KnowledgeBoat Logo
OPEN IN APP

Chapter 3

Laws of Motion

Class 9 - Concise Physics Selina Solutions


Exercise 3(A)

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:

  1. 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.
  2. 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:

  1. 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.
  2. Electrostatic Force — Two like charges repel, while two unlike charges attract each other. The force between the charges is called electrostatic force.

Question 2

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 3

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 4(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.

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. Laws of Motion, Concise Physics Solutions ICSE Class 9.

Question 4(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.

A spring is compressed against a rigid wall. Draw a neat labelled diagram showing the forces acting on the spring. Laws of Motion, Concise Physics Solutions ICSE Class 9.

Question 4(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.

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. Laws of Motion, Concise Physics Solutions ICSE Class 9.

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.

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

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 8

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.

Multiple Choice Type

Question 1

Which of the following is a contact force —

  1. electrostatic force
  2. gravitational force
  3. frictional force ✓
  4. magnetic force

Answer

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

One of the examples of contact force is 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.

Question 2

The non-contact force is —

  1. force of reaction
  2. force due to gravity ✓
  3. tension in string
  4. force of the friction

Answer

The forces experienced by bodies even without being physically in touch, are called non contact forces.

One of the examples of non contact force is the force due to gravity.

The earth, because of it's mass attracts all other masses around it. The force on a body due to earth's attraction is called the force of gravity or the weight of the body.

Exercise 3(B)

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.

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 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 4

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

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, when no force is applied on a ball moving on a perfectly smooth horizontal surface, it's speed will remain unchanged.

Question 5

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 6

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 7

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 8

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 9

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 10

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 11

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:

  1. 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).
  2. 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 12

'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 13

Name the two kinds of inertia.

Answer

The two kinds of inertia are —

  1. Inertia of rest
  2. Inertia of motion

Question 14

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 15

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 16

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 17

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.

Question 18

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 19

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 20

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 21(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 21(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 21(c)

Explain the following:

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

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.

Question 21(d)

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(e)

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(f)

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.

Multiple Choice Type

Question 1

The property of inertia is more in —

  1. a car
  2. a truck ✓
  3. a horse cart
  4. a toy car

Answer

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 2

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

  1. A less force is required for the cricket ball than for the tennis ball
  2. A less force is required for the tennis ball than for the cricket ball ✓
  3. Same force is required for both the balls
  4. Nothing can be said.

Answer

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 3

A force is needed to —

  1. change the state of motion or state of rest of the body ✓
  2. keep the body in motion
  3. keep the body stationary
  4. keep the velocity of body constant.

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

Exercise 3(C)

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:

  1. Mass
  2. It's Velocity

Question 2

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 3

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 106 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 = m01(vc)2\dfrac{m_0}{\sqrt {1-\big(\dfrac{v}{c}\big)^2}}

where, m0 is the mass of the particle when it is at rest (i.e., v = 0)

Then,

Change in momentum (Δp) = Δ(mv)

Question 4

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

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

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

And we know,

a = (v - u)t\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 5

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,

PA = mvA

For B,

PB = mvB

Ratio between the two is —

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

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

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

Substituting the values, we get,

PAPB=v2vPAPB=12\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 6

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,

Inertia of AInertia of B=mass of Amass of B\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,

Inertia of AInertia of B=m2mInertia of AInertia of B=12\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,

vA = 2v

vB = v

Ratio between the two is —

PAPB=(mv)A(mv)B\dfrac{{P}_A}{{P}_B} = \dfrac{(mv)_A}{(mv)_B} \\[0.5em]

Substituting the values, we get,

PAPB=m×2v2m×vPAPB=2mv2mvPAPB=11\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 7

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 8

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 9

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 10

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 —

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

Question 11

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 12

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 —

Draw graph to show the dependence of acceleration on force for a constant mass. Laws of Motion, Concise Physics Solutions ICSE Class 9.

(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 —

Draw graph to show the dependence of force on mass for a constant acceleration. Laws of Motion, Concise Physics Solutions ICSE Class 9.

Question 13

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

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

How does the acceleration produced by a given force depend on mass of the body? Draw a graph to show it. Laws of Motion, Concise Physics Solutions ICSE Class 9.

Question 14

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 15

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 16

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

= 105 g cm s-2

= 105 dyne.

Thus, 1 newton = 105 dyne.

Question 17

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 18

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 (t1) to stop the ball.

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

F = Change in momentumTime interval\dfrac{\text {Change in momentum}}{\text {Time interval}} = mut1\dfrac{mu}{t_1}

But if the cricketer pulls back his hands along with the ball, he takes a much longer time t2 to stop the ball.

The force now exerted by the ball on his hands is F2 = mut2\dfrac{mu}{t_2}

Since, t2 > t1, therefore, F2 < 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.

Multiple choice type

Question 1

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

  1. v/m
  2. m/v
  3. mv ✓
  4. 1/mv

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

The unit of linear momentum is —

  1. Ns ✓
  2. kg m s-2
  3. N s-1
  4. kg2 m s-1

Answer

As we know,

Value of N = kg m s-2 and

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

= (kg m s-2) x s

= N s

Question 3

The correct form of Newton's second law is —

  1. F = ΔpΔt\dfrac{Δp}{Δt}
  2. F = m ΔvΔt\dfrac{Δv}{Δt}
  3. F = v ΔmΔt\dfrac{Δm}{Δt}
  4. F = mv

Answer

As we know,

Force = Rate of change of momentum

F = ΔpΔt\dfrac{Δp}{Δt}

= Δ(mv)Δt\dfrac{Δ(mv)}{Δt}

= m(Δv)Δt\dfrac{m(Δv)}{Δt}

= ma

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

Question 4

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

  1. size of the body
  2. shape of the body
  3. mass of the body ✓
  4. None of these

Answer

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.

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=5×2p=10p = 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,

1000g=1kg50g=(11000)×5050g=0.05kg1000 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×vv=0.50.05v=0.10.01v=10m s10.5 = 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×aa=152a=7.5m s215 = 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×aa=105a=2m s210 = 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=0.5×5F=2.5NF = 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×aa=102a=5m s210 = 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 1st equation of motion states that;

v = u + at

Substituting the values in the formula above we get,

v=0+(5×3)v=15m s1v = 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) = distance (S)time (t)\dfrac{\text {distance (S)}}{\text {time (t)}}

Given,

s = 100 m in 5 s

Substituting the values in the formula above we get,

v=1005v=20v = \dfrac{100}{5} \\[0.5em] v = 20 \\[0.5em]

Hence, velocity acquired by the body = 20 m s-1

(ii) As we know,

v2 - u2 = 2as

Given,

m = 100 kg

u = 0

v = 20 m s-1

Substituting the values in the formula above we get,

20202=2×a×100400=200×aa=2m s220^2 - 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=100×2F=200NF = 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.

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. Laws of Motion, Concise Physics Solutions ICSE Class 9.

(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 = (11000\dfrac{1}{1000}) x 100 kg

100 g = 0.1 Kg

Hence, m = 0.1 Kg

As,

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

= 205\dfrac{20}{5}

Hence, a = 4 m s-2

Substituting the values in the formula above we get,

F=0.1×4F=0.4F = 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 = (11000\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=0.5×10F=5NF = 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×aa=55a=1m s25 = 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 1st equation of motion states that;

v = u + at

Given,

m = 150 g

Converting g to kg,

1000 g = 1 kg

150 g = (11000\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×0.03)a=250.03a=833.330 = 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×(833.33)F=125NF = 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 1st 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×0.1)a=20.1a=20m s22 = 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=2×20F=40NF = 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 + (12\dfrac{1}{2}) at2

Given,

m = 500 g

Converting g to kg,

1000 g = 1 kg

500 g = (11000\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=(0×2)+(12×a×22)4=0+(12×a×4)4=2aa=42a=2m s24 = (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=0.5×2F=1NF = 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 = vut\dfrac{v - u}{t}

Given,

m = 480 kg

u = 54 km h-1

converting km h-1 to m s-1

54km h1=54×(1000m60×60s)54km h1=54×(10m36s)54km h1=540m36s54km h1=15m s154 \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=01510a=1.5m s2a = \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×(1.5)F=720NF = 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 = vut\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=0302a=15m s2a = \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×(15)m=100kg1500 = 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 = (11000\dfrac{1}{1000}) x 50 kg

100 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 = (1100\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,

v2 - u2 = 2as

Substituting the values in the formula above we get,

021002=2×a×0.0210000=0.04×aa=100000.04a=10000004a=250000m s20^2 - 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 105 m s-2

(iv) Now,

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

Substituting the values in the formula above we get,

F=0.05×(2.5×105)F=0.125×105NF = 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)

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 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”.

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 3

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”.

Example

(a) 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.

(b) 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 4

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) as shown.

Name and state the action and reaction in firing a bullet from a gun. Laws of Motion, Concise Physics Solutions ICSE Class 9.

(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. As a result of which we observe that the hammer moves back a little after it hits the nail.

(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.

Name and state the action and reaction force when a book is lying on a table. Laws of Motion, Concise Physics Solutions ICSE Class 9.

(d) A moving rocket — In a rocket, fuel is burnt inside the rocket and the burnt gases at high pressure and high temperature are excelled out of the rocket through a nozzle. Thus, rocket exerts a force F (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.

Name and state the action and reaction force for a moving rocket. Laws of Motion, Concise Physics Solutions ICSE Class 9.

(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. 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.

Name and state the action and reaction force when a person is walking on the floor. Laws of Motion, Concise Physics Solutions ICSE Class 9.

(f) A moving train colliding with a stationary train — When a moving train collides with a stationary train, it exerts a force F (action) on the stationary train and the stationary train in turn exerts a force R (reaction) on the moving train.

Thus, the stationary train exerts an equal and opposite force on the moving train. As a result of which the moving train moves a little backwards.

Question 5

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.

Explain the motion of a rocket with the help of Newton’s third law. Laws of Motion, Concise Physics Solutions ICSE Class 9.

Hence, Newton's third law is obeyed.

Question 6

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.

When a shot is fired from a gun, the gun gets recoiled. Explain. Laws of Motion, Concise Physics Solutions ICSE Class 9.

Hence, Newton's third law is obeyed.

Question 7

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 8

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 FBA and the pull on the spring A by the spring B is the reaction FAB.

This demonstrates that "to every action, there is an equal and opposite reaction" (i.e., in magnitude FAB = FBA but they are in opposite directions. )

Question 9

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 10

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 11

“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 12

“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 FBA and the pull on the spring A by the spring B is the reaction FAB.

This demonstrates that "to every action, there is an equal and opposite reaction" (i.e., in magnitude FAB = FBA but they are in opposite directions. )

Question 13

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.

Question 14

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 FAB applied by the body A) is equal in magnitude to the action (i.e., the force FBA 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.

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 FBA and the pull on the spring A by the spring B is the reaction FAB.

This demonstrates that "to every action, there is an equal and opposite reaction" (i.e., in magnitude FAB = FBA but they are in opposite directions. )

Multiple choice type

Question 1

Newton’s third law —

  1. defines the force qualitatively
  2. defines the force quantitatively
  3. explains the way a force acts on a body ✓
  4. gives the direction of force.

Answer

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.

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 2

Action and reaction act on the —

  1. same body in opposite directions
  2. different bodies in opposite directions ✓
  3. different bodies, but in same direction
  4. same body in same direction.

Answer

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 FAB applied by the body A) is equal in magnitude to the action (i.e., the force FBA applied by the body A on the body B), but they are in directions opposite to each other.

Hence, the action and reaction never act on the same body, but they always act simultaneously on two bodies i.e., the forces of interaction are always present in pair.

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 —

In figure, a block of weight 15 N is hanging from a rigid support by a string. What force is exerted by block on the string, string on the block. Laws of Motion, Concise Physics Solutions ICSE Class 9.

(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)

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.

State Newton’s law of gravitation. Laws of Motion, Concise Physics Solutions ICSE Class 9.

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 m1 and m2 separated by a distance r.

Answer

Write an expression for the gravitational force of attraction between two bodies of masses m1 and m2 separated by a distance r. Laws of Motion, Concise Physics Solutions ICSE Class 9.

Let there be two bodies of masses m1 and m2 separated by a distance r. The magnitude of force of attraction F acting between them is

F ∝ m1 m2 (i.e. directly proportional to the product of masses)

F ∝ 1r2\dfrac{1}{r^2} (i.e. inversely proportional to the square of distance between them)

Combining the two relations,

F ∝ m1m2r2\dfrac{m_1 m_2}{r^2}

F = Gm1m2r2\text{G}\dfrac{m_1 m_2}{r^2}

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

Question 4

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 ∝ 1r2\dfrac{1}{r^2}

Question 5

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

Answer

As we know,

F = G m1m2r2\dfrac{m_1 m_2}{r^2}

If separation between two bodies is doubled then,

F = G m1m2(2r)2\dfrac{m_1 m_2}{(2r)^2}

Hence,

F = G m1m24r2\dfrac{m_1 m_2}{4r^2}

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

Question 6

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.

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 7

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

Answer

The numerical value of gravitational constant G is 6.67 x 10-11 N m2 kg-2

The S.I. unit of G is N m2 kg-2

Question 8

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 9

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 —

F=GMmr2\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 10

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

F=GMmR2\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 11

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.

The acceleration due to gravity is denoted by the letter g. It's S.I. unit is m s-2.

It is a vector quantity directed vertically downwards towards the centre of earth.

Question 12

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 s2.

As we know,

F=GMmR2\text{F} = \text{G}\dfrac{\text{M}\text{m}}{\text{R}^2}

Taking the mass of earth M = 5.96 x 1024 kg and the radius of earth R = 6.37 x 106 m, G = 6.67 x 10 -11 N m 2 kg-2, the force of gravity on a body of mass m = 1 kg on the surface of earth will be

F = (6.67×1011)×(5.96×1024)×1(6.37×106)2\dfrac{(6.67 \times 10 ^{-11}) \times (5.96 \times 10^{24}) \times 1}{(6.37 \times 10^{6})^{2}} = 9.8 N.

Thus, earth attracts a body of mass 1 kg by a force of 9.8 N towards it's centre.

Question 13

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 = GMmR2\text{G}\dfrac{\text{M}\text{m}}{\text{R}^2}     [Equation 2]

From eqn. 1 and 2 we get,

GMmR2=mg\text{G}\dfrac{\text{M}\text{m}}{\text{R}^2} = \text{m}\text{g}

⇒ acceleration due to gravity g = GMR2\dfrac{\text{G}\text{M}}{\text{R}^2}.

Question 14

How are g and G related ?

Answer

As we know,

g = GMR2\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 15

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 + 12\dfrac{1}{2} gt2, 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) + 12\dfrac{1}{2} gt2

= 0 + 12\dfrac{1}{2} gt2

= 12\dfrac{1}{2} gt2

Therefore, maximum height attained by the body = 12\dfrac{1}{2} gt2

Question 16

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,

v2 = u2 - 2gh

Initial velocity = u

Final velocity = 0

Substituting the values in the formula above we get,

02=u22ghmax2ghmax=u2hmax=u22g0^2 = u^2 - 2gh_{max} \\[0.5em] 2gh_{max} = u^2 \\[0.5em] \Rightarrow h_{max} = \dfrac{u^2}{2g} \\[0.5em]

Hence, hmax = u22g\dfrac{u^2}{2g}

Question 17

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 18

Distinguish between mass and weight.

Answer

MassWeight
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 19

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 20

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

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

Hence,

Weight of a body of mass m kg at the center of the earth = zero.

Question 21

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.

Question 22

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.

Multiple choice type

Question 1

The gravitational force between two bodies is —

  1. always repulsive
  2. always attractive ✓
  3. attractive only at large distances
  4. repulsive only at large distances.

Answer

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 value of G is —

  1. 9.8 N m2 kg-2
  2. 6.7 x 10-11 N m2 kg-2
  3. 6.7 x 10-11 m s-2
  4. 6.7 N kg-1

Answer

The numerical value of gravitational constant G is —

6.67 x 10 -11 N m 2 kg-2

Question 3

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

  1. 9.8 N
  2. 6.7 N
  3. 980 N
  4. 6.7 x 10-11 N ✓

Answer

As we know,

F = G m1m2r2\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=(6.7×1011)×1×112F=6.7×1011F = \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 4

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 —

  1. u/g
  2. ug
  3. 2u/g ✓
  4. u/2g

Answer

At the maximum height, v = 0

maximum height = u22g\dfrac{u^2}{2g} (from equation v2 = u2 - 2gh)

time taken by the body to reach the maximum height (t) = ug\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 = 2ug\dfrac{2u}{g}.

Hence, the time for which the body remains in air = 2ug\dfrac{2u}{g}

Question 5

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 —

  1. 9.8 m s-1
  2. 4.9 m s-1
  3. 19.6 m s-1
  4. Zero

Answer

As we know,

Velocity of the object = g x t

Given,

g = 9.8 m s-2

t = 2 s

Substituting the values in the formula above, we get,

Velocity of the object=9.8×2Velocity of the object=19.6m s1\text {Velocity of the object} = 9.8 \times 2 \\[0.5em] \Rightarrow \text{Velocity of the object} = 19.6 \text{m s}^{-1}\\[0.5em]

Hence, the velocity of the object on reaching the ground will be 19.6 m s-1

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,

F=GMmR2\text{F} = \text{G}\dfrac{\text{M}\text{m}}{\text{R}^2}

Given,

F = 10 N, when separation = R,

Hence,

F1 = GMmR2\dfrac{G M m}{R^2} = 10 N     [Equation 1]

If separation = R2\dfrac{R}{2} then,

Substituting the values in the formula above, we get,

F2=GMm(R2)2F2=4GMmR2F_2 = \dfrac{G M m}{\big(\dfrac{R}{2}\big)^2} \\[0.5em] F_2 = \dfrac{4 G M m}{R^2} \\[0.5em]

F2=4×(GMmR2)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,

F2 = 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=5×10W=50NW = 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=10×9.8W=98NW = 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=5×9.8F=49NF = 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×10m=210m=0.2kg2 = 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×9.8m=989.8m=10kg98 = 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,

me = mm = 10 kg

gmoon = 1.6 m s-2

Substituting the values in the formula above, we get,

W=10×1.6W=16NW = 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,

We = 600 N

Wm = ?

and

gm = 16\dfrac{1}{6}ge

Hence, Wm = 16\dfrac{1}{6}We

Substituting the values, we get,

Wm = 16\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=10.5×10F=105NF = 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=10.5×10F=105NW = 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 + 12\dfrac{1}{2}gt2

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×t)+(12×9.8×32)S=0+(12×9.8×9)S=44.1mS = (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,

v2 = u2 - 2gs

where, v = final velocity

Substituting the values in the formula, we get,

v2=02(2×9.8×44.1)v2=2×9.8×44.1v2=864.36v=29.4m s1v^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=5×9.8F=49NF = 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,

v2 = u2 - 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=u22×10×20u2=400u=20m s10 = 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,

v2 = u2 + 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,

v2=02+(2×10×20)v2=400v=20m s1v^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) = 2ug\dfrac{2u}{g} and

u = 20 m s-1

g = 10 m s-2

Substituting the values in the formula, we get,

t=2×2010t=4st = \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,

v2 = u2 + 2gs

u = 0

v = 20 m s-1

g = 10 m s-2

Substituting the values in the formula, we get,

202=02+(2×10×s)400=20ss=40020s=20m20^2 = 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 + 12\dfrac{1}{2} gt2

(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 = 62=3\dfrac{6}{2} = 3 s

g = 10 m s-2

u = 0

Substituting the values in the formula, we get,

h=(0×3)+(12×10×32)h=0+(5×9)h=45mh = (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,

v2 = u2 - 2gs

Substituting the values in the formula, we get,

02=u2(2×10×45)u2=(2×10×45)u2=900u=30m s10^2 = 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 - 12\dfrac{1}{2} gt2 (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=(20×2)(12×10×22)h=40(5×4)h=20mh = (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 + 12\dfrac{1}{2} gt2

Given,

g = 10 m s-2

h = 80 m

u = 0

Substituting the values in the formula, we get,

80=(0×t)+(12×10×t2)80=0+(5×t2)80=5×t2t2=805t2=16t=4s80 = (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,

v2 = u2 + 2gh

u = 0

g = 10 m s-2

h = 80 m

Substituting the values in the formula, we get,

v2=02+(2×10×80)v2=1600v=40ms1v^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 + 12\dfrac{1}{2} gt2

Given,

g = 9.8 m s-2

t = 2.5 s

u = 0

Substituting the values in the formula, we get,

h=(0×2.5)+(12×9.8×2.52)h=0+(4.9×2.52)h=4.9×(2.5)2h=30.6mh = (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,

v2 = u2 - 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,

02=4922×9.8×h2401=19.6×hh=240119.6h=122.5m0^2 = 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 = 2ug\dfrac{2u}{g}

Substituting the values in the formula, we get,

t=2×499.8t=989.8t=10st = \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 + 12\dfrac{1}{2} gt2

Given,

g = 10 m s-2

t = 4 s

u = 0

Substituting the values in the formula, we get,

h=(0×4)+(12×10×42)h=0+(5×16)h=5×16h=80mh = (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 + 12\dfrac{1}{2} gt2

Given,

g = 10 m s-2

t = 20 s

u = 0

Substituting the values in the formula, we get,

h=(0×20)+(12×10×202)h=0+(5×400)h=5×400h=2000mh = (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) = distancespeed\dfrac{\text{distance}}{\text{speed}}

or

t = depth of wellspeed of sound\dfrac{\text {depth of well}}{\text{speed of sound}}

Substituting the values in the formula, we get,

t=2000330t=6.1st = \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=19.69.8×t19.6=9.8×tt=19.69.8t=2s0 = 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 + 12\dfrac{1}{2} gt2

we get,

h=(19.6×1)+(12×9.8×12)h=19.6+4.9h=24.5mh = (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+(9.8×3)v=29.4ms1v = 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

PrevNext