Laws of Motion Class 9 Physics

Selina ICSE Solutions for Class 9 Physics Chapter 3 Laws of Motion

Exercise 3(A)

1S.Explain giving two examples each of :

a. Contact forces
b. Non - contact forces

Solution 1S.

a.The forces which act on bodies when they are in physical contact are called contact forces.

Example: frictional force and force exerted on two bodies during collision.

b. The forces experienced by bodies even without being physically touched are called non-contact forces.

Example: Gravitational force and Electrostatic force.

2S.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

Solution 2S.

Contact force: (a) frictional force (b) normal reaction force (c) force of tension in a string

Non-contact force: (d) gravitational force (e) electric force (f) magnetic force

3S.Give one example in each case where :
a. The force is of contact, and
b. Force is at a distance

Solution 3S.

a. Force exerted on two bodies during collision.

b. Magnetic force between magnetic poles.

4S.A. A ball is hanging by string from the ceiling of the roof. Draw a neat labelled diagram showing the forces acting on the ball and the string.

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

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.

Solution 4S.

C. The forces acting on the block are its weight in the downward direction and the normal reaction force due to the table on the upward direction.

5S.State one factor on which the magnitude of a non-contact force depends. How does it depend on the factor stated by you?

Solution 5S.

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

The force decreases as the distance of separation increases.

The force is inversely proportional to the square of the distance of separation.

6S.The separation between two masses is reduced to half. How is the magnitude of gravitational force between them affected?

Solution 6S.

The magnitude of gravitational force between two masses will become four times as gravitational force varies inversely as the square of distance of separation.

7S.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?

Solution 7S.

A force when applied on a non-rigid body changes the inter-spacing between its constituent particles and therefore causes a change in its dimensions and can also produce motion in it.

On the other hand, a force when applied on a rigid body, does not change the inter-spacing between its constituent particles and therefore it does not change the dimensions of the body but causes motion in it

8S.Give one example in each of the following cases where a force:



 i. Stops a moving body.
 ii. Moves a stationary body.
 iii. Changes the size of a body.
 iv. Changes the shape of a body.

Solution 8S.

i. A fielder on the ground stops a moving ball by applying a force with his hands.

ii. The pull exerted by horse makes a cart moves.

iii. In a cycle pump, when the piston is lowered, the air is compressed to occupy a less volume.

iv. On pressing a piece of rubber, its shape changes.

1M.Which of the following is a contact force:

a. Electrostatic force
b. Gravitational force
c. Frictional force
d. Magnetic force

Solution 1M.

Frictional force is a contact force.

2M.The non - contact force is :

a. Force of reaction
b. Force due to gravity
c. Tension in string
d. Force of friction

Solution 2M.

Force due to gravity is a non-contact force.

Exercise 3(B)

1S.Name the physical quantity that causes motion in a body.

Solution 1S.

Force causes motion in a body.

2S.Is force needed to keep a moving body in motion?

Solution 2S.

Force is not needed to keep a moving body in motion.

3S.A ball moving on a table eventually stops. Explain the reason.

Solution 3S.

The force of friction between the table and the ball opposes the motion of the ball.

4S.A ball is moving on a perfectly smooth horizontal surface. If no force is applied on it, then will its speed decrease, increase or remain unchanged?

Solution 4S.

In absence of any external force, its speed shall remain unchanged.

5S.What is Galileo's law of inertia?

Solution 5S.

Galileo’s law of inertia states that a body continues to be in its state of rest or of uniform motion

unless an external force is applied on it.

6S.State Newton's first law of motion.

Solution 6S.

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 the body 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.

7S.State and explain the law of inertia (or Newton's first law of motion).

Solution 7S.

Statement of Newton's first law: If a body is in a state of rest, it will remain in the state of rest, and if the body 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.

Explanation: Newton's first law can be explained in the following two parts:

(i) Definition of inertia: The 1st part of Newton's first law of motion gives the definition of inertia, according to which an object cannot change its state by itself.

Example: A book lying on a table will remain in its position unless it is displaced.

(ii) Definition of force: The 2nd 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 stop a moving object.

Example: A book lying on a table is displaced from its place when it is pushed.

8S.What is meant by the term inertia?

Solution 8S.

The property of an object by virtue of which it neither changes its state nor tends to change the state is called inertia.

9S.Give the qualitative definition of force based on Newton's first law of motion.

Solution 9S.

Force is that external cause which can move a stationary object or which can stop a moving object.

10S.Name the factor on which the inertia of a body depends and state how does it depend on the factor stated above.

Solution 10S.

Inertia of a body depends on its mass. Inertia is directly proportional to mass, i.e. greater the mass of a body, greater is its inertia.

11S.Give two examples to show that greater the mass, greater is the inertia of the body.

Solution 11S.

Examples to show that greater the mass, greater is the inertia of the body are as shown below:

1. If you want to start a car by pushing it, you find that it takes a very large force to overcome its inertia. On the other hand, only a small force is needed to start a child’s express wagon. The difference between the car and express wagon is the difference in mass. The car has a large mass, whereas the wagon has a small one.

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

12S.'More the mass, the more difficult it is to move the body from rest'. Explain this statement by giving an example.

Solution 12S.

It is difficult, i.e. a larger force is required to set a loaded trolley (which has more mass) in motion than an unloaded trolley (which has less mass).

13S.Name the two kinds of inertia.

Solution 13S.

Two kinds of inertia are as listed below:

1. Inertia of rest.

2. Inertia of motion.

14S.Give one example of each of the following cases of inertia:

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

Solution 14S.

Examples of inertia of rest: A coin placed on top of a card remains in place when the card is slightly and quickly jerked horizontally.

Example of inertia of motion: A ball thrown vertically upwards by a person in a moving train comes back to his hand.

15S.Two equal and opposite forces act on a stationary body. Will the body move? Give a reason to your answer.

Solution 15S.

No, the body will not move because the net force acting on it is zero. Hence, it will remain stationary due to inertia of rest.

16S.Two equal and opposite forces act on a moving object. How is its motion affected? Give reason.

Solution 16S.

The motion remains unaffected because the net force acting on it is zero.

17S.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?

Solution 17S.

The net force on the airplane is zero or the upward force is equal to the downward force.

18S.Why does a person tend to fall when he jumps out from a moving train and tries to stop immediately?

Solution 18S.

If a person jumps out of a moving train and tries to stop immediately, he falls due to inertia of motion. This is because his body tends to move forward with the velocity of the train while his feet are stationary.

19S.Why does a coin, placed on a card, drop into the tumbler when the card is rapidly flicked with a finger?

Solution 19S.

The reason is that when the card is flicked, a momentary force acts on the card, so it moves away. However, the coin kept on it does not share the motion at once and it remains stationary at its place due to the inertia of rest. The coin then falls down into the tumbler due to the pull of gravity.

20S.When a ball is thrown vertically upwards from a moving train it comes back to the thrower's hand. Why does this happen?

Solution 20S.

The reason is that when the ball is thrown, the ball is in motion along with the person and train. Due to the inertia of motion, during the time the ball remains in air, the person and ball move ahead by the same distance. This makes the ball fall back into the thrower’s hand.

21S.The reason is that when the ball is thrown, the ball is in motion along with the person and train. Due to the inertia of motion, during the time the ball remains in air, the person and ball move ahead by the same distance. This makes the ball fall back into the thrower's hand.

Solution 21S.

(a) When a train suddenly starts, the passengers tend to fall backwards. This is because the lower part of the body, which is in contact with the train, begins to move while the upper part of the body tends to maintain its position of rest. As a result, the upper part tends to fall backwards.

(b) The frame of the sliding door being in contact with the floor of the train also comes in motion on start of the train, but the sliding door remains in its position due to inertia. Thus, the frame moves ahead with the train, while the door slides opposite to the direction of motion of the train. Thus, the door may shut or open accordingly.

(c) When the branches of the tree are shaken, they come in motion, while the fruits due to inertia remain in a state of rest. Thus, the larger and weakly attached fruits get detached from the branches and fall down due to the pull of gravity.

(d) When people alight from a moving bus, they continue to run alongside the bus to avoid falling. If they were to stop at once, the feet would come to rest suddenly but the upper part of the body would still be in motion and they would tend to fall forward.

(e) The part of the carpet where the stick strikes comes in motion at once, while the dust particles settled on it remain in the state of rest due to inertia of rest. Thus, a part of the carpet moves ahead with the stick leaving behind the dust particles that fall down due to gravity.

(f) When running, the athlete brings his body in the state of motion. When the body is in motion, it becomes easier to take a long jump.

1M.The property of inertia is more in the following vehicle:

(a) A car (b) A truck
(c) A horse cart  (d) A toy car.

Solution 1M.

A truck

2M.A tennis ball and cricket ball are both stationary. To get them into the state of motion, the following action is required:

(a) Less force is required for the cricket ball than for the tennis ball.
(b) Less force is required for the tennis ball than for the cricket ball.
(c) Same force is required for both the balls.
(d) Nothing can be said.

Solution 2M.

Less force is required for the tennis ball than for the cricket ball.

3M.A force is needed to do the following actions:

(a) Change the state of motion or state of rest of the body.
(b) Maintain the body in motion. 
(c) Maintain the body at rest.
(d) Maintain constant velocity of the body.

Solution 3M.

Change the state of motion or state of rest of the body.

Exercise 3(C)

1S.Name the two factors on which the force needed to stop a moving body in a given time depends.

Solution 1S.

Force needed to stop a moving body in a given time depends on its mass and velocity.

2S.Define linear momentum and state its S.I. unit.

Solution 2S.

Linear momentum of a body is the product of its mass and velocity.

Its SI unit is kgms-1.

3S.A body of mass m moving with a velocity v is acted upon by a force. Write an 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.

Solution 3S.

4S.Show that the rate of change of momentum = mass × acceleration. Under what condition does this relation hold?

Solution 4S.

Let a force ‘F’ be applied on a body of mass m for a time ‘t’ due to which its velocity changes from u to v. Then,

Initial momentum of body = mu

Final momentum of body = mv

Change in momentum of the body in ‘t’ seconds = mv – mu = m (v – u)

Rate of change of momentum = Change in momentum/time

= [m (v – u)]/t

However, acceleration a = Change in velocity/time = (v – u)/t

Therefore, rate of change of momentum = ma = mass × acceleration

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

5S.Two bodies A and B of same mass are moving with velocities v and 2v, respectively. Compare their (i) inertia and (ii) momentum.

Solution 5S.

(i) Mass is the measure of inertia.

Let ‘m’ be the mass of the two bodies.

Inertia of body A:Inertia of body B :: m:m

Or, Inertia of body A:Inertia of body B :: 1:1

(ii) Momentum of body A = m (v)

Momentum of body B = m (2v) = 2mv

Momentum of body A:Momentum of body B :: mv:2mv

Or, Momentum of body A:Momentum of body B :: 1:2.

6S.Two balls A and B of masses m and 2 m are in motion with velocities 2v and v, respectively. Compare:

(i) Their inertia.
(ii) Their momentum.
(iii)  The force needed to stop them in the same time.

Solution 6S.

(i) Inertia of body A:Inertia of body B :: m:2m

Or, Inertia of body A:Inertia of body B :: 1:2.

(ii) Momentum of body A = m ( 2v) = 2mv

Momentum of body B = (2m) v = 2mv

Momentum of body A:Momentum of body B :: 2 mv:2mv

Or, Momentum of body A:Momentum of body B :: 1:1.

(iii) According to Newton’s 2nd law of motion, rate of change of momentum is directly proportional to the force applied on it. Therefore,

Force needed to stop A:Force needed to stop B :: 1:1.

7S.State the Newton's second law of motion. What information do you get from it?

Solution 7S.

According to Newton’s second law of motion, the rate of change of momentum is directly proportional to the force applied on it and the change of momentum takes place in the direction in which the force is applied.

It gives the quantitative value of force, i.e. it relates the force to the measurable quantities such as acceleration and mass.

8S.How does Newton's second law of motion differ from the first law of motion?

Solution 8S.

Newton’s first law of motion gives the qualitative definition of force. It explains the force as the cause of acceleration only qualitatively but Newton’s second law of motion gives the quantitative value of force. It states force as the product of mass and acceleration. Thus, it relates force to the measurable quantities such as acceleration and mass.

9S.Write the mathematical form of Newton's second law of motion. State the conditions if any.

Solution 9S.

Mathematical expression of Newton’s second law of motion is as shown below:

Force = Mass x Acceleration

Above relation holds for the following conditions:

(i) When the velocity of the body is much smaller than the velocity of light.

(ii) When the mass remains constant.

10S.State Newton's second law of motion. Under what condition does it take the form F = ma?

Solution 10S.

According to Newton’s second law of motion, the rate of change of momentum is directly proportional to the force applied on it, and the change of momentum takes place in the direction in which the force is applied.

The relation F=ma holds for the following conditions:

(i) When the velocity of the body is much smaller than the velocity of light.

(ii) When the mass remains constant.

11S.How can Newton's first law of motion be obtained from the second law of motion?

Solution 11S.

From Newton’s second law of motion, F = ma.

If F = 0, then a = 0.

This means that if no force is applied on the body, its acceleration will be zero. If the body is at rest, then it will remain in the state of rest and if it is moving, then it will remain moving in the same direction with the same speed. Thus, a body not acted upon by an external force, does not change its state of rest or motion. This is the statement of Newton’s first law of motion.

12S.Draw graphs to show the dependence of (i) acceleration on force for a constant mass and (ii) force on mass for a constant acceleration.

Solution 12S.

13S.How does the acceleration produced by a given force depend on the mass of the body? Draw a graph to show it.

Solution 13S.

If a given force is applied on bodies of different masses, then the acceleration produced in them is inversely proportional to their masses.

A graph plotted for acceleration (a) against mass (m) is a hyperbola.

14S.Name the S.I. unit of force and define it.

Solution 14S.

The S.I. unit of force is newton.

One newton is the force which acts on a body of mass 1kg and produces an acceleration of 1 m/s2, i.e. 1 N = 1 kg × 1 m/s2.

15S.What is the C.G.S. unit of force? How is it defined?

Solution 15S.

The C.G.S. unit of force is dyne.

One dyne is the force which acts on a body of mass 1 gramme and produces an acceleration of 1 cms-2, i.e. 1 dyne = 1 g × 1 cms-2.

16S.Name the S.I. and C.G.S. units of force. How are they related?

Solution 16S.

The S.I. unit of force is newton and the C.G.S. unit of force is dyne.

1 N = 105 dyne.

17S.Why does a glass vessel break when it falls on a hard floor, but it does not break when it falls on a carpet?

Solution 17S.

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

18S.Use Newton's second law of motion to explain the following instances:

(a) You pull your hands back while catching a fast moving cricket ball.

(b) You prefer to land on sand instead of hard floor while taking a high jump.

Solution 18S.

(a) We pull our hands back while catching a fast moving cricket ball, because by doing so, we increase the time of catch, i.e. increase the time to bring about a given change in momentum, and hence, the rate of change of momentum decreases. Thus, a small force is exerted on our hands by the ball.

(b) When an athlete lands from a height on a hard floor, his feet comes to rest instantaneously, so a very large force is exerted by the floor on his feet, but if 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.

1M.The linear momentum of a body of mass m moving with velocity v is given by the following relation:

(a) v/m   (b) m/v

(c) mv    (d) 1/mv

Solution 1M.

Mv

2M.The unit of linear momentum is :

(a) N s  (b) kg m s-2

(c) N s-1  (d) kg2 m s-1

Solution 2M.

N s

3M.The correct form of Newton's second law is given by the following equation:

(a)  (b) 

(c)  (d) 

Solution 3M.

4M.The acceleration produced in a body by a force of given magnitude depends on the following factor:

(a) Size of the body   (b) Shape of the body

(c) Mass of the body  (d) None of these.

Solution 4M.

Mass of the body

1N.A body of mass 5 kg is moving with velocity 2 m s-1. Calculate its linear momentum.

Solution 1N.

Mass of the body, m = 5kg

Velocity, v = 2 m/s

Linear momentum = mv = (5)(2) kg m/s

= 10 kg m/s

2N.The linear momentum of a ball of mass 50 g is 0.5 kg m s-1. Find its velocity.

Solution 2N.

Linear momentum = 0.5 kg m/s

Mass, m = 50 g = 0.05 kg

Velocity = Linear momentum/mass

= 0.5/0.05 m/s

= 10 m/s

3N.A force of 15 N acts on a body of mass 2 kg. Calculate the acceleration produced.

Solution 3N.

Force, F = 15 N

Mass, m = 2kg

Acceleration, a = F/m [ From Newton’s second law]

Or, a = (15/2) ms-2

Or, a = 7.5 ms-2

4N.A force of 10 N acts on a body of mass 5 kg. Find the acceleration produced.

Solution 4N.

Force, F = 10 N

Mass, m = 5kg

Acceleration, a = F/m [ From Newton’s second law]

Or, a = (10/5) ms-2

Or, a = 2 ms-2

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

Solution 5N.

Mass, m = 0.5 kg.

Acceleration, a = 5 ms-2

Force, F = ma   [ From Newton’s second law]

Or, F = (0.5) (5) N = 2.5 N.

6N.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.

Solution 6N.

Force, F = 10 N

Mass, m = 2 kg

Time, t = 3 s

Initial velocity, u = 0 m/s.

(i) Let v be the final velocity acquired.

From Newton’s second law,

F = ma.

Or, a = F/m = 10/2 = 5 ms-2.

From the 1st equation of motion,

a = (v – u)/t

Or, v = at + u.

Or, v = (5)(3) + 0 = 15 m/s.

(ii) Change in momentum = Final momentum – initial momentum

Δp = mv – mu.

Or, Δp = m (v – u).

Or, Δp = 2 ( 15 – 0) = 30 kg m/s.

7N.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.

Solution 7N.

Mass, m = 100 kg

Distance moved, s = 100 m

Initial velocity, u = 0

(i) Because the body moves through a distance of 100 m in 5 s,

Velocity of the body = Distance moved / time taken

Velocity = (100/5) = 20 m/s

(ii) From Newton’s third equation of motion,

v2  u2 = 2as.

Or, a = (v2  u2) /2s.

Or, a = [ (202  02)/ 2(100) ] ms-2.

Or, a = 2 ms-2.

(iii) Force, F = ma

Or, F = (100) (2) N.

Or, F = 200 N.

8N.Fig. 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 the v-t graph)

Solution 8N.

Slope of a velocity-time graph gives the value of acceleration.

Here, slope = 20/5 = 4 m/s2.

Or, acceleration, a = 4 m/s2.

Force = Mass × Acceleration.

Given mass, m = 100 g = 0.1 kg.

Force = (0.1) (4) = 0.4 N.

9N.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 the force. 

Solution 9N.

Mass, m = 2 kg

Initial velocity, u = 0

Final velocity, v = 2 m/s

Time, t = 0.1 s

Acceleration = Change in velocity/time

Or, a = (v – u) /t

Or, a = (2 – 0)/ 0.1 = 20 ms-2.

Force = Mass x Acceleration

Or, F = (2) (20) = 40 N.

10N.

Solution 10N.

Mass, m = 100g = 0.1 kg.

Initial velocity, u = 30 m/s.

Final velocity, v = 0.

Time, t = 0.03 s.

Acceleration = Change in velocity/time.

Or, a = (v – u)/t.

Or, a = (0 – 30)/ 0.03 = -1000 ms-2.

Here, negative sign indicates retardation.

Now, Force = Mass x Acceleration

Or, F = (0.1) (1000) = 100 N.

11N.A car of mass 480 kg moving at a speed of 54 km per hour is stopped in 10 s. Calculate the force applied by the brakes.

Solution 11N.

Mass, m = 480 kg.

Initial velocity, u = 54 km/hr = 15 m/s.

Final velocity, v = 0.

Time, t = 10 s.

Acceleration = Change in velocity/time.

Or, a = (v – u)/t.

Or, a = (0 – 15)/10 = -1.5 ms-2.

Here, negative sign indicates retardation.

Now, Force = Mass x Acceleration

Or, F = (480) (1.5) = 720 N.

12N.A bullet of mass 50 g moving with an initial velocity 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.

Solution 12N.

Mass, m = 50 gm = 0.05 kg.

Initial velocity, u = 100 m/s.

Final velocity, v = 0.

Distance, s = 2cm = 0.02 m.

(i) Initial momentum = mu = (0.05) (100) = 5 kg m/s.

(ii) Final momentum = mv = (0.05) (0) = 0 kg m/s.

(iii) Acceleration, a = (v2 – u2)/2s.

Or, a = (02 – 1002)/ 2(0.02).

Or, a = – 2.5 x 105 ms-2.

Therefore, retardation is 2.5 x 105 ms-2.

(iv) Force, F = ma

Or, F = (0.05 kg) (2.5 x 105 ms-2)

Or, F = 12500 N

13N.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?

Solution 13N.

Let the force be F.

Force F causes an acceleration, a = 10 m/s2 in a body of mass, m = 500 g or 0.5 kg

Thus, F = ma

Or, F = (0.5) (10) = 5 N

Let a’ be the acceleration which force F (=5N) cause on a body of mass, m’ = 5 kg.

Then, a’ = F/m’.

Or, a’ = (5/5) ms-2.

Or, a’ = 1 ms-2.

14N.A car is moving with a uniform velocity 30 ms-1. It is stopped in 2 s by applying a force of 1500 N through its brakes.

Calculate the following values:

(a) The change in momentum of car.
(b) The retardation produced in car.
(c) The mass of car.

Solution 14N.

Initial velocity, u = 30 m/s

Final velocity, v = 0

Time, t = 2s

Force, F = 1500 N

Here, a = (v – u)/t = (0 – 30)/ 2 = – 15 ms-2. Here, negative sign indicates retardation.

Now, F = ma.

Or, m = F/a = (1500/ 15) = 100 kg.

(a) Change in momentum = Final momentum – Initial momentum

Or, Δp = m (v – u)

Or, Δp = 100 (0 – 30)

Or, Δp = 3000 kg m/s

(b) Acceleration, a = (v – u)/t.

Or, a = (0 – 30)/ 2 = – 15 ms-2,

Here, negative sign indicates retardation.

Thus, retardation = 15 ms-2.

(c) From Newton’s second law of motion,

F = ma

Or, m = F/a = (1500/ 15) = 100 kg.

Exercise 3(D)

1S.State the purpose of Newton's third law of motion.

Solution 1S.

Newton’s third law explains how a force acts on an object.

2S.State Newton's third law of motion.

Solution 2S.

According to Newton’s third law of motion, to every action there is always an equal and opposite reaction. The action and reaction act simultaneously on two different bodies.

3S.State and explain the law of action and reaction. by giving two examples. 

Solution 3S.

Law of action and reaction: In an interaction of two bodies A and B, the magnitude of action, i.e. the force FAB applied by the body B on the body A, is equal in magnitude to the reaction, i.e., the force FBA applied by the body A on the body B, but they are in directions opposite to each other.

Examples:

1. When a book is placed on a table, it does not move downwards. It implies that the resultant force on the book is zero, which is possible only if the table exerts an upward force of reaction on the book, equal to the weight of the book.

2. 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 it easier for us to move.

Explanation: In the above stated example, there are two objects and two forces. In the first example, the weight of the book acts downwards (action) and the force of the table acts upwards (reaction).

In the second example, our feet exerts a force on the ground (action) and the ground exerts an equal and opposite force (reaction) on our feet.

4S.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 moving on the floor,

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

Solution 4S.

(a) Action: Force exerted on the bullet.

Reaction: Recoil experienced by the gun.

(b) Action: The force exerted by the hammer on the nail.

Reaction: The force applied by the nail on the hammer.

(c) Action: Weight of the book acting downwards.

Reaction: Force acted by the table upwards.

(d) Action: Force exerted by the rocket on the gases backwards.

Reaction: Force exerted by outgoing gases on the rocket in forward direction.

(e) Action: Force exerted by the feet on the ground in backward direction.

Reaction: Force exerted by the ground on feet in forward direction.

(f) Action: Force exerted by a moving train on a stationary train.

Reaction: Force exerted by a stationary train on a moving train.

5S.Explain the motion of a rocket with the help of Newton's third law. 

Solution 5S.

When a rocket moves in space, it pushes gases outside, i.e. the rocket applies force on the gases in the backward direction. As a reaction, the gases put equal amount of force on the rocket in the opposite direction and the rocket moves in the forward direction.

6S.When a shot is fired from a gun, the gun is recoiled. Explain. 

Solution 6S.

When a man fires a bullet from a gun, a force F is exerted on the bullet (action), and the gun experiences an equal and opposite recoil (reaction) and hence gets recoiled.

7S.When you step ashore from a stationary boat, it tends to leave the shore. Explain. 

Solution 7S.

When a man exerts a force (action) on the boat by stepping into it, its force of reaction makes him step out of the boat, and the boat tends to leave the shore due to the force exerted by the man (i.e. action).

8S.When two spring balances joined at their free ends are pulled apart, both show the same reading. Explain. 

Solution 8S.

Couple two spring balances A and B as shown in the figure. When we pull the balance B, both the balances show the same reading indicating that both the action and reaction forces are equal and opposite. In this case, the pull of either of the two spring balances can be regarded as action and that of the other balance as the reaction.

9S.To move a boat ahead in water, the boatman has to push the water backwards by his oar. Explain this statement. 

Solution 9S.

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

10S.A person pushing a wall hard is liable to fall backwards. Give reason. 

Solution 10S.

A person pushing a wall hard (action) by his palm, experiences a force (reaction) exerted by the wall on his palm in the opposite direction; thus, he is liable to fall backwards.

11S.The action and reaction both act simultaneously. Is this statement true? 

Solution 11S.

Yes, action and reaction act simultaneously.

12S.'Action and reaction are equal in magnitude'. Is this statement true? 

Solution 12S.

Yes, action and reaction are equal in magnitude.

13S.A light ball falling on ground, after striking the ground rises upwards. Explain the reason.

Solution 13S.

When a falling ball strikes the ground, it exerts a force on the ground. The ground exerts a force back at the ball in the opposite direction. This is the reason the ball rises upwards.

14S.Comment on the statement 'the sum of action and reaction on a body is zero'.

Solution 14S.

The given statement is wrong.

Reason: According to Newton’s third law of motion, the action and reaction act simultaneously on different bodies. Hence they do not cancel each other.

1M.Newton's third law does which of the following actions:

(a) Defines the force qualitatively. (b) Defines the force quantitatively.

(c) Explains the way the force acts on a body. (d) Gives the direction of force. 

Solution 1M.

Explains the way the force acts on a body.

2M.Action and reaction act on:



(a) Same body in opposite directions.

(b) Different bodies in opposite directions.

(c) Different bodies, but in the same direction.

(d) Same body in the same direction. 

Solution 2M.

Different bodies in opposite directions

1N.A boy pushes a wall with a force of 10 N towards east. What force is exerted by the wall on the boy? 

Solution 1N.

The wall exerts an equal force of 10 N on the boy in the opposite direction, i.e. west.

2N.In the Fig., a block of weight 15 N is hanging from a rigid support by a string. What force is exerted by (a) a block on the string and (b) a string on the block? Name and show them in the diagram.

Solution 2N.

(a) A block exerts 15 N force (weight) on the string downwards.

(b) The string exerts an equal force of 15 N on the block in the opposite direction, i.e. upward direction (tension).

Exercise 3(E)

1S.State Newton's law of gravitation.

Solution 1S.

Newton’s law of gravitation: Every particle in the universe attracts every other particle with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them, and the direction of the force is along the line joining the masses.

2S.State whether the gravitational force between two masses is attractive or repulsive?

Solution 2S.

Gravitational force is always attractive.

3S.Write an expression for the gravitational force of attraction between two bodies of masses m1 and m2 separated by a distance r.

Solution 3S.

Here G is a constant of proportionality called the universal gravitational constant.

4S.How does the gravitational force of attraction between two masses depend on the distance between them?

Solution 4S.

The gravitational force of attraction between two masses is inversely proportional to the square of distance between them.

5S.How is the gravitational force between two masses affected if the separation between them is doubled?

Solution 5S.

If the distance between the masses becomes half, the force reduces to one-fourth.

6S.Define gravitational constant G.

Solution 6S.

The gravitational constant is defined as the force of attraction between two bodies of unit mass separated by a unit distance.

7S.Write the numerical value of gravitational constant G with its S.I. unit.

Solution 7S.

The value of G in the S.I. system is 6.67 x 10-11Nm2kg-2.

8S.What is the importance of law of gravitation?

Solution 8S.

The gravitational force of attraction is significant to explain the motion of heavenly bodies, e.g. motion of planets around the Sun, motion of the Moon around the Earth etc.

9S.What do you understand by the term force due to gravity?

Solution 9S.

The force with which the Earth attracts a body towards its centre is called the force due to gravity.

10S.Write an expression for the force due to gravity on a body of mass m and explain the meaning of symbols used in it.

Solution 10S.

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

11S.Define the term acceleration due to gravity? Write its S.I. unit.

Solution 11S.

The rate at which the velocity of a freely falling body increases is called acceleration due to gravity. Its S.I. unit is m/s2.

12S.Write down the average value of g on Earth's surface.

Solution 12S.

The average value of ‘g’ on the Earth’s surface is 9.8 m/s2.

13S.How is the acceleration due to gravity on the Earth's surface related to the mass and radius of the Earth?

Solution 13S.

Let g be the acceleration due to gravity on the Earth’s surface (mass = M and radius = R).

14S.How are g and G related?

Solution 14S.

Acceleration due to gravity (g) is directly proportional to universal gravitational constant (G).

15S.A body falls freely under gravity from rest and reaches the ground in time t. Write the expression for the height fallen by the body.

Solution 15S.

16S.A body is thrown vertically upwards with an initial velocity u. Write the expression for the maximum height attained by the body.

Solution 16S.

If a body is thrown vertically up with an initial velocity u to a height h, then there will be retardation (a = – g).

At the highest point, the final velocity v = 0.

Thus, from the third equation,

17S.Define the terms mass and weight.

Solution 17S.

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

Weight: The weight of a body is the force with which the Earth attracts it.

18S.Distinguish between mass and weight.

Solution 18S.

Mass is a scalar quantity, but weight is a vector quantity. Mass is the measure of the quantity of matter contained in a body, but weight is the measure of force with which the Earth attracts the body. Mass of a body is always constant but weight varies from place to place.

19S.State the S.I. units of (a) mass and (b) weight.

Solution 19S.

The S.I. unit of mass is kg and that of weight is newton.

20S.The value of g at the centre of Earth is zero. What will be the weight of a body of mass m kg at the centre of the Earth?

Solution 20S.

W = mg

At the centre of Earth, g = 0.

Therefore, W = 0.

21S.Which of the quantity is constant for a body: mass or weight?

Solution 21S.

Mass of a body is always constant.

22S.Explain the meaning of the following statement '1 kgf = 9.8 N'.

Solution 22S.

1 kgf = 9.8 N.

One kilogramme force is the force due to gravity on a mass of 1 kilogramme.

1M.The gravitational force between two bodies has the following property:

(a) Always repulsive

(b) Always attractive

(c) Attractive only at large distances (d) Repulsive only at large distances.

Solution 1M.

Always attractive

2M.The value of G is one of the values listed below:

(a) 9.8 N m2 kg-2

(b) 6.7 x 10-11 N m2 kg-2 (c) 6.7 x 10-11 m s-2

(d) 6.7 N kg-1

Solution 2M.

6.7 x 10-11 N m2 kg-2

3M.The force of attraction between two masses each of 1 kg kept at a separation of 1 m is which one of the values shown below:

(a) 9.8 N  (b) 6.7 N

(c) 980 N  (d) 6.7 x 10-11 N 

Solution 3M.

6.7 x 10-11 N

4M.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 given by the following relation:
(a)  (b) ug

(c)  (d) 

Solution 4M.

METHOD: Let’t’ be the time in which the body reaches its maximum height.

Initial velocity = u.

Final velocity (at the highest point) = 0.

Acceleration due to gravity = g (negative sign indicates the body is moving against gravity).

Using the first equation of motion,

v = u + gt.

We get,

0 = u gt

Or t = u/g

Now total time for which the ball remains in air = Time of ascent + Time of descent

Because time of ascent = Time of descent,

Total time taken = u/g + u/g = 2u/g

5M.An object falling freely from rest reaches ground in 2 s. If acceleration due to gravity is 9.8 m s-2, then the velocity of object on reaching the ground will be as shown below:

(a) 9.8 m s-1 (b) 4.9 m s-1 (c) 19.6 m s-1 (d) Zero. 

Solution 5M.

19.6 m s-1

METHOD: Given, u = 0

g = 9.8 m/s2

Time t = 2s

Let ‘v’ be the velocity of object on reaching the ground.

Using the first equation of motion,

v = u + gt

We get,

v = 0 + (9.8) (2)

Or, v = 19.6 m/s.

1N.The force of attraction between two bodies at certain separation is 10 N. What will be the force of attraction between them if the separation between them is reduced to half? 

Solution 1N.

2N.Write the approximate weight of a body of mass 5 kg. What assumption have you made?

Solution 2N.

Weight = mg

W = (5) (9.8) = 50 N.

Assumption: Value of acceleration due to gravity = 9.8 m/s2.

3N.Calculate the weight of a body of mass 10 kg in (a) kgf and (b) newton. Take g = 9.8 m s-2.

Solution 3N.

Mass = 10 kg

(i) Weight (in kgf) = 10 x 1 kgf = 10 kgf

[1 kgf = 9.8 N]

(ii) Weight (in newton) = 10 x 9.8 = 98 N.

4N.State the magnitude and direction of the force of gravity acting on the body of mass 5 kg. Take g = 9.8 m s-2.

Solution 4N.

Mass = 5 kg.

g = 9.8 m/s2.

Let F be the force of gravity,

F = mg.

F = (5) (9.8) = 49 N.

Force of gravity always acts downwards.

5N.The weight of a body is 2.0 N. What is the mass of the body? (g = 10 m s-2)

Solution 5N.

Weight, W = 2.0 N

g = 9.8 m/s2

Let ‘m’ be the mass of the body.

W = mg

Or, m = W/g = (2/9.8) kg = 0.2 kg.

6N.The weight of a body on Earth is 98 N, where acceleration due to gravity is 9.8 m s-2. What will be its (a) mass and (b) weight on the Moon, where acceleration due to gravity is 1.6 m s-2?

Solution 6N.

Weight of the body on Earth = 98 N.

Acceleration due to gravity on Earth = 9.8 m/s2.

Let ‘m’ be the mass of the body on Earth.

m = W/g

m = (98/9.8) = 10 kg

Thus, the mass of the body is 10 kg, which always remains constant.

(a) Mass on moon = mass on Earth = 10 kg

(b) Let weight on moon is W’.

W’ = mass x acceleration due to gravity on the Moon.

[Given, acceleration due to gravity on the Moon = 1.6 m/s2]

W’ = 10 x 1.6 =16 N.

7N.A man weighs 600 N on the Earth. What would be his approximate weight on the Moon? Give a reason for your answer?

Solution 7N.

Man’s weight on Earth = 600 N

Man’s weight on the Moon = (1/6) man’s weight on Earth;

Because the acceleration due to gravity on the Moon is 1/6th that of Earth and w = mg.

Therefore, man’s weight on Moon = (600/6) = 100 N.

8N.What (a) force and (b) weight act on a mass of 10.5 kg being pulled by the Earth? Take g = 10 m s-2.

Solution 8N.

Mass, m = 10.5 kg

G = 10 m/s2

(a) Force, F = mg

F = (10.5) (10) = 105 N

(b) Weight, w = mg

w = (10.5) (10) = 105 N

9N.A ball is released from a height and it reaches the ground in 3 s. If g= 9.8 m s-2, find the value for the following variables:

(a) The height from which the ball was released.

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

Solution 9N.

Let ‘S’ be the height.

Time taken, t = 3s; g = 9.8 m/s2

Initial velocity, u = 0 (because the body starts from rest)

(a) Using the second equation of motion,

S = ut + (1/2) gt2

We get,

S = 0 + (1/2) (9.8) (3) (3)

S = 44.1 m

(b) Let ‘v’ be the velocity with which the ball strikes the ground.

Using the third equation of motion,

v2 – u2 = 2gs

or, v2 – 02 = 2(9.8) (44.1)

or, v2 = 864.36

or, v = 29.4 m/s

10N.What force, in newton, your muscles need to apply to hold a mass of 5 kg in your hand? State the assumption.

Solution 10N.

Mass, m = 5kg

Force, F = mg

F = (5) (9.8) = 49 N

Assumption: Value of acceleration due to gravity is 9.8 m/s2.

11N.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 the following variables:

(a) The initial velocity of the ball

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

(c) The total time of journey of ball.

Solution 11N.

Given, maximum height reached, s = 20 m

Acceleration due to gravity, g = 10 m/s2

(a) Let ‘u’ be the initial velocity.

At the highest point, velocity = 0

Using the third equation of motion,

v2   u2 = 2gs

or, 0  u2 = 2 (10) (20) m/s

or, u2 =  (400) m/s [Negative sign indicates that the motion is against gravity]

or, u = 20 m/s

(b) Let v’ be the final velocity of the ball on reaching the ground.

Considering the motion from the highest point to ground,

Velocity at highest point = 0 = Initial velocity for downward journey of the ball.

Distance travelled, s = 20m

Using the third equation of motion,

v2  u2 = 2gs

or, v2  0 = 2 (10) (20) m/s

or, v2 = 400 m/s

or, v = 420 m/s

(c) Now total time for which the ball remains in air, t = 2u/g.

Or, t = 2 (20)/(10).

Or, t = 4s.

12N.An object 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)

Solution 12N.

Initial velocity u = 0

Final velocity = 20 m/s

g = 10 m/s2

Let ‘h’ be the height of the tower.

Using the third equation of motion,

v2 – u2 = 2gs

or, (20)2 – 0 = 2 (10) h

or, h = 20 m

13N.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)

Solution 13N.

Total time of journey = 6 s

g = 10 m/s2

(i) Let ‘H’ be the greatest height.

Time of ascent, t = 6/2 = 3 s,

For ascent, initial velocity, u = 0

Using the second equation of motion,

H = ut + (1/2) gt2

H = 0 + (1/2) (10) (3) 2

H = 45 m

(ii) Let u’ be the initial velocity.

Final velocity, v = 0

Using the third equation of motion,

v2 – u2 = 2gH

or, v2 – 0 = 2(10) (45)

or, v2 = 900

or, v = 30 m/s

14N.A pebble is thrown vertically upwards with a speed of 20 m s-1. How high will it be after 2 s? (Take g = 10 m s-2)

Solution 14N.

Initial velocity, u = 20 m/s

Time, t = 2s

g = 10 m/s2

Maximum height reached in 2s, H = (1/2) gt2

Or, H = (1/2) (10) (2) 2

Or, H = 20 m

15N.(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)

Solution 15N.

(a) Height, s = 80m

g = 10 m/s2

Using the second equation of motion,

S = ut + (1/2) gt2

Or, 80 = 0+ (1/2) (10) (t) 2

Or, (t) 2 = 16

Or, t = 4s

(b) Let ‘v’ be the velocity on reaching the ground.

Using the third equation of motion,

v2 – u2 = 2gH

or, v2 – 0 = 2(10) (80)

or, v2 = 1600

or, v = 40 m/s

16N.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)

Solution 16N.

Given time t = 2.5, g = 9.8 m/s2

Height, H = (1/2) gt2

Or, H = (1/2) (9.8) (2.5)2

Or, H = 30.6 m

17N.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).

Solution 17N.

Initial velocity, u = 49 m/s

g = 9.8 m/s2

(i) Let H be the maximum height attained.

At the highest point, velocity = 0.

Using the third equation of motion,

v2 – u2 = 2gH

or, 0 – 492 = 2(-9.8) (H)

or, H = (492)/ 19.6

or, H = 122.5 m

(ii) Total time of flight is given by t = 2u/g

Or, t = 2(49)/ 9.8

Or, t = 10 s

18N.A stone is dropped freely from the top of a tower and it reaches the ground in 4 s. Taking g = 10m s-2, calculate the height of the tower.

Solution 18N.

Initial velocity u = 0

Time t = 4 s

g = 10 m/s2

Let ‘H’ be the height of the tower.

Using the second equation of motion,

H = ut + (1/2) gt2

Or, H = 0 + (1/2)(10)(4)2

Or, H = 80 m

19N.A pebble is dropped freely in a well from its 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.

Solution 19N.

(i) Time t =20 s

g = 10 m/s2

Let ‘D’ be the depth of the well.

Using the second equation of motion,

D = ut + (1/2) gt2

D = 0 + (1/2)(10)(20) 2

D = 2000 m

(ii) Speed of sound = 330 m/s

Depth of well = 2000 m

Time taken to hear the echo after the pebble reaches the water surface = Depth/speed

= (2000/330) s

= 6.1 s

Time taken for pebble to reach the water surface = 20 s.

Therefore, the total time taken to hear the echo after the pebble is dropped = 20 + 6.1 = 21.6 s.

20N.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 the ball on reaching the ground. Take g= 9.8 ms-2.

Solution 20N.

Let x be the height of the tower.

Let h be the distance from the top of the tower to the highest point as shown in the diagram.

Initial velocity u = 19.6 m/s

g = 9.8 m/s2

At the highest point, velocity = 0

Using the third equation of motion,

v2 – u2 = 2gh

Or, – (19.6) 2 = 2 (-9.8) h

Or, h = 19.6 m

If the ball takes time t1 to go to the highest point from the top of building, then for the upward journey from the relation, v = u – gt,

0 = 19.6 – (9.8) (t1)

Or, t1 = 2s

(ii) Let us consider the motion for the part (x+h)

Time taken to travel from highest point to the ground = (5 – 2) = 3s

Using the equation s = ut + (1/2) gt2

We get,

(x + h) = 0 + (1/2) (9.8) (3) 2

Or, (x + 19.6) = 44.1 m

Or, x = 44.1 – 19.6 = 24.5 m

Thus, height of the tower = 24.5 m

(iii) Let v be the velocity of the ball on reaching the ground.

Using the relation, v = u + gt

We get:

v = 0 + (9.8) (3)

Or, v = 29.4 m/s