,

W_{m} = $\dfrac{1}{6}$ x 600 = 100 N

Hence, **weight of the man on the moon = 100 N**.

#### Question 8

What is the (a) force of gravity and (b) weight of a block of mass 10.5 kg? Take g = 10 m s^{-2}

**Answer**

As we know,

**Force of gravity (F) = mass (m) x acceleration due to gravity (g)**

Given,

**m = 10.5 kg**

**g = 10 m s ^{-2}**

Substituting the values in the formula above, we get,

$F = 10.5 \times 10 \\[0.5em] \Rightarrow F = 105 N \\[0.5em]$

Hence, the **force of gravity = 105 N**

(b) As we know,

**Weight (W) = mass (m) x acceleration due to gravity (g)**

Given,

**m = 10.5 kg**

**g = 10 m s ^{-2}**

Substituting the values, we get,

$W = 10.5 \times 10 \\[0.5em] \Rightarrow F = 105 N \\[0.5em]$

Hence, **weight = 105 N**.

#### Question 9

A ball is released from a height and it reaches the ground in 3 s. If g = 9.8 m s^{-2}, find —

(a) the height from which the ball was released,

(b) the velocity with which the ball will strike the ground.

**Answer**

(a) As we know from the equation of motion;

**s = ut + $\dfrac{1}{2}$gt ^{2}**

where, s = height

Given,

**t = 3 s**

**g = 9.8 m s ^{-2}**

initial velocity (**u) = 0**

Substituting the values in the formula above, we get,

$S = (0 \times t) + (\dfrac{1}{2} \times 9.8 \times 3^2) \\[0.5em] S = 0 + (\dfrac{1}{2} \times 9.8 \times 9) \\[0.5em] S = 44.1 m \\[0.5em]$

Hence, **the height from which the ball was released = 44.1 m**

(b) From the equation of motion,

**v ^{2} = u^{2} - 2gs**

where, v = final velocity

Substituting the values in the formula, we get,

$v^2 = 0^2 - (2 \times 9.8 \times 44.1) \\[0.5em] v^2 = 2 \times 9.8 \times 44.1 \\[0.5em] \Rightarrow v^2 = 864.36 \\[0.5em] \Rightarrow v = 29.4 \text{m s}^{-1} \\[0.5em]$

Hence, **velocity with which the ball strikes the ground = 29.4 m s ^{-1}**

#### Question 10

What force, in newton, your muscles need to apply to hold a mass of 5 kg in your hand ? State the assumption.

**Answer**

As we know,

**Force (F) = mass (m) x acceleration due to gravity (g)**

Given,

**m = 5 kg**

**Assumption : g = 9.8 N kg ^{-1}**

Substituting the values in the formula above, we get,

$F = 5 \times 9.8 \\[0.5em] \Rightarrow F = 49 N \\[0.5em]$

Hence, **force = 49 N**

#### Question 11

A ball is thrown vertically upwards. It goes to a height 20 m and then returns to the ground. Taking acceleration due to gravity g to be 10 m s^{-2}

find —

(a) the initial velocity of the ball

(b) the final velocity of the ball on reaching the ground and

(c) the total time of journey of the ball.

**Answer**

As we know, from the equation of motion,

**v ^{2} = u^{2} - 2gs** (a = - g as movement is against gravity )

Given,

**s = 20 m**

**g = 10 m s ^{-2}**

**v = 0**

Substituting the values in the formula, we get,

$0 = u^2 - 2 \times 10 \times 20 \\[0.5em] u^2 = 400 \\[0.5em] \Rightarrow u = 20 \text{m s}^{-1} \\[0.5em]$

Hence, **initial velocity of the ball = 20 m s ^{-1}**

(b) As we know, from the equation of motion,

**v ^{2} = u^{2} + 2gs**

When the ball starts falling after reaching the maximum height, it's velocity (u) = 0

**s = 20 m**

**g = 10 m s ^{-2}**

Substituting the values in the formula, we get,

$v^2 = 0^2 + (2 \times 10 \times 20) \\[0.5em] v^2 = 400 \\[0.5em] \Rightarrow v = 20 \text{m s}^{-1} \\[0.5em]$

Hence, **final velocity of the ball on reaching the ground = 20 m s ^{-1}**

(c) As we know,

total time of of journey of ball (t) = $\dfrac{2u}{g}$ and

**u = 20 m s ^{-1}**

**g = 10 m s ^{-2}**

Substituting the values in the formula, we get,

$t = \dfrac{2 \times 20}{10} \\[0.5em] \Rightarrow t = 4 s \\[0.5em]$

Hence, total time for which the ball stays in air = **4 s**

#### Question 12

A body is dropped from the top of a tower. It acquires a velocity 20 m s^{-1} on reaching the ground. Calculate the height of the tower. (Take g = 10 m s^{-2})

**Answer**

As we know, from the equation of motion,

**v ^{2} = u^{2} + 2gs**

**u = 0**

**v = 20 m s ^{-1}**

**g = 10 m s ^{-2}**

Substituting the values in the formula, we get,

$20^2 = 0^2 + (2 \times 10 \times s) \\[0.5em] 400 = 20 s \\[0.5em] \Rightarrow s = \dfrac{400}{20} \\[0.5em] \Rightarrow s = 20 m \\[0.5em]$

Hence, **the height of the tower = 20 m**

#### Question 13

A ball is thrown vertically upwards. It returns 6 s later. Calculate (i) the greatest height reached by the ball, and (ii) the initial velocity of the ball. (Take g = 10 m s^{-2} )

**Answer**

From the equation of motion,

**h = ut + $\dfrac{1}{2}$ gt ^{2}**

(we consider motion of ball from highest point to the ground)

Since, total time for journey of the ball in air = 6 s.

Therefore, time for reaching maximum height = $\dfrac{6}{2} = 3$ s

**g = 10 m s ^{-2}**

**u = 0**

Substituting the values in the formula, we get,

$h = (0 \times 3) + (\dfrac{1}{2} \times 10 \times 3^2) \\[0.5em] h = 0 + (5 \times 9) \\[0.5em] h = 45 m \\[0.5em]$

Hence, **greatest height reached by the ball = 45 m**

(ii) From the equation of motion,

**v ^{2} = u^{2} - 2gs**

Substituting the values in the formula, we get,

$0^2 = u^2 - (2 \times 10 \times 45) \\[0.5em] u^2 = (2 \times 10 \times 45) \\[0.5em] u^2 = 900 \\[0.5em] u = 30 \text {m s}^{-1} \\[0.5em]$

Hence, the **initial velocity of the ball = 30 m s ^{-1}**

#### Question 14

A pebble is thrown vertically upwards with a speed of 20 m s^{-2}. How high will it be after 2 s ? (Take g = 10 m s^{-2})

**Answer**

As we know, the equation of motion

**h = ut - $\dfrac{1}{2}$ gt ^{2}** (a = -g, as movement is against gravity)

Given,

**t = 2 s**

**g = 10 m s ^{-2}**

u = 20 m s^{-2}

Substituting the values in the formula, we get,

$h = (20 \times 2) - (\dfrac{1}{2} \times 10 \times 2^2) \\[0.5em] h = 40 - (5 \times 4) \\[0.5em] h = 20 m \\[0.5em]$

Hence, **greatest height reached by the ball = 20 m**

#### Question 15

(a) How long will a stone take to fall to the ground from the top of a building 80 m high and (b) what will be the velocity of the stone on reaching the ground ? (Take g = 10 m s^{-2} )

**Answer**

From the equation of motion,

**h = ut + $\dfrac{1}{2}$ gt ^{2}**

Given,

**g = 10 m s ^{-2}**

**h = 80 m**

**u = 0**

Substituting the values in the formula, we get,

$80 = (0 \times t) + (\dfrac{1}{2} \times 10 \times t^2) \\[0.5em] 80 = 0 + (5 \times t^2) \\[0.5em] 80 = 5 \times t^2 \\[0.5em] t^2 = \dfrac{80}{5} \\[0.5em] t^2 = 16 \\[0.5em] t = 4 s \\[0.5em]$

Hence, **time taken = 4 s**

(b) From the equation of motion,

**v ^{2} = u^{2} + 2gh**

**u = 0**

**g = 10 m s ^{-2}**

**h = 80 m**

Substituting the values in the formula, we get,

$v^2 = 0^2 + (2 \times 10 \times 80) \\[0.5em] v^2 = 1600 \\[0.5em] \Rightarrow v = 40 m s ^{-1}\\[0.5em]$

Hence, **final velocity of the stone on reaching the ground = 40 m s ^{-1}**

#### Question 16

A body falls from the top of a building and reaches the ground 2.5 s later. How high is the building ? (Take g = 9.8 m s^{-2})

**Answer**

From the equation of motion,

**h = ut + $\dfrac{1}{2}$ gt ^{2}**

Given,

**g = 9.8 m s ^{-2}**

**t = 2.5 s**

**u = 0**

Substituting the values in the formula, we get,

$h = (0 \times 2.5) + (\dfrac{1}{2} \times 9.8 \times 2.5^2) \\[0.5em] h = 0 + (4.9 \times 2.5^2) \\[0.5em] h = 4.9 \times (2.5)^2 \\[0.5em] h = 30.6 m \\[0.5em]$

Hence, the **height of the building is 30.6 m**

#### Question 17

A ball is thrown vertically upwards with an initial velocity of 49 m s^{-1}. Calculate: (i) the maximum height attained, (ii) the time taken by it before it reaches the ground again. (Take g = 9.8 m s^{-2})

**Answer**

From the equation of motion,

**v ^{2} = u^{2} - 2gh** (a = -g , as the movement is against gravity)

**u = 49 m s ^{-1}**

**v = 0**

**g = 9.8 m s ^{-2}**

Substituting the values in the formula, we get,

$0^2 = 49^2 - 2 \times 9.8 \times h \\[0.5em] 2401 = 19.6 \times h \\[0.5em] \Rightarrow h = \dfrac{2401}{19.6} \\[0.5em] \Rightarrow h = 122.5 m\\[0.5em]$

Hence, **maximum height attained = 122.5 m**

(ii) As we know, total time of journey is,

t = $\dfrac{2u}{g}$

Substituting the values in the formula, we get,

$t = \dfrac{2 \times 49}{9.8} \\[0.5em] t = \dfrac{98}{9.8} \\[0.5em] t = 10 s \\[0.5em]$

Hence, the **time taken by the ball before it reaches the ground again = 10 s**

#### Question 18

A stone is dropped freely from the top of a tower and it reaches the ground in 4 s. Taking g = 10 m s^{-2}, calculate the height of the tower.

**Answer**

From the equation of motion,

**h = ut + $\dfrac{1}{2}$ gt ^{2}**

Given,

**g = 10 m s ^{-2}**

**t = 4 s**

**u = 0**

Substituting the values in the formula, we get,

$h = (0 \times 4) + (\dfrac{1}{2} \times 10 \times 4^2) \\[0.5em] h = 0 + (5 \times 16) \\[0.5em] h = 5 \times 16 \\[0.5em] h = 80 m \\[0.5em]$

Hence, the **height of the tower is 80 m**

#### Question 19

A pebble is dropped freely in a well from it's top. It takes 20 s for the pebble to reach the water surface in the well. Taking g = 10 m s^{-2} and speed of sound = 330 m s^{-1}, find: (i) the depth of water surface, and (ii) the time when echo is heard after the pebble is dropped.

**Answer**

(i) From the equation of motion,

**h = ut + $\dfrac{1}{2}$ gt ^{2}**

Given,

**g = 10 m s ^{-2}**

**t = 20 s**

**u = 0**

Substituting the values in the formula, we get,

$h = (0 \times 20) + (\dfrac{1}{2} \times 10 \times 20^2) \\[0.5em] h = 0 + (5 \times 400) \\[0.5em] h = 5 \times 400 \\[0.5em] h = 2000 m \\[0.5em]$

Hence, the **height of the well is 2000 m**

(ii) As we know,

time (t) = $\dfrac{\text{distance}}{\text{speed}}$

or

t = $\dfrac{\text {depth of well}}{\text{speed of sound}}$

Substituting the values in the formula, we get,

$t = \dfrac{2000}{330} \\[0.5em] \Rightarrow t = 6.1 s \\[0.5em]$

As **the pebble reaches the ground after 20 s, hence echo will be heard after (20 + 6.1) s = 26.1 s**

#### Question 20

A ball is thrown vertically upwards from the top of a tower with an initial velocity of 19.6 m s^{-1}. The ball reaches the ground after 5 s. Calculate: (i) the height of the tower, (ii) the velocity of ball on reaching the ground. Take g = 9.8 m s^{-2}

**Answer**

(i) As we know,

**v = u - gt** (a = -g as the movement is against gravity)

Given,

**u = 19.6 m s ^{-1}**

**v = 0 (velocity on reaching the maximum height)**

**g = 9.8 m s ^{-2}**

**t = 5 s**

Substituting the values in the formula, we get the time taken to reach the maximum height.

$0 = 19.6 - 9.8 \times t \\[0.5em] 19.6 = 9.8 \times t \\[0.5em] \Rightarrow t = \dfrac{19.6}{9.8} \\[0.5em] \Rightarrow t = 2 s \\[0.5em]$

Hence, the time taken to reach the maximum height is 2 s and from maximum height back to the top of the tower = 2s.

Therefore, time taken from the top of the tower to the ground = 5 - (2+2) = 1 s.

So, with the help of the formula:

**h = ut + $\dfrac{1}{2}$ gt ^{2}**

we get,

$h = (19.6 \times 1) + (\dfrac{1}{2} \times 9.8\times 1^2) \\[0.5em] h = 19.6 + 4.9 \\[0.5em] h = 24.5 m \\[0.5em]$

Hence, **height of the tower = 24.5 m**

(ii) Initial velocity (u) = 0 (on falling from maximum height)

Time taken in falling from maximum height to ground = Total Time - Time taken to reach maximum height

= (5 - 2) s

= 3 s

As we know, **v = u + gt**

Substituting the values in the formula, we get,

$v = 0 + (9.8 \times 3) \\[0.5em] \Rightarrow v = 29.4 m s ^{-1} \\[0.5em]$

Hence, the **velocity of ball on reaching the ground = 29.4 m s ^{-1}**